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diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER11_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER11_3.ipynb new file mode 100644 index 00000000..f46ad2ec --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER11_3.ipynb @@ -0,0 +1,561 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 11 - Carburetion and Carburettors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.1 PAGE 399" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Suction at throat = 4447.61 N/m**2 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "d=0.1#..................#Cylinder bore in m\n", + "l=0.12#................#Cylinder stroke in m\n", + "N=1800#..................#Engine rpm\n", + "d2=0.028#................#Throat diameter in m\n", + "Cda=0.8#................#Co efficient of air flow\n", + "etaV=0.75#..................#Volumetric efficiency\n", + "rhoa=1.2#................#Density of air in kg/m**3\n", + "n=4#.......................#No of cylinders\n", + "#Calculations\n", + "Vs=(pi/4)*d*d*l*n#.................#Stroke Volume in m**3\n", + "Va=etaV*Vs#.......................#Actual volume per stroke in m**3\n", + "Vas=Va*(N/2)*(1/60)#.............#Actual volume sucked per second\n", + "ma=Vas*rhoa#.........................#Air consumed in kg/sec\n", + "delp=((ma/(Cda*(pi/4)*d2*d2))**2)/(2*rhoa)#.............#Suction at throat in N/m**2\n", + "print \"Suction at throat = %0.2f N/m**2 \"%delp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.2 PAGE 400" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Suction at the throat = 539.13 N/m**2 \n", + "Throat area = 7.72 cm**2 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "# Initialisation of Variables\n", + "cp=5#.................#Consumption of petrol in kg/h\n", + "afr = 16#...............#Air fuel ratio\n", + "Af=2*10**(-6)#..............#Fuel orifice area in m**2\n", + "z=0.005#................#Distance between tip of jet and level of petrol in float chamber in m\n", + "spgrp=0.75#..............#Specific gravity of petrol\n", + "rhow=1000#.................#Density of water in kg/m**3\n", + "rhoa=1.2#....................#Density of air in kg/m**3\n", + "Cda=0.8#...............#Coefficient of discharge for venturi throat\n", + "g=9.81#...............#Acceleration due to gravity in m/sec**2\n", + "#Calculations\n", + "mf=cp/3600#.................#Fuel consumed in kg/sec\n", + "delp=(((mf/(Af*Cda))**2)*(1/(2*spgrp*rhow)))+(g*z*spgrp*rhow)#\n", + "print \"Suction at the throat = %0.2f N/m**2 \"%delp\n", + "ma=mf*afr#................#Air flow rate\n", + "Atsqr=((ma/Cda)**2)*(1/(2*rhoa*delp))#....................#Throat area in m**2\n", + "print \"Throat area = %0.2f cm**2 \"%(sqrt(Atsqr)*10**4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.3 PAGE 401" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diameter of the fuel jet = 1.32 mm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi,sqrt\n", + "# Initialisation of Variables\n", + "pc=7.2#.................#Petrol consumed in kg/h\n", + "spgrp=0.75#................#Specific gravity of fuel\n", + "rhow=1000#.................#Density of water in kg/m**3\n", + "t1=300#...................#Temperature of air in Kelvin\n", + "afr=15#....................#Air fuel ratio\n", + "d2=0.024#....................#Diameter of choke tube in m\n", + "z=0.0042#...................#The height of the jet above petrol level in float chamber in m\n", + "Cda=0.8#....................#Coefficient of discharge for air\n", + "Cdf=0.7#.....................#Coefficient of discharge for fuel\n", + "p1=1.013#.....................#Atmospheric pressure in bar\n", + "g=9.81#.......................#Acceleration due to gravity in m/s**2\n", + "R=287#........................#Gas constant in J/kg.K\n", + "#calculations\n", + "mf=pc/3600#....................#Rate of fuel consumption in kg/sec\n", + "rhof=spgrp*rhow#...............#Density of fuel in kg/m**3\n", + "rhoa=(p1*10**5)/(R*t1)#............#Density of air in kg/m**3\n", + "ma=mf*afr#.......................#Air flow rate \n", + "delpa=((ma/(Cda*(pi/4)*d2**2))**2)*(1/(2*rhoa))#....................#Suction in N/m**2\n", + "df=sqrt((mf/sqrt(2*rhof*(delpa-(g*z*rhof))))*(1/(Cdf*(pi/4))))#.................#Diameter of fuel jet in m\n", + "print \"Diameter of the fuel jet = %0.2f mm\"%(df*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.4 PAGE 402" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Suction at the throat = 7.67 cm of Water \n", + "Throat area = 2.86 cm \n", + "Velocity of air across the venturi throat = 8.34 m/s \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "# Initialisation of Variables\n", + "pc=5.45#......................#Petrol consumption in kg/h\n", + "afr=15#......................#Air fuel ratio\n", + "af=2*10**(-6)#................#Fuel jet orifice area in m**2\n", + "z=0.00635#...................#Distance between tip of fuel jet and level of petrol in the float chamber in m\n", + "Cda=0.8#............................#Coefficient of discharge of venturi throat\n", + "rhoa=1.29#........................#Density of air in kg/m**3\n", + "spgrp=0.72#........................#Specific gravity of fuel\n", + "rhow=1000#.........................#Density of water in kg/m**3\n", + "g=9.81#..............................#Acceleration due to gravity in m/s**2\n", + "Cdf=0.75#........................#Coefficient of discharge of the fuel\n", + "#calculations\n", + "mf=pc/3600#....................#Fuel consumed in kg/sec\n", + "rhof=spgrp*rhow#...............#Density of fuel in kg/m**3\n", + "delp=(((mf/(af*Cdf))**2)*(1/(2*rhof)))+(g*z*rhof)#......................#Depression in venturi throat in N/m**2\n", + "h2odep=delp/(g*1000)#................................#Depression in venturi throat in cm of Water\n", + "print \"Suction at the throat = %0.2f cm of Water \"%(h2odep*100)\n", + "ma=mf*afr#................#Air flow rate\n", + "At=sqrt(((ma/Cda)**2)*(1/(2*rhoa*delp)))#....................#Throat area in m**2\n", + "dt=sqrt(At/(pi/4))#........................................#Throat diameter in m\n", + "print \"Throat area = %0.2f cm \"%(dt*100)\n", + "Ct=sqrt((2*g*z*rhof)/rhoa)#..........................#Velocity of air across the venturi throat in m/sec\n", + "print \"Velocity of air across the venturi throat = %0.2f m/s \"%(Ct)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.5 PAGE 403" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Throat pressure = 0.76 bar \n", + "Air fuel ratio when the air cleaner is fitted : 13.69\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "# Initialisation of Variables\n", + "afr=15#.....................#Air fuel ratio\n", + "p1=1#.........................#Atmospheric pressure in bar\n", + "p2=0.8#.......................#Pressure at venturi throat in bar\n", + "pd=30#....................#Pressure drop to air cleaner in mm of Hg\n", + "rhohg=13600#....................#Density of Hg in kg/m**3\n", + "af=240#........................#Air flow at sea level in kg/h\n", + "g=9.81#.....................#Acceleration due to gravity in m/s**2\n", + "#calculations\n", + "delpa=p1-p2#........................#When there is no air cleaner\n", + "pt=1-(rhohg*g*(pd/1000)*10**(-5))-delpa#..........................#Throat pressure in bar\n", + "print \"Throat pressure = %0.2f bar \"%pt\n", + "afrn=afr*sqrt(delpa/(p1-pt))#...............................#Air fuel ratio when the air cleaner is fitted\n", + "print \"Air fuel ratio when the air cleaner is fitted : %0.2f\"%afrn" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.6 PAGE 403" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Throat diameter = 0.03 m\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "# Initialisation of Variables\n", + "As=4.6#........................#Air supply in kg/min\n", + "p1=1.013#.......................#Atmospheric pressure in bar\n", + "t1=298#......................#Atmospheric temperature in Kelvin\n", + "C2=80#........................#Air flow velocity in m/s\n", + "Cv=0.8#....................#Velocity co efficient\n", + "ga=1.4#........................#Degree of freedom of gas\n", + "R=0.287#........................#Gas constant in kJ/kgK\n", + "#Calculations\n", + "cp=R*(ga/(ga-1))#.......................#Specific heat capacity of air in kJ/kgK\n", + "p2=((1-(((C2/Cv)**2)*(1/(2*cp*1000*t1))))**(ga/(ga-1)))*p1#...................#Throat pressure in bar\n", + "rho1=(p1*10**5)/(R*1000*t1)#\n", + "rho2=rho1*(p2/p1)**(1/ga)#\n", + "ma=As/60#...................#Air flow in kg/s\n", + "A2=ma/(rho2*C2)#.................#Throat area in m**2\n", + "d2=sqrt((4*A2)/pi)#................#Throat diameter in m\n", + "print \"Throat diameter = %0.2f m\"%d2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.7 PAGE 404" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Throat diameter = 3.53 cm \n", + "Orifice diameter = 0.0023 cm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "# Initialisation of Variables\n", + "As=6#........................#Air supply in kg/min\n", + "fs=0.45#..........................#Fuel supply in kg/min\n", + "p1=1.013#.......................#Atmospheric pressure in bar\n", + "t1=300#......................#Atmospheric temperature in Kelvin\n", + "rhof=740#......................#Density of fuel in kg/m**3\n", + "C2=92#........................#Air flow velocity in m/s\n", + "Cda=0.8#....................#Velocity co efficient\n", + "Cdf=0.6#.........................#Coefficient of discharge for fuel\n", + "ga=1.4#........................#Degree of freedom of gas\n", + "r=0.75#......................#ratio of pressure drop across venturi and of that of choke\n", + "R=0.287#........................#Gas constant in kJ/kgK\n", + "#Calculations\n", + "ma=As/60#.................................#Air flow in kg/s\n", + "mf=fs/60#.................................#Fuel flow in kg/s\n", + "cp=R*(ga/(ga-1))#.......................#Specific heat capacity of air in kJ/kgK\n", + "p2=((1-(((C2/Cda)**2)*(1/(2*cp*1000*t1))))**(ga/(ga-1)))*p1#...................#Throat pressure in bar\n", + "v1=(R*t1*1000)/(p1*10**5)#\n", + "v2=v1*(p1/p2)**(1/ga)#................#specific volume in m**3/kg\n", + "A2=(ma*v2)/(C2)#.................#Throat area in m**2\n", + "d2=sqrt((4*A2)/pi)#................#Throat diameter in m\n", + "print \"Throat diameter = %0.2f cm \"%(d2*100)\n", + "pdv=p1-p2#..........#Pressure drop at venturi in bar\n", + "pdj=r*pdv#.............#Pressure drop at jet in bar\n", + "Af=((mf/Cdf)*(1/sqrt(2*rhof*pdj*10**5)))#.............#Area of orifice in m**2\n", + "df=sqrt((4*Af)/pi)#................#Orifice diameter in m\n", + "print \"Orifice diameter = %0.4f cm \"%(df)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.8 PAGE 404" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diameter of choke = 29.61 mm \n", + "Diameter of the fuel jet = 1.68 mm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "# Initialisation of Variables\n", + "Vs=1489*10**(-6)#.......................#Capacity of engine in m**3\n", + "N=4200#...............#Engine rpm at which max speed is developed\n", + "etaV=0.75#.....................#Volumetric efficiency\n", + "afr=13#........................#air fuel ratio\n", + "Ct=85#..........................#Theoretical air speed at peak power in m/s\n", + "C2=Ct#\n", + "Cda=0.82#....................#Coefficient of discharge for the venturi\n", + "Cdf=0.65#....................#Coefficient of discharge of main petrol jet\n", + "spgr=0.74#..................#Specific gravity of petrol\n", + "z=0.006#.................................#Level of petrol surface below choke\n", + "p1=1.013#......................#Atmospheric pressure in bar\n", + "t1=293#.........................#Atmospheric temperature in Kelvin\n", + "r=0.4#.............................#Ratio of diameter of emulsion tube to choke diameter\n", + "R=0.287#.............................#Gas constant in kJ/kgK\n", + "ga=1.4#..............................#Degree of freedom for air\n", + "g=9.81#..............................#Acceleration due to gravity in m/s**2\n", + "rhow=1000#...........................#Density of water in kg/m**3\n", + "#calculations\n", + "rhof=rhow*spgr#............................#Density of fuel in kg/m**3\n", + "Va=(etaV*Vs*N)/(60*2)#.....................#Volume of air induced in m**3/s\n", + "ma=(p1*10**5*Va)/(R*t1*1000)#...............#mass flow of air in kg/s\n", + "cp=R*(ga/(ga-1))#.......................#Specific heat capacity of air in kJ/kgK\n", + "p2=((1-(((C2)**2)*(1/(2*cp*1000*t1))))**(ga/(ga-1)))*p1#...................#Throat pressure in bar\n", + "pt=p2#\n", + "vt=Va*(p1/p2)**(1/ga)#.....................#Volume flow of air at choke in m**3/s\n", + "At=vt/(Ct*Cda)#...................#Area of emulsion tube in m\n", + "D=sqrt((4*At*10**6)/(pi*(1-r**2)))#...................#Diameter of choke in mm\n", + "print \"Diameter of choke = %0.2f mm \"%D\n", + "mf=ma/afr#..................#Mass flow of fuel in kg/s\n", + "delpa=(p1-p2)*10**5#\n", + "df=sqrt((mf/sqrt(2*rhof*(delpa-(g*z*rhof))))*(1/(Cdf*(pi/4))))#.................#Diameter of fuel jet in m\n", + "print \"Diameter of the fuel jet = %0.2f mm \"%(df*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.9 PAGE 405" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air fuel ratio when the nozzle lip is neglected : 11.35\n", + "Air fuel ratio when nozzle lip is taken into account : 11.39\n", + "Minimum velocity of air = 8.58 m/s\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "# Initialisation of Variables\n", + "da=0.018#..........................#Throat Diameter in m\n", + "df=0.0012#......................#Diameter of fuel orifice in m\n", + "Cda=0.82#.................#Coefficient of air flow\n", + "Cdf=0.65#......................#Coefficient of fuel flow\n", + "z=0.006#........................#Level of petrol surface below the throat\n", + "rhoa=1.2#.......................#density of air in kg/m**3\n", + "rhof=750#.........................#density of fuel in kg/m**3\n", + "g=9.81#........................#Acceleration due to gravity in m/s**2\n", + "delp=0.065*10**5#...................#Pressure drop in N/m**2\n", + "#Calculations\n", + "afr1=(Cda/Cdf)*((da/df)**2)*sqrt(rhoa/rhof)#..................#Air fuel ratio when the nozzle lip is neglected\n", + "print \"Air fuel ratio when the nozzle lip is neglected : \",round(afr1,2)\n", + "afr2=afr1*sqrt(delp/(delp-(g*z*rhof)))#.....................#Air fuel ratio when nozzle lip is taken into account\n", + "print \"Air fuel ratio when nozzle lip is taken into account : \",round(afr2,2)\n", + "C2=sqrt((2*g*z*rhof)/rhoa)#.........................#Minimum velocity of air in m/s\n", + "print \"Minimum velocity of air = %0.2f m/s\"%C2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.10 PAGE 406" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Fuel consumption = 46.66 kg/h \n", + "Velocity of air at the throat = 132.52 m/s \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "# Initialisation of Variables\n", + "d=0.11#..................#Engine bore in m\n", + "l=0.11#..................#Engine length in m\n", + "da=0.042#................#Throat diameter of the choke tube in m\n", + "N=3000#..................#Engine rpm\n", + "etaV=0.75#...............#Volumetric efficiency\n", + "Ra=287#..................#Gas constant for air in J/kgK\n", + "Rv=97#...................#Gas constant for fuel vapour in J/kgK\n", + "t=273#....................#Temperature in Kelvin\n", + "p=1.013#...................#Pressure in bar\n", + "delpa=0.12#.................#Pressure depression in bar\n", + "t2=273+15#...................#Temperature at throat\n", + "n=8#........................#No of cylinders\n", + "mO=32#.......................#Mass of Oxygen molecule in amu\n", + "mC=12#........................#Mass of Carbon molecule in amu\n", + "mH=1#.......................#Mass of Hydrogen molecule in amu\n", + "cC=84#......................#Composition of carbon in %\n", + "cH2=16#.....................#Composition of Hydrogen in % \n", + "#Calculations\n", + "Vfm=(pi/4)*d*d*l*n*(N/2)*etaV#.....................#Volume of fuel mixture supplied in m**3/min\n", + "afr=((cC*(mO/mC))+(cH2*(mO/(4*mH))))/23#..................#Air fuel ratio\n", + "va=(Ra*t)/(p*10**5)#.....................#Volume of 1 kg of air in m**3/kg\n", + "vf=(Rv*t)/(p*10**5)#......................#Volume of 1 kg of fuel vapour in m**3/kg\n", + "fc=(Vfm/((afr*va)+vf))*60#...............#Fuel consumption in kg/h\n", + "print \"Fuel consumption = %0.2f kg/h \"%fc\n", + "rhoa=((p-delpa)*10**5)/(Ra*t2)#...............#Density of air at the throat in kg/m**3\n", + "Ca=(afr*(fc/3600))/((pi/4)*da*da*rhoa)#................#Velocity of air at the throat in m/s\n", + "print \"Velocity of air at the throat = %0.2f m/s \"%Ca" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 11.11 PAGE 407" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air fuel ratio at the altitude : 11.26\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "# Initialisation of Variables\n", + "a=4500#.................#Altitude\n", + "afr=14#...............#Air fuel ratio at sea level\n", + "t1=25#...........#Temperature at sea level in Celsius\n", + "p1=1.013#...........#Pressure at sea level in bar\n", + "#Calculations\n", + "t2=t1-(0.0064*a)#.........................#Temperature at the given altitude using the given formula in Celsius\n", + "p2=p1/(10**(a/19300))#....................#Pressure at the given altitude using the given formula in bar\n", + "afr2=afr*sqrt((p2*(t1+273))/(p1*(t2+273)))#...................#Air fuel ratio at the altitude\n", + "print \"Air fuel ratio at the altitude : %0.2f\"%afr2" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER12_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER12_3.ipynb new file mode 100644 index 00000000..36e5f2df --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER12_3.ipynb @@ -0,0 +1,445 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER-12 Fuel injection systems for C.I. Engines " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 12.1 PAGE 421" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volume of Fuel Injected per Cycle = 0.05 cc \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "\n", + "n=6#...............#No of cylinders\n", + "BP=125#...............#Brake Power in kW\n", + "N=3000#..............#Engine rpm\n", + "bsfc=200#............#Brake Specific Fuel Consumption g/kWh\n", + "spgr=0.85#.............#Specific Gravity\n", + "\n", + "#Calculations\n", + "\n", + "fc=(bsfc/1000)*BP#.........#Fuel consumption in kg/h\n", + "fcpc=fc/n#..................#Fuel consumption per cylinder\n", + "FCPC=(fcpc/60)/(N/2)#................#Fuel Consumption per cycle in kg\n", + "VFIC = (FCPC*1000)/spgr#...................#Volume of fuel injected per cycle in cc\n", + "print \"Volume of Fuel Injected per Cycle = %0.2f cc \"%VFIC" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 12.2 PAGE 421" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diameter of Nozzle Orifice = 8.13e-04 m \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# Initialisation of Variables\n", + "n=6#...............#No of cylinders\n", + "N=1500#............#Engine rpm\n", + "BP=220#.............#Brake Power in kW\n", + "bsfc=0.273#..........#Brake Specific Fuel Consumption in kg/kWh\n", + "theta=30#.............#The Period of Injection in degrees of crank angle\n", + "spgr=0.85#............#Specific Gravity of fuel\n", + "Cf=0.9#................#Orifice discharge co-efficient\n", + "ip=160#...............#Injection pressure in bar\n", + "cp=40#.................#Pressure in combustion chamber in bar\n", + "rhow=1000#................#Density of water in kg/m**3\n", + "#Calculations\n", + "vf = Cf*sqrt((2*(ip-cp)*10**5)/(spgr*rhow))#.............#Actual fuel velocity of injection in m/sec\n", + "qf=(bsfc*BP)/(spgr*rhow*3600)#..................# Volume of fuel injected per sec in m**3\n", + "d=sqrt (qf/((pi/4)*n*vf*(theta/360)*(60/N)*(N/120)))#...........#Diameter of nozzle orifice\n", + "print \"Diameter of Nozzle Orifice = %0.2e m \"%d" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 12.3 PAGE 422" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volume of Fuel Injected per Cycle = 0.13 cm**3 \n", + "Diameter of orifice = 0.40 mm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# Initialisation of Variables\n", + "n=1#............#No of cylinders\n", + "N=650#............#Engine rpm\n", + "theta=28#...........#Crank Travel in degree\n", + "fc=2.2#...........#Fuel consumption in kg/h\n", + "spgr=0.875#............#Specific Gravity\n", + "ip=150#................#Injection Pressure in bar\n", + "cp=32#.................#Combustion chamber Pressure in bar\n", + "Cd=0.88#...............#co-efficient of discharge of orifice\n", + "rhow=1000#...........#Density of water in kg/m**3\n", + "#Calculation\n", + "fcpc = fc/60#..............#Fuel consumption per cylinder\n", + "fipc = fcpc/(N/2)#.........#Fuel Injected per cycle in kg\n", + "vfpc = fipc/(spgr*rhow)#....#volume of fuel injected per cycle\n", + "print \"Volume of Fuel Injected per Cycle = %0.2f cm**3 \"%(vfpc*10**6)\n", + "tfic=(theta/360)*(60/N)#....#Time for Fuel Injection per Cycle in sec\n", + "mf = fipc/tfic#...............#Mass of fuel injected per cycle in kg/s\n", + "vf = Cd*sqrt((2*(ip-cp)*10**5)/(spgr*rhow))#.............#Actual fuel velocity of injection in m/sec\n", + "d=sqrt((mf*4)/(pi*vf*spgr*rhow))\n", + "print \"Diameter of orifice = %0.2f mm \"%(d*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 12.4 PAGE 423" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diameter of orifice = 0.56 mm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# Initialisation of Variables\n", + "N=2000#............#Engine rpm\n", + "theta=30#...........#Crank Travel in degree\n", + "sfc=0.272#...........#Fuel consumption in kg/kWh\n", + "ip=120#................#Injection Pressure in bar\n", + "cp=30#.................#Combustion chamber Pressure in bar\n", + "Cd=0.9#...............#co-efficient of discharge of orifice\n", + "rhow=1000#...........#Density of water in kg/m**3\n", + "api = 32#..............#API in degree\n", + "pw=15#..................#Power Output in kW\n", + "#Calculation\n", + "spgr= 141.5/(131.5+api)#............#Specific Gravity\n", + "fcpc = (sfc*pw)/((N/2)*60)#..............#Fuel consumption per cycle in kg\n", + "tfic=(theta/360)*(60/N)#....#Time for Fuel Injection per Cycle in sec\n", + "mf = fcpc/tfic#...............#Mass of fuel injected per cycle in kg/s\n", + "vf = Cd*sqrt((2*(ip-cp)*10**5)/(spgr*rhow))#.............#Actual fuel velocity of injection in m/sec\n", + "d=sqrt((mf*4)/(pi*vf*spgr*rhow))\n", + "print \"Diameter of orifice = %0.2f mm \"%(d*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 12.5 PAGE 424" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Fuel consumption = 0.28 kg/kWh \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# Initialisation of Variables\n", + "N=1800#............#Engine rpm\n", + "theta=32#...........#Crank Travel in degree\n", + "ip=118.2#................#Injection Pressure in bar\n", + "cp=31.38#.................#Combustion chamber Pressure in bar\n", + "Cd=0.9#...............#co-efficient of discharge of orifice\n", + "rhow=1000#...........#Density of water in kg/m**3\n", + "api = 32#..............#API in degree\n", + "pw=11#..................#Power Output in kW\n", + "d=0.47#...................#Fuel Injection orifice diameter in mm\n", + "#Calculation\n", + "spgr= 141.5/(131.5+api)#............#Specific Gravity\n", + "tfic=(theta/360)*(60/N)#....#Time for Fuel Injection per Cycle in sec\n", + "vf = Cd*sqrt((2*(ip-cp)*10**5)/(spgr*rhow))#.............#Actual fuel velocity of injection in m/sec\n", + "mf=vf*spgr*rhow*(pi/4)*(d/1000)**2#\n", + "tncp=(N/2)*60#...............#Total no of cycles per hour\n", + "FIPC=mf*tfic#.................#Mass of fuel injected per cycle in kg/cycle\n", + "fc=FIPC*tncp*(1/pw)#...................#Fuel consumption in kg/kWh\n", + "print \"Fuel consumption = %0.2f kg/kWh \"%fc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 12.6 PAGE 424" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Orifice Area Required per injector = 0.0136 cm**2 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# Initialisation of Variables\n", + "n=8#......#No of cylinders\n", + "pw=386.4#...........#Power output in kW\n", + "N=800#.............#Engine rpm\n", + "fc=0.25#.............#Fuel Consumption in kg/kWh\n", + "theta=12#..............#Crank Travel in degree (for injection)\n", + "spgr=0.85#...........#Specific Gravity\n", + "patm=1.013#............#Atmospheric pressure\n", + "cf=0.6#................#Co-efficient of discharge for injector\n", + "pcB=32#................#Pressure in cylinder in beginning in bar\n", + "piB=207#...............#Pressure in injector in beginning in bar\n", + "pcE=55#...............#Pressure in cylinder at the end in bar\n", + "piE=595#................#Pressure in injector at the end in bar\n", + "rhow=1000#..............#density of water in kg/m**3\n", + "#calculations\n", + "pwpc = pw/n#......................#Output per cylinder\n", + "fcpc = (pwpc*fc)/60#.............#Fuel consumption per cylinder in kg/min\n", + "fipc = fcpc/(N/2)#................#Fuel injected per cycle in kg\n", + "tfic = (theta*60)/(360*N)#...........#Time for fuel Injection per cycle\n", + "mf = fipc/tfic#......................#Mass of fuel injected per second\n", + "pdb = piB-pcB#....................#Pressure difference at beginning\n", + "pde = piE-pcE#...................#Pressure difference at end\n", + "apd = (pdb+pde)/2#\n", + "Ao=mf/(cf*sqrt(2*apd*10**5*spgr*rhow))#\n", + "print \"Orifice Area Required per injector = %0.4f cm**2 \"%(Ao*10000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 12.7 PAGE 425" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mass of fuel injected into each cylinder per second = 1.10 kg/s\n", + "Diameter of the fuel orifice = 0.55 mm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# Initialisation of Variables\n", + "n=6#...............#No of cylinders\n", + "afr=20#...........#Air fuel ratio\n", + "d = 0.1#...............#cylinder bore in mm\n", + "l=0.14#..............#Cylinder length in mm\n", + "etav=0.8#............#Volumetric Efficiency\n", + "pa=1#.................#Pressure at the beginning of the compression in bar\n", + "ta = 300#.............#Temperature at the beginning of the compression in Kelvin\n", + "theta = 20#...............#Crank travel in degree for injection\n", + "N = 1500#...................#engine rpm\n", + "rhof=960#.................#Fuel density in kg/m**3\n", + "cf=0.67#................#Co efficient of discharge for injector\n", + "pi=150#....................#injection pressure in bar\n", + "pc=40#....................#combustion pressure in bar\n", + "R=287#........................#gas constant for air in kJ/kg.K\n", + "#calculations\n", + "V=(pi/4)*d**2*l*etav#......................#Volume of air supplied per cylinder per cycle in m**3\n", + "ma=(pa*10**5*V)/(R*ta)#.....................#Mass of this air at suction conditions in kg/cycle\n", + "mf=ma/afr#............................#Mass of fuel in kg/cycle\n", + "fipc = (theta*60)/(360*N)#...........#Time taken for fuel injection per cycle in seconds\n", + "MF = mf/fipc#........................#Mass of fuel injected into each cylinder per second\n", + "print \"The mass of fuel injected into each cylinder per second = %0.2f kg/s\"%MF\n", + "vf=cf*sqrt((2*(pi-pc)*10**5)/rhof)#.............#fuel velocity injection in m/s\n", + "d0=sqrt((MF*4)/(pi*vf*rhof))#..................#diameter of fuel orifice in m\n", + "print \"Diameter of the fuel orifice = %0.2f mm \"%(d0*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 12.8 PAGE 425" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total displacement of plunger = 0.26 cc\n", + "Effective stroke of plunger = 0.69 mm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# Initialisation of Variables\n", + "Vpbes=7#.................#Volume of fuel in the pump barrel before commencement of effective stroke in cc\n", + "df=3#.................#Diameter of fuel line from pump to injector in mm\n", + "lf=700#.................#Length of fuel line from pump to injector in mm\n", + "Vfiv=2#................#Volume of fuel in the injection valve in cc\n", + "Vfd=0.1#.................#Volume of fuel to be delivered in cc\n", + "p1=150#..............#Pressure at which fuel is delivered in bar\n", + "p2=1#.................#atmospheric pressure in bar\n", + "cc=78.8*10**(-6)#..........#Co - efficient of compressibility per bar\n", + "dp=7#..............#Diameter of plunger in mm\n", + "#calculations\n", + "V1=Vpbes+(pi/4)*((df/10)**2)*(lf/10)+Vfiv#...................#Total initial fuel volume\n", + "delV=cc*(p1-p2)*V1#................#Change in volume due to compression\n", + "displu=delV+Vfd#.....................#Total displacement of plunger\n", + "print \"Total displacement of plunger = %0.2f cc\"%displu\n", + "lp=(displu*4)/(pi*(dp/10)**2)#.............#Effective stroke of plunger\n", + "print \"Effective stroke of plunger = %0.2f mm \"%lp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 12.9 PAGE 426" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time required for spray penetration at an injection pressure of 235 bar = 11.98 milliseconds:\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# Initialisation of Variables\n", + "p1=145#...........#injection pressure in bar\n", + "p2=235#.........#Injection pressure in bar (2nd case)\n", + "t1=16#.............#spray penetration time in milliseconds\n", + "s1=22#................#spray penetration length in cm\n", + "s2=22#.................#spray penetration length in cm (2nd case)\n", + "pc=30#.................#combustion chamber pressure in bar\n", + "#calculations\n", + "delp1=p1-pc#\n", + "delp2=p2-pc#\n", + "t2=(s2/s1)*t1*sqrt(delp1/delp2)#..........#Spray time in seconds for 2nd case\n", + "#Given that s=t*sqrt(delp)\n", + "print \"The time required for spray penetration at an injection pressure of 235 bar = %0.2f milliseconds:\"%t2" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER15_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER15_3.ipynb new file mode 100644 index 00000000..2a5367b8 --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER15_3.ipynb @@ -0,0 +1,76 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 15 - Engine Cooling" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 15.1 PAGE 499" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mass of cooling water required = 3668.50 kg/h for petrol engine \n", + "The mass of cooling water required = 2674.95 kg/h for diesel engine \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "BP=90#.................#Brake Power in kW\n", + "deltw=27#.................#Raise in temperature of water \n", + "etaP=0.25#...................#Efficiency of petrol engine\n", + "etaD=0.3#....................#Efficiency od diesel engine\n", + "Pec=32#......................#Percentage of energy going to coolant in petrol engine\n", + "Dec=28#......................#Percentage of energy going to coolant in diesel engine\n", + "cp=4.187#..........#specific heat of water at constant pressure\n", + "#Calculations\n", + "hsP = BP/etaP#............#Heat supplied in kW or kJ/s\n", + "ecP=hsP*(Pec/100)#.............#Energy going to cooling water in kg/s\n", + "mwP=ecP/(cp*deltw)#.............#Mass of cooling water required\n", + "hsD = BP/etaD#............#Heat supplied in kW or kJ/s\n", + "ecD=hsD*(Dec/100)#.............#Energy going to cooling water in kg/s\n", + "mwD=ecD/(cp*deltw)#.............#Mass of cooling water required\n", + "print \"The mass of cooling water required = %0.2f kg/h for petrol engine \"%(mwP*3600)\n", + "print \"The mass of cooling water required = %0.2f kg/h for diesel engine \"%(mwD*3600)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER16_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER16_3.ipynb new file mode 100644 index 00000000..d86cda95 --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER16_3.ipynb @@ -0,0 +1,306 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 16 - Supercharging of IC Engines" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 16.1 PAGE 525" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power required to run the supercharger = 90.47 kW \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "pwu=735#............#Power developed by naturally aspirated engine in kW\n", + "afru=12.8#.............#Air fuel ratio for naturally aspirated engine\n", + "bsfc=0.350#......#Brake specific fuel consumption in kg/kWh\n", + "metau=0.86#...........#Mechanical efficiency of naturally aspirated engine\n", + "pi=730#...........#Inlet pressure in mm of Hg absolute\n", + "tm=325#...........#Mixture temperature in Kelvin\n", + "pr=1.6#.............#Pressure ratio of supercharged engine\n", + "etaa=0.7#.............#Adiabatic efficiency of supercharged engine\n", + "metas=0.9#..............#Mechanical efficiency of supercharged engine\n", + "afrs=12.8#.............#Air fuel ratio for supercharged engine\n", + "rhohg=13600#.............#Density of mercury in kg/m**3\n", + "R=0.287#...................#Gas constant in kJ/kgK\n", + "ga=1.4#................#Degree of freedom for gas\n", + "cp=1.005#..................#Specific heat of the fuel\n", + "g=9.81#................#Acceleration due to gravity in m/s**2\n", + "#calculations\n", + "t2=tm*(pr)**((ga-1)/ga)#..............#Ideal temperature for the supercharged engine\n", + "t2a=tm+(t2-tm)/etaa#................#Actual temperature for the supercharged engine\n", + "wa=cp*(t2a-tm)#.....................#Work of the supercharger\n", + "wsup=cp*(t2a-tm)/metas#..............#Work required to drive the supercharger in kJ/kg of air\n", + "#When unsupercharged\n", + "p1=(pi/1000)*((g*rhohg)/1000)#..............#Inlet pressure in kN/m**2\n", + "rhounsup=p1/(R*tm)#\n", + "maunsup=(bsfc*pwu*afrs)/3600#...................#Air consumption in kg/s for unsupercharged engine\n", + "#When supercharged\n", + "rhosup=(pr*p1)/(R*t2a)#\n", + "masup=maunsup*(rhosup/rhounsup)#..................#Air consumption in kg/s\n", + "Psup=masup*wsup#...............#Power required to run the supercharger in kW\n", + "print \"The Power required to run the supercharger = %0.2f kW \"%Psup" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 16.2 PAGE 526" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Engin Capacity = 0.02 m**3 \n", + "The Brake mean effective pressure = 8.34 bar\n", + "The Increase of pressure required = 0.47 bar\n" + ] + } + ], + "source": [ + "# Initialisation of Variables\n", + "p1=1.0132#..............#Mean pressure at sea level in bar\n", + "t1=283#................#Mean temperature at sea level in Kelvin\n", + "BP=260#....................#Brake Power output in kW\n", + "etaV=0.78#..................#Volumetric efficiency at sea level free air condition\n", + "sfc=0.247#............#Specific Fuel consumption in kg/kW.h\n", + "afr=17#...................#Air fuel ratio\n", + "N=1500#...................#Engine rpm\n", + "at=2700#.................#Altitude in mts\n", + "p2=0.72#................#Pressure in bar at the given altitude\n", + "Psup=0.08#.................#8% power of engine is taken by the supercharger\n", + "R=287#...................#Gas constant in J/kgK\n", + "t2=32+273#..............#Temperature in Kelvin at the given altitude\n", + "#calculations\n", + "mf=(sfc*BP)/60#.............#Fuel consumption in kg/min\n", + "ma = mf*afr#..................#Air consumption in ig/min\n", + "acps = ma/(N/2)#............#Air consumption per stroke in kg\n", + "Vs=(acps*R*t1)/(etaV*p1*10**5)#................#Engine capacity in m**3\n", + "print \"The Engin Capacity = %0.2f m**3 \"%Vs\n", + "pmb=(BP*6)/(Vs*10*(N/2))#........#Brake Mean Effective Pressure in bar\n", + "print \"The Brake mean effective pressure = %0.2f bar\"%pmb\n", + "gp=BP/(1-Psup)#.................#Gross power produced by supercharged engine in kW\n", + "masup=ma*gp/BP#......................#Mass of air required for supercharged engine in kg\n", + "matc=masup/(N/2)#..............#Mass of air taken per cycle\n", + "pressure=(matc*R*t2)/(etaV*10**5*Vs)#\n", + "print \"The Increase of pressure required = %0.2f bar\"%(pressure-p2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 16.3 PAGE 527" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Net increase in Brake Power = 17.60 kW \n" + ] + } + ], + "source": [ + "# Initialisation of Variables\n", + "ec=3600*10**(-6)#.............#Engine capacity in m**3\n", + "pw=13#...............#Power developed in kW per m**3 of free air induced per minute\n", + "etaV=0.82#............#Volumetric Efficiency\n", + "N=3000#................#Engine rpm\n", + "p1=1.0132#...........................#Initial Air pressure in bar\n", + "t1=298#........................#Initial Temperature in Kelvin\n", + "pr=1.8#.....................#Pressure ratio in rotary compressor\n", + "etaC=0.75#.................#Isentropic efficiency of compressor\n", + "etaM=0.8#....................#Mechanical efficiency\n", + "ga=1.4#.....................#Degree of freedom for the gas\n", + "td=4#.......................#The amount by which the temperature is kess than delivery temperature from compressor\n", + "R=287#......................#Gas constant in J/kg.K\n", + "cp=1.005#.....................#Specific heat capacity\n", + "#Calculations\n", + "Vs=(ec*N)/2#....................#Swept volume in m**3/min\n", + "Vu=Vs*etaV#....................#Unsupercharged volume induced per min\n", + "rcdp=pr*p1#........#Rotary compressor delivery pressure\n", + "t2=t1*(pr)**((ga-1)/ga)#..............#Ideal temperature for the supercharged engine\n", + "t2a=t1+(t2-t1)/etaC#................#Actual temperature for the supercharged engine\n", + "ta=t2a-td#............................#Temperature of air at intake to the engine cylinder\n", + "V1=(rcdp*Vs*t1)/(p1*ta)#.................#Equivalent volume at 1.0132 bar and 298 K\n", + "Vinc=V1-Vs#...........................#Increase in induced Volume of air in m**3/min\n", + "ipincai=pw*Vinc#.......................#Increase in IP from air induced in kW\n", + "ipinciip=((rcdp-p1)*10**5*Vs)/(60*1000)#...........#Increase in IP due to increased induction pressure kW\n", + "ipinctot=ipincai+ipinciip#...............#Total increase in Input Power in kW\n", + "bpinc=ipinctot*etaM#....................#Increase in Brake Power of the engine in kW\n", + "ma=(rcdp*10**5*Vs)/(60*R*ta)#...................#Mass of air delivered by the compressor kg/s\n", + "pc=(ma*cp*(t2a-t1))/etaM#....................#Power required by the compressor\n", + "bpincnet=bpinc-pc#..........................#Net Increase in BP\n", + "print \"The Net increase in Brake Power = %0.2f kW \"%bpincnet" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 16.4 PAGE 528" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Engine Capacity = 0.02 m**3:\n", + "The Brake mean effective pressure = 8.41 bar\n", + "The Increase of pressure required = 0.47 bar\n" + ] + } + ], + "source": [ + "# Initialisation of Variables\n", + "p1=1.0132#..............#Mean pressure at sea level in bar\n", + "t1=283#................#Mean temperature at sea level in Kelvin\n", + "BP=250#....................#Brake Power output in kW\n", + "etaV=0.78#..................#Volumetric efficiency at sea level free air condition\n", + "sfc=0.245#............#Specific Fuel consumption in kg/kW.h\n", + "afr=17#...................#Air fuel ratio\n", + "N=1500#...................#Engine rpm\n", + "at=2700#.................#Altitude in mts\n", + "p2=0.72#................#Pressure in bar at the given altitude\n", + "Psup=0.08#.................#8% power of engine is taken by the supercharger\n", + "R=287#...................#Gas constant in J/kgK\n", + "t2=32+273#..............#Temperature in Kelvin at the given altitude\n", + "#calculations\n", + "mf=(sfc*BP)/60#.............#Fuel consumption in kg/min\n", + "ma = mf*afr#..................#Air consumption in ig/min\n", + "acps = ma/(N/2)#............#Air consumption per stroke in kg\n", + "Vs=(acps*R*t1)/(etaV*p1*10**5)#................#Engine capacity in m**3\n", + "print \"The Engine Capacity = %0.2f m**3:\"%Vs\n", + "pmb=(BP*6)/(Vs*10*(N/2))#........#Brake Mean Effective Pressure in bar\n", + "print \"The Brake mean effective pressure = %0.2f bar\"%pmb\n", + "gp=BP/(1-Psup)#.................#Gross power produced by supercharged engine in kW\n", + "masup=ma*gp/BP#......................#Mass of air required for supercharged engine in kg\n", + "matc=masup/(N/2)#..............#Mass of air taken per cycle\n", + "pressure=(matc*R*t2)/(etaV*10**5*Vs)#\n", + "print \"The Increase of pressure required = %0.2f bar\"%(pressure-p2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 16.5 PAGE 529" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The indicated mean effective pressure = 4.55 bar \n", + "The total aspirated air mass flow into the engine = 6.73 kg/min \n", + "Air flow into the compressor = 16.79 kg/min\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "t1=298#.................#Temperature of the air while entering the compressor in Kelvin\n", + "qrej=1210#..............#Amount of heat rejected in cooler in kJ/min\n", + "t2=273+65#...............#Temperature of the air leaving the cooler in Kelvin\n", + "p2=1.75#.................#Pressure of the air leaving the cooler in bar\n", + "n=6#.....................#No of cylinders\n", + "d=0.1#...................#Bore of the cylinder in m\n", + "l=0.11#...................#Stroke of the cylinder in m\n", + "etaV=0.72#................#volumetric efficiency\n", + "N=2000#...............#Engine rpm\n", + "Tout=150#..................#Torque Output in Nm\n", + "etaM=0.8#..................#Mechanical efficiency\n", + "R=287#.......................#Gas constant for air in J/kgK\n", + "cp=1.005#...................#Specific capacity of air\n", + "#calculations\n", + "BP=(2*pi*N*Tout)/(60*1000)#...........#Brake power in kW\n", + "IP=BP/etaM#..........#Input Power in kW\n", + "Vc=(pi/4)*d*d*l#...................#Cylinder Volume in m**3\n", + "pmi=(6*IP)/(n*Vc*(N/2)*10)#................#Indicated mean effective pressure\n", + "print \"The indicated mean effective pressure = %0.2f bar \"%(pmi)\n", + "Vs=Vc*6*(N/2)#.........................#Engine Swept Volume in m**3/min\n", + "Vaa=Vs*etaV#..........................#Aspirated volume of air into engine in m**3/min\n", + "maa=(p2*10**5*Vaa)/(R*t2)#..............#Aspirated air mass flow into the engine in kg/min\n", + "print \"The total aspirated air mass flow into the engine = %0.2f kg/min \"%maa\n", + "t2a=((((BP/cp)/(qrej/(60*cp)))*t2)-t1)/(((BP/cp)/(qrej/(60*cp)))-1)#\n", + "mc=((BP/cp)/(t2a-t1))*60#........................#Air flow into the compressor in kg/min\n", + "print \"Air flow into the compressor = %0.2f kg/min\"%mc" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER17_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER17_3.ipynb new file mode 100644 index 00000000..af20eb27 --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER17_3.ipynb @@ -0,0 +1,2535 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 17 - Testing and performance of I.C. engines" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.1 PAGE 570" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power developed = 13.30 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "Pmi=6#.....................#Mean effective pressure in bar\n", + "N=1000#....................#Engine rpm\n", + "d=0.11#.....................#Diameter of piston in m\n", + "l=0.14#.....................#Stroke length in m\n", + "n=1#........................#No of cylinders\n", + "k=1#........................#k=1 for two stroke engine\n", + "#Calculations\n", + "V=l*(pi/4)*d*d#.............#Volume of the cylinder in m**3\n", + "IP=(n*Pmi*V*k*10*N)/6#.........#Indicated Power developed in kW\n", + "print \"Indicated power developed = %0.2f kW\"%IP" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.2 PAGE 571" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The bore of the engine = 87.96 mm\n", + "The stroke of the engine = 131.94 mm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=4#........................#No of cylinders\n", + "P=14.7#....................#Power developed in kW\n", + "N=1000#....................#Engine speed in rpm\n", + "Pmi=5.5#....................#Mean effective pressure in bar\n", + "lbyd=1.5#...................#Ratio of stroke to bore\n", + "k=0.5#.......................#For four stroke engine\n", + "#Calculations\n", + "d=((P*6)/(n*Pmi*N*k*10*(pi/4)*lbyd))**(1/3)#......................#Calculation of bore in m\n", + "l=lbyd*d#................................#Calculation of stroke in m\n", + "print \"The bore of the engine = %0.2f mm\"%(d*1000)\n", + "print \"The stroke of the engine = %0.2f mm\"%(l*1000)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.3 PAGE 572" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Brake Power = 2.51 KW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "Db=0.6#.....................#Diameter of the brake wheel in m\n", + "d=0.026#......................#Diameter of the rope in m\n", + "W=200#.......................#Dead load on the brake in N\n", + "S=30#......................#Spring balance reading in N\n", + "N=450#......................#Engine speed in rpm\n", + "#Calculations\n", + "BP=((W-S)*pi*(Db+d)*N)/(60*1000)#...............#Brake Power in KW\n", + "print \"Brake Power = %0.2f KW\"%(BP)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.4 PAGE 572" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Since it is given that the stroke is equal to bore, their value is = 138.67 mm\n", + "The engine displacement = 0.00 m**3\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=4#........................#No of cylinders\n", + "k=0.5#......................#For four stroke engine\n", + "Tb=160#.....................#Max brake torque in Nm\n", + "N=3000#......................#Engine rpm\n", + "Pm=9.6#....................#Brake mean effective pressure in bar\n", + "#Calculations\n", + "D=((2*pi*N*Tb*6)/(60*1000*Pm*(pi/4)*N*k*10))**(1/3)#.....................#Bore of engine in m\n", + "L=D#...................#Given that the stroke is equal to bore\n", + "Disp=(pi/4)*D*D*L#....................................#Displacement in m**3\n", + "print \"Since it is given that the stroke is equal to bore, their value is = %0.2f mm\"%(D*1000)\n", + "print \"The engine displacement = %0.2f m**3\"%(Disp)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.5 PAGE 572" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Since the stroke and bore are equal, their value is = 290.90 mm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=6#.....................#No of cylinders\n", + "Pmb=6#....................#Brake mean effective pressure in bar\n", + "N=1000#..................#Engine rpm\n", + "k=0.5#.......................#For four stroke engine\n", + "Wce=820#.................#Work during compression and expansion in kW\n", + "Wie=50#...................#Work during intake and exhaust in kW\n", + "f=150#......................#Rubbing friction in engiine in kW\n", + "WnetT=40#...................#Net work done by turbine in kW\n", + "#Calculations\n", + "BP=Wce-(Wie+f+WnetT)#.....................#Net work available or brake power in kW\n", + "D=((BP*6)/(n*Pmb*(pi/4)*N*k*10))**(1/3)#......................#Bore of engine in m\n", + "L=D#.........................................#Given that bore is equal to stroke\n", + "print \"Since the stroke and bore are equal, their value is = %0.2f mm\"%(D*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.6 PAGE 573" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ratio of the power input of the engine with methane fuel to that with octane : 1.12\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "Cm=50150#............................#Heating value of methane in kJ/kg\n", + "Co=44880#............................#Heating value of octane in kJ/kg\n", + "#Calculations\n", + "#Since Energy supplied is proportional to mass of fuel supplied time calorific value of the fuel supplied\n", + "ratioP=Cm/Co#.........................#Ratio of the power input of the engine with methane fuel to that with octane\n", + "print \"Ratio of the power input of the engine with methane fuel to that with octane : %0.2f\"%ratioP" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.7 PAGE 573" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Brake power = 250.00 kW\n", + "Brake thermal efficiency = 33.07 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "N=2000#...........................#Engine rpm\n", + "k=0.5#..............................#Four stroke engine\n", + "Disp=0.025#........................#Engine displacement in m**3\n", + "Pmb=6#..............................#Brake mean effective pressure in bar\n", + "mf=0.018#............................#Fuel consumption in kg/s\n", + "Cf=42000#............................#Calorific value of fuel in kJ/kg\n", + "#Calcuations \n", + "BP=(Pmb*Disp*N*k*10)/(6)#................#Brake power in kW\n", + "etaBT=BP/(mf*Cf)#.................#Brake thermal efficiency\n", + "print \"The Brake power = %0.2f kW\"%(BP)\n", + "print \"Brake thermal efficiency = %0.2f %%\"%(etaBT*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.8 PAGE 574" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Brake power developed by engine = 9.16 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "T=175#.......................#Torque due to brake load in Nm\n", + "N=500#.........................#Engine speed in rpm\n", + "#calcuations\n", + "BP=(2*pi*N*T)/(60*1000)#.......................#Brake power developed by engine in kW\n", + "print \"Brake power developed by engine = %0.2f kW\"%(BP)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.9 PAGE 575" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 47.71 kW\n", + "Brake power = 42.88 kW\n", + "Mechanical efficiency = 89.88 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "D=0.3#.............................#Bore of engine cylinder in m\n", + "L=0.45#............................#Stroke of engine cylinder in m\n", + "N=300#.............................#Engine rpm\n", + "Pmi=6#.............................#Indicated mean effective pressure in bar\n", + "Nbl=1.5#...........................#Net brake load in kN\n", + "Db=1.8#............................#Diameter of brake drum in m\n", + "d=0.02#............................#Brake rope diameter\n", + "k=0.5#.............................#Four stroke engine\n", + "n=1#...............................#No of cylinders\n", + "#Calculations\n", + "IP=(n*Pmi*L*(pi/4)*D*D*N*k*10)/6#......................#Indicated power in kW\n", + "BP=(Nbl*pi*(Db+d)*N)/60#...............................#Brake power in kW\n", + "etam=BP/IP#.............................................#Mechanical efficiency\n", + "print \"Indicated power = %0.2f kW\"%(IP)\n", + "print \"Brake power = %0.2f kW\"%(BP)\n", + "print \"Mechanical efficiency = %0.2f %%\"%(etam*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.10 PAGE 576" + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Brake specific fuel consumption = 0.27 kJ/hW-h\n", + "Brake thermal efficiency : 30.78 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "Db=0.7#.............................#Diameter of brake pulley in m\n", + "d=0.025#............................#Diameter of the rope in m\n", + "W=50#...............................#Load on the tight side of the rope in kg\n", + "S=50#...............................#Spring balance reading in N\n", + "N=900#..............................#Engine rpm\n", + "mf=4#...............................#Rate of fuel consumption in kg/h\n", + "C=44000#............................#Calorific value of fuel in kJ/kg\n", + "g=9.81#.............................#Acceleration due to gravity in m/s**2\n", + "#Calculations\n", + "BP=(((W*g)-S)*pi*(Db+d)*N)/(60*1000)#........................#Brake power in kW\n", + "bsfc=mf/BP#...................................................#Brake specific fuel consumption in kJ/hW-h\n", + "print \"Brake specific fuel consumption = %0.2f kJ/hW-h\"%bsfc\n", + "etathB=(BP*3600)/(mf*C)#.............................................#Brake thermal efficiency\n", + "print \"Brake thermal efficiency : %0.2f %%\"%(etathB*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.11 PAGE 576" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated thermal efficiency = 42.52 %\n", + "Brake power = 100.53 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=4#.......................#No of cylinders\n", + "k=0.5#.....................#Four stroke engine\n", + "r=8#.......................#Compression ratio\n", + "d=0.1#.....................#Engine bore in m\n", + "l=0.1#.....................#Engine stroke in m\n", + "etaV=0.75#.................#Volumetric efficiency\n", + "N=4800#....................#Engine rpm\n", + "afr=15#....................#Air fuel ratio\n", + "C=42000000#................#Calorific value of fuel\n", + "rhoa=1.12#.................#Atmospheric density in kg/m**3\n", + "Pmi=10#....................#Mean effective pressure in bar\n", + "etamech=0.8#...............#Mechanical efficiency\n", + "#Calculations \n", + "IP=(n*Pmi*l*(pi/4)*d*d*N*k*10)/6#.................#Indicated power in kW\n", + "Ac=n*(pi/4)*d*d*l*(N/2)*(etaV/60)#.....................#Air consumption in m**3/s\n", + "ma=Ac*rhoa#........................................#Mass flow of air in kg/s\n", + "mf=ma/afr#.........................................#Mass flow of fuel in kg/s\n", + "etath=(IP*1000)/(mf*C)#...................................#Indicated thermal efficiency\n", + "print \"Indicated thermal efficiency = %0.2f %%\"%(etath*100)\n", + "BP=IP*etamech#.....................................#Brake Power in kW\n", + "print \"Brake power = %0.2f kW\"%BP" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.12 PAGE 577" + ] + }, + { + "cell_type": "code", + "execution_count": 45, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volumetric efficiency = 84.48 %\n", + "Brake specific fuel consumption = 0.28 kg/kWh\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "N=1800#...................#Engine rpm\n", + "l=0.11#...................#Engine stroke in m\n", + "d=0.085#..................#Engine bore in m\n", + "ma=0.56#..................#Air flow rate in kg/min\n", + "BP=6#.....................#Brake power developed in kW\n", + "afr=20#...................#Air fuel ratio\n", + "C=42550#..................#Calorific value of fuel in kJ/kg\n", + "rhof=1.18#................#Density of fuel in kg/m**3\n", + "#calculations\n", + "V=(pi/4)*d*d*l*(N/2)#.....................#Volume displacemt in m**3/min\n", + "Ma=V*rhof#.................................#Mass of air in kg/min\n", + "etaV=ma/Ma#................................#Volumetric efficiency\n", + "fc=ma/afr#.................................#Fuel concumption\n", + "bsfc=(fc*60)/BP#...........................#Brake specific fuel consumption in kg/kWh\n", + "print \"The volumetric efficiency = %0.2f %%\"%(etaV*100)\n", + "print \"Brake specific fuel consumption = %0.2f kg/kWh\"%(bsfc)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.13 PAGE 578" + ] + }, + { + "cell_type": "code", + "execution_count": 46, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mechanical efficiency = 63.59 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "pmicover=6.5#....................#Mean effective pressure on cover side in bar\n", + "pmicrank=7#......................#Mean effective pressure on crank side in bar\n", + "D=0.2#...........................#Engine bore in m\n", + "l=0.35#..........................#Engine stroke in m\n", + "drod=0.02#.........................#Diameter of piston rod in m\n", + "W=1370#............................#Dead load on the brake in N\n", + "S=145#.............................#Spring balance reading in N\n", + "Db=1.2#............................#Brake wheel diameter in m\n", + "d=0.02#............................#Brake rope diameter in m\n", + "k=0.5#.............................#Four stroke engine\n", + "N=420#.......................#Engine rpm\n", + "#calculations\n", + "Acover=(pi/4)*D*D#.......................#Area of cylinder on the cover side in m**2\n", + "Acrank=(pi/4)*((D**2)-(drod**2))#..........#Effective area of cylinder on the crank end side in m**2\n", + "IPcover=(pmicover*l*Acover*N*k*10)/6#................#Indicated power on the cover end side in kW\n", + "IPcrank=(pmicrank*l*Acrank*N*k*10)/6#................#Indicated power on the crank end side in kW\n", + "IPtotal=IPcover+IPcrank#....................#TOtal\n", + "BP=((W-S)*pi*(Db+d)*N)/(60*1000)#...................#Brake power in kW\n", + "etamech=BP/IPtotal#..................................#Mechanical efficiency\n", + "print \"The mechanical efficiency = %0.2f %%\"%(etamech*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.14 PAGE 579" + ] + }, + { + "cell_type": "code", + "execution_count": 47, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Engine bore = 152.98 mm\n", + "Engine stroke = 229.47 mm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "BP=14.7#........................#Brake power in kW\n", + "p1=0.9#.........................#Suction pressure in bar\n", + "etamech=0.8#....................#Mechanical efficiency\n", + "r=5#............................#Compression ratio\n", + "p3=24#..........................#maximum explosion pressure in bar\n", + "N=1000#.........................#Engine rpm\n", + "rld=1.5#........................#Ratio of length and stroke\n", + "ic=1.35#........................#Index of compression curve\n", + "ie=1.3#.........................#Index of expansion curve\n", + "k=0.5#..........................#Four stroke engine\n", + "#calculations\n", + "p2=(r**ic)*p1#......................#intermediate pressure = %0.2f bar) during compression\n", + "p4=p3/(r**ie)#......................#Intermediate pressure = %0.2f bar) during expansion\n", + "pm=((((p3-r*p4)/(ie-1))-((p2-p1*r)/(ic-1)))*(10**5))/(r-1)#...........#Mean effective pressure in N/m**2\n", + "pmb=pm/100000#........................................#Mean effective pressure in bar\n", + "IP=BP/etamech#........................................#Indicated power in kW\n", + "D=((IP*6*4)/(pmb*rld*(pi)*N*k*10))**(1/3)#............#Engine bore in m\n", + "L=rld*D#..............................................#Engine stroke in m\n", + "print \"Engine bore = %0.2f mm\"%(D*1000)\n", + "print \"Engine stroke = %0.2f mm\"%(L*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.15 PAGE 580" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated thermal efficiency = 27.03 %\n", + "Indicated brake efficiency = 23.43 %\n", + "Mechanical efficiency = 86.67 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "#Initialisation of Variables\n", + "IP=30#........................#Indicated power in kW\n", + "BP=26#........................#Brake power in kW\n", + "N=1000#.......................#Engine rpm\n", + "fpbph=0.35#...................#Fuel per brake power hour in kg/B.P.h\n", + "C=43900#......................#Calorific value of fuel used in kJ/kg\n", + "#Calculations\n", + "mf=BP*fpbph#.............#Fuel consumption per hour in kg/h\n", + "etaIth=IP/((mf/3600)*C)#.................#Indicated thermal efficiency\n", + "etaBth=BP/((mf/3600)*C)#.................#Indicated brake efficiency\n", + "etamech=BP/IP#...........................#Mechanical efficiency\n", + "print \"Indicated thermal efficiency = %0.2f %%\"%(etaIth*100)\n", + "print \"Indicated brake efficiency = %0.2f %%\"%(etaBth*100)\n", + "print \"Mechanical efficiency = %0.2f %%\"%(etamech*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.16 PAGE 581" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Brake specific fuel consumption = 0.29 kg/kWh\n", + "Brake thermal efficiency = 28.62 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "#Initialisation of Variables\n", + "Db=0.75#.....................#Diameter of brake pulley in m\n", + "d=0.05#......................#Rope diameter in m\n", + "W=400#.......................#Dead load in N\n", + "S=50#........................#Spring balance reading in N\n", + "cf=4.2#......................#Consumption of fuel in kg/h\n", + "N=1000#......................#Engine rpm\n", + "C=43900#.....................#Calorific value of fuel in kJ/kg\n", + "#Calculations\n", + "BP=((W-S)*pi*(Db+d)*N)/(60*1000)#...............#Brake power in kW\n", + "bsfc=cf/BP#......................................#Brake specific fuel consumption in kg/kWh\n", + "etabth=BP/((cf/3600)*C)#.........................#Brake thermal efficiency\n", + "print \"Brake specific fuel consumption = %0.2f kg/kWh\"%(bsfc)\n", + "print \"Brake thermal efficiency = %0.2f %%\"%(etabth*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.17 PAGE 581" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The calorific value of the fuel used = 40341.03 kJ/kg\n", + "Fuel consumption = 20.52 (kg/h\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "#Initialisation of Variables\n", + "n=6#.............................#No of cylinders\n", + "D=0.09#..........................#Bore of cylinder in m\n", + "L=0.1#...........................#Stroke length in m\n", + "r=7#.............................#Compression ratio\n", + "etarel=0.55#.....................#Relative efficiency\n", + "isfc=0.3#........................#Indicated specific fuel consumption in kg/kWh\n", + "imep=8.6#........................#Indicated mean effective pressure in bar\n", + "N=2500#..........................#Engine speed\n", + "ga=1.4#..........................#Degree of freedom for air\n", + "k=0.5#...........................#Four stroke engine\n", + "#calculations\n", + "etastan=1-1/(r**(ga-1))#...................#Air standard efficiency\n", + "etath=etarel*etastan#.....................#Indicated thermal efficiency \n", + "C=3600/(etath*isfc)#.......................#Calorific value of fuel in kJ/kg\n", + "IP=(n*imep*L*D*D*(pi/4)*N*k*10)/6#................#Indicated power in kW\n", + "fc=IP*isfc#.............................#Fuel consumption in kg/h\n", + "print \"The calorific value of the fuel used = %0.2f kJ/kg\"%(C)\n", + "print \"Fuel consumption = %0.2f (kg/h\"%(fc)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.18 PAGE 582" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Engine bore = 62.04 mm\n", + "Engine stroke = 93.05 mm\n", + "Fuel consumption = 8.79 kg/h\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "BP=30#........................#Brake power in kW\n", + "pmi=8#.........................#Mean effective pressure in bar\n", + "etamech=0.8#....................#Mechanical efficiency\n", + "n=4#............................#No of cylinders\n", + "N=2500#.........................#Engine rpm\n", + "rld=1.5#........................#Ratio of length and stroke\n", + "etabth=0.28#......................#Brake thermal efficiency\n", + "k=1#..........................#Two stroke engine\n", + "C=43900#.........................#Calorific value of fuel in kJ/kg\n", + "#calculations\n", + "IP=BP/etamech#........................................#Indicated power in kW\n", + "D=((IP*6*4)/(pmi*n*rld*(pi)*N*k*10))**(1/3)#............#Engine bore in m\n", + "L=rld*D#..............................................#Engine stroke in m\n", + "print \"Engine bore = %0.2f mm\"%(D*1000)\n", + "print \"Engine stroke = %0.2f mm\"%(L*1000)\n", + "mf=BP/(etabth*C)#..............................#Fuel consumption in kg/s\n", + "print \"Fuel consumption = %0.2f kg/h\"%(mf*3600)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.19 PAGE 583" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volumetric efficiency of the engine = 78.16 %\n", + "The brake thermal efficiency of the engine = 23.64 %\n", + "Brake torque = 0.23 kNm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=6#......................#No of cylinders\n", + "pdpc=700*10**(-6)#.................#Piston displacement per cylinder in m**3\n", + "P=78#............................#Power developed in kW\n", + "N=3200#.............................#Engine rpm\n", + "mf=27#.............................#Fuel consumption in kg/h\n", + "C=44000#...........................#Calorific value of fuel in kJ/kg\n", + "afr=12#..............................#Air fuel ratio\n", + "p1=0.9#..........................#Intake air pressure\n", + "pa=p1#\n", + "t1=305#...............................#Intake air temperature\n", + "ta=t1#\n", + "R=0.287#.....................#Gas constant in kJ/kgK\n", + "#Calculations\n", + "ma=afr*mf#............................#maaa of air in kg/h\n", + "Va=(ma*R*t1)/(p1*100)#.............#Volume of air intake in m**3/h\n", + "Vs=pdpc*n*(N/2)*60#.....................#Swept volume in m**3/h\n", + "etaV=Va/Vs#.............................#Volumetric efficiency\n", + "print \"The volumetric efficiency of the engine = %0.2f %%\"%(etaV*100)\n", + "etabt=P/(mf*(C/3600))#...................#Brake thermal efficiency\n", + "print \"The brake thermal efficiency of the engine = %0.2f %%\"%(etabt*100)\n", + "Tb=(P*60)/(2*pi*N)#..........................#Brake torque in kNm\n", + "print \"Brake torque = %0.2f kNm\"%(Tb)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.20 PAGE 583" + ] + }, + { + "cell_type": "code", + "execution_count": 53, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average no of times each cylinder misfires in one min : 2.00\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi,floor\n", + "# Initialisation of Variables\n", + "n=6#.......................#No of cylinders\n", + "Vs=1.75*10**(-3)#..............#Stroke volume in m**3\n", + "IP=26.3#.....................#Indicated power in kW\n", + "Ne=504#.......................#Expected Engine rpm\n", + "Pmi=6#........................#Mean effective pressure in bar\n", + "k=0.5#.........................#Four stroke engine\n", + "#Calculations\n", + "Na=floor((IP*6)/(n*Pmi*Vs*k*10))#.......................#Actual Engine rpm\n", + "af=(Na*n)/2#.......................#Actual no of fires in min\n", + "ef=(Ne*n)/2#.......................#Expected no of fires in min\n", + "Nm=ef-af#........................#No of misfires/min\n", + "nm=Nm/n#....................#Average no of times each cylinder misfires in one min\n", + "print \"Average no of times each cylinder misfires in one min : %0.2f\"%nm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.21 PAGE 584" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 14.24 kW\n", + "Indicated thermal efficiency = 34.63 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=4#.......................#No of cylinders\n", + "D=0.075#.......................#Engine bore in m\n", + "L=0.09#........................#Engine length in m\n", + "err=39/8#.......................#Engine to rear axle ratio\n", + "Dw=0.65#.........................#Wheel diameter in m\n", + "pc=0.227#.......................#Petrol consumption in kg\n", + "pmi=5.625#.........................#Mean effective pressure in bar\n", + "C=43470#..............................#Calorific value of petrol in kJ/kg\n", + "k=0.5#.............................#Four stroke engine\n", + "sc=48#............................#Speed of the car in km/h\n", + "d=3.2#.............................#Distance covered by car in km\n", + "#Calculations\n", + "sc1=sc*(1000/60)#...................#Speed of the car in m/min\n", + "Nt=sc1/(pi*Dw)#......................#Revolutions made by tire per min\n", + "Ne=Nt*err#............................#Speed of engine shaft\n", + "IP=(n*pmi*L*(pi/4)*D*D*Ne*k*10)/6#........#Indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%(IP)\n", + "sc2=sc/60#.......................#Speed of the car in km/min\n", + "t=d/sc2#..........................#Time for covering 3.2 km in min\n", + "fc=pc/(t*60)#.....................#Fuel consumed per second in kg\n", + "etait=IP/(fc*C)#...............#Indicated thermal efficiency\n", + "print \"Indicated thermal efficiency = %0.2f %%\"%(etait*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.22 PAGE 584" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 26.59 kW\n", + "Brake power = 21.21 kW\n", + "Mechanical efficiency = 79.75 %\n", + "Indicated thermal efficiency = 21.61 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=1#.................#No of cylinders\n", + "D=0.25#................#Engine bore in m\n", + "L=0.4#.................#Engine stroke in m\n", + "pmg=7#.................#Gross mean effective pressure in bar\n", + "pmp=0.5#...............#Pumping mean effective pressure in bar\n", + "N=250#..................#Engine rpm\n", + "Db=1.5#................#Effective diameter of the brake in m\n", + "Nl=1080#..............#Net load on the brake in N\n", + "fh=10#.................#Fuel used per hour in kg\n", + "C=44300#...............#Calorific value of fuel in kJ/kg\n", + "k=0.5#.................#Four stroke engine\n", + "#Calculations\n", + "mf=fh/3600#.........................#Fuel used per second in kg\n", + "pm=pmg-pmp#.......................#Net pressure\n", + "IP=(n*pm*L*(pi/4)*D*D*N*k*10)/6#..........#/Indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%(IP)\n", + "BP=((Nl)*pi*Db*N)/(60*1000)#...............#Brake power in kW\n", + "print \"Brake power = %0.2f kW\"%(BP)\n", + "etamech=BP/IP#...........................#Mechanical efficiency\n", + "print \"Mechanical efficiency = %0.2f %%\"%(etamech*100)\n", + "etath=IP/(mf*C)#.........................#Indicated thermal efficiency\n", + "print \"Indicated thermal efficiency = %0.2f %%\"%(etath*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.23 PAGE 585" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Brake mean effective pressure = 553.38 bar\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "etabth=0.3#....................#Brake thermal efficiency\n", + "afrw=20#........................#Air fuel ratio by weight\n", + "C=41800#.........................#Calorific value of fuel used in kJ/kg\n", + "R=287#........................#Gas constant in J/kg\n", + "#Calculations\n", + "Wp=etabth*C#...................#Work produced per kg of fuel in kJ\n", + "p1=1.0132#\n", + "t=273+15#............#STP conditions in bar and Kelvin\n", + "V=(afrw*t*R)/(p1*10**5)#.......#Volume of air used in m**3\n", + "pmb=(Wp*1000)/(V*10**5)#........#Brake mean effective pressure in bar\n", + "print \"Brake mean effective pressure = %0.2f bar\"%(pmb)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.24 PAGE 585" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volumetric efficiency = 85.67 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "v1=0.216#.....................#Gas consumption in m**3/min\n", + "pw=75#........................#Pressure of gas in mm of water\n", + "t1=290#......................#Temperature of gas in K\n", + "ac=2.84#....................#Air consumption in kg/min\n", + "br=745#......................#Barometer reading in m of Hg\n", + "D=0.25#.....................#Engine bore in m\n", + "L=0.475#......................#Engine stroke in m\n", + "N=240#........................#Engine rpm\n", + "R=287#......................#Gas constant for air in J/kgK\n", + "#Calculations\n", + "p1=br+(pw/13.6)#...................#Pressure of gas in mm of mercury\n", + "p2=760\n", + "t2=273#.....................#NTP conditions in mm of Hg and Kelvin\n", + "v2=(p1*v1*t2)/(t1*p2)#...............#Volume of gas used at NTP in m**3\n", + "gs=v2/(N/2)#.........................#Gas used per stroke in m**3\n", + "v=(ac*R*t2)/(1.0132*10**5)#...........#Volume occupied by air at NTP in m**3/min\n", + "aps=v/(N/2)#...........................#Air used per stroke\n", + "Va=gs+aps#.....................#Actual volume of mixture in m**3 drawn per stroke at NTP\n", + "Vs=(pi/4)*D*D*L#...............#Swept volume in mm**3\n", + "etaV=(Va/Vs)#...................#Volumetric efficiency\n", + "print \"Volumetric efficiency = %0.2f %%\"%(etaV*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.25 PAGE 586" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 26.59 kW\n", + "Brake power = 22.87 kW\n", + "Mechanical efficiency = 86.00 %\n", + "Indicated thermal efficiency = 22.79 %\n", + "Relative efficiency = 43.24 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "t=1#...................#Duration of trial in hrs\n", + "Rev=14000#.............#Revolutions\n", + "nmc=500#...............#Number of missed cycles\n", + "bl=1470#................#Net Brake load in N\n", + "mep=7.5#................#Mean effective pressure in bar\n", + "gc=20000#...............#Gas consumption in litres\n", + "lcv=21#.................#LCV of gas at supply condition in kJ/litre\n", + "D=0.25#.................#Engine bore in m\n", + "L=0.4#.................#Engine stroke in m\n", + "r=6.5#..................#Compression ratio\n", + "n=1#......................#No of cylinders\n", + "Cb=4#......................#Effective brake Circumference \n", + "k=0.5#....................#Four stroke engine\n", + "ga=1.4#......................#Degree of freedom\n", + "#Calculations\n", + "N=Rev/60#..............#Engine rpm\n", + "Vg=gc/3600#.............#Fuel consumption in litres/s\n", + "Na=((Rev/2)-nmc)/60#................#Working cycles per min\n", + "IP=(n*mep*L*(pi/4)*D*D*Na*10)/6#............#indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%(IP)\n", + "BP=((bl)*Cb*N)/(60*1000)#...............#Brake power in kW\n", + "print \"Brake power = %0.2f kW\"%(BP)\n", + "etamech=BP/IP#...........................#Mechanical efficiency\n", + "print \"Mechanical efficiency = %0.2f %%\"%(etamech*100)\n", + "etath=IP/(Vg*lcv)#.........................#Indicated thermal efficiency\n", + "print \"Indicated thermal efficiency = %0.2f %%\"%(etath*100)\n", + "etast=1-(1/r**(ga-1))#............#Air standard efficiency \n", + "etarel=etath/etast#............#Relative efficiency\n", + "print \"Relative efficiency = %0.2f %%\"%(etarel*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.26 PAGE 587" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The compression ratio : 5.58\n", + "Thermal efficiency = 19.89 %\n", + "Gas consumption = 0.96 m**3/IP hour\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=1.3#...................#Index of compression\n", + "pa=1.4#\n", + "pb=3.6#\n", + "posa=(1/4)#..........#Point a - the position 1/4 of the stroke\n", + "posb=(3/4)#..........#Point b - the position 3/4 of the stroke\n", + "ga=1.4#...............#Degree of freedom for gas\n", + "etarel=0.4#...................#Relative efficiency\n", + "C=18800#....................#Calorific value of fuel in kJ/m**3\n", + "#Calculations\n", + "r=1+((((pb/pa)**(1/n))-1)/(posb-(((pb/pa)**(1/n))*(posa))))#.........#Compression ratio\n", + "print \"The compression ratio : %0.2f\"%r\n", + "etast=1-(1/r**(ga-1))#............#Air standard efficiency \n", + "etath=etarel*etast#............#Thermal efficiency\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)\n", + "v=1/(etath*C)#...............#Gas consumption per IP sec\n", + "print \"Gas consumption = %0.2f m**3/IP hour\"%(v*3600)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.27 PAGE 587" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mean effective pressure developed = 6.54 bar\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=6#.....................#No of cylinders\n", + "r=5#..................#Compression ratio\n", + "Vc=0.000115#................#Clearance volume of each cylinder in m**3\n", + "fc=10.5#.....................#Fuel consumed in kg/h\n", + "C=41800#......................#Calorific value of fuel in kJ/kg\n", + "N=2500#.......................#Engine speed in rpm\n", + "er=0.65#.......................#Efficiency ratio\n", + "ga=1.4#........................#Degree of freedom\n", + "#calculations\n", + "etast=1-(1/r**(ga-1))#...............................#Air standard efficiency\n", + "etath=etast*er#.................................#Thermal efficiency\n", + "IP=etath*(fc/3600)*C#..........................#Indicated power in kW\n", + "Wnet=(IP*(10**3)*60)/(n*(N/2))#..............#Net work froom one cycle per cylinder in N-m\n", + "Vs=(r-1)*Vc#......................#Swept volume in m**3\n", + "pm=Wnet/(Vs*10**5)#...................#Mean effective pressure developed\n", + "print \"Mean effective pressure developed = %0.2f bar\"%(pm)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.28 PAGE 588" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mean effective pressure = 8.11 bar\n", + "Brake mean effective pressure = 6.16 bar\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "D=0.2#.................#Engine bore in m\n", + "L=0.25#...............#Engine stroke in m\n", + "n=2#......................#No of cylinders\n", + "r=13#......................#Compression ratio\n", + "fc=14#..................#Fuel consumption in kg/h\n", + "N=300#....................#Engine rpm\n", + "etarel=0.65#..............#Relative efficiency\n", + "etamech=0.76#.............#Mechanical efficiency\n", + "co=0.05#.....................#Cut off of the stroke\n", + "C=41800#.....................#Calorific value of fuel in kJ/kg\n", + "k=1#........................#Two stroke engine\n", + "ga=1.4#.......................#Degree of freedom\n", + "#calculations\n", + "rho=1+(co*(r-1))#\n", + "etast=1-((1/(r**(ga-1)))*(1/ga)*((rho**ga)-1)*(1/(rho-1)))#............#Air standard efficiency\n", + "etath=etarel*etast#........................#Thermal efficiency\n", + "IP=etath*(fc/3600)*C#........................#Indicated power in kW\n", + "BP=etamech*IP#................................#Brake power in kW\n", + "pmi=(6*IP)/(n*N*L*(pi/4)*D*D*k*10)#............#mean effective pressure in bar\n", + "print \"Mean effective pressure = %0.2f bar\"%(pmi)\n", + "pmb=pmi*etamech#...........................#Brake mean effective pressure in bar\n", + "print \"Brake mean effective pressure = %0.2f bar\"%(pmb)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.29 PAGE 589" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The compression ratio : 6.26\n", + "Indicated thermal efficiency : 36.40 %\n", + "Brake specific fuel consumption = 0.27 kg/kWs\n", + "Engine bore = 108.20 mm\n", + "Engine stroke = 135.25 mm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=4#.................#No of cylinders\n", + "C=45200#..................#calorific value of fuel in kJ/kg\n", + "etamech=0.82#...............#Mechanical efficiency\n", + "etarel=0.7#.................#Relative efficiency\n", + "etast=0.52#...............#Air standard efficiency\n", + "etav=0.78#...............#Volumetric efficiency\n", + "sbr=1.25#...................#Stroke bore ratio\n", + "N=2400#...................#Engine rpm\n", + "p=1#.......................#Suction pressure in bar\n", + "t=298#....................#Suction temperature in bar\n", + "BP=72#...................#Brake power in kW\n", + "ga=1.4#......................#Degree of freedom\n", + "afr=16#.................#Air fuel ratio\n", + "R=287#.......................#Gas constant in J/kg\n", + "#calculations\n", + "r=(1/(1-etast))**(1/(ga-1))#............#Compression ratio\n", + "print \"The compression ratio : %0.2f\"%r\n", + "etath=etast*etarel#......................#Indicated thermal efficiency\n", + "print \"Indicated thermal efficiency : %0.2f %%\"%(etath*100)\n", + "IP=BP/etamech#....................#Indicated power in kW\n", + "mf=IP/(etath*C)#......................#Fuel consumption in kg/s\n", + "bsfc=mf/BP#......................#Brake specific fuel consumption in kg/kWs\n", + "print \"Brake specific fuel consumption = %0.2f kg/kWs\"%(bsfc*3600)\n", + "mafm=afr+1#......................#Mass of air fuel mixture in kg/kg of fuel\n", + "mafm1=mafm*mf#....................#Mass of air fuel mixture when mf amount of fuel is supplied to engine per second\n", + "v=(mafm1*R*t)/(p*10**5)#.......................#/Volume of air fuel mixture supplied to the engine in m**3\n", + "Vs=v/etav#..............................#Swept volume in m**3\n", + "D=((Vs)/((pi/4)*sbr*n*(N/(2*60))))**(1/3)#............#Engine bore in m\n", + "print \"Engine bore = %0.2f mm\"%(D*1000)\n", + "print \"Engine stroke = %0.2f mm\"%(D*1000*sbr)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.30 PAGE 589" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Brake power = 14.81 kW\n", + "Mechanical efficiency = 85.00 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=1#.......................#No of cylinders\n", + "D=0.18#...................#Engine bore in m\n", + "L=0.34#....................#Engine stroke in m\n", + "N=400#......................#Engine rpm\n", + "mepw=6.4#.................#Mean effective pressure of working loop in bar\n", + "mepp=0.36#..................#Mean effective pressure of pumping loop in bar\n", + "mepd=0.64#.................#Mean effective pressure (dead cycle) iin bar\n", + "fs=46#................#Firing strokes per min\n", + "#calculations\n", + "pminet=mepw-mepp#..........#Net indicated mean effective pressure in bar\n", + "dc=(N/2)-fs#...............#Dead cycles per min\n", + "IPnet=(n*pminet*(pi/4)*L*D*D*fs*4*10)/6#.............#Net indicated power output in kW\n", + "ppdc=(n*pminet*L*(pi/4)*D*D*10*dc)/6#.............#Pumping power of dead cycles in kW\n", + "FP=IPnet-ppdc#...........................#Frictional power in kW\n", + "IP=(n*pminet*L*(pi/4)*D*D*(N/2)*10)/6#...............#Indicated power in kW\n", + "BP=IP-FP#..................#Brake power in kW\n", + "print \"Brake power = %0.2f kW\"%(BP)\n", + "etamech=BP/IP#.................#Mechanical efficiency\n", + "print \"Mechanical efficiency = %0.2f %%\"%(etamech*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.31 PAGE 590" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mechanical efficiency = 81.61 %\n", + "Brake thermal efficiency = 39.54 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=1#..............#No of cylinders\n", + "B=0.32#...............#Engine bore in m\n", + "L=0.42#..............#Engine stroke in m\n", + "N=200#................#Engine rpm\n", + "Nk=90#..................#No of explosions per min\n", + "v1=11.68#............#Gas used in m**3/h\n", + "pg=170#................#Pressure of gas in mm of water \n", + "br=755#................#Barometer reading in mm of Hg\n", + "pmi=6.2#.................#Mean effective pressure in bar\n", + "C=21600#.......................#Calorific value of gas in kJ/kg\n", + "bl=2040#......................#Net load on brake in N\n", + "Db=1.2#......................#Brake drum diameter in m\n", + "t1=298#.....................#Ambient temperature in Kelvin\n", + "#Calculations\n", + "IP=(n*pmi*L*(pi/4)*B*B*Nk*10)/6#..........................#Indicated power in kW\n", + "BP=(bl*pi*Db*N)/(60*1000)#.........................#Brake power in kW\n", + "etamech=(BP/IP)#...................#Mechanical efficiency\n", + "print \"Mechanical efficiency = %0.2f %%\"%(etamech*100)\n", + "p1=br+(pg/13.6)#.................#In mm of Hg\n", + "p2=760#\n", + "t2=273#...................#NTP conditions in mm of Hg and Kelvin\n", + "v2=(p1*v1*t2)/(p2*t1)#\n", + "etabth=BP/((v2/3600)*C)#..............#Brake thermal efficiency\n", + "print \"Brake thermal efficiency = %0.2f %%\"%(etabth*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.32 PAGE 591" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volumetric efficiency = 66.45 %\n", + "Air fuel ratio : 13.20\n", + "Mean effective pressure = 7.55 bar\n", + "Relative efficiency = 53.79 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi,sqrt\n", + "# Initialisation of Variables\n", + "n=1#...................#No of cylinders\n", + "d=0.032#................#Diameter of circular orifice in m\n", + "Cd=0.62#.............#Co efficient of discharge\n", + "hw=150#.................#Pressure across orfice in mm of water\n", + "t=20+273#..............#Temperature of air in the room in Kelvin\n", + "p=1.0132#.................#Ambient pressure in bar\n", + "pd=0.00178#............#Piston displacement in m**3\n", + "R=287#....................#Gas constant in J/kg\n", + "r=6.5#..................#Compression ratio\n", + "fc=0.135#................#Fuel consumption in kg/min\n", + "C=43900#.................#Calorific value of fuel in kJ/kg\n", + "BP=28#................#Brake power in kW\n", + "N=2500#...................#Engine rpm\n", + "k=0.5#....................#Four stroke engine\n", + "g=9.81#.......................#Acceleration due to gravity in m/s**2\n", + "rhow=1000#....................#Density of water in kg/m**3\n", + "ga=1.4#........................#Degree of freedom\n", + "#calculations\n", + "mbyv=(p*10**5)/(R*t)#\n", + "pw=(hw/rhow)*rhow#....................#Pressure across orifice in kg/m**2\n", + "H=pw/mbyv#........................#Head of air column causing the flow in m\n", + "ma=Cd*(pi/4)*d*d*sqrt(2*g*H)#................#Air flow through orifice in m**3/s\n", + "maps=(ma*60)/(N/2)#........................#Air consumption per stroke\n", + "etav=maps/pd#.................#Volumetric efficiency\n", + "print \"Volumetric efficiency = %0.2f %%\"%(etav*100)\n", + "ac=ma*60*mbyv#...............#Mass of air drawn into cylinder per min in kg\n", + "afr=ac/fc#...................#Air fuel ratio\n", + "print \"Air fuel ratio : %0.2f\"%afr\n", + "pmb=(6*BP)/(n*pd*N*k*10)#...................#Mean effective pressure in bar\n", + "print \"Mean effective pressure = %0.2f bar\"%(pmb)\n", + "etast=1-(1/(r**(ga-1)))#...............#Air standard efficiency\n", + "etabth=BP/((fc/60)*C)#...............#Brake thermal efficiency\n", + "etarel=etabth/etast#.................#Relative efficiency\n", + "print \"Relative efficiency = %0.2f %%\"%(etarel*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.33 PAGE 591" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 7.50 kW\n", + "Brake power = 8.04 kW\n", + "Brake mean effective pressure = 6.00 bar\n", + "Brake specific fuel consumption = 0.35 kg/BP h\n", + "Brake thermal efficiency = 24.74 %\n", + "Indicated thermal efficiency = 30.92 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "N=400#.................#Engine rpm\n", + "n=1#....................#no of cylinders\n", + "W=370#.................#Load on the brake in N\n", + "S=50#..................#Spring balance readin in N\n", + "Db=1.2#.................#Diameter of the brake drum\n", + "mf=2.8#.................#Fuel consumption in kg/h\n", + "C=41800#..................#Calorific value of fuel in kJ/kg\n", + "D=0.16#...................#Engine bore in m\n", + "L=0.2#....................#Engine stroke in m\n", + "k=0.5#.....................#Four stroke engine\n", + "Sc=1#....................#Spring constant in bar/mm\n", + "l=40#....................#Length of diagram in mm\n", + "aic=300#.................#Area of indicator card in mm**2\n", + "#Calculations\n", + "pmi=aic*(Sc/l)#..................#Mean effective pressure in bar\n", + "IP=(n*pmi*L*(pi/4)*D*D*k*N*10)/6#..............#Indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%(pmi)\n", + "BP=((W-S)*pi*Db*N)/(60*1000)#............#Brake power in kW\n", + "print \"Brake power = %0.2f kW\"%(BP)\n", + "pmb=(BP*6)/(n*L*D*D*(pi/4)*k*N*10)#...........#Brake mean effective pressure in bar\n", + "print \"Brake mean effective pressure = %0.2f bar\"%(pmb)\n", + "bsfc=mf/BP#.................#Brake specific fuel consumption in kg/BP h\n", + "print \"Brake specific fuel consumption = %0.2f kg/BP h\"%bsfc\n", + "etabth=BP/((mf/3600)*C)#..................#Brake thermal efficiency\n", + "print \"Brake thermal efficiency = %0.2f %%\"%(etabth*100)\n", + "etaith=IP/((mf/3600)*C)#....................#Indicated thermal efficiency\n", + "print \"Indicated thermal efficiency = %0.2f %%\"%(etaith*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.34 PAGE 592" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated thermal efficiency = 27.96 %\n", + "Mean effective pressure = 8.06 bar\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "R=287#................#Gas constant in J/kg K\n", + "n=4#...................#No of cylinders\n", + "D=0.0825#..............#Engine bore in m\n", + "L=0.13#................#Engine stroke in m\n", + "BP=28#..................#Brake power in kW\n", + "N=1500#.................#Engine rpm\n", + "afrth=14.8#...............#theoretical air fuel ratio \n", + "C=45980#..................#Calorific value of fuel in kJ/kg\n", + "etamech=0.9#.............#Mechanical efficiency\n", + "ap=70#..................#Percentage of Volume of air in he cylinder\n", + "fr=20#..................#Percentage richness of the fuel\n", + "p1=1.0132#.................#Ambient pressure in bar\n", + "pc=762#...................#Pressure in the cylinder in mm of Hg \n", + "tc=273+15.5#...............#Temperature in the cylinder in Kelvin\n", + "k=0.5#..................#Four stroke engine\n", + "#Calculations\n", + "Vs=(pi/4)*D*D*L#.......................#Swept volume in m**3\n", + "va=(ap/100)*Vs#.....................#Volume of air drawn in m**3\n", + "p=(pc/760)*p1#\n", + "m=(p*(10**5)*va)/(R*tc)#...................#Mass of air per stroke per cylinder\n", + "tmau=m*(N/2)*n#...................#Theoretical mass of air used per minute in kg\n", + "tmfu=tmau/afrth#..................#Theoretical mass of fyel used per min in kg\n", + "mf=(tmfu/60)*((100+fr)/100)#...............#Mass of fuel burnt per second in kg\n", + "IP=BP/etamech#.........................#Indicated power in kW\n", + "etaith=IP/(mf*C)#.....................#Indicated thermal efficiency \n", + "print \"Indicated thermal efficiency = %0.2f %%\"%(etaith*100)\n", + "pmb=(BP*6)/(n*L*D*D*(pi/4)*N*10*k)#...............#Mean effective pressure in bar\n", + "print \"Mean effective pressure = %0.2f bar\"%(pmb)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.35 PAGE 593" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 13.44 kW\n", + "Brake power = 10.76 kW\n", + "Indicated thermal efficiency = 23.87 %\n", + "Brake thermal efficiency = 19.11 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=1#..................#No of cylinders\n", + "D=0.2#............#Engine bore in m\n", + "L=0.4#..............#Engine stroke in m\n", + "Nt=9400#...............#Total no of revolutions \n", + "Ne=4200#...............#Total no of explosions\n", + "t=40#...................#Duration of testing in min\n", + "Nk=Ne/t#...............#No of explosions\n", + "bl=540#...............#Brake load in N\n", + "Db=1.6#.................#Diameter of brake wheel in m\n", + "d=0.02#................#Diameter of rope in m\n", + "gu=8.5#..................#Gas used in m**3/sec\n", + "C=15900#...............#Calorific value of fuel in kJ/kg\n", + "Vg=(gu/(t*60))#.................#Volume of gas used in m**3/sec\n", + "aic=550#.....................#Area of indicator diagram mm**2\n", + "l=72#.......................#Length of indicator diagram in mm\n", + "s=0.8#.....................#Spring number in bar/mm\n", + "#calculations\n", + "pmi=(aic*s)/l#................#Mean effective pressure in bar\n", + "IP=(n*pmi*L*D*D*(pi/4)*Nk*10)/6#............#Indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%(IP)\n", + "BP=(bl*pi*(Db+d)*(Nt/t))/(60*1000)#...............#Brake power in kW\n", + "print \"Brake power = %0.2f kW\"%(BP)\n", + "etaith=IP/(Vg*C)#...............#Indicated thermal efficiency\n", + "print \"Indicated thermal efficiency = %0.2f %%\"%(etaith*100)\n", + "etabth=BP/(Vg*C)#...............#Brake thermal efficiency\n", + "print \"Brake thermal efficiency = %0.2f %%\"%(etabth*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.36 PAGE 594" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Brake mean effective pressure = 3.97 bar\n", + "Brake specific fuel consumption = 0.30 kg/kWh\n", + "Brake thermal efficiency = 26.38 %\n", + "Volumetric efficiency = 61.01 %\n", + "Percentage of excess air supplied : 43.76\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi,sqrt\n", + "# Initialisation of Variables\n", + "n=6#....................#No of cylinders\n", + "D=0.125#................#Engine bore in m\n", + "L=0.125#...............#Engine stroke in m\n", + "N=2400#.................#Engine rpm\n", + "W=490#...............#Load on the dynamometer in N\n", + "CD=16100#...............#Dynamometer constant\n", + "d0=0.055#...................#Air orifice diameter in m\n", + "Cd=0.66#...................#Co efficient of discharge\n", + "hw=310#.................#Head causing flow through prifice in mm of water\n", + "br=760#................#Barometer reading in mm of Hg\n", + "t=298#..................#Ambient temperature in Kelvin\n", + "fc=22.1#..................#Fuel consumption per hour in kg\n", + "C=45100#..................#Calorific value of fuel used in kJ/kg\n", + "perc=85#...................#Percentage of carbon in the fuel\n", + "perh=15#...................#Percentage of hydrogen in the fuel\n", + "p1=1.013#....................#Pressure of air at the end of suction stroke in bar\n", + "t1=298#......................#Temperature of air the the end of suction stroke in Kelvin\n", + "k=0.5#.......................#Four stroke engine\n", + "R=287#.......................#Gas constant in J/kgK\n", + "#calculations\n", + "BP=W*(N/CD)#................#Brake power in kW\n", + "pmb=(BP*6)/(L*D*D*k*10*N*n*(pi/4))#................#Brake mean effective pressure in bar\n", + "print \"Brake mean effective pressure = %0.2f bar\"%(pmb)\n", + "bsfc=fc/BP#.......................#Brake specific fuel consumption in kg/kWh\n", + "print \"Brake specific fuel consumption = %0.2f kg/kWh\"%(bsfc)\n", + "etathb=BP/((fc/3600)*C)#......................#Brake thermal efficiency\n", + "print \"Brake thermal efficiency = %0.2f %%\"%(etathb*100)\n", + "Vst=(pi/4)*D*D*L#..............#Stroke volume in m**3\n", + "Val=840*(pi/4)*d0*d0*Cd*sqrt((hw/10)/((p1*10**5)/(R*t1)))#............#Volume of air passing through orifice of air box per min\n", + "Vac=Val/n#.........................#Actual volume of air per cylinder in m**3/min\n", + "asps=Vac/(N/2)#.......................#Air supplied per stroke per cylinder in m**3\n", + "etav=asps/Vst#....................#Volumetric efficiency\n", + "print \"Volumetric efficiency = %0.2f %%\"%(etav*100)\n", + "Qa=(100/23)*(((perc/100)*(8/3))+((perh/100)*(8/1)))#.....................#Quantity of air required per kg of fuel combustion\n", + "aqas=(Val*((p1*10**5)/(R*t1))*60)/fc#....................#Actual quantity of air supplied per kg of fuel\n", + "pe=(aqas-Qa)/Qa#....................#Fraction of excess air supplied to engine\n", + "print \"Percentage of excess air supplied : %0.2f\"%(pe*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.37 PAGE 595" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mechanical efficiency = 77.28 %\n", + "Brake thermal efficiency = 23.26 %\n", + "HEAT BALANCE TABLE : \n", + "_______________________________________________________________________\n", + "Item kJ Percent\n", + "_______________________________________________________________________\n", + "Heat supplied by fuel 367840 100.00\n", + "Heat absorbed in IP 110694 30.09\n", + "Heat taken away by cooling water 59774 16.25\n", + "Heat carried away by exhaust gases 197371 53.66\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=1#......................#No of cylinders\n", + "D=0.3#....................#Engine bore in m\n", + "L=0.45#....................#Engine stroke in m\n", + "mf=8.8#...................#Fuel consumption in kg/h\n", + "C=41800#...................#Calorific value of fuel in kJ/kg\n", + "N=200#....................#Engine rpm\n", + "pmi=5.8#....................#Mean effective pressure in bar\n", + "bl=1860#....................#Brake load in N\n", + "Db=1.22#...................#Diameter of brake drum in m\n", + "k=0.5#........................#four stroke engine\n", + "mw=650#......................#Mass of cooling water in kg\n", + "cpw=4.18#....................#Specific heat capacity of water\n", + "delt=22#......................#Temperature rise\n", + "#Calculations\n", + "IP=(n*L*D*D*k*10*pmi*N*(pi/4))/6#...............#Indicated power in kW\n", + "BP=(bl*pi*Db*N)/(60*1000)#..................#Brake power in kW\n", + "etamech=BP/IP#............#Mechanical efficiency\n", + "print \"Mechanical efficiency = %0.2f %%\"%(etamech*100)\n", + "etathb=BP/((mf/3600)*C)#...................#Brake thermal efficiency\n", + "print \"Brake thermal efficiency = %0.2f %%\"%(etathb*100)\n", + "#Heat supplied\n", + "hip=IP*3600#...........#Heat equivalent of IP in kJ/h\n", + "hcw=mw*cpw*delt#..........#Heat carried away by cooling water\n", + "hf=mf*C#................#heat supplied by fuel\n", + "hex=hf-hip-hcw#..........#Heat carried by exhaust gasses\n", + "pf=100#\n", + "pip=(hip/hf)*100#\n", + "pcw=(hcw/hf)*100#\n", + "pex=(hex/hf)*100\n", + "print \"HEAT BALANCE TABLE : \"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Item kJ Percent\"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Heat supplied by fuel %d %0.2f\"%(hf,pf)\n", + "print \"Heat absorbed in IP %d %0.2f\"%(hip,pip)\n", + "print \"Heat taken away by cooling water %d %0.2f\"%(hcw,pcw)\n", + "print \"Heat carried away by exhaust gases %d %0.2f\"%(hex,pex)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.38 PAGE 596" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Brake power = 37.01 kW\n", + "Brake specific fuel consumption = 0.28 kg/kWh:\n", + "Brake thermal efficiency = 29.76 %\n", + "HEAT BALANCE TABLE:\n", + "_______________________________________________________________________\n", + "Item kJ Percent\n", + "_______________________________________________________________________\n", + "Heat supplied by fuel 7461 100.00\n", + "Heat absorbed in BP 2220 29.76\n", + "Heat taken away by cooling water 2332 31.26\n", + "Heat carried away by exhaust gases 1811 24.28\n", + "Unaccounted heat 1097 14.70\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "r=15#................#Compression ratio\n", + "n=1#...................#No of cylinders\n", + "mf=10.2#..................#Fuel consumption in kg/h\n", + "C=43890#.................#Calorific value of fuel in kJ/kg\n", + "ma=3.8#.................#Consumption of air in kg/min\n", + "N=1900#...................#Engine rpm\n", + "T=186#....................#Torque on brake drum in Nm\n", + "mw=15.5#.................#Mass of cooling water used in kg/min\n", + "delt=36#..................#temperature rise\n", + "tg=410#..................#Exhaust gas temperature in Celsius\n", + "tr=20#...................#Room temperature in Celsius\n", + "cp=1.17#.................#Specific heat capacity for exhaust gases kJ/kgK\n", + "cpw=4.18#..................#Specific heat capacity for water in kJ/kgK\n", + "#calculations\n", + "BP=(2*pi*N*T)/(60*1000)#................#Brake power in kW\n", + "print \"Brake power = %0.2f kW\"%BP\n", + "bsfc=mf/BP#.............................#Brake specific fuel consumption in kg/kWh\n", + "print \"Brake specific fuel consumption = %0.2f kg/kWh:\"%bsfc\n", + "etabth=BP/((mf/3600)*C)#....................#Brake thermal efficiency\n", + "print \"Brake thermal efficiency = %0.2f %%\"%(etabth*100)\n", + "#Heat supplied\n", + "mg=(mf/60)+ma#....................#Mass of exhaust gases in kg/min\n", + "hbp=BP*60#...........#Heat equivalent of BP in kJ/min\n", + "hcw=mw*cpw*delt#..........#Heat carried away by cooling water\n", + "hf=(mf/60)*C#................#heat supplied by fuel\n", + "hex=mg*cp*(tg-tr)#..........#Heat carried by exhaust gasses\n", + "ha=round(hf)-round(hbp+hex+hcw)#............#Unaccounted heat\n", + "pf=100#\n", + "pbp=(hbp/hf)*100#\n", + "pcw=(hcw/hf)*100#\n", + "pex=(hex/hf)*100#\n", + "pa=(ha/hf)*100#\n", + "print \"HEAT BALANCE TABLE:\"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Item kJ Percent\"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Heat supplied by fuel %d %0.2f\"%(hf,pf)\n", + "print \"Heat absorbed in BP %d %0.2f\"%(hbp,pbp)\n", + "print \"Heat taken away by cooling water %d %0.2f\"%(hcw,pcw)\n", + "print \"Heat carried away by exhaust gases %d %0.2f\"%(hex,pex)\n", + "print \"Unaccounted heat %d %0.2f\"%(ha,pa)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.39 PAGE 597" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 15.91 kW\n", + "Brake power = 11.73 kW\n", + "HEAT BALANCE TABLE\n", + "_______________________________________________________________________\n", + "Item kJ Percent\n", + "_______________________________________________________________________\n", + "Heat supplied by fuel 66728 100.000000\n", + "Heat absorbed in IP 19088 28.606158\n", + "Heat taken away by cooling water 16929 25.370159\n", + "Heat carried away by dry exhaust gases 13448 20.154897\n", + "Heat carried away by steam in exhaust gases 6606 9.901049\n", + "Unaccounted heat 10655 15.967810\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "Cpw=4.18#..............#Specific heat of water in kJ/kgK\n", + "n=1#................#No of cylinders\n", + "N=350#.......#Engine rpm\n", + "pmi=3.1#..........#Mean effective pressure in bar\n", + "bl=640#..........#Brake load in N\n", + "mf=1.52#............#Fuel consumption in kg\n", + "mw=162#..............#Mass of cooling water\n", + "tw1=30#...............#Water inlet temperature in C\n", + "tw2=55#................#Water outlet temperature in C\n", + "ma=32#..................#Mass of air used per kg of fuel in kg\n", + "tr=25#.................#Room temperature in C\n", + "tg=305#.................#Exhaust temperature in C\n", + "D=0.2#.................#Engine bore in m\n", + "L=0.28#.................#Engine stroke in m\n", + "Db=1#......................#Brake drum diameter in m\n", + "ms=1.4#......................#Mass of steam formed per kg of fuel exhaust in kg\n", + "C=43900#...................#Calorirfic value of fuel in kJ/kg\n", + "Cps=2.09#..................#Specific heat of steamm in exhaust in kJ/kgK\n", + "Cpg=1.0#...................#Specific heat of dry exhaust gases in kJ/kgK\n", + "k=1#....................#Two stroke engiine\n", + "t=20#.....................#Duration of testing in min\n", + "#Calculations\n", + "IP=(n*pmi*N*D*D*L*k*10*(pi/4))/6#...................#Indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%IP\n", + "BP=(bl*pi*Db*N)/(60*1000)#......................#Brake power in kW\n", + "print \"Brake power = %0.2f kW\"%BP\n", + "#Heat supplied\n", + "hf=mf*C#................#heat supplied by fuel\n", + "hip=IP*60*t#...........#Heat equivalent of BP in kJ/min\n", + "hcw=mw*Cpw*(tw2-tw1)#..........#Heat carried away by cooling water\n", + "mg=mf+(ma*mf)#....................#Mass of exhaust gases in kg/min\n", + "mst=mf*ms#..................#Mass of steam formed\n", + "hg=(mg-mst)*Cpg*(tg-tr)#..........#Heat carried by exhaust gasses\n", + "hst=mst*(417.5+2257.9+(Cps*(305-99.6)))#....................#Heat carried by exhaust steam, the obtained values are from steam table and hence are constants at NTP\n", + "ha=round(hf)-round(hip+hg+hst+hcw)#............#Unaccounted heat\n", + "pf=100#\n", + "pip=(hip/hf)*100#\n", + "pcw=(hcw/hf)*100#\n", + "pg=(hg/hf)*100#\n", + "pa=(ha/hf)*100#\n", + "pst=(hst/hf)*100#\n", + "print \"HEAT BALANCE TABLE\"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Item kJ Percent\"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Heat supplied by fuel %d %f\"%(hf,pf)\n", + "print \"Heat absorbed in IP %d %f\"%(hip,pip)\n", + "print \"Heat taken away by cooling water %d %f\"%(hcw,pcw)\n", + "print \"Heat carried away by dry exhaust gases %d %f\"%(hg,pg)\n", + "print \"Heat carried away by steam in exhaust gases %d %f\"%(hst,pst)\n", + "print \"Unaccounted heat %d %f\"%(ha,pa)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.40 PAGE 598" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volumetric efficiency = 68.49 %\n", + "Brake mean effective pressure = 5.14 bar\n", + "Engine torque = 269.63 N-m\n", + "Brake specific fuel consumption = 0.27 kg/kWh\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi,sqrt\n", + "# Initialisation of Variables\n", + "n=6#................#No of cylinders\n", + "D=0.1#...............#Engine bore in m\n", + "L=0.14#...............#Engine stroke in m\n", + "N=2500#...............#Engine rpm\n", + "k=0.5#..................#Four stroke\n", + "bl=480#.................#Brake load in N\n", + "br=76#..................#Barometer reading in cm of Hg\n", + "d0=3.3/100#..............#Orifice diameter in m\n", + "Cd=0.62#.................#Co efficient of discharge of orifice\n", + "pd=14#...................#Pressure drop across orifice in cm of Hg\n", + "tr=25#...............#Room temperature in C\n", + "mf=0.32#................#Fuel consumption in kg/min\n", + "rhohg=13600#.................#Density of Hg in kg/m**3\n", + "R=0.287#...................#gas constant in kJ/kgK\n", + "g=9.81#.................#Acceleration due to gravity in m/s**2\n", + "CD=17000#....................#dynamometer constant\n", + "#Calculations\n", + "Vs=(pi/4)*D*D*L*(N/2)*(n/60)#..............#Swept volume in m**3\n", + "br1=(br/100)*rhohg*g*(10**-3)#.............#Barometer reading into kN/m**2\n", + "rhoa=br1/(R*(tr+273))#...............#Density of air\n", + "pd1=(pd/100)*rhohg*g#......................#Conversion of pd into N/m**2\n", + "ha=pd1/(rhoa*g)#.......................#Head of air causing flow in m\n", + "Va=Cd*(pi/4)*d0*d0*sqrt(2*g*ha)#............#Volume of air passing through orifice of air box per min\n", + "etav=Va/Vs#....................#Volumetric efficiency\n", + "print \"Volumetric efficiency = %0.2f %%\"%(etav*100)\n", + "BP=bl*(N/CD)#................#Brake power in kW\n", + "pmb=(BP*6)/(L*D*D*k*10*N*n*(pi/4))#................#Brake mean effective pressure in bar\n", + "print \"Brake mean effective pressure = %0.2f bar\"%(pmb)\n", + "T=(BP*60*1000)/(2*pi*N)#....................#Engine torque in N-m\n", + "print \"Engine torque = %0.2f N-m\"%T\n", + "bsfc=(mf*60)/BP#.......................#Brake specific fuel consumption in kg/kWh\n", + "print \"Brake specific fuel consumption = %0.2f kg/kWh\"%bsfc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.41 PAGE 598" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 14.06 kW\n", + "Indicated thermal efficiency = 26.85 %\n", + "HEAT BALANCE TABLE\n", + "_______________________________________________________________________\n", + "Item kJ Percent\n", + "_______________________________________________________________________\n", + "Heat supplied by fuel 3141.79 100.00\n", + "Heat equivalent of BP 659.73 21.00\n", + "Heat taken away by cooling water 862.12 27.44\n", + "Heat carried away by dry exhaust gases 782.78 24.92\n", + "Heat carried away by steam in exhaust gases 294.85 9.38\n", + "Unaccounted heat 543.00 17.28\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "Cpw=4.18#..............#Specific heat of water in kJ/kgK\n", + "n=1#................#No of cylinders\n", + "N=350#.......#Engine rpm\n", + "pmi=2.74#..........#Mean effective pressure in bar\n", + "bl=600#..........#Brake load in N\n", + "mf=4.22#............#Fuel consumption in kg\n", + "mw=495#..............#Mass of cooling water\n", + "tw1=13#...............#Water inlet temperature in C\n", + "tw2=38#................#Water outlet temperature in C\n", + "ma=135#..................#Mass of air used in kg/h\n", + "tr=20#.................#Room temperature in C\n", + "tg=370#.................#Exhaust temperature in C\n", + "D=0.2#.................#Engine bore in m\n", + "L=0.28#.................#Engine stroke in m\n", + "Db=1#......................#Brake drum diameter in m\n", + "C=44670#...................#Calorirfic value of fuel in kJ/kg\n", + "Cps=2.093#..................#Specific heat of steamm in exhaust in kJ/kgK\n", + "Cpg=1.005#...................#Specific heat of dry exhaust gases in kJ/kgK\n", + "k=1#....................#Two stroke engiine\n", + "t=60#.....................#Duration of testing in min\n", + "perh=15#.................#Percentage of H2 in the fuel\n", + "#Calculations\n", + "IP=(n*pmi*N*D*D*L*k*10*(pi/4))/6#...................#Indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%IP\n", + "BP=(bl*pi*Db*N)/(60*1000)#......................#Brake power in kW\n", + "etaith=(IP)/((mf/3600)*C)#.................#Indicated thermal efficiency\n", + "print \"Indicated thermal efficiency = %0.2f %%\"%(etaith*100)\n", + "#Heat supplied\n", + "hf=(mf/t)*C#................#heat supplied by fuel\n", + "hbp=BP*t#...........#Heat equivalent of BP in kJ/min\n", + "hcw=(mw/60)*Cpw*(tw2-tw1)#..........#Heat carried away by cooling water\n", + "mg=(mf+ma)/t#....................#Mass of exhaust gases in kg/min\n", + "mst=9*(perh/100)*(mf/60)#..................#Mass of steam formed\n", + "mdg=mg-mst#..............................#Mass of dry exhaust gases per min\n", + "hg=(mdg)*Cpg*(tg-tr)#..........#Heat carried by exhaust gasses\n", + "hst=mst*(417.5+2257.9+(Cps*(305-99.6)))#....................#Heat carried by exhaust steam, the obtained values are from steam table and hence are constants at NTP\n", + "ha=round(hf)-round(hbp+hg+hst+hcw)#............#Unaccounted heat\n", + "pf=100#\n", + "pbp=(hbp/hf)*100#\n", + "pcw=(hcw/hf)*100#\n", + "pg=(hg/hf)*100#\n", + "pa=(ha/hf)*100#\n", + "pst=(hst/hf)*100#\n", + "print \"HEAT BALANCE TABLE\"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Item kJ Percent\"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Heat supplied by fuel %0.2f %0.2f\"%(hf,pf)\n", + "print \"Heat equivalent of BP %0.2f %0.2f\"%(hbp,pbp)\n", + "print \"Heat taken away by cooling water %0.2f %0.2f\"%(hcw,pcw)\n", + "print \"Heat carried away by dry exhaust gases %0.2f %0.2f\"%(hg,pg)\n", + "print \"Heat carried away by steam in exhaust gases %0.2f %0.2f\"%(hst,pst)\n", + "print \"Unaccounted heat %0.2f %0.2f\"%(ha,pa)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.42 PAGE 599" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 17.38 kW\n", + "Brake power = 10.81 kW\n", + "HEAT BALANCE TABLE : \n", + "_______________________________________________________________________\n", + "Item kJ Percent\n", + "_______________________________________________________________________\n", + "Heat supplied by fuel 3146 100.00\n", + "Heat absorbed in IP 1043 33.15\n", + "Heat taken away by cooling water 870 27.68\n", + "Heat carried away by dry exhaust gases 877 27.89\n", + "Heat carried away by steam in exhaust gases 319 10.16\n", + "Unaccounted heat 35 1.11\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "Cpw=4.18#..............#Specific heat of water in kJ/kgK\n", + "n=1#................#No of cylinders\n", + "N=350#.......#Engine rpm\n", + "pmi=2.8#..........#Mean effective pressure in bar\n", + "bl=590#..........#Brake load in N\n", + "mf=4.3#............#Fuel consumption in kg\n", + "mw=500#..............#Mass of cooling water\n", + "tw1=25#...............#Water inlet temperature in C\n", + "tw2=50#................#Water outlet temperature in C\n", + "ma=33#..................#Mass of air used per kg of fuel in kg\n", + "tr=25#.................#Room temperature in C\n", + "tg=400#.................#Exhaust temperature in C\n", + "D=0.22#.................#Engine bore in m\n", + "L=0.28#.................#Engine stroke in m\n", + "Db=1#......................#Brake drum diameter in m\n", + "C=43900#...................#Calorirfic value of fuel in kJ/kg\n", + "Cps=2.09#..................#Specific heat of steamm in exhaust in kJ/kgK\n", + "Cpg=1.0#...................#Specific heat of dry exhaust gases in kJ/kgK\n", + "k=1#....................#Two stroke engiine\n", + "perh=15#...................#Percentage of hydrogen\n", + "#Calculations\n", + "IP=(n*pmi*N*D*D*L*k*10*(pi/4))/6#...................#Indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%IP\n", + "BP=(bl*pi*Db*N)/(60*1000)#......................#Brake power in kW\n", + "print \"Brake power = %0.2f kW\"%BP\n", + "#Heat supplied\n", + "hf=(mf/60)*C#................#heat supplied by fuel\n", + "hip=IP*60#...........#Heat equivalent of BP in kJ/min\n", + "hcw=(mw/60)*Cpw*(tw2-tw1)#..........#Heat carried away by cooling water\n", + "mg=(mf+(mf*ma))/60#....................#Mass of exhaust gases in kg/min\n", + "mst=9*(perh/100)*(mf/60)#..................#Mass of steam formed\n", + "mdg=mg-mst#..............................#Mass of dry exhaust gases per min\n", + "hg=(mdg)*Cpg*(tg-tr)#..........#Heat carried by exhaust gasses\n", + "hst=mst*(417.5+2257.9+(Cps*(400-99.6)))#....................#Heat carried by exhaust steam, the obtained values are from steam tables at NTP\n", + "mg=mf+(ma*mf)#....................#Mass of exhaust gases in kg/min\n", + "ha=round(hf)-round(hip+hg+hst+hcw)#............#Unaccounted heat\n", + "pf=100#\n", + "pip=(hip/hf)*100#\n", + "pcw=(hcw/hf)*100#\n", + "pg=(hg/hf)*100#\n", + "pa=(ha/hf)*100#\n", + "pst=(hst/hf)*100#\n", + "print \"HEAT BALANCE TABLE : \"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Item kJ Percent\"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Heat supplied by fuel %d %0.2f\"%(hf,pf)\n", + "print \"Heat absorbed in IP %d %0.2f\"%(hip,pip)\n", + "print \"Heat taken away by cooling water %d %0.2f\"%(hcw,pcw)\n", + "print \"Heat carried away by dry exhaust gases %d %0.2f\"%(hg,pg)\n", + "print \"Heat carried away by steam in exhaust gases %d %0.2f\"%(hst,pst)\n", + "print \"Unaccounted heat %d %0.2f\"%(ha,pa)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.43 PAGE 600" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "HEAT BALANCE TABLE: \n", + "_______________________________________________________________________\n", + "Item kJ Percent\n", + "_______________________________________________________________________\n", + "Heat supplied by fuel 7951 100.00\n", + "Heat absorbed in BP 3073 21.00\n", + "Heat taken away by cooling water 2544 27.68\n", + "Heat carried away by exhaust gases 1745 27.89\n", + "Unaccounted heat 589 1.11\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "I=210#.....................#Output of generator in A\n", + "V=200#.....................#Generator voltage in V\n", + "etag=0.82#.................#Generator efficiency \n", + "mf=11.2#....................#Fuel used in kg/h\n", + "C=42600#.......................#Calorific value of fuel in kJ/kg\n", + "afr=18#....................#Air fuel ratio\n", + "mc=580#.....................#Mass of water through calorimeter in kg/h\n", + "delt=36#....................#Temperature raise in C\n", + "tg=98#........................#Temperature of exhaust in C\n", + "ta=20#.......................#Ambient temperature in C\n", + "phcw=0.32#.....................#Heat lost to cooling jacket is 32% of heat supplied\n", + "cpe=1.05#...................#Specific heat of exhause gases in kJ/kgK\n", + "cpw=4.18#...................#Specific heat of water in kJ/kgK\n", + "#Calculations\n", + "pow=V*I#......................#Total power generated in W\n", + "BP=(pow/1000)/etag#..................#Brake power in kW\n", + "hf=(mf/60)*C#...................#Heat supplied to the engine\n", + "hbp=BP*60#.........................#Heat equivalent of BP\n", + "mg=(mf/60)*(afr+1)#...............#Mass of exhaust gases formed per min in kg\n", + "hg=((mc/60)*cpw*(delt))+(mg*cpe*(tg-ta))#..........#Heat carried by exhaust gases per min\n", + "hcw=phcw*hf#...................#Heat lost to cooling jacket\n", + "ha=hf-(hcw+hg+hbp)#...................#Unaccounted heat\n", + "pf=100#pbp=(hbp/hf)*100#pcw=(hcw/hf)*100#pg=(hg/hf)*100#pa=(ha/hf)*100\n", + "print \"HEAT BALANCE TABLE: \"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Item kJ Percent\"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Heat supplied by fuel %d %0.2f\"%(hf,pf)\n", + "print \"Heat absorbed in BP %d %0.2f\"%(hbp,pbp)\n", + "print \"Heat taken away by cooling water %d %0.2f\"%(hcw,pcw)\n", + "print \"Heat carried away by exhaust gases %d %0.2f\"%(hg,pg)\n", + "print \"Unaccounted heat %d %0.2f\"%(ha,pa)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.44 PAGE 601" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "HEAT BALANCE TABLE : \n", + "_______________________________________________________________________\n", + "Item kJ Percent\n", + "_______________________________________________________________________\n", + "Heat supplied by fuel 8745 100.00\n", + "Heat absorbed in BP 3056 34.95\n", + "Heat lost by FP 658 7.53\n", + "Heat taken away by cooling water 2612 29.87\n", + "Heat carried away by dry exhaust gases 1569 17.94\n", + "Heat carried away by steam in exhaust gases 795 9.10\n", + "Unaccounted heat 52 0.60\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "D=0.34#..............#Engine bore in m\n", + "k=0.5#...............#Four stroke engine\n", + "n=1#..................#No of cylinders\n", + "L=0.44#................#Engine stroke in m\n", + "Ne=400#................#Engine rpm\n", + "aic=465#..............#Area of indicator diagram in mm**2\n", + "l=60#..................#Length of diagram in mm\n", + "s=0.6#...............#Spring constant in bar/mm\n", + "W=950#.................#Load of dynamometer in N\n", + "CD=7460#................#Dynamometer constant\n", + "mf=10.6#................#Fuel used in kg/h\n", + "Ca=49500#.................#Calorific value of fuel in kJ/kg\n", + "mw=25#...................#Cooling water circulated in kg/min\n", + "cpw=4.18#..................#Specific heat capacity of water in kJ/kgC\n", + "delt=25#..................#Rise in temperature of water\n", + "#Mass analysis of fuel\n", + "C=84#..................#Percentage of carbon\n", + "H=15#..................#Percentage of hydrogen\n", + "In=1#..................#Percentage of incombustible\n", + "#Volume analysis of exhaust gases\n", + "CO2=9#..................#Percentage of caron dioxide\n", + "O=10#..................#Percentage of oxygen\n", + "N=81#..................#Percentage of nitrogen\n", + "tg=400#................#Temperature of exhaust gases in C\n", + "cpg=1.05#..............#Specific heat of exhaust gases in kJ/kgC\n", + "tr=25#.................#Temperature of room in C\n", + "ppst=0.03#..............#Partial pressure of steam in exhaust gases in bar\n", + "cpst=2.1#.................#Specific heat of superheated steam in kJ/kgC\n", + "#Calculations\n", + "pmi=(aic*s)/l#................#Mean effective pressure in bar\n", + "IP=(n*pmi*L*D*D*k*10*Ne*(pi/4))/6#...............#Indicated power in kW\n", + "BP=(W*Ne)/CD#.......................#Brake power in kW\n", + "FP=IP-BP#...........................#Frictional power in kW\n", + "hf=(mf/60)*Ca#...................#Heat supplied in kJ per min\n", + "hbp=BP*60#.....................#Heat equivalent of Brake power in kW\n", + "hfp=FP*60#......................#heat equivalent of frictional power in kW\n", + "hcw=mw*cpw*delt#..................#Heat carried away by cooling water\n", + "ma1=(N*C)/(33*(CO2))#...........#Mass of air supplied per kg of fuel\n", + "mg1=ma1+1#.......................#Mass of exhaust gases per kg of fuel\n", + "mg=mg1*(mf/60)#.................#Mass of exhaust gas formed per min\n", + "mst1=9*(H/100)#.................#Mass of steam formed per kg of fuel\n", + "mst=mst1*(mf/60)#..................#Mass of steam formed per min\n", + "mdg=mg-mst#.......................#Mass of dry exhaust gas \n", + "Hex=mdg*cpg*(tg-tr)#...............#Heat carried by exhaust gases\n", + "hst=(2545.5+(cpst*(tg-24.1)))*mst#................#Heat carried by steam in exhaust gases in kJ/kg.....The values are from steam tables corresponding to the partial pressure 0.03 and temperature 400 Celsius\n", + "ha=hf-(hbp+hfp+hcw+Hex+hst)#.....................#Unaccounted heat\n", + "pf=100#\n", + "pbp=(hbp/hf)*100#\n", + "pfp=(hfp/hf)*100#\n", + "pcw=(hcw/hf)*100#\n", + "pex=(Hex/hf)*100#\n", + "pa=(ha/hf)*100#\n", + "pst=(hst/hf)*100#\n", + "print \"HEAT BALANCE TABLE : \"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Item kJ Percent\"\n", + "print \"_______________________________________________________________________\"\n", + "print \"Heat supplied by fuel %d %0.2f\"%(hf,pf)\n", + "print \"Heat absorbed in BP %d %0.2f\"%(hbp,pbp)\n", + "print \"Heat lost by FP %d %0.2f\"%(hfp,pfp)\n", + "print \"Heat taken away by cooling water %d %0.2f\"%(hcw,pcw)\n", + "print \"Heat carried away by dry exhaust gases %d %0.2f\"%(Hex,pex)\n", + "print \"Heat carried away by steam in exhaust gases %d %0.2f\"%(hst,pst)\n", + "print \"Unaccounted heat %d %0.2f\"%(ha,pa)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.45 PAGE 602" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mechanical efficiency = 83.29 %\n", + "Indicated thermal efficiency = 13.44 %\n", + "Air standard efficiency = 49.13 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=4#........................#No of cylinders\n", + "ga=1.4#...................#Degree of freedom\n", + "D=0.075#..................#Engine bore in m\n", + "L=0.1#...................#Engine stroke in m\n", + "mf=6#.......................#Fuel consumption in kg/h\n", + "C=83600#..................#Calorific value of fuel used\n", + "Vc=0.0001#.................#Clearence volume in m**3\n", + "BP=15.6#.................#Brake power wilh all cylinder working in kW\n", + "BP1=11.1#...................#Brake power wilh cylinder no 1 cutout in kW\n", + "BP2=11.03#...................#Brake power wilh cylinder no 2 cutout in kW\n", + "BP3=10.88#...................#Brake power wilh cylinder no 3 cutout in kW\n", + "BP4=10.66#...................#Brake power wilh cylinder no 4 cutout in kW\n", + "#Calculations\n", + "IP1=BP-BP1#...........................#Indicated power produced in cylinder 1 in kW\n", + "IP2=BP-BP2#...........................#Indicated power produced in cylinder 2 in kW\n", + "IP3=BP-BP3#...........................#Indicated power produced in cylinder 3 in kW\n", + "IP4=BP-BP4#...........................#Indicated power produced in cylinder 4 in kW\n", + "IP=IP1+IP2+IP3+IP4#.............................#Total Indicated power produced in kW\n", + "etamech=BP/IP#..............................#Mechanical efficiency\n", + "print \"Mechanical efficiency = %0.2f %%\"%(etamech*100)\n", + "etaith=IP/((mf/3600)*C)#.....................#Indicated thermal efficiency\n", + "print \"Indicated thermal efficiency = %0.2f %%\"%(etaith*100)\n", + "Vs=(pi/4)*D*D*L#..........................#Stroke volume in m**3\n", + "r=(Vs+Vc)/Vc#................................#Compression ratio\n", + "etast=1-(1/(r**(ga-1)))#...........................#Air standard efficiency\n", + "print \"Air standard efficiency = %0.2f %%\"%(etast*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 17.46 PAGE 603" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Engine torque = 59.20 Nm:\n", + "Brake mean effective pressure = 7.31 bar\n", + "Brake thermal efficiency = 21.31 %\n", + "Specific fuel consumption = 0.38 kg/kWh\n", + "Mechanical efficiency = 76.56 %\n", + "Indicated mean effective pressure = 9.55 bar \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "n=4#..........................#No of cylinders\n", + "D=0.06#......................#Engine bore in m\n", + "L=0.09#.......................#Engine stroke in m\n", + "N=2800#.......................#Engine rpm\n", + "Ta=0.37#.......................#Length of torque arm in m\n", + "spgr=0.74#.....................#Specific graviy of fuel \n", + "fc=8.986#......................#Fuel consumption in ltrs/h\n", + "mf=fc*spgr#.....................#Fuel consumed in kg/h\n", + "C=44100#......................#Calorific value of fuel in kJ/kg\n", + "BPnl=160#.................#Net brake load in N\n", + "BP1=110#...................#Brake load with cylinder no 1 cutout in N\n", + "BP2=107#...................#Brake load with cylinder no 2 cutout in N\n", + "BP3=104#...................#Brake load with cylinder no 3 cutout in N\n", + "BP4=110#...................#Brake load with cylinder no 4 cutout in N\n", + "k=0.5#.....................#Four stroke engine\n", + "#Calculations\n", + "T=BPnl*Ta#.......................#Engine torque in N-m\n", + "print \"Engine torque = %0.2f Nm:\"%T\n", + "BP=(2*pi*N*T)/(60*1000)#..........................#Brake power in kW\n", + "pmb=(BP*6)/(n*D*D*L*N*10*(pi/4)*k)#...................#Brake mean effective pressure in bar\n", + "print \"Brake mean effective pressure = %0.2f bar\"% pmb\n", + "etabth=BP/((mf/3600)*C)#...........................#Brake thermal efficiency\n", + "print \"Brake thermal efficiency = %0.2f %%\"%(etabth*100)\n", + "sfc=mf/BP#.......................#Specific fuel consumption in kg/kWh\n", + "print \"Specific fuel consumption = %0.2f kg/kWh\"%sfc\n", + "IP1=BPnl-BP1#...........................#Indicated power produced in cylinder 1 in kW\n", + "IP2=BPnl-BP2#...........................#Indicated power produced in cylinder 2 in kW\n", + "IP3=BPnl-BP3#...........................#Indicated power produced in cylinder 3 in kW\n", + "IP4=BPnl-BP4#...........................#Indicated power produced in cylinder 4 in kW\n", + "IP=IP1+IP2+IP3+IP4#.............................#Total Indicated power produced in kW\n", + "etamech=BPnl/IP#..............................#Mechanical efficiency\n", + "print \"Mechanical efficiency = %0.2f %%\"%(etamech*100)\n", + "pmi=pmb/etamech#............................#Indicated mean effective pressure in bar\n", + "print \"Indicated mean effective pressure = %0.2f bar \"%pmi" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER20_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER20_3.ipynb new file mode 100644 index 00000000..25da35d7 --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER20_3.ipynb @@ -0,0 +1,2554 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 20 - Air Compressors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.1 PAGE 750" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 4.24 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "v1=1#.........#Volume of air taken in m**3/min\n", + "p1=1.013#...............#Intake pressure in bar\n", + "t1=288#...............#Intake temperature in K\n", + "p2=7#......................#Delivery pressure in bar\n", + "n=1.35#..............#Adiabatic index\n", + "R=287#..............#Gas constant in kJ/kgK\n", + "#Calculations\n", + "m=(p1*v1*10**5)/(R*t1)#..............#Mass of air delivered per min in kg\n", + "t2=t1*((p2/p1)**((n-1)/n))#...........#Delivery temperature in K\n", + "iw=(n/(n-1))*m*R*(t2-t1)#............#Indicated work in kJ/min\n", + "IP=iw/(60*1000)#....................#Indicated power\n", + "print \"Indicated power = %0.2f kW\"%IP" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.2 PAGE 750" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cylinder bore = 141.44 mm\n", + "Motor power = 5.54 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "N=300#............#Compressor rpm\n", + "afr=15#.........#Air fuel ratio\n", + "etamech=0.85#....#Mechanical efficiency\n", + "etamt=0.9#.......#Motor transmission efficiency\n", + "v=1#............#Volume dealt with per min at inlet in m**3/min\n", + "rld=1.5#........#Ratio of stroke to diameter\n", + "v1=1#.........#Volume of air taken in m**3/min\n", + "p1=1.013#...............#Intake pressure in bar\n", + "t1=288#...............#Intake temperature in K\n", + "p2=7#......................#Delivery pressure in bar\n", + "n=1.35#..............#Adiabatic index\n", + "R=287#..............#Gas constant in kJ/kgK\n", + "#Calculations\n", + "m=(p1*v1*10**5)/(R*t1)#..............#Mass of air delivered per min in kg\n", + "t2=t1*((p2/p1)**((n-1)/n))#...........#Delivery temperature in K\n", + "iw=(n/(n-1))*m*R*(t2-t1)#............#Indicated work in kJ/min\n", + "IP=iw/(60*1000)#....................#Indicated power in kW\n", + "vdc=v/N#........#Volume drawn in per cycle in m**3\n", + "D=(vdc/((pi/4)*rld))**(1/3)#..........#Cylinder bore in m\n", + "print \"Cylinder bore = %0.2f mm\"%(D*1000)\n", + "pc=IP/etamech#.........#Power input to the compressor in kW\n", + "mp=pc/etamt#..........#Motor power in kW\n", + "print \"Motor power = %0.2f kW\"%mp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.3 PAGE 751" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at the end of compression = 430.07 K\n", + "Work done during compression per kg of air = 236.03 kJ\n", + "Heat transferred during compression per kg of air = -98.28 kJ\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "p1=1#......#Suction pressure in bar\n", + "t1=293#.....#Suction temperature in K\n", + "n=1.2#......#Compression index\n", + "p2=10#......#Delivery pressure in bar\n", + "R=0.287#....#Gas constant in kJ/kgK\n", + "cv=0.718#...#Specific heat at constant volume in kJ/kgK\n", + "#Calculations\n", + "t2=t1*((p2/p1)**((n-1)/n))#.....#Temperature at the end of compression in K\n", + "print \"Temperature at the end of compression = %0.2f K\"%t2\n", + "W=1*R*t1*(n/(n-1))*(((p2/p1)**((n-1)/n))-1)#.......#Work done during compression of air in kJ\n", + "print \"Work done during compression per kg of air = %0.2f kJ\"%W\n", + "Q=(t2-t1)*(cv-((R)/(n-1)))#..........#Heat transferred during compression of air in kJ/kg\n", + "print \"Heat transferred during compression per kg of air = %0.2f kJ\"%Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.4 PAGE 752" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volumetric Efficiency = 77.39 %\n", + "Indicated power = 5.35 kW\n", + "Isothermal efficiency = 79.79 %\n", + "Mechanical efficiency = 85.60 %\n", + "Overall efficiency = 68.30 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt,log\n", + "# Initialisation of Variables\n", + "p1=1#........#Suction pressure in bar\n", + "t1=293#.......#Suction temperature in K\n", + "p2=6#..........#Discharge pressure in bar\n", + "t2=453#.......#Discharge temperature in K\n", + "N=1200#.........#Compressor rpm\n", + "Ps=6.25#........#Shaft power in kW\n", + "ma=1.7#........#Mass of air delivered in kg/min\n", + "D=0.14#......#Engine bore in m\n", + "L=0.10#.......#Engine stroke in m\n", + "R=287#..........#Gas constant in kJ/kgK\n", + "#Calculations\n", + "Vd=(pi/4)*D*D*L*N#.........#Displlacement volume in m**3/min\n", + "FAD=ma*R*t1/(p1*10**5)#........#Free air delivered\n", + "etav=FAD/Vd#.....#Volumetric efficiency \n", + "print \"Volumetric Efficiency = %0.2f %%\"%(etav*100)\n", + "n=1/(1-((log(t2/t1))/(log(p2/p1))))#........#Index of compression\n", + "IP=(n/(n-1))*(ma/60)*(R/1000)*t1*(((p2/p1)**((n-1)/n))-1)#..........#Indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%IP\n", + "Piso=((ma/60)*(R/1000)*t1*(log(p2/p1)))#..........#Isothermal power\n", + "etaiso=Piso/IP#..............#Isothermal efficiency\n", + "print \"Isothermal efficiency = %0.2f %%\"%(etaiso*100)\n", + "etamech=IP/Ps#...........#Mechanical efficiency\n", + "print \"Mechanical efficiency = %0.2f %%\"%(etamech*100)\n", + "etao=Piso/Ps#........#Overall efficiency\n", + "print \"Overall efficiency = %0.2f %%\"%(etao*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.5 PAGE 753" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work output = 148.95 kW\n", + "Mass of cooling water = 2.12 kg/min\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "ma=6.75#.........#Mass of air compressed in kg/min\n", + "p1=1#............#Initial pressure in atm\n", + "cp=1.003#.........#Specifc heat at constant vpressure in kJ/kgK\n", + "t1=21#...........#Initial temperature in Celsius\n", + "t2=43#..........#Final temperature in Celsius\n", + "rp=1.35#.........#Pressure ratio\n", + "ga=1.4#.........#Ratio os specific heats\n", + "delt=3.3#.......#Change in temperature \n", + "cpw=4.18#.......#Specific heat for water in kJ/kgK\n", + "#Calculations\n", + "W=ma*cp*(t2-t1)#............#Work output in kJ\n", + "print \"Work output = %0.2f kW\"%W\n", + "t21=(t1+273)*(rp**((ga-1)/ga))#...........#Final temperature if the compression had been isentropic\n", + "Qr=ma*cp*(t21-(t2+273))#............#Heat rejected in kJ\n", + "mw=Qr/(cpw*delt)#........#Mass of cooling water in kg/min\n", + "print \"Mass of cooling water = %0.2f kg/min\"%mw" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.6 PAGE 754" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Swept volume = 0.03 m**3\n", + "Delivery temperature = 449.90 K\n", + "Indicated Power = 57.58 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "ma=14#........#Quantity of air delivered in kg/min\n", + "p1=1.013#......#Intake pressure in bar\n", + "t1=288#.........#Intake temperature in K\n", + "p2=7#...........#Delivery pressure in bar\n", + "N=300#..........#Compressor rpm\n", + "pervc=0.05#.......#Percentage of clearance volume in the total stroke volume\n", + "n=1.3#............#Compressor and expansion index\n", + "#Calculations\n", + "V1byVs=pervc+1#\n", + "v1minv4=ma/(N*2)#\n", + "v4byv3=((p2/p1)**(1/n))#\n", + "v4byvs=v4byv3*pervc#\n", + "Vs=v1minv4/(V1byVs-v4byvs)#.....#Swept volume in m**3\n", + "print \"Swept volume = %0.2f m**3\"%Vs\n", + "t2=t1*((p2/p1)**((n-1)/n))#........#Delivery Temperature in K\n", + "print \"Delivery temperature = %0.2f K\"%t2\n", + "IP=((n)/(n-1))*p1*(10**5)*((ma)/(60*1000))*(((p2/p1)**((n-1)/n))-1)#\n", + "print \"Indicated Power = %0.2f kW\"%IP" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.7 PAGE 754" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 63.51 kW\n", + "Volumetric efficiency = 2.07 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "FAD=14#...........#Free air delivered in m**3/min\n", + "p1=0.95#.........#Induction pressure in bar\n", + "t1=305#........#Induction temperature in K\n", + "p2=7#...........#Delivery pressure in bar\n", + "n=1.3#...........#Adiabatic index\n", + "VcbyVs=0.05#........#Ratio of clearance volume and swept volume\n", + "R=287#...........#Gas constant in J/kgK\n", + "t=288#...........#free air temperature in K\n", + "p=1.013#.........#free air pressure in bar\n", + "#Calculations\n", + "m=(p*100000*FAD)/(R*t)#..........#Mass delivered per min in kg\n", + "t2=t1*((p2/p1)**((n-1)/n))#\n", + "IP=((n/(n-1))*m*(R/1000)*(t2-t1))/60#.........#Indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%IP\n", + "v4byv3=(p2/p1)**(1/n)#v4byvs=v4byv3*VcbyVs#v1minv4=(1+VcbyVs)-v4byvs#\n", + "Vbyvs=v1minv4*(t/t1)*(p1/p)#\n", + "etav=Vbyvs/1#.............#Volumetric efficiency\n", + "print \"Volumetric efficiency = %0.2f %%\"%(etav*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.8 PAGE 755" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power input to compressor = 83.65 kW\n", + "Compressor stroke = 300.00 mm\n", + "Compressor bore = 291.43 mm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "FAD=16#.......#Free air delivered in m**3/min\n", + "p1=0.96#......#Suction pressure in bar\n", + "t1=303#.......#Suction temperature in K\n", + "n=1.3#........#Compression index\n", + "k=0.04#........#Clearance ratio\n", + "p2=6#.........#Delivery pressure in bar\n", + "etamech=0.9#...#Mechanical efficiency\n", + "vp=300#.......#Piston speed in m/min\n", + "N=500#........#Compressor rpm\n", + "p=1#.....#Ambient pressure in bar\n", + "t=288#.....#Ambient temperature in K\n", + "etac=0.85#...#Compressor efficiency\n", + "R=0.287#......#Universal gas constant\n", + "#Calculations\n", + "m=(p*10**5*FAD)/(R*1000*t)#...........#Mass flow rate of compressor in kg/min\n", + "t2=t1*((p2/p1)**((n-1)/n))#.....#Temperature at the end of compression in K\n", + "P=(n/(n-1))*(m/60)*R*(t2-t1)*(1/etamech)*(1/etac)#..........#Power input to compressor in kW\n", + "print \"Power input to compressor = %0.2f kW\"%P\n", + "L=vp/(2*N)#.........#Stroke in m\n", + "print \"Compressor stroke = %0.2f mm\"%(L*1000)\n", + "etav=((t/t1)*(p1/p)*(1+k-(k*((p2/p1)**(1/n)))))#........#Volumetric efficiency\n", + "D=sqrt(FAD/((pi/4)*L*N*2*etav))#...........#Compressor bore in m\n", + "print \"Compressor bore = %0.2f mm\"%(D*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.9 PAGE 756" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volumetric efficiency = 91.10 %\n", + "Power required to drive the compressor = 2.27 kW\n", + "Compressor rpm : 486.0\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "m=0.6#..........#Mass of air delivered in kg/min\n", + "p2=6#.........#Delivery pressure in bar\n", + "p1=1#..........#Induction pressure in bar\n", + "t1=303#........#Induction temperature in K\n", + "D=0.1#.........#Compressor bore in m\n", + "L=0.15#........#Compressor stroke in m\n", + "k=0.03#........#Clearance ratio\n", + "etamech=0.85#....#Mechanical efficiency\n", + "R=0.287#.......#Gas constant in kJ/kgK\n", + "n=1.3#........#Compression index\n", + "#Calculations\n", + "etav=(1+k)-(k*((p2/p1)**(1/n)))#..........#Volumetric efficiency\n", + "print \"Volumetric efficiency = %0.2f %%\"%(etav*100)\n", + "IP=(n/(n-1))*(m/60)*R*t1*(((p2/p1)**((n-1)/n))-1)#.........#Indicated power in kW\n", + "P=IP/etamech#...........#Power required to drive the compressor in kW\n", + "print \"Power required to drive the compressor = %0.2f kW\"%P\n", + "FAD=(m*R*t1*1000)/(p1*10**5)#...........#Free air delivery in m**3/min\n", + "Vd=FAD/etav#........#Displacement volume in m**3/min\n", + "N=Vd/((pi/4)*D*D*L)#.........#Compressor rpm\n", + "print \"Compressor rpm : \",round(N)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.10 PAGE 756" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Percentage change in free air delivery = 2.68 %\n", + "Percentage change in power delivered = 2.68 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "L=88#...........#Compressor stroke in cm\n", + "k=0.02#...........#Clearance ratio\n", + "p3=8.2#...........#Delivery pressure in bar\n", + "p4=1.025#.......#Suction pressure in bar\n", + "p1=p4#.......#Suction pressure in bar \n", + "n=1.3#.........#Compression index\n", + "lo=0.55#...#Length of distance piece fitted after overhaul in cm\n", + "#Calculations\n", + "pcfa=(((L+(L*k))-((L*k)*((p3/p4)**(1/n))))-(((k*L)+lo+L)-(((k*L)+lo)*((p3/p4)**(1/n)))))/((L+L*k)-((L*k)*((p3/p4)**(1/n))))\n", + "print \"Percentage change in free air delivery = %0.2f %%\"%(pcfa*100)\n", + "pcpa=pcfa#......#Percentage change in power delivered\n", + "print \"Percentage change in power delivered = %0.2f %%\"%(pcpa*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.11 PAGE 757" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Motor power = 241.07 kW\n", + "Compressor bore = 262.71 mm\n", + "Compressor stroke = 315.25 mm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "v=30#.............#Suction volume in m**3/min\n", + "p1=1#...........#Suction pressure in bar\n", + "t1=300#.........#Suction temperature in K\n", + "p2=16#...........#Delivery pressure in bar\n", + "N=320#..........#Compressor rpm\n", + "k=0.04#.........#Clearance ratio\n", + "rld=1.2#.........#Ratio of stroke to bore\n", + "etamech=0.82#....#Mechanical efficiency\n", + "n=1.32#.........#Compression index\n", + "ti=39+273#......#Temperature inside the suction chamber in K\n", + "nc=4#.........#No of cylineders\n", + "#Calculations\n", + "W=(n/(n-1))*(p1/1000)*10**5*(v/60)*(((p2/p1)**((n-1)/n))-1)#........#Work done in kW\n", + "mp=W/etamech#..........#Motor power in kW\n", + "print \"Motor power = %0.2f kW\"%mp\n", + "etav=((1+k)-(k*((p2/p1)**(1/n))))*(t1/ti)#........#Volumetric efficiency\n", + "Vs=(v/nc)*(1/(2*N))*(1/etav)#............#Swept volume of cylinder in m**3\n", + "D=(Vs/((pi/4)*rld))**(1/3)#.............#Compressor bore in m\n", + "L=D*rld#..............#Compresor stroke in m\n", + "print \"Compressor bore = %0.2f mm\"%(D*1000)\n", + "print \"Compressor stroke = %0.2f mm\"%(L*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.12 PAGE 758" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diameter of compressor cylinder = 159.70 mm\n", + "Stroke of compressor cylinder = 191.64 mm\n", + "Diameter of engine cylinder = 98.49 mm\n", + "Stroke of engine cylinder = 118.18 mm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "n=2#........#No of cylinders \n", + "ma=16#........#Mass of air supplied per min in kg\n", + "p1=1#........#Suction pressure in bar\n", + "t1=288#.......#Suction temperature in K\n", + "k=0.04#.......#Clearance ratio\n", + "ni=1.3#........#Compression index\n", + "R=0.287#........#Gas constant in kJ/kgK\n", + "N=2000#........#Engine rpm\n", + "p3=7#...........#Delivery pressure in bar\n", + "rld=1.2#...........#Ratio of stroke to bore for compressor cylinder and engine cylinder\n", + "etamech=0.82#.........#Mechanical efficiency of engine\n", + "pmb=5.5#..........#Mean effective pressure in bar in engine\n", + "ne=4#.............#No of engine cylinders\n", + "#Calculations\n", + "Vs=(((ma/n)*R*1000*t1)/(p1*10**5*N))/((1+k)-(k*((p3/p1)**(1/ni))))#\n", + "Dc=(Vs/((pi/4)*rld))**(1/3)#.........#Diameter of compressor cylinder in m\n", + "Lc=rld*Dc#.............#Stroke of the compressor cylinder in m\n", + "print \"Diameter of compressor cylinder = %0.2f mm\"%(Dc*1000)\n", + "print \"Stroke of compressor cylinder = %0.2f mm\"%(Lc*1000)\n", + "IP=(ni/(ni-1))*(ma/60)*R*t1*(((p3/p1)**((ni-1)/ni))-1)#......#Indicated power of the compressor in kW\n", + "BP=IP/etamech#...............#Brake power of the engine in kW\n", + "De=((BP*60*1000)/(ne*pmb*10**5*rld*(pi/4)*N))**(1/3)#......#Diameter of the engine cylinder in m\n", + "Le=rld*De#...........#Stroke of the engine cylinder in m\n", + "print \"Diameter of engine cylinder = %0.2f mm\"%(De*1000)\n", + "print \"Stroke of engine cylinder = %0.2f mm\"%(Le*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.13 PAGE 759" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volumetric efficiency = 81.44 %\n", + "Diameter of the cylinder = 17.98 cm\n", + "Stroke of the cylinder = 17.98 cm\n", + "Brake power = 11.10 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "nc=1.25#......#Index of compression\n", + "ne=1.3#......#Index of expansion\n", + "etamech=0.85#.......#Mechanical efficiency\n", + "p1=1#.........#Suction pressure in bar\n", + "p2=7.5#.......#Delivery pressure in bar\n", + "t1=25+273#....#Suction temperature in bar\n", + "Vamb=2.2#.....#Volume of free air delivered in m**3\n", + "N=310#........#Engine rpm\n", + "k=0.05#.......#Clearance ratio\n", + "pamb=1.03#.....#Ambient pressure in bar\n", + "tamb=293#......#Ambient temperature in K\n", + "#Calculations\n", + "etav=(1+k-(k*((p2/p1)**(1/ne))))#........#Volumetric efficiency\n", + "print \"Volumetric efficiency = %0.2f %%\"%(etav*100)\n", + "v1=(pamb*Vamb*t1)/(p1*tamb)#.......#Volume of air delivered at suction condition in m**3\n", + "vs=(v1/(etav*N*2))#.........#Swept volume in m**3\n", + "D=(vs/(pi/4))**(1/3)#........#Diameter of the cylinder in m\n", + "L=D#\n", + "print \"Diameter of the cylinder = %0.2f cm\"%(D*100)\n", + "print \"Stroke of the cylinder = %0.2f cm\"%(L*100)\n", + "W=2*vs*10**5*(((nc)/(nc-1))*p1*(1+k)*(((p2/p1)**((nc-1)/(nc)))-1)-((ne)/(ne-1))*p1*(k*((p2/p1)**(1/ne)))*(((p2/p1)**((ne-1)/(ne)))-1))#..........#Work done per cycle of operation in Nm/cycle\n", + "IP=W*N/(60*1000)#...............#Indicated power in kW\n", + "BP=IP/etamech#.............#Brake power in kW\n", + "print \"Brake power = %0.2f kW\"%BP" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.14 PAGE 760" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Compressor diameter = 35.75 cm\n", + "Compressor stroke = 53.63 cm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "v=14#..........#Volume of air delivered in m**3\n", + "p1=1#........#Suction pressure in bar\n", + "p2=7#........#Delivery pressure in bar\n", + "N=310#........#Compressor rpm\n", + "n=1.35#........#Compression index\n", + "k=0.05#........#Clearance ratio\n", + "rld=1.5#.........#Ratio of cylinder length and diameter\n", + "#Calculations\n", + "etav=(1+k)-(k*((p2/p1)**(1/n)))#..........#Volumetric efficiency\n", + "Vs=v/(etav*N)#.............#Swept volume in m**3\n", + "D=((Vs)/((pi/4)*rld))**(1/3)#......#Compressor diameter in m\n", + "L=rld*D#......................#Compressor stroke in m\n", + "print \"Compressor diameter = %0.2f cm\"%(D*100)\n", + "print \"Compressor stroke = %0.2f cm\"%(L*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.15 PAGE 761" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Free air delivered = 13.89 m**3/min\n", + "Heat rejected during compression = 748.46 kJ/min\n", + "Power needed to drive the compressor = 52.01 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "D=0.33#.........#Cylinder diameter in m\n", + "L=0.35#.........#Cylinder stroke in m\n", + "k=0.05#.........#Clearance ratio \n", + "N=300#..........#Compressor rpm\n", + "psuc=0.95#........#Suction pressure in bar\n", + "tsuc=298#.........#Suction temperature in K\n", + "pamb=1.013#......#Ambient pressure in bar\n", + "tamb=293#.........#Ambient temperature in K\n", + "p2=4.5#...........#Delivery pressure in bar\n", + "n=1.25#..........#Compression index\n", + "cv=0.717#...........#Specific heat at constant volume in kJ/kgK\n", + "ga=1.4#..........#Ratio of specific heats\n", + "etamech=0.8#......#Mechanical efficiency\n", + "R=0.287#.........#Gas constant in kJ/kgK\n", + "#Calculations\n", + "Vs=(pi/4)*D*D*L*N*2#............#Swept volume in m**3\n", + "p1=psuc#\n", + "etav=1-(k*(((p2/p1)**(1/n))-1))#........#Volumetric efficiency\n", + "Vad=Vs*etav#................#Actual air drawn per min in m**3\n", + "FAD=(psuc/pamb)*(tamb/tsuc)*Vad#............#Free air delivered in m**3/min\n", + "print \"Free air delivered = %0.2f m**3/min\"%FAD\n", + "t1=tsuc#\n", + "ma=(p1*10**5*Vad)/(R*1000*t1)#.......#Mass of air delivered per min in kg\n", + "t2=t1*((p2/p1)**((n-1)/n))#..........#Delivery temperature in K\n", + "Qr=ma*cv*((ga-n)/(n-1))*(t2-t1)#..........#Heat rejected during compression in kJ/min\n", + "print \"Heat rejected during compression = %0.2f kJ/min\"%Qr\n", + "P=((n)/(n-1))*R*t1*(ma/60)*(((p2/p1)**((n-1)/(n)))-1)*(1/etamech)#..........#Power needed to drive the compressor in kW\n", + "print \"Power needed to drive the compressor = %0.2f kW\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.16 PAGE 762" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power required to run the compressor = 4.19 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "#Initialisation of variables\n", + "p1=1.03#...........#Intake pressure in bar\n", + "t1=300#............#Intake temperature in K\n", + "p2=7#.............#Intake pressure for High pressure cylinder in bar\n", + "t2=310#..............#Temperature of air entering high pressure cylinder in K\n", + "p3=40#............#Pressure of air after compression in bar\n", + "V=30#.........#volume of air delivered in m**3/h\n", + "R=0.287#............#Gas constant for air in kJ/kgK\n", + "ga=1.4#...........#Ratio of specific heats\n", + "#Calculations\n", + "m=p1*10**5*V/(R*1000*t1)#..........#Mass of air compressed in kg/h\n", + "t21=t1*((p2/p1)**((ga-1)/ga))#.......#Actual temperature of air entering high pressure cylinder in K\n", + "t3=t2*((p3/p2)**((ga-1)/ga))#........#Actual temperature of air after compression in K\n", + "W=((ga)/(ga-1))*m*(R/3600)*(t21-t1+t3-t2)#..........#Power required to run compressor in kW\n", + "print \"Power required to run the compressor = %0.2f kW\"%W" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.17 PAGE 762" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diameter of low pressure cylinder = 267.30 mm\n", + "Diameter of high pressure cylinder = 109.13 mm\n", + "Power required to run the compressor = 57.49 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "FAD=6#.......#Free air delivered in m**3/min\n", + "p1=1#........#suction pressure in bar\n", + "t1=300#......#Suction temperature in K\n", + "p3=40#.......#Delivery pressure in bar\n", + "p2=6#........#Intermediate pressure in bar\n", + "t3=300#........#Temperature at the inlet to 2nd stage in K\n", + "n=1.3#.........#Compression index\n", + "etamech=0.8#.....#Mechanical efficiency\n", + "N=400#............#Compressor rpm\n", + "R=0.287#.........#Gas constant in kJ/kgK\n", + "#Calculations \n", + "dlp=(FAD/(N*(pi/4)))**(1/3)#...............#Diameter of the low pressure cylinder in m\n", + "dhp=sqrt(1/(dlp*N*(pi/4)))#............#Diameter of high pressure cylinder in m\n", + "print \"Diameter of low pressure cylinder = %0.2f mm\"%(dlp*1000)\n", + "print \"Diameter of high pressure cylinder = %0.2f mm\"%(dhp*1000)\n", + "m=(p1*FAD*10**5)/(R*t1*1000*60)#........#Mass flow of air in kg/s\n", + "W=n*(1/(n-1))*m*R*t1*(((p2/p1)**((n-1)/n))+((p3/p2)**((n-1)/n))-2)#........#Indicated work in kJ/s\n", + "P=W/etamech#...............#Power required in kW\n", + "print \"Power required to run the compressor = %0.2f kW\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.18 PAGE 763" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Intermediate pressure = 0.26 Mpa\n", + "Volume of low pressure cylinder = 0.02 m**3\n", + "Volume of high pressure cylinder = 0.01 m**3\n", + "Power required to drive the compressor = 42.96 kW\n", + "Heat rejected in the intercooler = 15.04 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "ns=2#.............#No of stages\n", + "v1=0.2#.........#Intake volume in m**3/s\n", + "p1=1#..........#Intake pressure in bar\n", + "t1=289#...............#Intake temperature in K\n", + "p3=7#.........#Final pressure in bar\n", + "n=1.25#.......#Compression index\n", + "N=600#........#Compressor rpm\n", + "cp=1.005#.....#Specific heat at constant pressure in kJ/kgK\n", + "R=0.287#......#Gas constant in kJ/kgK\n", + "#Calculations\n", + "p2=sqrt(p1*p3)#......#Intermediate pressure in bar\n", + "print \"Intermediate pressure = %0.2f Mpa\"%(p2/10)\n", + "vslp=60*v1/N#..........#Volume of low pressure cylinder in m**3\n", + "vshp=p1*vslp/p2#..........#Volume of high pressure cylinder in m**3\n", + "print \"Volume of low pressure cylinder = %0.2f m**3\"%vslp\n", + "print \"Volume of high pressure cylinder = %0.2f m**3\"%vshp\n", + "W=(ns*(n/(n-1)))*p1*10**5*(v1/1000)*(((p3/p1)**((n-1)/(ns*n)))-1)#...........#Power required to drive the compressor in kW\n", + "print \"Power required to drive the compressor = %0.2f kW\"%W\n", + "m=p1*10**5*v1/(R*t1*1000)#.........#Mass of air handled in kg/s\n", + "t2=t1*((p2/p1)**((n-1)/n))#.........#Temperature at the end of first stage compression in K\n", + "Qr=m*cp*(t2-t1)#.......#Heat rejected in the intercooler in kW\n", + "print \"Heat rejected in the intercooler = %0.2f kW\"%Qr" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.19 PAGE 764" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ratio of cylinder diameters : 2.340\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "#initialisation of variables\n", + "p3=30#..........#delivery pressure in bar\n", + "p1=1#.........#suction pressure in bar\n", + "t1=273+15#.......#suction temperature in K\n", + "n=1.3#.........#adiabatic index\n", + "#calculation\n", + "p2=sqrt(p1*p3)#.....#Pressure before entering High pressure cylinder in bar\n", + "t21=t1*((p2/p1)**((n-1)/n))#.........#Actual temperature before entering the high pressure turbine in K\n", + "r=sqrt((p2**(1/n))*(t21/t1))#............#Ratio of cylinder diameters\n", + "print \"Ratio of cylinder diameters : %0.3f\"%r" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.20 PAGE 765" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work done in compressing 1 kg of gas = 388.53 kJ\n", + "Pressure in the intercooler would rise.\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "ns=2#.........#No of stages\n", + "p1=1#........#Suction pressure in bar\n", + "p2=7.4#.......#Intercooler pressure in bar\n", + "p3=42.6#.......#Delivery pressure in bar\n", + "t1=15+273#......#Suction temperature in K\n", + "n=1.3#........#Compression index\n", + "R=0.287#.......#Gas constant in kJ/kgK\n", + "dlp=0.09#.......#Diameter of low pressure cylinder in m\n", + "dhp=0.03#.......#Diameter of high pressure cylinder in m\n", + "etav=0.9#.....#Volumetric efficiency\n", + "#Calculations\n", + "W=n*(1/(n-1))*R*t1*(((p2/p1)**((n-1)/n))+((p3/p2)**((n-1)/n))-2)#\n", + "print \"Work done in compressing 1 kg of gas = %0.2f kJ\"%W\n", + "#Given that stroke length is same in both cases\n", + "rV=p2/p1#.........#Ratio of volumes\n", + "rECV=((dlp/dhp)**2)*etav#.........#Ratio of effective cylinder volumes\n", + "if (rECV>rV):\n", + " print \"Pressure in the intercooler would rise.\"\n", + "elif (rECV<rV):\n", + " print \"Pressure in the intercooler would fall\"\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.21 PAGE 765" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Stroke length = 0.36 m\n", + "Bore = 0.27 m\n", + "Work required = 49.20 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "V=4#...........#Volume of air handled in m**3/min\n", + "p1=1.016#.....#Suction pressure in bar\n", + "t1=288#........#Suction temperature in K\n", + "N=250#.........#Compressor rpm\n", + "p3=78.65#.....#Delivery pressure in bar\n", + "vp=3#........#Piston speed in m/s\n", + "etamech=0.75#.....#mechanical efficiency\n", + "etav=0.8#........#Volumetric efficiency\n", + "n=1.25#..........#Compression index\n", + "R=287#...........#Gas constant in J/kgK\n", + "ns=2#............#No of stages\n", + "#Calculations\n", + "l=(vp*60)/(2*N)#..........#Stroke length in m\n", + "d=sqrt(V/((pi/4)*l*N*etav))#.......#Bore in m\n", + "print \"Stroke length = %0.2f m\"%l\n", + "print \"Bore = %0.2f m\"%d\n", + "m=(p1*10**5*V)/(R*t1)#.......#Mass of air handled by the compressor in kg/min\n", + "p2=sqrt(p1*p3)#.............#Intermediate pressure in bar\n", + "t2=t1*((p2/p1)**((n-1)/n))#.........#Temperature at the end of first stage compression in K\n", + "W=ns*(n/(n-1))*(m/60)*(R/1000)*(t2-t1)*(1/etamech)#..........#Work required in kW\n", + "print \"Work required = %0.2f kW\"%W" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.22 PAGE 766" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Indicated power = 15.48 kW\n", + "Swept volume for low pressure cylinder = 0.0131 m**3\n", + "Swept volume of high pressure cylinder = 0.0044 m**3\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "m=4.5#.........#Amount of air compressed in kg/min\n", + "ps=1.013#.......#Suction pressure in bar\n", + "ts=288#.........#Suction temperature in K\n", + "rp=9#...........#Pressure ratio\n", + "n=1.3#.........#Compression index\n", + "k=0.05#........#Clearance ratio\n", + "N=300#.........#Compressor rpm\n", + "R=287#.........#Gas constant in J/kgK\n", + "ns=2#............#No of stages\n", + "#Calculations\n", + "ti=round(ts*((sqrt(rp))**((n-1)/n)))#......#Intermediate temperature in K\n", + "W=round(ns*n*(1/(n-1))*m*(R/1000)*(ti-ts))#..........#Work required per min in kJ\n", + "IP=W/60#.........#Indicated power in kW\n", + "print \"Indicated power = %0.2f kW\"%IP\n", + "mc=m/N#...........#Mass induced per cycle in kg\n", + "etav=(1+k)-(k*(sqrt(rp)**(1/n)))#.......#Volumetric efficiency\n", + "Vs=(mc*R*ts)/(ps*10**5*etav)#........#Swept volume for low pressure cylinder in m**3\n", + "print \"Swept volume for low pressure cylinder = %0.4f m**3\"%Vs\n", + "vdhp=(mc*ts*R)/(sqrt(rp)*ps*10**5)#............#Volume of air drawn in high pressure cylinder per cycle in m**3\n", + "vshp=vdhp/etav#...............#Swept volume ofhigh pressure cylinder in m**3\n", + "print \"Swept volume of high pressure cylinder = %0.4f m**3\"%vshp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.23 PAGE 767" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power required to run the compressor = 18.68 kW\n", + "Bore of low pressure cylinder = 19.06 cm\n", + "Stroke of low pressure cylinder = 19.06 cm\n", + "Bore of high pressure cylinder = 7.00 cm\n", + "Stroke of high pressure cylinder = 19.06 cm\n", + "Ratio of cylinder volumes : 7.4162\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "v1=2.2#...........#free air delivered by the compressor in m**3/min\n", + "p1=1#........#Suction pressure in bar\n", + "t1=298#.......#Suction temperature in K\n", + "pd=55#........#Delivery pressure in bar\n", + "N=210#.......#Compressor rpm\n", + "n=1.3#........#Compression index\n", + "k=0.05#.......#Clearance ratio for high pressure and low pressure cylinders\n", + "R=287#.......#Gas constant in J/kgK\n", + "ns=2#.......#No of stages\n", + "#Calculations\n", + "ps=p1#\n", + "m =(p1*v1*10**5)/(R*t1)#.........#Mass of air deivered in m**3/min\n", + "W=(ns*(n/(n-1)))*m*R*t1*(((pd/ps)**((n-1)/(ns*n)))-1)#...........#Work done by compressor in Nm/min\n", + "P=W/(60*1000)#...........#Power required to run the compressor\n", + "print \"Power required to run the compressor = %0.2f kW\"%P\n", + "pi=sqrt(ps*pd)#........#Intermediate pressure in bar\n", + "etav1=(1+k)-(k*((pi/p1)**(1/n)))#...........#Volumetric efficiency of the low pressure cylinder \n", + "Vs=(v1*10**6)/(etav1*N)#............#Swept volume in cm**3\n", + "dlp=(Vs/((pi/4)))**(1/3)#..........#Diameter of low pressure cylinder in cm\n", + "llp=dlp#.................#Stroke of low pressure cylinder in cm\n", + "print \"Bore of low pressure cylinder = %0.2f cm\"%dlp\n", + "print \"Stroke of low pressure cylinder = %0.2f cm\"%llp\n", + "dhp=sqrt(dlp*dlp/pi)#.........#Diameter of high pressure cylinder in cm\n", + "lhp=llp#\n", + "print \"Bore of high pressure cylinder = %0.2f cm\"%dhp\n", + "print \"Stroke of high pressure cylinder = %0.2f cm\"%lhp\n", + "rcv=pi/ps#.....#Ratio of cylinder volumes\n", + "print \"Ratio of cylinder volumes : %0.4f\"%rcv" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.24 PAGE 768" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat rejected to the intercooler = 2179.62 kJ/min\n", + "Diameter of high pressure cylinder = 184.68 mm\n", + "Power required for high pressure cylinder = 45.02 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "p1=1#........#Suction pressure in bar\n", + "p2=4#.....#Intermediate pressure in bar\n", + "p5=3.8#.......#Pressure of air leaving the interooler in bar\n", + "p6=15.2#........#Delivery pressure in bar\n", + "t1=300#..........#Suction temperature in K\n", + "dlp=0.36#........#Diameter of low pressure cylinder in m\n", + "llp=0.4#........#Stroke of low pressure cylinder in m\n", + "N=220#........#Compressor rpm\n", + "k=0.04#........#Clearance ratio\n", + "cp=1.0035#.........#Specific heat at constant pressure in kJ/kgK\n", + "n=1.3#........#Compression index\n", + "R=0.287#........#Gas constant in kJ/kgK\n", + "p8=p5#\n", + "p3=p2#\n", + "p7=p6#\n", + "t5=t1#\n", + "#Calculations\n", + "Vslp=(pi/4)*dlp*dlp*llp*N*2#.......#Swept volume in m**3\n", + "etavlp=(1+k)-(k*((p2/p1)**(1/n)))#.....#Volumetric efficiency\n", + "valp=Vslp*etavlp#................#Volume of air drawn in low pressure cylinder in m**3\n", + "m=(p1*10**5*valp)/(R*1000*t1)#........#Mass of air drawin in kg/min\n", + "t2=round(t1*((p2/p1)**((n-1)/n)))#\n", + "Qr=m*cp*(t2-t5)#........#Heat rejected to the intercooler in kJ/min\n", + "print \"Heat rejected to the intercooler = %0.2f kJ/min\"%Qr\n", + "vahp=(m*R*t5*1000)/(p5*10**5)#...#Volume of air drawn into high pressure cylinder per min in m**3\n", + "Vshp=vahp/etavlp#.........#Swept volume of high pressure cylinder in m**3/min\n", + "dhp=sqrt(Vshp/((pi/4)*2*N*llp))#..........#Diameter of high pressure cylinder in m\n", + "print \"Diameter of high pressure cylinder = %0.2f mm\"%(dhp*1000)\n", + "P=(n/(n-1))*m*(1/60)*R*(t2-t1)#.......#Power required for high pressure cylinder in kW\n", + "print \"Power required for high pressure cylinder = %0.2f kW\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.25 PAGE 769" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power required to run the compressor = 85.38 kW\n", + "Heat rejected in intercooler = 1742.36 kJ/min\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "ps=1#........#Suction pressure in bar\n", + "pi=4.2#.....#Intermediate pressure in bar\n", + "pi1=4#.......#Pressure of air leaving the interooler in bar\n", + "pd=18#........#Delivery pressure in bar\n", + "t1=298#..........#Suction temperature in K\n", + "t5=t1#\n", + "dlp=0.4#........#Diameter of low pressure cylinder in m\n", + "llp=0.5#........#Stroke of low pressure cylinder in m\n", + "N=200#........#Compressor rpm\n", + "k=0.05#........#Clearance ratio\n", + "cp=1.004#.........#Specific heat at constant pressure in kJ/kgK\n", + "n=1.25#........#Compression index\n", + "R=0.287#........#Gas constant in kJ/kgK\n", + "#Calculations\n", + "Vslp=(pi/4)*dlp*dlp*llp#..........#Swept volume of low pressure cylinder in m**3\n", + "etavlp=(1+k)-(k*((pi/ps)**(1/n)))#.....#Volumetric efficiency\n", + "t2=round(t1*((pi/ps)**((n-1)/n)))#\n", + "m=(ps*10**5*etavlp*Vslp)/(R*1000*t1)#...#Mass of air in kg\n", + "wlp=((n)/(n-1))*R*1000*t1*m*(((pi/ps)**((n-1)/(n)))-1)#..........#Work done per min in Nm in low pressure cylinder\n", + "whp=((n)/(n-1))*R*t5*m*1000*(((pd/pi1)**((n-1)/(n)))-1)#..........#Work done per min in Nm in high pressure cylinder\n", + "W=wlp+whp#.........#Net work done in Nm\n", + "IP=(W*N)/(60*1000)#............#Power required to run the compressor in kW\n", + "print \"Power required to run the compressor = %0.2f kW\"%IP\n", + "Qr=m*N*cp*(t2-t1)#...........#Heat rejected in intercooler in kJ/min\n", + "print \"Heat rejected in intercooler = %0.2f kJ/min\"%Qr" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.26 PAGE 771" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power required = 49.23 kW\n", + "Isothermal work done = 41.78 kW\n", + "Isothermal efficiency = 84.86 %\n", + "Free air delivered = 9.04 m**3/min\n", + "Heat transferred in intercooler = 19.89 kW\n", + "\n", + "Swept volume for low pressure stage = 0.02 m**3\n", + "\n", + "\n", + "Clearance volume for low pressure stage = 8.90e-04 m**3\n", + "\n", + "\n", + "Swept volume for high pressure stage = 0.01 m**3\n", + "\n", + "\n", + "Clearance volume for high pressure stage = 3.48e-04 m**3\n", + "\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt,log\n", + "# Initialisation of Variables\n", + "p1=1#............#Intake pressure in bar\n", + "p2=4#..............#Pressure after first stage in bar\n", + "p3=16#............#Final pressure in bar\n", + "ns=2#............#No of stages\n", + "t1=300#............#Intake temperature in K\n", + "n=1.3#............#Compression index\n", + "klp=0.04#.........#Clearance ratio for low pressure cylinder\n", + "khp=0.06#........#Clearance ratio for high pressure cylinder\n", + "N=440#............#Engine rpm\n", + "R=0.287#..........#Gas constant in kJ/kgK\n", + "m=10.5#.............#Mass of air delivered in kg/min\n", + "cp=1.005#.........#Specific heat at constant pressure in kJ/kgK\n", + "#Calculations\n", + "rp=sqrt(p1*p3)#...........#Pressure ratio per stage\n", + "P=((ns*n)/(n-1))*R*t1*(m/60)*(((p3/p1)**((n-1)/(ns*n)))-1)#..........#Work done per min in Nm\n", + "print \"Power required = %0.2f kW\"%P\n", + "isoWd=(m/60)*R*t1*log(p3/p1)#..........#Isothermal work done in Nm\n", + "print \"Isothermal work done = %0.2f kW\"%isoWd\n", + "etaiso=isoWd/P#...............#Isothermal efficiency\n", + "print \"Isothermal efficiency = %0.2f %%\"%(etaiso*100)\n", + "FAD=(m*R*t1*1000)/(p1*10**5)#.............#Free air delivered in m**3/min\n", + "print \"Free air delivered = %0.2f m**3/min\"%FAD\n", + "t2=t1*((p2/p1)**((n-1)/n))#.....#Temperature at the end of compression in K\n", + "Qt=(m/60)*cp*(t2-t1)#..............#Heat transferred in intercooler in kW\n", + "print \"Heat transferred in intercooler = %0.2f kW\"%Qt\n", + "etavlp=(1+klp)-(klp*((p2/p1)**(1/n)))#..........#Volumetric efficiency of low pressure stage\n", + "etavhp=(1+khp)-(khp*((p2/p1)**(1/n)))#..........#Volumetric efficiency of high pressure stage\n", + "vslp=FAD/(N*etavlp)#......#Swept volume for low pressure stage in m**3\n", + "vclp=klp*vslp#..............#Clearance volume for low pressure stage in m**3\n", + "print \"\\nSwept volume for low pressure stage = %0.2f m**3\\n\"%(vslp)\n", + "print \"\\nClearance volume for low pressure stage = %0.2e m**3\\n\"%(vclp)\n", + "vshp=FAD/(N*rp*etavhp)#......#Swept volume for high pressure stage in m**3\n", + "vchp=khp*vshp#..............#Clearance volume for high pressure stage in m**3\n", + "print \"\\nSwept volume for high pressure stage = %0.2f m**3\\n\"%(vshp)\n", + "print \"\\nClearance volume for high pressure stage = %0.2e m**3\\n\"%(vchp)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.27 PAGE 771" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work done = 1297727.20 Nm\n", + "Isothermal work done = 1146628.13 Nm\n", + "Isothermal efficiency = 88.36 %\n", + "Single stage work done per min = 1686780.23 Nm\n", + "Percentage of work saved : 23.06 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "ns=3#......#No of stages\n", + "p1=1.05#......#Intake pressure in bar\n", + "pd=40#..........#Delivery pressure in bar\n", + "V=3#..........#Volume of air xupplied per min in m**3\n", + "n=1.25#........#Compression index\n", + "#Calculations\n", + "Wd=((ns*n)/(n-1))*p1*V*10**5*(((pd/p1)**((n-1)/(ns*n)))-1)#..........#Work done per min in Nm\n", + "print \"Work done = %0.2f Nm\"%Wd\n", + "isoWd=10**5*p1*V*log(pd/p1)#..........#Isothermal work done in Nm\n", + "print \"Isothermal work done = %0.2f Nm\"%isoWd\n", + "etaiso=isoWd/Wd#...............#Isothermal efficiency\n", + "print \"Isothermal efficiency = %0.2f %%\"%(etaiso*100)\n", + "wdss=((n)/(n-1))*p1*V*10**5*(((pd/p1)**((n-1)/(n)))-1)#..........#Single stage Work done per min in Nm\n", + "print \"Single stage work done per min = %0.2f Nm\"%wdss\n", + "perws=(wdss-Wd)/wdss#.......#Percentage of work saved\n", + "print \"Percentage of work saved : %0.2f %%\"%(perws*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.28 PAGE 772" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + "\n", + "Intermediate pressures: \n", + " p2 = 3.30\n", + " p3 = 10.90\n", + "\n", + "Indicated power required = 101.19 kW\n" + ] + } + ], + "source": [ + "# Initialisation of Variables\n", + "p1=1#.............#Intake pressure in bar\n", + "p4=36#........#Final pressure in bar\n", + "n=1.25#.........#Compression index\n", + "R=0.287#.......#Gas constant in kJ/kgK\n", + "t1=300#..........#Intake temperature in K\n", + "ns=3#...........#No of stages\n", + "v=15#..........#Volume of air delivered in m**3\n", + "#Calculations\n", + "p2=p1*((p4/p1)**(1/ns))#\n", + "p3=p2*((p4/p1)**(1/ns))#\n", + "print \"\\n\\nIntermediate pressures: \\n p2 = %0.2f\\n p3 = %.2f\\n\"%(p2,p3)\n", + "t2=t1*((p4/p1)**(((n-1)/n)*(1/ns)))#....#Delivery temperature in K\n", + "m=p1*10**5*v/(R*1000*t1)#...........#Mass of air handled per min in kg\n", + "Wt=((n/(n-1))*m*R*(1/60)*(t2-t1)*ns)#........#Total work done in three stages \n", + "print \"Indicated power required = %0.2f kW\"%Wt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.29 PAGE 772" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat rejected in intercooler after first stage = 1276.59 kJ/min\n", + "Heat rejected in intercooler after second stage = 1639.45 kJ/min\n", + "Heat rejected in intercooler after third stage = 1875.18 kJ/min\n", + "Diameter of the intermediate cylinder = 179.48 mm\n", + "Diameter of the intermediate cylinder = 87.59 mm\n", + "Shaft power = 142.68 kW\n", + "\n", + "Heat transferred during first stage = 455.70 kJ/min\n", + "\n", + "Heat transferred during second stage = 438.71 kJ/min\n", + "\n", + "Heat transferred during third stage = 334.22 kJ/min\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "ns=3#........#No of stages\n", + "N=200#.......#Compressor rpm\n", + "p1=1#.......#Intake pressure in bar\n", + "t1=20+273#....#Intake temperature in K\n", + "D=0.35#......#Engine bore in m\n", + "L=0.4#.......#Engine stroke in m\n", + "p2=4#........#Discharge pressure from first stage in bar\n", + "p6=16#........#Discharge pressure from second stage in bar\n", + "p10=64#........#Discharge pressure from third stage in bar\n", + "pd=0.2#........#Loss of pressure between intercoolers in bar\n", + "R=0.287#......#Gas constant in kJ/kgK\n", + "k=0.04#.......#Clearence volume in 4% of the stroke volume\n", + "n1=1.2#.....#Compressor index for first stage\n", + "n2=1.25#.....#Compressor index for second stage\n", + "n3=1.3#.....#Compressor index for third stage\n", + "cp=1.005#......#Specific heat at constant pressure in kJ/kgK\n", + "etamech=0.8#.....#Mechanical efficiency\n", + "#Calculations\n", + "p5=p2-pd#\n", + "p9=p6-pd#\n", + "t5=t1#\n", + "t9=t1#\n", + "Vs=(pi/4)*D*D*L*N*2#............#Swept volume of low pressure cylinder per min in m**3\n", + "etav1=(1+k)-(k*((p2/p1)**(1/n1)))#.....#Volumetric efficiency in first stage\n", + "etav2=(1+k)-(k*((p6/p5)**(1/n2)))#.....#Volumetric efficiency in second stage\n", + "etav3=(1+k)-(k*((p10/p9)**(1/n3)))#.....#Volumetric efficiency in third stage\n", + "vain1=Vs*etav1#.................#Volume of air taken in first stage in m**3/min\n", + "m=(p1*10**5)*vain1/(R*t1*1000)#...........#Mass of air intake in kg/min in first stage\n", + "t2=round(t1*((p2/p1)**((n1-1)/n1)))#\n", + "t6=t5*((p6/p5)**((n2-1)/n2))#\n", + "t10=t9*((p10/p9)**((n3-1)/n3))#\n", + "Qr1=m*cp*(t2-t5)#........#Heat rejected in intercooler after first stage in kJ/min\n", + "Qr2=m*cp*(t6-t9)#........#Heat rejected in intercooler after second stage in kJ/min\n", + "Qr3=m*cp*(t10-t1)#........#Heat rejected in intercooler after third stage in kJ/min\n", + "print \"Heat rejected in intercooler after first stage = %0.2f kJ/min\"%Qr1\n", + "print \"Heat rejected in intercooler after second stage = %0.2f kJ/min\"%Qr2\n", + "print \"Heat rejected in intercooler after third stage = %0.2f kJ/min\"%Qr3\n", + "vainip=m*R*t5*1000/(p5*10**5)#.........#Volume drawn in intermediate pressure cylinder/min\n", + "Vsip=vainip/etav2#.............#Swept volume of intermediate cylinder in m**3/min\n", + "Dip=sqrt(Vsip/(2*N*L*(pi/4)))#............#Diameter of the intermediate cylinder in m\n", + "print \"Diameter of the intermediate cylinder = %0.2f mm\"%(Dip*1000)\n", + "vainhp=m*R*t9*1000/(p9*10**5)#.........#Volume drawn in high pressure cylinder/min\n", + "Vshp=vainhp/etav3#.............#Swept volume of high pressure cylinder in m**3/min\n", + "Dhp=sqrt(Vshp/(2*N*L*(pi/4)))#............#Diameter of the intermediate cylinder in m\n", + "print \"Diameter of the intermediate cylinder = %0.2f mm\"%(Dhp*1000)\n", + "Ps=(((n1/(n1-1))*m*R*(t2-t1))+((n2/(n2-1))*m*R*(t6-t5))+((n3/(n3-1))*m*R*(t10-t9)))*(1/(60*etamech))#...#Shaft power in kW\n", + "print \"Shaft power = %0.2f kW\"%Ps\n", + "cv=cp-R#..........#Specific heat at constant volume in kJ/kgK\n", + "ga=cp/cv#...........#Ratio of specific heats\n", + "Qt1=cv*((ga-n1)/(ga-1))*(t2-t1)*m#............#Heat transfer during first stage in kJ/min\n", + "Qt2=cv*((ga-n2)/(ga-1))*(t6-t1)*m#............#Heat transfer during second stage in kJ/min\n", + "Qt3=cv*((ga-n3)/(ga-1))*(t10-t1)*m#............#Heat transfer during third stage in kJ/min\n", + "print\"\\nHeat transferred during first stage = %0.2f kJ/min\"%(Qt1)\n", + "print \"\\nHeat transferred during second stage = %0.2f kJ/min\"%(Qt2)\n", + "print \"\\nHeat transferred during third stage = %0.2f kJ/min\"%(Qt3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.30 PAGE 773" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "No of stages : 4\n", + "Exact stage pressure ratio : 3.34\n", + "\n", + "Intermediate pressures\n", + "p4=37.38\n", + "p3=10.90\n", + "p2=4.00\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "p1=1#.....#Intake pressure in bar\n", + "p5=125#.....#Pressure of the compressed air in bar\n", + "rpr=4#.........#Pressure ratio is restricted to 4\n", + "#Calculations\n", + "X=(log(p5/p1)/log(rpr))#\n", + "if(X>round(X)):\n", + " x=round(X)+1#\n", + "else:\n", + " x=round(X)#\n", + "print \"No of stages : %d\"%x\n", + "esrp=(p5/p1)**(1/x)#\n", + "print \"Exact stage pressure ratio : %0.2f\"%esrp\n", + "p4=p5/esrp#p3=p4/esrp#p2=p3/esrp#......#Intermediate pressures in bar\n", + "print \"\\nIntermediate pressures\\np4=%.2f\\np3=%.2f\\np2=%.2f\"%(p4,p3,p2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.31 PAGE 774" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "No of stages: 4\n", + "Total work done = 512.24 kJ/kg\n", + "Heat rejected in intercoolers = 254.23 kJ/kg\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "ps=1#.........#Suction pressure in bar\n", + "t1=273+125#.......#Delivery temperature in K\n", + "pd=160#...........#Delivery pressure in bar\n", + "tm=40+273#........#Min temperature\n", + "ts=298#........#Suction temperature in K\n", + "n=1.25#......#Adiabatic index\n", + "cv=0.71#.......#Specific heat at constant volume in kJ/kgK\n", + "R=0.287#......#Gas constant in kJ/kgK\n", + "ns=3#.......#No of stages\n", + "#Calculations\n", + "p1=ps*((t1/ts)**(n/(n-1)))#\n", + "x=(log(pd/p1))/(((n/(n-1))*(log(t1/tm))))#\n", + "print \"No of stages: %d\"%(round(x)+1)\n", + "rp1=p1#...........#Pressure ratio in 1st stage\n", + "rp=(pd/rp1)**(1/ns)#.........#Pressure ratio in the following stage\n", + "W=(n/(n-1))*R*ts*(((rp1)**((n-1)/n))-1)#.........#Work done in first stage in kJ\n", + "Wf=ns*(n/(n-1))*R*tm*(((rp)**((n-1)/n))-1)#.........#Work done in next three stages in kJ\n", + "wt=W+Wf#............#Total work done per kg in kJ\n", + "print \"Total work done = %0.2f kJ/kg\"%wt\n", + "cp=cv+R#..............#Specific heat at constant pressure in kJ/kgK\n", + "Qr=ns*cp*(t1-tm)#.............#Heat rejected in intercoolers in kJ/kg\n", + "print \"Heat rejected in intercoolers = %0.2f kJ/kg\"%Qr" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.32 PAGE 775" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Swept capacity of low pressure cylinder = 0.12 m**3\n", + "Swept capacity of intermediate pressure cylinder = 0.03 m**3\n", + "Swept capacity of intermediate pressure cylinder = 0.01 m**3\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "Vamb=10.5#........#Free air volume in m**3\n", + "Pamb=1.013#...........#Free air presssure in bar\n", + "Tamb=273+15#..........#Free air temperature in K\n", + "T1=(273+25)#...........#Temperature at the end of suction in all cylinders in K\n", + "P1=1#............#Pressure at the suction in bar\n", + "pd=95#...........#Delivery presssure in bar\n", + "N=100#.........#Compressor rpm\n", + "n=1.25#..........#Adiabatic index\n", + "k=0.04#.........#Fractional clearances for LP\n", + "k1=0.07#.........#Fractional clearances for HP\n", + "#Calculations\n", + "z=(pd/P1)**(1/3)#.........#Pressure ratio\n", + "pi1=z*P1#\n", + "pi2=z*pi1#\n", + "etavollp=1+k-(k*(z**(1/n)))#\n", + "etavolhp=1+k1-(k1*(z**(1/n)))#\n", + "v1=(Pamb*Vamb*T1)/(Tamb*P1)#\n", + "sclp=(round(v1))/(etavollp*N)#.........#Swept capacity of LP cylinder in m**3\n", + "print \"Swept capacity of low pressure cylinder = %0.2f m**3\"%sclp\n", + "vip=(Pamb*Vamb*T1)/(pi1*Tamb)#.........#Volume of free air reduced to suction conditions of IP cylinder\n", + "scip=vip/(etavolhp*N)#.........#Swept capacity of IP cylinder in m**3\n", + "print \"Swept capacity of intermediate pressure cylinder = %0.2f m**3\"%scip\n", + "vhp=(Pamb*Vamb*T1)/(pi2*Tamb)#.........#Volume of free air reduced to suction conditions of HP cylinder\n", + "schp=vhp/(etavolhp*N)#.........#Swept capacity of HP cylinder in m**3\n", + "print \"Swept capacity of intermediate pressure cylinder = %0.2f m**3\"%schp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.34 PAGE 776" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at the end of expansion = 244.63 K\n", + "Indicated power of the motor = 0.75 kW\n", + "Mass of air supplied per min = 0.42 kg\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "D=0.0635#.......#Engine bore in m\n", + "L=0.114#.........#Engine stroke in m\n", + "p1=6.3#..........#Supply pressure in bar\n", + "t1=273+24#.........#Supply temperature in K\n", + "p4=1.013#..........#Exhaust pressure in bar\n", + "cv=0.05#............#Clearance volume is 5% of the swept volume\n", + "cr=0.5#..........#Cut off ratio\n", + "n=1.3#...........#Adiabatic index\n", + "R=287#...............#gas constant in kJ/kgK\n", + "N=300#.............#Engine rpm\n", + "ga=1.4#...........#Ratio of specific heats\n", + "#Calculations\n", + "Vs=(pi*D*D*L)/4#........#Swept volume in m**3\n", + "Vc=cv*Vs#..........#Clearance volume in m**3\n", + "v6=Vc#\n", + "v5=v6#\n", + "v1=(Vs/2)+Vc#\n", + "v2=Vs+Vc#\n", + "v3=v2#\n", + "p3=p4#\n", + "v4=v5+(cv*Vs)#\n", + "p2=p1*((v1/v2)**n)#.......#Pressure at the end of expansion\n", + "t2=t1*((v1/v2)**(n-1))#........#Temperature at the end of expansion in K\n", + "print \"Temperature at the end of expansion = %0.2f K\"%t2\n", + "p5=p4*((v4/v5)**n)#\n", + "w=((p1*(v1-v6))+(((p1*v1)-(p2*v2))/(n-1))-(p3*(v3-v4))-(((p5*v5)-(p4*v4))/(n-1)))*10**5#.......#Workk done per cycle in Nm\n", + "IP=(w*N)/(60*1000)#..........#Indicated power in kW\n", + "print \"Indicated power of the motor = %0.2f kW\"%IP\n", + "t3=t2*((p3/p2)**((ga-1)/ga))#\n", + "t4=t3#\n", + "m4=(p4*v4*10**5)/(R*t4)#\n", + "m1=(p1*v1*10**5)/(R*t1)#\n", + "ma=(m1-m4)*N#..........#Mass of air supplied per min\n", + "print \"Mass of air supplied per min = %0.2f kg\"%ma" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.35 PAGE 777" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work input for roots compressor = 1.52 kJ/rev\n", + "Work input for vane compressor = 1.35 kJ/rev\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "v=0.03#..............#Induced volume in m**3/rev\n", + "p1=1.013#...........#Inlet pressure in bar\n", + "rp=1.5#............#Pressure ratio\n", + "ga=1.4#...........#Ratio of specific heats\n", + "#Calculations\n", + "p2=rp*p1#\n", + "wr=(p2-p1)*(10**5)*v/1000#.....#Work input for roots compressor in kJ\n", + "print \"Work input for roots compressor = %0.2f kJ/rev\"%wr\n", + "pi=(p2+p1)/2#\n", + "wv=((p2-pi)*(10**5)*v*((p1/pi)**(1/ga))*(1/1000))+((ga/(ga-1))*p1*(10**5)*(v/1000)*(((pi/p1)**((ga-1)/ga))-1))#...#Work input required for vane type in kJ/rev\n", + "print \"Work input for vane compressor = %0.2f kJ/rev\"%wv" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.36 PAGE 778" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Compressor efficiency = 85.98 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "v=0.08#.........#Volume of air compressed in m**3\n", + "p1=1#..........#Intake pressure in bar\n", + "p2=1.5#........#Pressure after compression in in bar\n", + "ga=1.4#.........#Ratio of specific heats\n", + "#Calculations\n", + "wac=v*(p2-p1)*10**5#........#Actual work done in Nm\n", + "wid=(ga/(ga-1))*p1*v*(10**5)*(((p2/p1)**((ga-1)/ga))-1)#...........#Ideal work done per revolution in Nw\n", + "etac=wid/wac#................#Compressor efficiency\n", + "print \"Compressor efficiency = %0.2f %%\"%(etac*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.37 PAGE 778" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Isentropic efficiency = 74.46 %\n", + "Power required to drive the coompressor = 190.69 kW\n", + "Overall efficiency = 72.92 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "m=2.5#..........#Air flow rate in kg/s\n", + "p1=1#........#Inlet pressure in bar\n", + "t1=290#............#Inlet temperature in bar\n", + "C1=80#..........#Inlet Velocity in m/s\n", + "p2=1.5#........#pressure after compression in bar\n", + "t2=345#............#temperature after compression in bar\n", + "C2=220#..........#Velocity after compression in m/s\n", + "cp=1.005#...........#Specific heat at constant pressure in kJ/kgK\n", + "ga=1.4#............#Ratio of specific heats\n", + "R=287#..............#Gas constant for air in kJ/kgK\n", + "#Calculations\n", + "t21=t1*((p2/p1)**((ga-1)/ga))#\n", + "wisen=cp*(t21-t1)+((C2*C2)-(C1*C1))/(2*1000)#.....#Isentropic work done in kJ/kg\n", + "w=cp*(t2-t1)+((C2*C2)-(C1*C1))/(2*1000)#.....#Actual work done (in impeller) in kJ/kg\n", + "etaisen=wisen/w#...............#Isentropic efficiency\n", + "print \"Isentropic efficiency = %0.2f %%\"%(etaisen*100)\n", + "P=m*w#..........#Power required to drive the coompressor in kW\n", + "print \"Power required to drive the coompressor = %0.2f kW\"%P\n", + "t3=(((C2*C2)-(C1*C1))/(2*1000*cp))+t2#....#Temperature of air after leaving the diffuser in K\n", + "p3=p2*((t3/t2)**(ga/(ga-1)))#..........#Pressure of air after leaving the diffuser in bar\n", + "t31=t1*((p3/p1)**((ga-1)/ga))#...........#Delivery temperature from diffuser in K\n", + "etao=(t31-t1)/(t3-t1)#...............#Overall efficiency \n", + "print \"Overall efficiency = %0.2f %%\"%(etao*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.38 PAGE 779" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Actual rise in temperature = 165.00\n", + "Tip diameter of the impeller = 40.99 cm\n", + "Power required = 1459.26 kW\n", + "Diameter of the eye = 28.17 cm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "ma=528#.............#Air flow in kg/min\n", + "m=ma/60#.............#Air flow in kg/s\n", + "p1=1#........#Inlet pressure in bar\n", + "t1=293#............#Inlet temperature in bar\n", + "N=20000#..............#Compressor rpm\n", + "etaisen=0.8#.........#Isentropic efficiency\n", + "po1=1#.........#Static pressure in bar\n", + "p02=4#...........#Final total pressure in bar\n", + "C1=145#.........#Velocity of air when entering the impeller in m/s\n", + "rwt=0.9#..........#Ratio of whirl speed to tip speed \n", + "dh=0.12#........#Hub diameter in m\n", + "cp=1.005#...........#Specific heat at constant pressure in kJ/kgK\n", + "ga=1.4#............#Ratio of specific heats\n", + "R=287#..............#Gas constant for air in kJ/kgK\n", + "#Calculations\n", + "t01=t1+((C1*C1)/(2*cp*1000))#..........#Stagnation temperature at the inlet to the machine in K\n", + "p01=p1*((t01/t1)**(ga/(ga-1)))#.....#Stagnation pressure at the inlet to the machine in bar\n", + "t021=t01*((p02/p01)**((ga-1)/ga))#\n", + "deltisen=t021-t01#.........#Isentropic rise in temperature in K\n", + "delt=round(deltisen/etaisen)#........#Actual rise in temperature \n", + "print \"Actual rise in temperature = %0.2f\"%delt\n", + "wc=cp*delt#.........#Work consumed by compressor in kJ/kg\n", + "Cbl2=sqrt(wc*1000/rwt)#\n", + "d2=Cbl2*60/(pi*N)#..........#Tip diameter of the impeller in m\n", + "print \"Tip diameter of the impeller = %0.2f cm\"%(d2*100)\n", + "P=m*wc#............#Power required in kW\n", + "print \"Power required = %0.2f kW\"%P\n", + "rho1=(p1*10**5)/(R*t1)#.......#Density at entry in kg/m**3\n", + "d1=sqrt(((m*4)/(C1*rho1*pi))+(dh**2))#.......#Eye diameter in m\n", + "print \"Diameter of the eye = %0.2f cm\"%(d1*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.39 PAGE 780" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Final temperature of air = 466.65 Kevin\n", + "Theoretical power = 2282.94 kW\n", + "Impeller diameter at outlet = 84.10 cm\n", + "Impeller diameter at inlet = 42.05 cm\n", + "Breadth of impeller at inlet = 14.92 cm\n", + "Impeller blade angle at inlet = 15.73 degrees\n", + "Diffuser blade angle at inlet = 8.89 degrees\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt,atan\n", + "# Initialisation of Variables\n", + "N=10000#.................#Compressor rpm\n", + "v=660#............#Volume of air delivered in m**3/min\n", + "p1=1#.................#Inlet pressure in bar\n", + "t1=293#.............#Inlet temperature in K\n", + "rp=4#.............#Pressure ratio\n", + "etaisen=0.82#........#Isentropic efficiency\n", + "Cf2=62#...............#Flow velocity in m/s\n", + "rr=2#.............#Ratio of outer radius of impeller to inner radius of impeller\n", + "ka=0.9#..............#Blade area co efficient\n", + "fis=0.9#...........#Slip factor\n", + "cp=1.005#..............#Specific heat at constant pressure in kJ/kgK\n", + "ga=1.4#............#Ratio of specific heats\n", + "R=287#..............#Gas constant for air in kJ/kgK\n", + "#Calculations\n", + "t21=t1*(rp**((ga-1)/ga))#\n", + "Cf1=Cf2#\n", + "t2=t1+((t21-t1)/etaisen)#..........#Final temperature of air\n", + "m=(p1*10**5*v/60)/(R*t1)#...............#Mass flow rate in m**3/s\n", + "P=m*cp*(t2-t1)#.........#Theoretical power in kW\n", + "print \"Final temperature of air = %0.2f Kevin\"%t2\n", + "print \"Theoretical power = %0.2f kW\"%P\n", + "Cbl2=sqrt(1000*cp*(t2-t1)/fis)#\n", + "d2=60*Cbl2/(pi*N)#..........#Impeller diameter at outlet in m\n", + "d1=d2/rr#...............#Impeller diameter at inlet in m\n", + "print \"Impeller diameter at outlet = %0.2f cm\"%(d2*100)\n", + "print \"Impeller diameter at inlet = %0.2f cm\"%(d1*100)\n", + "b1=(v/60)/(2*pi*(d1/2)*Cf1*ka)#.........#Breadth of impeller at inlet in m\n", + "print \"Breadth of impeller at inlet = %0.2f cm\"%(b1*100)\n", + "Cbl1=Cbl2/rr#\n", + "beta1=(atan(Cf1/Cbl1))*180/pi#\n", + "al2=(atan(Cf2/(fis*Cbl2)))*180/pi#\n", + "print \"Impeller blade angle at inlet = %0.2f degrees\"%(beta1)\n", + "print \"Diffuser blade angle at inlet = %0.2f degrees\"%(al2,)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.40 PAGE 780" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + "Blade angles \n", + "\n", + "\t Blade angle at the inlet of the impeller: beta 1=25.869916 \n", + "\t Blade angle at the outlet of the impeller: beta 2=34.206408 \n", + "\t Absolute angle at the tip of impeller: alpha 2=21.610499\n", + "\n", + "\n", + "Breadth of the blade at inlet = 7.35 cm\n", + "Breadth of the blade at outlet = 2.89 cm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt,atan\n", + "# Initialisation of Variables\n", + "v1=4.8#......#Volume of air compressed in m**3/s\n", + "p1=1#....#Inlet pressure in bar\n", + "t1=293#........#Inlet pressure in K\n", + "n=1.5#........#Compression index\n", + "Cf1=65#......#Air flow velocity at inlet in m/s\n", + "Cf2=Cf1#......#Flow velocity is same at inlet and outlet\n", + "d1=0.32#..........#Inlet impeller diameter in m\n", + "d2=0.62#........#Outlet impeller diameter in m\n", + "N=8000#........#Blower rpm\n", + "cp=1.005#......#Specific heat at constant pressure in kJ/kgK\n", + "#Calculations\n", + "t21=t1*((n/p1)**((n-1)/n))#....#Temperature at the outlet of compressor in K\n", + "Cbl1=(pi*d1*N)/60#......#Peripheral velocity at inlet in m/s\n", + "Cbl2=(pi*N*d2)/60#......#Tip peripheral velocity at outlet in m/s\n", + "Cw2=(cp*(t21-t1)*1000)/Cbl2#\n", + "be1=(atan(Cf1/Cbl1))*180/pi#\n", + "be2=(atan(Cf2/(Cbl2-Cw2)))*180/pi#......#Blade angles at the tip of the impeller\n", + "al2= (atan(Cf2/Cw2))*180/pi# \n", + "print \"\\nBlade angles \\n\\n\\t Blade angle at the inlet of the impeller: beta 1=%f \\n\\t Blade angle at the outlet of the impeller: beta 2=%f \\n\\t Absolute angle at the tip of impeller: alpha 2=%f\\n\\n\"%(be1,be2,al2)\n", + "b1=v1/(2*pi*(d1/2)*Cf1)#........#Breadth of blade at inlet in m\n", + "print \"Breadth of the blade at inlet = %0.2f cm\"%(b1*100)\n", + "v2=(v1*t21*p1)/(n*t1)#..............#Discharge at the outlet in m**3/s\n", + "b2=v2/(2*pi*(d2/2)*Cf2)#........#Breadth of blade at outlet in m\n", + "print \"Breadth of the blade at outlet = %0.2f cm\"%(b2*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.41 PAGE 781" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diameter of the impeller = 0.56 m\n", + "Power input to compressor = 2951.61 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "m=16.5#.......#Air flow in kg/s\n", + "rp=4#........#Pressure ratio\n", + "N=15000#.........#Compressor rpm\n", + "t01=293#.........#Inlet head temperature\n", + "fis=0.9#...........#Slip factor\n", + "fiw=1.04#.........#Power input factor\n", + "etaisen=0.8#.......#Isentropic efficiency\n", + "cp=1.005#........#Specific heat at constant pressure in kJ/kgK\n", + "ga=1.4#......#Ratio of specific heats\n", + "#Calculations\n", + "t021=t01*(rp**((ga-1)/ga))#\n", + "delt=(t021-t01)/etaisen#\n", + "Cbl2=sqrt((1000*cp*delt)/(fiw*fis))#\n", + "D=(60*Cbl2)/(pi*N)#..............#Diameter of impeller\n", + "print \"Diameter of the impeller = %0.2f m\"%D\n", + "P=m*cp*delt#\n", + "print \"Power input to compressor = %0.2f kW\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.42 PAGE 781" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total head temperature = 447.09 K\n", + "Static temperature at outlet = 439.93 K\n", + "Static pressure at exit = 3.45 bar\n", + "Static temperature at inlet = 278.25 K\n", + "Static pressure at inlet = 0.90 bar\n", + "Static pressure ratio : 3.8237\n", + "Work done on air = 159.89 kJ/kg\n", + "Power required to drive the compressor = 1918.67 kW\n" + ] + } + ], + "source": [ + "\n", + "# Initialisation of Variables\n", + "rp=3.6#..........#Pressure ratio\n", + "die=0.35#.......#Diameter of inlet eye of compressor in m\n", + "Cf=140#..........#Axial velocity in m/s\n", + "m=12#.............#Mass flow in kg/s\n", + "Cbl2=120#.........#Velocity in the delivery duct in m/s\n", + "Ci=460#..........#The tip speed of the impeller in m/s\n", + "N=16000#............#Speed of impeller in rpm\n", + "etaisen=0.8#.......#Isentropic efficiency\n", + "pc=0.73#........#Pressure co efficient\n", + "pa=1.013#..........#Ambient pressure in bar\n", + "ta=273+15#................#Ambient temperature in K\n", + "ga=1.4#..........#Ratio of specific heats\n", + "cp=1.005#.........#Specific heat at constant pressure in kJ/kgK\n", + "R=0.287#........#Gas constant in kJ/kgK\n", + "#Calculations\n", + "delt=((ta*((rp**((ga-1)/ga))-1))/etaisen)#.......#Rise in temperature\n", + "t02=ta+delt#............#Total head temperature in K\n", + "print \"Total head temperature = %0.2f K\"%t02\n", + "t2=t02-((Cbl2*Cbl2)/(2*cp*1000))#..........#Static temperature at outlet in K\n", + "print \"Static temperature at outlet = %0.2f K\"%t2\n", + "p02=pa*rp#\n", + "p2=p02/(1+((Cbl2*Cbl2)/(2*R*t2*1000)))#...........#Static pressure at exit in bar\n", + "print \"Static pressure at exit = %0.2f bar\"%p2\n", + "t1=ta-((Cf*Cf)/(2*cp*1000))#.............#Static temperature at inlet in K\n", + "print \"Static temperature at inlet = %0.2f K\"%t1\n", + "p1=pa/(1+((Cf*Cf)/(2*R*t1*1000)))#...........#Static pressure at inlet in bar\n", + "print \"Static pressure at inlet = %0.2f bar\"%p1\n", + "rp=p2/p1#.....#Static pressure ratio\n", + "print \"Static pressure ratio : %0.4f\"%rp\n", + "W=cp*delt#...........#Work done on air in kJ/kg of air\n", + "print \"Work done on air = %0.2f kJ/kg\"%W\n", + "P=m*cp*delt#..........#Power required to drive the compressor in kW\n", + "print \"Power required to drive the compressor = %0.2f kW\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.43 PAGE 782" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature rise of air while passing through compressor: 159.09 \n", + "Pressure ratio : 3.8237\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Initialisation of Variables\n", + "t1=300#.........#Inlet temperature in K\n", + "N=18000#.............#Compressor rpm\n", + "etaisen=0.76#.......#Isentropic efficiency\n", + "od=0.55#......#Outer diameter of blade tip\n", + "sf=0.82#......#Slip factor\n", + "cp=1.005#.........#Specific heat capacity at constant pressure in kJ/kgK\n", + "ga=1.4#.............#Ratio of specific heats\n", + "#Calculations\n", + "Cbl2=(pi*od*N)/60#W=Cbl2*Cbl2*sf/1000#...........#Work done per kg of air in kW\n", + "delt=W/cp#..............#Temperature rise of air while passing through compressor \n", + "print \"Temperature rise of air while passing through compressor: %0.2f \"%delt\n", + "t21=(etaisen*delt)+t1#rp=((t21/t1)**(ga/(ga-1)))#.....#Pressure ratio\n", + "print \"Pressure ratio : %0.4f\"%rp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.44 PAGE 782" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pressure rise = 0.15 bar\n", + "Work done per kg of air = 27.54 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt,tan\n", + "# Initialisation of Variables\n", + "Cbl=240#........#Mean blade velocity in m/s\n", + "Cf=190#.........#Air flow velocity in m/s\n", + "al1=45#\n", + "al2=14#.........#Blade angels in degrees\n", + "rho=1#.........#Density of air in kg/m**3\n", + "#Calculations \n", + "pr=(1/2)*(rho*Cf*Cf/(10**5))*(((tan(al1*pi/180))**2)-((tan(al2*pi/180))**2))#.......#Pressure rise in bar\n", + "print \"Pressure rise = %0.2f bar\"%pr\n", + "W=Cbl*Cf/1000*((tan(al1*pi/180))-(tan(al2*pi/180)))#.............#Work done per kg of air in kW\n", + "print \"Work done per kg of air = %0.2f kW\"%W" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.45 PAGE 783" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work done by the machine = 174.52 kJ/kg\n", + "\n", + "Blade angels are as follows (In degrees)\n", + "\n", + "alpha1=18.091622\tbeta1=59.136727\n", + "\n", + "alpha2=59.136727\tbeta2=18.091622\n", + "\n", + "\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt,atan\n", + "# Initialisation of Variables\n", + "etaisen=0.82#.......#Overall isentropic efficiency \n", + "N=8#............#No of stages\n", + "t1=293#...........#Inlet temperature in K\n", + "ga=1.4#............#Ratio of specific heats\n", + "rp=4#.............#Pressure ratio\n", + "Rd=0.5#................#Reaction factor\n", + "Cbl=180#.................#Mean blade speed in m/s\n", + "Cf=90#...............#Air flow velocity in m/s\n", + "cp=1.005#.........#Specific heat at constant pressure in kJ/kgK\n", + "#Calculations\n", + "t21=t1*(rp**((ga-1)/ga))#\n", + "t2=((t21-t1)/etaisen)+t1#\n", + "wrt=cp*(t2-t1)#.........#Work done by the machine in kJ/kg\n", + "print \"Work done by the machine = %0.2f kJ/kg\"%wrt\n", + "be1=atan(((cp*(t2-t1)*1000/(Cf*Cbl*N))+(Cbl/Cf))/2)*180/pi#\n", + "al1=atan((Cbl/Cf)-tan(be1*pi/180))*180/pi#\n", + "print \"\\nBlade angels are as follows (In degrees)\\n\\nalpha1=%f\\tbeta1=%f\\n\\nalpha2=%f\\tbeta2=%f\\n\\n\"%(al1,be1,be1,al1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.46 PAGE 784" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Flow velocity = 95.45 m/s\n", + "No of stages : 8.0\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt,tan\n", + "# Initialisation of Variables\n", + "etaisen=0.85#.......#Overall isentropic efficiency\n", + "t1=293#...........#Inlet temperature in K\n", + "rp=4#.............#Pressure ratio\n", + "Rd=0.5#................#Reaction factor\n", + "Cbl=180#.................#Mean blade speed in m/s\n", + "wip=0.82#..............#Work input factor\n", + "al1=12#\n", + "be1=42#......#Blade angels in degrees\n", + "ga=1.4#............#Ratio of specific heats \n", + "cp=1.005#.........#Specific heat at constant pressure in kJ/kgK\n", + "#Calculations\n", + "t21=t1*(rp**((ga-1)/ga))#\n", + "t2=((t21-t1)/etaisen)+t1#\n", + "wrt=cp*(t2-t1)#.........#Theoretical work required in kJ/kg\n", + "Cf=Cbl/(tan(al1*pi/180)+tan(be1*pi/180))#\n", + "Cw1=Cf*tan(al1*pi/180)#Cw2=Cf*tan(be1*pi/180)#\n", + "wcps=Cbl*(Cw2-Cw1)*wip/1000#.............#Work consumed per stage in kJ/kg\n", + "N=round(wrt/wcps)#.......#No of stages\n", + "print \"Flow velocity = %0.2f m/s\"%Cf\n", + "print \"No of stages : \",N" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.47 PAGE 784" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + "Blade angels are as follows (In degrees)\n", + "\n", + "alpha1=4.924070\t\tbeta1=59.410637\n", + "\n", + "alpha2=59.410637\tbeta2=4.924070\n", + "\n", + "\n", + "Power required by the compressor = 184.95 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt,atan,tan\n", + "# Initialisation of Variables\n", + "rp=5#..........#Stagnation pressure ratio ga\n", + "etaisen=0.92#.......#Overall isentropic efficiency\n", + "t1=290#.............#Inlet stagnation temperature in K\n", + "p1=1#...............#Inlet stagnation pressure in bar\n", + "Cbl=160#...........#Mean blade speed in m/s\n", + "ga=1.4#...........#Ratio of specific heats\n", + "Rd=0.5#............#Degree of reaction\n", + "Cf=90#................#Axial velocity of air through compressor in m/s\n", + "N=8#.............#No of stages\n", + "m=1#.........#Mass flow in kg/s\n", + "cp=1.005#............#Specific heat at constant pressure in kJ/kgK\n", + "#Calculations\n", + "tN1=t1*(rp**((ga-1)/ga))#......#Temperature at the end of compression stage due to isentropic expansion in K\n", + "tN=((tN1-t1)/etaisen)+t1#\n", + "be1=atan(((cp*(tN-t1)*1000/(Cf*Cbl*N))+(Cbl/Cf))/2)*180/pi#\n", + "al1=atan((Cbl/Cf)-tan(be1*pi/180))*180/pi#\n", + "print \"\\nBlade angels are as follows (In degrees)\\n\\nalpha1=%f\\t\\tbeta1=%f\\n\\nalpha2=%f\\tbeta2=%f\\n\\n\"%(al1,be1,be1,al1)\n", + "P=m*cp*(tN-t1)#..........#Power required by the compressor in kW\n", + "print \"Power required by the compressor = %0.2f kW\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.48 PAGE 785" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Stagnation polytropic efficiency = 87.59 %\n", + "No of stages : 27.0\n", + "Inlet temperature = 295.35 K\n", + "Inlet pressure = 0.95 bar\n", + "Height of the blade in the first stage = 28.23 cm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt,cos,log,tan,atan\n", + "# Initialisation of Variables\n", + "rp=4#........#Stagnation pressure ratio\n", + "etaisen=0.85#.....#Stagnation isentropic efficiency\n", + "p1=1#.............#Inlet stagnation pressure in bar\n", + "t1=300#...........#Inlet stagnation temperature in K\n", + "Rd=0.5#............#Degree of reaction\n", + "Cu=180#...........#Mean blade speed in m/s\n", + "Wd=0.9#...........#Work done factor\n", + "htr=0.42#.......#Hub tip ratio\n", + "al1=12#\n", + "be2=al1#.......#Relative air angle at rotor inlet in degrees\n", + "al2=32#\n", + "be1=al2#........#Relative air angle at rotor at outlet in degrees\n", + "ga=1.4#...........#Ratio of specific heats\n", + "cp=1.005#..........#Specific heat capacity at constant pressure in kJ/kgK\n", + "R=287#..........#Gas constant in J/kgK\n", + "m=19.5#..........#Mass flow in kg/s\n", + "#Calculations\n", + "tN1=t1*(rp**((ga-1)/ga))#......#Temperature at the end of compression stage due to isentropic expansion in K\n", + "tN=((tN1-t1)/etaisen)+t1#\n", + "etap=log(rp**((ga-1)/ga))/log(tN/t1)#...........#Stagnation polytropic efficiency\n", + "print \"Stagnation polytropic efficiency = %0.2f %%\"%(etap*100)\n", + "Cf=Cu/(tan(al1*pi/180)+tan(be1*pi/180))#\n", + "Cw1=Cf*tan(al1*pi/180)\n", + "Cw2=Cf*tan(al2*pi/180)#\n", + "wcps=Cu*(Cw2-Cw1)*Wd/1000#.............#Work consumed per stage in kJ/kg\n", + "wc=cp*(tN-t1)#...............#Work consumed by compressor in kJ/kg\n", + "N=round(wc/wcps)#.......#No of stages\n", + "print \"No of stages : \",N\n", + "C1=Cf/cos(al1*pi/180)#.......#Absolute velocity at exit from guide vanes in m/s\n", + "ti=t1-((C1*C1)/(2*cp*1000))#..........#Inlet temperature in K\n", + "print \"Inlet temperature = %0.2f K\"%ti\n", + "pi=p1*((ti/t1)**(ga/(ga-1)))#......#Inlet pressure in bar\n", + "print \"Inlet pressure = %0.2f bar\"%pi\n", + "rho1=(pi*10**5)/(R*ti)#.............#Density of air approaching the first stage\n", + "r1=sqrt(m/(rho1*pi*Cf*(1-(htr**2))))#\n", + "rh=r1*htr#\n", + "l=r1-rh#............#Height of the blade in the first stage in m\n", + "print \"Height of the blade in the first stage = %0.2f cm\"%(l*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 20.49 PAGE 785" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Delivery pressure = 5.57 bar\n", + "No of stages : 11.0\n", + "Overall efficiency = 84.86 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt,cos,log,tan,atan,exp\n", + "# Initialisation of Variables\n", + "ma=20#..........#Air flow rate in kg/s\n", + "p1=1#.........#Inlet stagnation pressure in bar\n", + "t1=290#.........#Inlet stagnation temperature in Kelvin\n", + "t2=305#.........#Temperature at the end of first stage in K\n", + "etapc=0.88#.....#Polytropic efficiency of compression\n", + "P=4350#......#Power consumed by compressor in kW\n", + "ga=1.4#.....#Ratio of specific heats\n", + "cp=1.005#......#Specific heat at constant pressure \n", + "#Calculations\n", + "p2byp1=(exp(etapc*log(t2/t1)))**(ga/(ga-1))#\n", + "tN=(P/(ma*cp))+t1#\n", + "pN=p1*((tN/t1)**((etapc*ga)/(ga-1)))#......#Delivery pressure in bar\n", + "print \"Delivery pressure = %0.2f bar\"%pN\n", + "N=log(pN/p1)/log(p2byp1)#...........#No of stages \n", + "print \"No of stages : \",round(N)\n", + "tN1=t1*((pN/p1)**((ga-1)/ga))#\n", + "etao=(tN1-t1)/(tN-t1)#...............#Overall efficiency\n", + "print \"Overall efficiency = %0.2f %%\"%(etao*100)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER21_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER21_3.ipynb new file mode 100644 index 00000000..2de19a92 --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER21_3.ipynb @@ -0,0 +1,1275 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 21 - Gas turbines and jet propulsion" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.1 PAGE 848" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power developed = 250.22 kW/kg of air\n", + "Thermal efficiency = 17.96 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#.........#Pressure of air while entering the turbine in bar\n", + "t1=293#........#Temperature of air entering the turbine in K\n", + "p2=4#.........#Pressure of air after compression in bar\n", + "etac=0.8#....#Efficiency of compressor\n", + "etat=0.85#.....#Efficiency of turbine\n", + "afr=90#........#Air fuel ratio\n", + "ma=3#...........#Mass of air in kg/s\n", + "ga=1.4#........#Ratio of specific heats\n", + "cp=1#.............#Specific heat at constant pressure in kJ/kgK\n", + "C=41800#.............#Calorific value of fuel in kJ/kg\n", + "#Calculations\n", + "t2=t1*((p2/p1)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K\n", + "t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K\n", + "t3=round((C/((afr+1)*cp))+t21)#..............#Temperature before expansion in turbine in K\n", + "p4=p1#\n", + "p3=p2#\n", + "t4=t3*((p4/p3)**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K\n", + "t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K\n", + "wt=((afr+1)/afr)*cp*(t3-t41)#........#Work done by turbine in kJ/kg of air\n", + "wc=round(1*cp*(t21-t1))#.................#Work done by compression in kJ/kg of air\n", + "wnet=wt-wc#..........#Net work done in kJ/kg\n", + "P=wnet*ma#.................#Power developed in kW/kg of air\n", + "print \"Power developed = %0.2f kW/kg of air\"%P\n", + "qs=(1/afr)*C#................#Heat supplied in kJ/kg of air\n", + "etath=wnet/qs#................#Thermal efficiency\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.2 PAGE 848" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power developed = 770.31 kW/kg of air\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=288#........#Temperature of air entering the turbine in K\n", + "t3=883#..............#Temperature before expansion in turbine in K\n", + "etac=0.8#....#Efficiency of compressor\n", + "etat=0.82#.....#Efficiency of turbine\n", + "rp=6#...........#Pressure ratio\n", + "ma=16#...........#Mass of air in kg/s\n", + "gac=1.4#........#Ratio of specific heats for compression process\n", + "gae=1.333#............#Ratio of specific heats for expansion process\n", + "cpc=1.005#.............#Specific heat at constant pressure in kJ/kgK during compression process\n", + "cpe=1.11#.............#Specific heat at constant pressure in kJ/kgK during expansion process\n", + "C=41800#.............#Calorific value of fuel in kJ/kg\n", + "#Calculations\n", + "t2=t1*((rp)**((gac-1)/gac))#...............#Ideal temperature of air after compression in K\n", + "t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K\n", + "t4=t3/((rp)**((gae-1)/gae))#............#Ideal temperature after expansion in turbine in K\n", + "t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K\n", + "wt=cpe*(t3-t41)#........#Work done by turbine in kJ/kg of air\n", + "wc=(1*cpc*(t21-t1))#.................#Work done by compression in kJ/kg of air\n", + "wnet=wt-wc#..........#Net work done in kJ/kg\n", + "P=wnet*ma#.................#Power developed in kW/kg of air\n", + "print \"Power developed = %0.2f kW/kg of air\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.3 PAGE 849" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal efficiency = 32.65 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#.........#Pressure of air while entering the turbine in bar\n", + "t1=300#........#Temperature of air entering the turbine in K\n", + "p2=6.2#.........#Pressure of air after compression in bar\n", + "etac=0.88#....#Efficiency of compressor\n", + "etat=0.9#.....#Efficiency of turbine\n", + "far=0.017#........#Fuel air ratio\n", + "ga=1.4#........#Ratio of specific heats for compression\n", + "gae=1.333#........#Ratio of specific heats for expansion\n", + "cp=1.147#.............#Specific heat at constant pressure in kJ/kgK during expansion\n", + "cpc=1.005#.............#Specific heat at constant pressure in kJ/kgK during compression\n", + "C=44186#.............#Calorific value of fuel in kJ/kg\n", + "#Calculations\n", + "t2=t1*((p2/p1)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K\n", + "t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K\n", + "t3=(((C*far)/((far+1)*cpc))+t21)#..............#Temperature before expansion in turbine in K\n", + "p4=p1#\n", + "p3=p2#\n", + "t4=t3*((p4/p3)**((gae-1)/gae))#............#Ideal temperature after expansion in turbine in K\n", + "t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K\n", + "wt=(cp*(t3-t41))#........#Work done by turbine in kJ/kg of air\n", + "wc=round(1*cpc*(t21-t1))#.................#Work done by compression in kJ/kg of air\n", + "wnet=wt-wc#..........#Net work done in kJ/kg\n", + "qs=(far)*C#................#Heat supplied in kJ/kg of air\n", + "etath=wnet/qs#................#Thermal efficiency\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.4 PAGE 850" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air fuel ratio is 56:1\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=300#........#Temperature of air entering the turbine in K\n", + "t3=1148#..............#Temperature before expansion in turbine in K\n", + "etac=0.8#....#Efficiency of compressor\n", + "etat=0.852#.....#Efficiency of turbine\n", + "rp=4#...........#Pressure ratio\n", + "p1=1#...............#Pressure of air before entering compressor\n", + "ga=1.4#........#Ratio of specific heats\n", + "cp=1.0#.............#Specific heat at constant pressure in kJ/kgK \n", + "C=42000#.............#Calorific value of fuel in kJ/kg\n", + "perlcc=10#............#Percent loss of calorific value of fuel in combustion chamber\n", + "#Calculations\n", + "p2=p1*rp#.................#Pressure of air after compression in bar\n", + "etacc=(100-perlcc)/100#.......#Efficiency of combustion chamber\n", + "t2=t1*((rp)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K\n", + "t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K\n", + "afr=((C*etacc)/(cp*(t3-t21)))-1#........#Air fuel ratio\n", + "print \"Air fuel ratio is %d:1\"%round(afr)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.5 PAGE 851" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal efficiency = 15.92 %\n", + "The work ratio is 0.259290\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=300#........#Temperature of air entering the turbine in K\n", + "t3=883#..............#Temperature before expansion in turbine in K\n", + "etac=0.8#....#Efficiency of compressor\n", + "etat=0.852#.....#Efficiency of turbine\n", + "rp=4#...........#Pressure ratio\n", + "p1=1#...............#Pressure of air before entering compressor\n", + "ga=1.4#........#Ratio of specific heats\n", + "cp=1.11#.............#Specific heat at constant pressure in kJ/kgK \n", + "C=42000#.............#Calorific value of fuel in kJ/kg\n", + "perlcc=10#............#Percent loss of calorific value of fuel in combustion chamber\n", + "#Calculations\n", + "p2=p1*rp#.................#Pressure of air after compression in bar\n", + "etacc=(100-perlcc)/100#.......#Efficiency of combustion chamber\n", + "t2=t1*((rp)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K\n", + "t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K\n", + "qs=cp*(t3-t21)#...................#Heat supplied in kJ/kg\n", + "t4=t3/((rp)**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K\n", + "t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K\n", + "wt=cp*(t3-t41)#........#Work done by turbine in kJ/kg of air\n", + "wc=(1*cp*(t21-t1))#.................#Work done by compression in kJ/kg of air\n", + "wnet=wt-wc#..........#Net work done in kJ/kg\n", + "etath=wnet/qs#................#Thermal efficiency\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)\n", + "wrr=wnet/wt#...................#Work ratio\n", + "print \"The work ratio is %f\"%wrr" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.6 PAGE 851" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Quantity of air in circulation = 13.49 kg\n", + "Actual heat supplied per kg of air circulation = 552.66 kJ\n", + "Thermal efficiency = 14.28 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#.........#Pressure of air while entering the turbine in bar\n", + "t1=293#........#Temperature of air entering the turbine in K\n", + "p2=5#.........#Pressure of air after compression in bar\n", + "plcc=0.1#.....#Pressure loss in combustion chamber in bar\n", + "t3=953#............#Temperature before expansion in turbine in K\n", + "etac=0.85#....#Efficiency of compressor\n", + "etat=0.8#.....#Efficiency of turbine\n", + "etacc=0.85#......#Efficiency of combustion chamber\n", + "ga=1.4#........#Ratio of specific heats\n", + "cp=1.024#.............#Specific heat at constant pressure in kJ/kgK \n", + "P=1065#.............#Power developed by the plant in kW\n", + "\n", + "#Calculations\n", + "p3=p2-plcc#........................#Pressure before expansion in turbine in bar\n", + "p4=p1#\n", + "t2=t1*((p2/p1)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K\n", + "t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K\n", + "t4=t3*((p4/p3)**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K\n", + "t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K\n", + "wt=(cp*(t3-t41))#........#Work done by turbine in kJ/kg of air\n", + "wc=round(1*cp*(t21-t1))#.................#Work done by compression in kJ/kg of air\n", + "wnet=wt-wc#..........#Net work done in kJ/kg\n", + "ma=P/wnet#.............#Quantity of air in circulation in kg\n", + "print \"Quantity of air in circulation = %0.2f kg\"%(ma)\n", + "qs=cp*(t3-t21)/etac#..................#Actual heat supplied per kg of air circulation in kJ\n", + "print \"Actual heat supplied per kg of air circulation = %0.2f kJ\"%qs\n", + "etath=wnet/qs#.............#Thermal efficiency\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.7 PAGE 852" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pressure ratio of low pressure turbine : 2.3042\n", + "Temperature of the exhaust from the unit = 672.95 K\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "ma=20#..............#Air flow rate in kg/s\n", + "t1=300#........#Temperature of air entering the turbine in K\n", + "t3=1000#............#Temperature before expansion in turbine in K\n", + "rp=4#...............#Pressure ratio\n", + "cp=1#.............#Specific heat at constant pressure in kJ/kgK \n", + "ga=1.4#........#Ratio of specific heats\n", + "#Calculations\n", + "t2=t1*((rp)**((ga-1)/ga))#...............#Temperature of air after compression in K\n", + "t4=t3-t2+t1#............#Temperature after expansion in turbine in K\n", + "prlp=rp/((t3/t4)**(ga/(ga-1)))#.............#Pressure ratio of low pressure turbine\n", + "print \"Pressure ratio of low pressure turbine : %0.4f\"%prlp\n", + "t5=t4/((prlp)**((ga-1)/ga))#............#Temperature of the exhaust from the unit in K\n", + "print \"Temperature of the exhaust from the unit = %0.2f K\"%t5" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.8 PAGE 853" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cycle efficiency = 31.04 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#.........#Pressure of air while entering the turbine in bar\n", + "t1=300#........#Temperature of air entering the turbine in K\n", + "t21=490#........#Actual temperature of air after compression in K\n", + "t3=1000#............#Temperature before expansion in turbine in K\n", + "rp=5#.............#Pressure ratio\n", + "etac=0.8#....#Efficiency of compressor\n", + "etat=0.8#.....#Efficiency of turbine\n", + "ga=1.4#........#Ratio of specific heats\n", + "cp=1.005#.............#Specific heat at constant pressure in kJ/kgK \n", + "#Calculations\n", + "t4=t3/((rp)**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K\n", + "t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K\n", + "t5=((t41-t21)*etac)+t21#...........#Temperature of the exhaust from the unit in K\n", + "wc=cp*(t21-t1)#.............#Work consumed by compressor in kJ/kg\n", + "wt=cp*(t3-t41)#........#Work done by turbine in kJ/kg\n", + "qs=cp*(t3-t5)#..........#Heat supplied in kJ/kg\n", + "etac=(wt-wc)/qs#.........#Cycle efficiency\n", + "print \"Cycle efficiency = %0.2f %%\"%(etac*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.9 PAGE 853" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal efficiency of power turbine = 21.73 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#.........#Pressure of air while entering the turbine in bar\n", + "t1=288#........#Temperature of air entering the turbine in K\n", + "p2=8#.........#Pressure of air after compression in bar\n", + "t3=1173#.............#Temperature before expansion in turbine in K\n", + "etac=0.76#....#Efficiency of compressor\n", + "etat=0.86#.....#Efficiency of turbine\n", + "ma=23#.........#Quantity of air circulation in kg/s\n", + "ga=1.4#........#Ratio of specific heats for compression\n", + "gag=1.34#........#Ratio of specific heats for expansion\n", + "cp=1.005#.............#Specific heat at constant pressure in kJ/kgK \n", + "cpg=1.128#.............#Specific heat at constant pressure in kJ/kgK\n", + "C=4200#.............#Calorific value of fuel in kJ/kg\n", + "etamech=0.95#........#Mechanical efficiency\n", + "etagen=0.96#.........#Generator efficiency\n", + "#Calculations\n", + "t2=t1*((p2/p1)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K\n", + "t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K\n", + "p4=p1#\n", + "p3=p2#.............#Isobaric processes\n", + "t4=t3*((p4/p3)**((gag-1)/gag))#............#Ideal temperature after expansion in turbine in K\n", + "t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K\n", + "wc=cp*(t21-t1)#................#Work dony by compressor\n", + "m1=(wc)/(cpg*(t3-t41))#.............#Flow through compressor turbine in kg\n", + "m2=1-m1#..............#Flow through power turbine in kg\n", + "wpt=m2*(cpg*(t3-t41))#.........#turbine work in kJ/kg\n", + "P=ma*wpt*etamech*etagen#.........#Power output in kW\n", + "qi=cpg*t3-cp*t21#.............#Input heat in kJ/kg of air\n", + "etath=wpt/qi#.............#Thermal efficiency of power turbine\n", + "print \"Thermal efficiency of power turbine = %0.2f %%\"%(etath*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.10 PAGE 854" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of gasses entering the turbine = 654.75 K\n", + "Pressure of gasses entering the power turbine = 1.65 bar\n", + "Net power output = 73.95 kW\n", + "Work ratio : 0.2198\n", + "Thermal efficiency of the unit : 19.26 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=288#........#Temperature of air entering the turbine in K\n", + "t3=883#..............#Temperature before expansion in turbine in K\n", + "etac=0.82#....#Efficiency of compressor\n", + "etathp=0.85#.....#Efficiency of high pressure turbine\n", + "etatlp=0.85#.....#Efficiency of low pressure turbine\n", + "rp=7#...........#Pressure ratio\n", + "p1=1.01#...............#Pressure of air before entering compressor\n", + "ga=1.4#........#Ratio of specific heats for compression\n", + "gag=1.333#........#Ratio of specific heats for expansion\n", + "cp=1.005#.............#Specific heat at constant pressure in kJ/kgK \n", + "cpg=1.15#.............#Specific heat at constant pressure in kJ/kgK in generator\n", + "#Calculations\n", + "p2=p1*rp#\n", + "t2=t1*((p2/p1)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K\n", + "t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K\n", + "wc=cp*(t21-t1)#............#Compressor work in kJ/kg\n", + "t41=t3-(wc/cpg)#..........#Temperature of gasses entering the turbine in K\n", + "print \"Temperature of gasses entering the turbine = %0.2f K\"%t41\n", + "t4=round(t3-((t3-t41)/etathp))#.........#Ideal temperature of gases entering the turbine in K\n", + "p3=p2#.........#Isobaric processes\n", + "p4=p3/((t3/t4)**(1/((gag-1)/gag)))#....#Pressure of gasses entering the power turbine in bar\n", + "print \"Pressure of gasses entering the power turbine = %0.2f bar\"%p4\n", + "t5=t41*((((t3/t4)**(1/((gag-1)/gag)))/(rp))**((gag-1)/gag))#\n", + "t51=t41-(etatlp*(t41-t5))#\n", + "wlp=cpg*(t41-t51)#............#Net power output in kW\n", + "print \"Net power output = %0.2f kW\"%wlp\n", + "wr=wlp/(wlp+wc)#............#Work ratio\n", + "print \"Work ratio : %0.4f\"%wr\n", + "qs=cpg*(t3-t21)#...........#Heat supplied in kJ/kg\n", + "etath=wlp/qs#..........#Thermal efficiency\n", + "print \"Thermal efficiency of the unit : %0.2f %%\"%(etath*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.11 PAGE 855" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net work output = 173.98 kW\n", + "Power plant efficiency = 25.00 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "# Initialisation of Variables\n", + "rp=5.6#................#Pressure ratio\n", + "t1=303#.............#Temperature of intake air in K\n", + "p1=1#............#Pressure of intake air in bar\n", + "t5=973#............#Highest temperature of the cycle in K\n", + "etac=0.85#..........#Effeciency of compressor\n", + "etat=0.9#..........#Efficiency of turbine\n", + "ma=1.2#..........#Rate of air flow in kg/s\n", + "cp=1.02#...........#Specific heat at constant volume in kJ/kgK\n", + "ga=1.41#.............#Ratio of specific heats\n", + "#Calculations\n", + "t2=t1*((sqrt(rp))**((ga-1)/ga))#\n", + "t21=((t2-t1)/etac)+t1#\n", + "wc=2*ma*cp*(t21-t1)#............#Work input for the two stage compressor in kJ/s\n", + "t6=t5/(rp**((ga-1)/ga))#\n", + "t61=t5-etat*(t5-t6)#\n", + "wt=ma*cp*(t5-t61)#...............#Work output from turbine in kJ/s\n", + "wnet=wt-wc#....................#Net work available in kJ/s\n", + "print \"Net work output = %0.2f kW\"%wnet\n", + "qs=ma*cp*(t5-t21)#.................#Heat supplied in kJ/s\n", + "etath=wnet/qs#\n", + "print \"Power plant efficiency = %0.2f %%\"%(etath*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.13 PAGE 856" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cycle efficiency = 11.81 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=288#.............#Temperature of intake air in K\n", + "rp=4#.............#Pressure ratio\n", + "etac=0.82#.........#Compressor efficiency\n", + "etahe=0.78#...........#Efficiency of heat exchanger\n", + "etat=0.7#...........#Turbine efficiency\n", + "t3=873#............#Temperature before expansion in turbine in K\n", + "R=0.287#............#Gas constant for air in kJ/kgK\n", + "ga=1.4#...........#Ratio of specific heats\n", + "#Calculations\n", + "t2=t1*((rp)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K\n", + "t21=((t2-t1)/etac)+t1#...............#Actual temperature of air after compression in K\n", + "t4=t3/(rp**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K\n", + "t41=t3-etat*(t3-t4)#............#Actual temperature after expansion in turbine in K\n", + "cp=R*(ga/(ga-1))#..............#Specific heat at constant pressure in kJ/kgK\n", + "wc=cp*(t21-t1)#.............#Compressor work in kJ/kg\n", + "wt=cp*(t3-t41)#....................#Turbine work in kJ/kg\n", + "wnet=wt-wc#....................#Net work available in kJ/s\n", + "t5=(etahe*(t41-t21))+t21#\n", + "qs=cp*(t3-t5)#.................#Heat supplied in kJ/kg\n", + "etac=wnet/qs#...............#Cycle efficiency\n", + "print \"Cycle efficiency = %0.2f %%\"%(etac*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.14 PAGE 857" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal efficiency in simple cycle = 23.25 %\n", + "Thermal efficiency in heat exchanger cycle = 20.94 %\n", + "Increase in thermal efficiency = -2.31 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "etahe=0.72#.................#Efficiency of heat exchanger\n", + "p1=1.01#.........#Pressure of air while entering the turbine in bar\n", + "t1=293#........#Temperature of air entering the turbine in K\n", + "p2=4.04#.........#Pressure of air after compression in bar\n", + "etat=0.85#..........#Turbine efficiency\n", + "pdhe=0.05#............#Pressure drop on each side of heat exchanger in bar\n", + "pdcc=0.14#...........#Pressure drop in combustion chamber in bar\n", + "etac=0.8#...........#Compressor efficiency\n", + "ga=1.4#.............#Ratio of specific heats\n", + "C=41800#.............#Calorific value of fuel in kJ/kg\n", + "cp=1.024#...........#Specific heat at constant pressure in kJ/kgK\n", + "afrc=90#..............#Air fuel ratio for simple cycle\n", + "#Calculations\n", + "t2=(t1*((p2/p1)**((ga-1)/ga)))#...............#Ideal temperature of air after compression in K\n", + "t21=round(((t2-t1)/etac)+t1)#...............#Actual temperature of air after compression in K\n", + "t3=((1*C)/(cp*(afrc+1)))+t21#............#Temperature before expansion in turbine in K\n", + "p4=p1#\n", + "p3=p2-pdcc#\n", + "t4=round(t3*((p4/p3)**((ga-1)/ga)))#............#Ideal temperature after expansion in turbine in K\n", + "t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K\n", + "etath=(t3-t41-t21+t1)/(t3-t21)#...........#Thermal efficiency in simple cycle\n", + "print \"Thermal efficiency in simple cycle = %0.2f %%\"%(etath*100)\n", + "p3he=p2-pdhe-pdcc#..........#Pressure before expansion in turbine in bar in heat exchanger cycle\n", + "p4he=p1+pdhe#................#Pressure after expansion in turbine in bar in heat exchanger cycle\n", + "t4he=t3*((p4he/p3he)**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K in heat exchanger cycle\n", + "t41he=round(t3-(etat*(t3-t4he)))#.................#Actual temperature after expansion in turbine in K in heat exchanger cycle\n", + "t5=(etahe*(t41he-t21))+t21#\n", + "etathhe=(t3-t41he-t21+t1)/(t3-t5)#.............#Thermal efficiency for heat exchanger cycle\n", + "print \"Thermal efficiency in heat exchanger cycle = %0.2f %%\"%(etathhe*100)\n", + "inc=etathhe-etath#\n", + "print \"Increase in thermal efficiency = %0.2f %%\"%(inc*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.15 PAGE 858" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal efficiency = 23.44 %\n", + "The workk ratio is 0.283\n", + "Mass flow = 39.12 kg/s\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "# Initialisation of Variables\n", + "t1=293#........#Temperature of air entering the turbine in K\n", + "rp=9#............#Overall pressure ratio\n", + "etac=0.8#........#Efficiency of compressor\n", + "t6=898#..........#Reheat remperature\n", + "t8=t6#\n", + "etat=0.85#.......#Efficiency of turbine\n", + "etamech=0.95#..........#Mechanical efficiency\n", + "etahe=0.8#...............#Heat exchanger thermal efficiency\n", + "cpg=1.15#.............#Specific heat capacity for gases in heat exchanger in kJ/kgK\n", + "cpa=1.005#............#Specific heat capacity for normal air in kJ/kgK\n", + "gag=1.333#.............#Ratio of specific heats for gases in heat exchanger \n", + "ga=1.4#...............#Ratio of specific heats for normal gases\n", + "P=4500#.................#Power output of turbine in kW\n", + "#Calculations\n", + "t2=t1*((sqrt(rp))**((ga-1)/ga))#\n", + "t21=((t2-t1)/etac)+t1#\n", + "wc=cpa*(t21-t1)#............#Work input per compressor stage\n", + "whp=(2*wc)/etamech#.........#Work output of HP turbine in kJ/kg\n", + "t71=t6-(whp/cpg)#\n", + "t7=round(t6-((t6-t71)/etat))#\n", + "k=(rp/((t6/t7)**((gag)/(gag-1))))**((gag-1)/gag)#\n", + "k1=((round((k/2)*100))*2)/100#..............#Rounding off upto 2 decimals\n", + "t9=t8/(k1)#\n", + "t91=t8-((t8-t9)*etat)#\n", + "wout=cpg*(t8-t91)*etamech#..............#Net work output in kJ/kg\n", + "t5=etahe*(t91-t21)+t21#\n", + "qs=cpg*(t6-t5)+cpg*(t8-t71)#...............#Heat supplied\n", + "etath=wout/qs#.................#Thermal efficiency\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)\n", + "wgross=whp+(wout/etamech)#.........#Gross work output in kJ/kg\n", + "wr=wout/wgross#................#Work ratio\n", + "print \"The workk ratio is %0.3f\"%wr\n", + "m1=P/wout#...............#Mass flow in kg/s\n", + "print \"Mass flow = %0.2f kg/s\"%m1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.16 PAGE 859" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal efficiency of the turbine without regenerator = 27.01 %\n", + "Thermal efficiency of the turbine with regenerator = 38.96 %\n", + "Mass of fluid circulated = 2.40 kg/s\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "# Initialisation of Variables\n", + "#Conditions of the closed gas turbine\n", + "t1=293#.............#Temperature at the inlet of first stage compressor in K\n", + "t5=1023#.................#Maximum temperature in K\n", + "p1=1.5#................#Inlet pressure in bar\n", + "p2=6#.................#Pressure in bar\n", + "etac=0.82#..............#Compressor efficiency\n", + "etat=0.82#..............#Turbine efficiency\n", + "etare=0.70#................#Regenerator efficiency\n", + "P=350#....................#Power developed by the plant in kW\n", + "ga=1.4#................#Ratio of specific heats\n", + "cp=1.005#..............#Specific heat at constant pressure in kJ/kgK\n", + "t3=t1#\n", + "#Calculations\n", + "t2=t1*((sqrt(p2/p1))**((ga-1)/ga))#\n", + "t21=((t2-t1)/etac)+t1#t41=t21#\n", + "t6=t5/((p2/sqrt(p1*p2))**((ga-1)/ga))#\n", + "t61=t5-(etat*(t5-t6))#\n", + "t81=t61#\n", + "t7=t5#\n", + "ta=(etare*(t81-t41))+t41#.......#Temperature of air coming out of regenerator in K\n", + "wnet=2*cp*(t5-t61-t21+t1)#........#Net work done in kJ/kg of air\n", + "qs=cp*(t5-t41+t7-t61)#...........#Heat supplied without regenerator in kJ/kg of air\n", + "qsr=cp*(t5-ta+t7-t61)#............#Heat supplied with regenerator in kJ/kg of air\n", + "etath=wnet/qs#.............#Thermal efficiency (without regenerator)\n", + "etathr=wnet/qsr#.........#Thermal efficiency (with regenerator)\n", + "mfl=P/wnet#..........#mass of fluid circulated in kg/s\n", + "print \"Thermal efficiency of the turbine without regenerator = %0.2f %%\"%(etath*100)\n", + "print \"Thermal efficiency of the turbine with regenerator = %0.2f %%\"%(etathr*100)\n", + "print \"Mass of fluid circulated = %0.2f kg/s\"%mfl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.17 PAGE 860" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net power output = 2042.40 kW \n", + "Thermal efficiency = 30.95 %\n", + "Specific fuel consumption = 0.28 kg/kWh\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=293#............#Temperature of inlet air into low pressure compressor in K\n", + "p1=1.05#.........#Pressure of inlet air into low pressure compressor in bar\n", + "t3=300#...........#Temperature of air after passing it through intercooler in K\n", + "t6=1023#..........#temperature of air in combustion chamber in K\n", + "rp=2#...........#Pressure ratio of each compressor \n", + "etac=0.82#........#Compressor efficiency\n", + "etat=0.82#..........#Turbine efficiency\n", + "etaht=0.72#............#Heat exchanger efficiency\n", + "ma=16#...........#Air flow in kg/s\n", + "ga=1.4#...........#Ratio of specific heats for air\n", + "gag=1.33#..........#Ratio of specific heats for gases\n", + "cpa=1.0#...........#Specific heat at constant pressure in kJ/kgK for air\n", + "cpg=1.15#.........#Specific heat at constant pressure in kJ/kgK for gases\n", + "C=42000#.........#Calorific value of fuel in kJ/kg\n", + "#Calculations\n", + "t2=round(t1*(rp**((ga-1)/ga)))#\n", + "t21=round(((t2-t1)/etac)+t1)#\n", + "t4=t3*(rp**((ga-1)/ga))#\n", + "t41=round(((t4-t3)/etac)+t3)#\n", + "t71=round(((cpg*t6)-cpa*(t21-t1+t41-t3))/cpg)#\n", + "t7=t6-((t6-t71)/etat)#\n", + "p6=p1*rp*rp#\n", + "p7=p6/((t6/t7)**((gag)/(gag-1)))#\n", + "t8=round(t71/((p7/p1)**((gag-1)/gag)))#\n", + "t81=round(t71-(etat*(t71-t8)))#\n", + "P=cpg*(t71-t81)#...........#Net power output in kJ/kg\n", + "print \"Net power output = %0.2f kW \"%(P*ma)\n", + "t5=etaht*(t81-t41)+t41#\n", + "qs=ma*cpg*(t6-t5)#......#Heat supplied in combustion chamber in kJ/s\n", + "etath=P*ma/qs#.........#Thermal efficiency\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)\n", + "afr=C/(cpg*(t6-t5))#......#Air fuel ratio\n", + "mf=ma*3600/afr#..............#Fuel supplied per hour in kg\n", + "sfc=mf/(P*ma)#...........#Specific fuel consumption in kg/kWh\n", + "print \"Specific fuel consumption = %0.2f kg/kWh\"%sfc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.18 PAGE 861" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Overall efficiency = 30.89 %\n", + "Work ratio = 0.363 Mass flow rate = 34.32 kg/s\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=293#............#Temperature of inlet air into low pressure compressor in K\n", + "p1=1.1#.........#Pressure of inlet air into low pressure compressor in bar\n", + "p2=3.3#..........#Pressure of air in the low pressure compressor in bar\n", + "t3=300#.............#Intercooled temperature in K\n", + "pli=0.15#..........#Loss in pressure due to intercooling in bar\n", + "p3=p2-pli#...........#Pressure after intercooling in bar\n", + "p4=9.45#............#Pressure of air after high pressure compressor in bar\n", + "p6=p4#\n", + "t6=973#.........#Temperature of gases supplied to high pressure turbine in K\n", + "t8=943#.........#Reheat temperature in K\n", + "plr=0.12#...........#Loss of pressure after reheating in bar\n", + "p7=3.62#............#Pressure of gases at the end of expansion in high pressure turbine in bar\n", + "p8=p7-plr#...........#Pressure of outlet gases in bar\n", + "ga=1.4#...........#Ratio of specific heats for air\n", + "gag=1.33#..........#Ratio of specific heats for gases\n", + "cpa=1.005#...........#Specific heat at constant pressure in kJ/kgK for air\n", + "cpg=1.15#.........#Specific heat at constant pressure in kJ/kgK for gases\n", + "etac=0.82#........#Compressor efficiency\n", + "etat=0.85#..........#Turbine efficiency\n", + "etaht=0.65#.........#Efficiency of heat exchanger\n", + "P=6000#..................#Power generated in kW\n", + "p9=p1#\n", + "#Calculations\n", + "t2=round(t1*((p2/p1)**((ga-1)/ga)))#\n", + "t21=round(((t2-t1)/etac)+t1)#\n", + "t4=round(t3*((p4/p3)**((ga-1)/ga)))#\n", + "t41=round(((t4-t3)/etac)+t3)#\n", + "t7=round(t6/((p6/p7)**((gag-1)/gag)))#\n", + "t71=round(t6-(etat*(t6-t7)))#\n", + "t9=round(t8/((p8/p9)**((gag-1)/gag)))#\n", + "t91=round(t8-(etat*(t8-t9)))#\n", + "t5=round(etaht*(t91-t41)+t41)#\n", + "wthp=cpg*(t6-t71)#.......#Work done by high pressure turbine in kJ/kg of gas\n", + "wtlp=cpg*(t8-t9)#.......#Work done by low pressure turbine in kJ/kg of gas\n", + "wchp=cpg*(t21-t1)#.......#Work done by high pressure compressor in kJ/kg of gas\n", + "wclp=cpg*(t41-t3)#.......#Work done by low pressure compressor in kJ/kg of gas\n", + "qs=cpg*(t6-t5+t8-t71)#.........#Heat supplied in kJ/kg of gas\n", + "etath=(wthp+wtlp-wchp-wclp)/qs#..#Overall efficiency\n", + "print \"Overall efficiency = %0.2f %%\"%(etath*100)\n", + "wr=(wthp+wtlp-wchp-wclp)/(wthp+wtlp)#......#Work ratio\n", + "print \"Work ratio = %0.3f \"%wr,\n", + "m=P/(wthp+wtlp-wchp-wclp)#.....#Mass flow rate\n", + "print \"Mass flow rate = %0.2f kg/s\"%m" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.19 PAGE 862" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Exit velocity of jet = 651.10 m/s\n", + "Rate of fuel consumption = 0.86 kg/s\n", + "Thrust specific fuel consumption = 0.00 kg/N\n", + "Thermal efficiency = 31.41 %\n", + "Propulsive power = 10437.94 kW\n", + "Propulsive efficiency = 59.81 %\n", + "Overall efficiency = 18.79 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "# Initialisation of Variables\n", + "ma=60.2#...........#Rate of air consumption in kg/s\n", + "delh=230#.......#Enthalpy change for nozzle in kJ/kg\n", + "z=0.96#..........#Velocity co efficient \n", + "afr=70#............#Air fuel ratio\n", + "etaco=0.92#...............#Combustion eficiency\n", + "CV=42000#..............#Calorific value of fuel in kJ/kg\n", + "v=1000#............#Velocity of aircraft in km/h\n", + "Ca=v*(5/18)#............#Aircraft velocity in m/s\n", + "#Calculations\n", + "Cj=z*sqrt(2*delh*v)#...........#Exit velocity of jet\n", + "print \"Exit velocity of jet = %0.2f m/s\"%Cj\n", + "mf=ma/afr#.........#Rate of fuel consumption\n", + "print \"Rate of fuel consumption = %0.2f kg/s\"%mf\n", + "tp=ma*(Cj-Ca)#......#Thrust produced in N\n", + "tsfc=mf/tp#.........#Thrust specific fuel consumption in kg/N\n", + "print \"Thrust specific fuel consumption = %0.2f kg/N\"%tsfc\n", + "etath=((Cj**2)-(Ca**2))/(2*(1/afr)*CV*etaco*1000)#.........#Thermal efficiency\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)\n", + "pp=(ma/1000)*((Cj**2)-(Ca**2))/2#................#Propulsive power in kW\n", + "print \"Propulsive power = %0.2f kW\"%pp\n", + "etapp=(2*Ca)/(Cj+Ca)#......................#Propulsive efficiency\n", + "print \"Propulsive efficiency = %0.2f %%\"%(etapp*100)\n", + "etao=((Cj-Ca)*Ca)/((1/afr)*CV*etaco*1000)#............#Overall efficiency\n", + "print \"Overall efficiency = %0.2f %%\"%(etao*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.20 PAGE 863" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Absolute velocity of jet = 363.64 m/s\n", + "Volume of air compressed = 5920.59 kg/min\n", + "Diameter of the jet = 463.09 mm\n", + "Turbine output = 2464.65 kW\n", + "\n", + "Air fuel ratio is 53.225:1\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi,sqrt\n", + "# Initialisation of Variables\n", + "v=800#.............#Speed of the turbojet in km/h\n", + "etapp=0.55#......#Propulsive efficiency\n", + "etao=0.17#.........#Overall efficiency\n", + "al=9500#...............#Altitude in m\n", + "rhoa=0.17#............#Density of air at the given altitude in kg/m**3\n", + "dr=6100#...........#Drag on the plane in N\n", + "CV=46000#.........#Calorific value of fuel in kJ/kg\n", + "#Calculations\n", + "Ca=v*(1000/3600)#.........#Velocity of jet in m/s\n", + "Cj=((2*Ca)/etapp)-Ca#........#Velocity of gases at nozzle exit relative to the aircraft in m/s\n", + "print \"Absolute velocity of jet = %0.2f m/s\"%(Cj-Ca)\n", + "ma=dr/(Cj-Ca)#............#Rate of air flow in kg/s\n", + "Va=(ma/rhoa)*60#..........#Volume of air compresssed per min in kg\n", + "print \"Volume of air compressed = %0.2f kg/min\"%Va\n", + "d=sqrt((Va*4)/(60*pi*Cj))#..........#Diameter of the jet in m\n", + "print \"Diameter of the jet = %0.2f mm\"%(d*1000)\n", + "tp=dr*(Ca/1000)#...........#Thrust power in kW\n", + "wt=tp/etapp#................#Turbine output in kW\n", + "print \"Turbine output = %0.2f kW\"%wt\n", + "mf=wt/(etao*CV)#...........#Rate of fuel consumption in kg/s\n", + "afr=ma/mf#..........#Air fuel ratio\n", + "print \"\\nAir fuel ratio is %0.3f:1\"%(afr)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.21 PAGE 864" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power required to drive the compressor = 171.46 kW\n", + "\n", + "\n", + "Air fuel ratio 73.094:1\n", + "\n", + "Pressure of gases leaving the turbine = 1.74 bar\n", + "Thrust per kg of air per second = 465.69 N\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi,sqrt\n", + "# Initialisation of Variables\n", + "t1=288#..........#Temperature of the inlet air into compressor in K\n", + "p1=1.01#......#Pressure of the inlet air into compressor in bar\n", + "t3=1023#.........#Maximum temperature in K\n", + "p2=4.04#.........#Pressure of air at the end of compression in bar\n", + "etac=0.82#.......#compressor efficiency\n", + "etat=0.78#......#Turbine efficiency\n", + "etan=0.88#........#Nozzle efficiency\n", + "R=0.287#.........#Gas constant for air in kJ/kgK\n", + "ga=1.4#............#Ratio of specific heats\n", + "C=42000#..........#Calorific value of fuel in kJ/kg\n", + "#Calculations\n", + "t2=t1*((p2/p1)**((ga-1)/ga))#........#Ideal temperature at the end of compression in K\n", + "t21=((t2-t1)/etac)+t1#...........#Actual temperature at the end of compression in K\n", + "cp=R*(ga/(ga-1))#..............#Specific heat at constant pressure in kJ/kgK\n", + "Pc=cp*(t21-t1)#.............#Power required to drive the compressor in kW\n", + "print \"Power required to drive the compressor = %0.2f kW\"%Pc\n", + "afr=((C)/(cp*(t3-t21)))-1#....#Air fuel ratio\n", + "print \"\\n\\nAir fuel ratio %0.3f:1\\n\"%(afr)\n", + "t41=t1+t3-t21#......#Actual temperatur of gases leaving the turbine in K\n", + "t4=t3-((t3-t41)/etat)#......#Ideal temperature of gases leaving the turbine in K\n", + "p3=p2#\n", + "p4=p3*((t4/t3)**(ga/(ga-1)))#.......#Pressure of gases leaving the turbine in bar\n", + "print \"Pressure of gases leaving the turbine = %0.2f bar\"%p4\n", + "p5=p1#\n", + "t5=t41/((p4/p5)**((ga-1)/ga))#\n", + "t51=t41-(etan*(t41-t5))#\n", + "Cj=sqrt(2*cp*(t41-t51)*1000)#..............#Jet velocity in m/s\n", + "th=Cj*1#..................#Thrust per kg per second in N\n", + "print \"Thrust per kg of air per second = %0.2f N\"%th" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.22 PAGE 865" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air fuel ratio : 48.348\n", + "Specific thrust = 822.74 N/kg\n", + "Total thrust = 21805.03 N\n" + ] + } + ], + "source": [ + "from scipy import sqrt\n", + "# Initialisation of Variables\n", + "Ca=216#................#Speed of aircraft in m/s\n", + "t1=265.8#...............#Intake air temperature in K\n", + "p1=0.78#...............#Intake air pressure in bar\n", + "rp=5.8#..................#Pressure ratio in compressor \n", + "t4=1383#.................#Temperature of gases entering the gas turbine in K\n", + "pd=0.168#...............#Pressure drop in combustion chamber in bar\n", + "etad=0.9#..............#Diffuser efficiency\n", + "etan=0.9#............#Nozzle efficiency\n", + "etac=0.9#............#Compressor efficiency\n", + "etat=0.8#.............#Turbine efficiency\n", + "C=44150#............#Calorific value of fuel in kJ/kg\n", + "cp=1.005#.............#Specific heat at constant pressure in kJ/kgK\n", + "ga=1.4#...............#Ratio of specific heats\n", + "cin=0.12#...............#Inlet cross sectio of the diffuser in m**3\n", + "R=0.287#............#Gas constant in kJ/kgK\n", + "#Calculations\n", + "t2=t1+((Ca*Ca)/(2*cp*1000))#......#For ideal diffuser\n", + "t21=t1+((Ca*Ca)/(2*cp*etad*1000))#......#For actual diffuser\n", + "p2=p1*((t2/t1)**(ga/(ga-1)))#\n", + "t3=t21*(rp**((ga-1)/ga))#\n", + "t31=t21+((t3-t21)/etac)#\n", + "afr=(C-(cp*t4))/(cp*(t4-t31))#............#Air fuel ratio\n", + "print \"Air fuel ratio : %0.3f\"%afr\n", + "p3=p2*rp#\n", + "p4=p3-pd#...............#Pressure of gases entering the turbine in bar\n", + "t51=t4-(t31-t21)#\n", + "t5=round(t4-((t4-t51)/etat))#\n", + "p5=p4/((t4/t5)**(ga/(ga-1)))#\n", + "p6=p1#\n", + "t6=t51/((p5/p6)**((ga-1)/ga))#\n", + "t61=t51-(etac*(t51-t6))#\n", + "Cj=44.72*sqrt(cp*(t51-t61))#........#Velocity at the exit of the nozzle in m/s\n", + "st=(1+(1/afr))*Cj#............#Specific thrust in N/kg\n", + "print \"Specific thrust = %0.2f N/kg\"%abs(st)\n", + "v1=Ca*cin#...........#Volume of flowing air in m**3/s\n", + "ma=(p1*v1*10**5)/(R*t1*1000)#.........#Mass flow of air\n", + "tt=ma*st#..............#Total thrust in N\n", + "print \"Total thrust = %0.2f N\"%abs(tt)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 21.23 PAGE 866" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Overall efficiency = 12.91 %\n", + "Rate of air consumption = 9.23 kg/s\n", + "Power developed by turbine = 1496.76 kW\n", + "The outlet area of jet tube = 0.10 m**2\n", + "Specific fuel consumption = 0.14 kg per kg of thrust\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Initialisation of Variables\n", + "al=9000#..........#Altitude in m\n", + "Ca=215#...........#Speed of aircraft in m/s\n", + "TP=750#.............#Thrust power developed in kW\n", + "p1=0.32#...........#Inlet pressure of air in bar\n", + "t1=231#.............#Inlet temperature of air in K\n", + "t3=963#.............#Temperature of gases leaving the combustion chamber in K\n", + "rpc=5.2#............#Pressure ratio\n", + "C=42500#..........#Calorific value of fuel in kJ/kg\n", + "C41=195#.........#Velocity in ducts\n", + "etac=0.86#..........#Compressor efficiency\n", + "ga=1.4#............#Ratio of specific heats for air\n", + "gag=1.33#............#Ratio of specific heats for gases\n", + "etat=0.86#..........#Turbine efficiency\n", + "etajt=0.9#..........#Jet tube efficiency\n", + "cp=1.005#............#Specific heat at constant pressure in kJ/kgK for air\n", + "cpg=1.087#............#Specific heat at constant pressure in kJ/kgK for gases\n", + "R=0.29#..................#Gas constant for exhaust gases in kJ/kgK\n", + "#Calculations\n", + "t2=t1*(rpc**((ga-1)/ga))#\n", + "t21=t1+((t2-t1)/etac)#\n", + "mf=(cpg*(t3-t21))/(C-(cpg*(t3-t21)))#\n", + "afr=1/mf#..........#Air fuel ratio\n", + "t41=round(t3-((cp*(t21-t1))/(cpg*(1+mf))))#\n", + "t4=t3-((t3-t41)/etat)#\n", + "p4=rpc#\n", + "rpt=(t3/t4)**(gag/(gag-1))#.............#Expansion pressure ratio in turbine\n", + "rpj=p4/rpt#....................#Expansion pressure ratio in jet tube\n", + "t5=t41/(rpj**((gag-1)/gag))#\n", + "Cj=sqrt(etajt*2*((cpg*1000*(t41-t5))+((C41*C41)/2)))#\n", + "etao=((((1+mf)*Cj)-Ca)*Ca)/(1000*mf*C)#......#Overall efficiency\n", + "print \"Overall efficiency = %0.2f %%\"%(etao*100)\n", + "ma=(TP*1000)/((((1+mf)*Cj)-Ca)*Ca)#........#Rate of air consumption in kg/s\n", + "print \"Rate of air consumption = %0.2f kg/s\"%ma\n", + "P=ma*(1+mf)*cpg*(t3-t41)#..............#Power developed by the turbine in kW\n", + "print \"Power developed by turbine = %0.2f kW\"%P\n", + "t51=t41-(((Cj**2)-(C41**2))/(2*1000*cpg))#\n", + "rhoe=(p1*10**5)/(R*1000*t51)#..........#Density of exhaust gases\n", + "Ajt=(ma*(1+mf))/(Cj*rhoe)#.......#Discharge of jet area in m**2\n", + "print \"The outlet area of jet tube = %0.2f m**2\"%Ajt\n", + "sfc=(mf*ma*3600)/(1000*(TP/Ca))#..........#Specific fuel consumption in kg/thrust-hour\n", + "print \"Specific fuel consumption = %0.2f kg per kg of thrust\"%sfc" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER3_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER3_3.ipynb new file mode 100644 index 00000000..2eca0c90 --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER3_3.ipynb @@ -0,0 +1,1877 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 3 - Air Standard Cycles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.1 PAGE 87" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Engine thermal efficiency = 53.49 %\n", + "Head added = 243.03 kJ\n", + "Change in entropy = 0.361 kJ/K\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=673#.....................#Max temp in Kelvin\n", + "t3=313##...................#Min temp in Kelvin\n", + "W=130#.................#Work produced in kJ\n", + "#calculations\n", + "etath=(t1-t3)/t1#................#Engine thermal efficiency\n", + "print \"Engine thermal efficiency = %0.2f %%\"%(etath*100)\n", + "ha=W/etath#.................#Heat added in kJ\n", + "print \"Head added = %0.2f kJ\"%ha\n", + "dels=(ha-W)/t3#...........#Change in entropy\n", + "print \"Change in entropy = %0.3f kJ/K\"%dels" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.2 PAGE 88" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum temperature = 585.37 Kelvin\n", + "The minimum temperature = 292.68 Kelvin\n", + "Volume at the end of isothermal expansion = 0.19 m**3\n", + "\n", + "\n", + "\n", + "Process Heat transfer\n", + "\n", + "_______________________________________________________________\n", + "\n", + "Isothermal expansion 40 kJ\n", + "\n", + "Adiabatic reversible expansion 0 kJ\n", + "\n", + "Isothermal compression -40 kJ\n", + "\n", + "Adiabatic reversible compressions 0 kJ\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "m=0.5#.....................#Mass of air in kg\n", + "etath=0.5#.................#Thermal efficiency of engine\n", + "hie=40#...................#Heat transferred during isothermal expansion in kJ\n", + "p1=7#....................#Pressure in bar at the beginning of expansion\n", + "v1=0.12#..................#Volume in m**3 at the beginning of expansion\n", + "cv=0.721#...................#Specific heat at constant volume in kJ/kgK\n", + "cp=1.008#..................#Specific heat at constant pressure in kJ/kgK\n", + "R=287#......................#Gas constant in J/kgK\n", + "#Calculations\n", + "t1=(p1*10**5*v1)/(R*m)#....................#Max temp in K\n", + "t2=t1*(1-etath)#.......................#Min temp in K\n", + "print \"The maximum temperature = %0.2f Kelvin\"%t1\n", + "print \"The minimum temperature = %0.2f Kelvin\"%t2\n", + "from math import exp\n", + "v2=(exp((hie*1000)/(m*R*t1)))*v1#..................#Volume at the end of isothermal expansion in m**3\n", + "print \"Volume at the end of isothermal expansion = %0.2f m**3\"%v2\n", + "print \"\\n\\n\"\n", + "print \"Process Heat transfer\\n\"\n", + "print \"_______________________________________________________________\\n\"\n", + "print \"Isothermal expansion %d kJ\\n\"%(hie)\n", + "print \"Adiabatic reversible expansion %d kJ\\n\"%(0)\n", + "print \"Isothermal compression %d kJ\\n\"%(-hie)\n", + "print \"Adiabatic reversible compressions %d kJ\"%(0)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.3 PAGE 89" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "p1=18 bar \n", + "p2=12 bar \n", + "p3=0.977 bar \n", + "p4=1.465 bar \n", + "t1=t2=683 Kelvin \n", + "t3=t4=334 Kelvin \n", + "\n", + "Change in entropy = 0.19 kJ/K\n", + "Mean effective pressure of the cycle = 0.47 bar\n", + "Mean effective pressure of the cycle = 235.25 bar\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "# Initialisation of Variables\n", + "p1=18#..................#Maximum pressure in bar\n", + "t1=410+273#.............#Maximum temperature in Kelvin\n", + "ric=6#.....................#Ratio of isentropic compression\n", + "rie=1.5#.................#Ratio of isothermal expansion\n", + "v1=0.18#..................#Volume of air at the beginning of expansion\n", + "ga=1.4#...................#Degree of freedom of gas\n", + "R=287#.....................#Gas constant in J/kgK\n", + "nc=210#..................#no of working cycles\n", + "#Calculations\n", + "\n", + "t4=t1/(ric**(ga-1))#.............#Min temp in K\n", + "t3=t4#\n", + "p4=p1/(ric**ga)#..................#Min pressure in bar\n", + "p2=p1/rie#.......................#pressure of gas before isentropic expansion in bar\n", + "p3=p2*((1/6)**ga)#.................#Pressure of gas after isentropic expansion in bar\n", + "print \"p1=%.f bar \\np2=%.f bar \\np3=%.3f bar \\np4=%.3f bar \\nt1=t2=%0.f Kelvin \\nt3=t4=%.f Kelvin \\n\"%(p1,p2,p3,p4,t1,t3)\n", + "dels=(p1*10**5*v1*log(rie))/(1000*t1)#....................#Change in entropy\n", + "print \"Change in entropy = %0.2f kJ/K\"%dels\n", + "qs=t1*dels#.......................#Heat supplied in kJ\n", + "Qr=t4*dels#.......................#Heat rejected in kJ\n", + "eta=(qs-Qr)/qs#............#Efficiency of the cycle\n", + "v3byv1=ric*rie#\n", + "Vs=(v3byv1-1)*v1#.................#Stroke volume\n", + "pm=((qs-Qr)*10**3)/(Vs*10**5)#........#Mean effective pressure of the cycle in bar\n", + "print \"Mean effective pressure of the cycle = %0.2f bar\"%pm\n", + "P=(qs-Qr)*(nc/60)#.........................#Power of engine\n", + "print \"Mean effective pressure of the cycle = %0.2f bar\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.4 PAGE 92" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of the sink = 1442 Celsius\n", + "temperature of source = 1785 Celsius:\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "eta=1/6#...................#Efficiency of the engine\n", + "rts=70#.................#The amount of temp which is reduced in the sink in C\n", + "#Calculation\n", + "t1byt2=1/(1-eta)#\n", + "t2=(rts+273)/((2*eta*t1byt2)-t1byt2+1)#............#Temperature of the sink in K\n", + "print \"Temperature of the sink = %0.f Celsius\"%(t2-273)\n", + "t1=t1byt2*t2#...............#Temperature of source in K\n", + "print \"temperature of source = %0.f Celsius:\"%(t1-273)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.5 PAGE 92" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Since the efficiency of the given engine is more than efficiency of carnot engine, the claim is not true.\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=1990#....................#Temperature of the heat source in K\n", + "t2=850#..................#Temperature of the sink in K\n", + "Q=32.5#...................#Heat supplied in kJ/min\n", + "P=0.4#....................#Power developed by the engine in kW\n", + "#Calculations\n", + "eta=1-(t2/t1)#..........#Efficiency of carnot engine\n", + "etath=P/(Q/60)#..................#Efficiency of the given engine\n", + "if (etath>eta):\n", + " print \"Since the efficiency of the given engine is more than efficiency of carnot engine, the claim is not true.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.7 PAGE 96" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The compression ratio of the engine is: 6.25\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "etaotto=0.6#............#Efficiency of otto engine\n", + "ga=1.5#.................#Ratio of specific heats\n", + "#Calculations\n", + "r=(1/(1-etaotto))**(1/(ga-1))#................#Compression ratio\n", + "print \"The compression ratio of the engine is:\",r" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.8 PAGE 96" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The air standard efficiency of otto cycle = 56.47 %\n", + "Mean effective pressure = 1.34 bar\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "D=0.25#......................#Engine bore in m\n", + "L=0.375#...................#Engine stroke in m\n", + "Vc=0.00263#................#Clearence volume in m**3\n", + "p1=1#..................#Initial pressure in bar\n", + "t1=323#...................#Initial temperature in K\n", + "p3=25#....................#Max pressure in bar\n", + "ga=1.4#....................#Ratio of specific heats\n", + "#Calculations\n", + "Vs=(pi/4)*D*D*L#................#Swept volume in m**3\n", + "r=round((Vs+Vc)/Vc)#..........................#Compression ratio\n", + "etaotto=1-(1/(r**(ga-1)))#..................#Air standard efficiency of otto cycle\n", + "print \"The air standard efficiency of otto cycle = %0.2f %%\"%(etaotto*100)\n", + "p2=p1*((r)**ga)#\n", + "rp=p3/p2#..........................#Pressure ratio\n", + "pm=(p1*r*((r**(ga-1))-1)*(rp-1))/((ga-1)*(r-1))#................#Mean effective pressure in bar\n", + "print \"Mean effective pressure = %0.2f bar\"%pm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.9 PAGE 98" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature after adiabatic compression = 689.22 K\n", + "Pressure after adiabatic compression = 18.38 bar\n", + "Temperature after isochoric compression = 2772.55 K\n", + "Pressure after isochoric compression = 73.93 bar\n", + "Temperature after adiabatic expansion = 1206.82 K\n", + "Pressure after adiabatic expansion = 4.02 bar\n", + "Specific work = 847.09 kJ/kg\n", + "Thermal efficiency = 56.47 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#.....................#Pressure in bar\n", + "t1=300#......................#Temperature in K\n", + "Q=1500#.....................#Heat added in kJ/kg\n", + "r=8#.......................#Compression ratio\n", + "Cv=0.72#....................#Specific heat at constant volume\n", + "ga=1.4#......................#Ratio of specific heats\n", + "#Calculations\n", + "t2=t1*(r)**(ga-1)#..........#Temperature after adiabatic compression in K\n", + "p2=p1*(r**ga)#..............#Pressure after adiabatic compression in bar\n", + "t3=(Q/Cv)+t2#.................#Temperature after isochoric compression in K\n", + "p3=(p2*t3)/t2#................#Pressure after isochoric compression in bar\n", + "t4=t3/(r**(ga-1))#.......................#Temperature after adiabatic expansion in K\n", + "p4=p3*(1/(r**(ga)))#................#Pressure after adiabatic expansion in bar\n", + "Ws=Cv*(t3-t2-t4+t1)#.........#Specific work in kJ/kg\n", + "etath=1-(1/(r**(ga-1)))#............#Thermal efficiency\n", + "print \"Temperature after adiabatic compression = %0.2f K\"%t2\n", + "print \"Pressure after adiabatic compression = %0.2f bar\"%p2\n", + "print \"Temperature after isochoric compression = %0.2f K\"%t3\n", + "print \"Pressure after isochoric compression = %0.2f bar\"%p3\n", + "print \"Temperature after adiabatic expansion = %0.2f K\"%t4\n", + "print \"Pressure after adiabatic expansion = %0.2f bar\"%p4\n", + "print \"Specific work = %0.2f kJ/kg\"%Ws\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.10 PAGE 99" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature after adiabatic compression = 614.30 K\n", + "Pressure after adiabatic compression = 12.29 bar\n", + "Temperature after isochoric compression = 1842.00 K\n", + "Pressure after isochoric compression = 36.84 bar\n", + "Temperature after adiabatic expansion = 899.56 K\n", + "Pressure after adiabatic expansion = 3.00 bar\n", + "Thermal efficiency = 56.47 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "r=6#..............#Compression ratio\n", + "p1=1#................#Pressure after isochoric expansion in bar\n", + "t1=300#................#Temperature after isochoric expansion in K\n", + "t3=1842#...............#Temperature after isochoric compression in K\n", + "ga=1.4#...............#Ratio of specific heats\n", + "#Calculations\n", + "p2=p1*(r**ga)#...............#Pressure after adiabatic compression in bar\n", + "t2=t1*(r**(ga-1))#.............#Temperature after adiabatic compression in K\n", + "p3=p2*(t3/t2)#..................#pressure after isochoric compression in bar\n", + "t4=t3/(r**(ga-1))#..............#Temperature after adiabatic expansion in K\n", + "p4=p3*(1/(r**(ga)))#...........#Pressure after adiabatic expansion in bar\n", + "etaotto=1-(1/(r**(ga-1)))#............#Efficiency of otto cycle\n", + "p5=p1#\n", + "t5=((p5/p3)**((ga-1)/ga))*t3#................#Atkinson cycle temp after further adiabatic expansion in K\n", + "etatk=1-((ga*(t5-t1))/(t3-t2))#...........#Efficiency of atkinson cycle\n", + "print \"Temperature after adiabatic compression = %0.2f K\"%t2\n", + "print \"Pressure after adiabatic compression = %0.2f bar\"%p2\n", + "print \"Temperature after isochoric compression = %0.2f K\"%t3\n", + "print \"Pressure after isochoric compression = %0.2f bar\"%p3\n", + "print \"Temperature after adiabatic expansion = %0.2f K\"%t4\n", + "print \"Pressure after adiabatic expansion = %0.2f bar\"%p4\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.11 PAGE 101" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Compression ratio: 3.95\n", + "Temperature at the end of compression = 607.79 K\n", + "Temperature at the end of heat addition = 1266.43 K\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#...................#Initial pressure in bar\n", + "t1=343#..................#Initial temperature in K\n", + "p2=7#....................#Pressure after adiabatic compression\n", + "Qs=465#...............#Heat addition at constant volume in kJ/kg\n", + "cp=1#.....................#Specific heat at constant pressure in kJ/kg\n", + "cv=0.706#..................#Specific heat at constant volume in kJ/kg\n", + "ga=cp/cv#.................#Ratio of specific heats\n", + "#Calculations\n", + "r=(p2/p1)**(1/ga)#...............#Compression ratio\n", + "t2=t1*(r**(ga-1))#.....................#Temperature at the end of compression in K\n", + "t3=t2+(Qs/cv)#.............#Temperature at the end of heat addition in K\n", + "print \"Compression ratio:\",round(r,2)\n", + "print \"Temperature at the end of compression = %0.2f K\"%t2\n", + "print \"Temperature at the end of heat addition = %0.2f K\"%t3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.12 PAGE 102" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Compression ratio: 6.92\n", + "Thermal efficiency = 53.87 %\n", + "Work done = 598.65 kJ\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "ga=1.4#..............#Ratio of specific heats\n", + "p2byp1=15#...............#Ratio pressure at the end of compression to that of pressure at the start\n", + "t1=311#................#Initial temperature in K\n", + "t3=2223#...............#Maximum temperature in K\n", + "R=0.287#...............#Gas constant in kJ/kg K\n", + "#Calculations\n", + "r=p2byp1**(1/ga)#...............#Compression ratio\n", + "etath=1-(1/(r**(ga-1)))#.............#Thermal efficiency\n", + "t2=t1*(r**(ga-1))#............#Temperature at the end of compression in K\n", + "t4=t3/(r**(ga-1))#...........#Temperature at the end of isothermal expansion in K\n", + "cv=R/(ga-1)#................#Specific heat at constant volume in kJ/kg\n", + "Q=cv*(t3-t2)#..............#Heat supplied in kJ/kg of air\n", + "Qr=cv*(t4-t1)#.................#Heat rejected in kJ/kg of air\n", + "W=Q-Qr#.................#Work done\n", + "print \"Compression ratio:\",round(r,2)\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)\n", + "print \"Work done = %0.2f kJ\"%W" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.13 PAGE 104" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature after adiabatic compression: 601.15 K\n", + "\n", + "Pressure after adiabatic compression: 11.00 bar\n", + "\n", + "Volume after adiabatic compression: 0.08 m**3\n", + "\n", + "Temperature after isochoric compression: 1172.73 K\n", + "\n", + "Pressure after isochoric compression: 21.46 bar\n", + "\n", + "Volume after isochoric compression: 0.08 m**3\n", + "\n", + "Temperature after adiabatic expansion: 591.09 K\n", + "\n", + "Pressure after adiabatic expansion: 1.95 bar\n", + "\n", + "Volume after adiabatic expansion: 0.45 m**3\n", + "\n", + "Percentage clearance: 22.01\n", + "\n", + "Efficiency of otto cycle: 49.60\n", + "\n", + "Mean effective pressure: 2.82 bar:\n", + "\n", + "Power developed: 364.54 kW\n", + "\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "v1=0.45#.............#Initial volume in m**3 \n", + "p1=1#...............#Initial pressure in bar\n", + "t1=303#.............#Initial temperature in K\n", + "p2=11#...................#Pressure at the end of compression stroke in bar\n", + "Q=210#...................#heat added at constant volume in kJ\n", + "N=210#.................#No of working cycles per min\n", + "ga=1.4#.............#Ratio of specific heats\n", + "R=287#.................#Gas constant in kJ/kgK\n", + "cv=0.71#.................#Specific heat at constant volume in kJ/kg\n", + "#Calculations\n", + "r=(p2/p1)**(1/ga)#...................#Compression ratio\n", + "t2=t1*(r**(ga-1))#...................#Temperature at the end of adiabatic compression\n", + "v2=(t2*p1*v1)/(t1*p2)#.................#Volume at the end of adiabatic compression in m**3\n", + "m=(p1*v1*10**5)/(R*t1)#................#Mass of engine fluid in kg\n", + "t3=(Q/(m*cv))+t2#...................#Temperature at the end of isochoric compression in K\n", + "p3=(t3/t2)*p2#................#Pressure at the end of isochoric compression in bar\n", + "v3=v2#\n", + "t4=t3*(1/r)**(ga-1)#...................#Temperature at the end of adiabatic expansion in K\n", + "p4=p3*(1/r)**ga#......................#Pressure at the end of adiabatic expansion in bar\n", + "v4=v1#\n", + "pc=(v2*100)/(v1-v2)#..................#Percentage clearence\n", + "etaotto=1-(1/(r**(ga-1)))#........................#Efficiency of otto cycle\n", + "Qr=m*cv*(t4-t1)#...............................#Heat rejected in kJ/kg\n", + "pm=((Q-Qr)*1000)/((v1-v2)*100000)#......#Mean effective pressure in bar\n", + "P=(Q-Qr)*(N/60)#........................#Power developed in kW\n", + "print \"Temperature after adiabatic compression: %0.2f K\\n\"%(t2)\n", + "print \"Pressure after adiabatic compression: %0.2f bar\\n\"%(p2)\n", + "print \"Volume after adiabatic compression: %0.2f m**3\\n\"%(v2)\n", + "print \"Temperature after isochoric compression: %0.2f K\\n\"%(t3)\n", + "print \"Pressure after isochoric compression: %0.2f bar\\n\"%(p3)\n", + "print \"Volume after isochoric compression: %0.2f m**3\\n\"%(v3)\n", + "print \"Temperature after adiabatic expansion: %0.2f K\\n\"%(t4)\n", + "print \"Pressure after adiabatic expansion: %0.2f bar\\n\"%(p4)\n", + "print \"Volume after adiabatic expansion: %0.2f m**3\\n\"%(v4)\n", + "print \"Percentage clearance: %0.2f\\n\"%(pc)\n", + "print \"Efficiency of otto cycle: %0.2f\\n\"%(etaotto*100)\n", + "print \"Mean effective pressure: %0.2f bar:\\n\"%(pm)\n", + "print \"Power developed: %0.2f kW\\n\"%(P)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.14 PAGE 106" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air standard efficiency of the engine = 49.59 %\n", + "There is no change in efficiency when Helium is used as working fluid instead of air\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=310#................#Min temperature in K\n", + "t3=1220#................#Max temperature in K\n", + "ga=1.4#................#Ratio of specific heats for air\n", + "cph=5.22#............#Specific heat at constant volume for helium in kJ/kg\n", + "cvh=3.13#...............#Specific heat at constant pressure for helium in kJ/kg\n", + "#Calculations \n", + "r=(t3/t1)**(1/((ga-1)*2))#..............#Compression ratio\n", + "etaotto=1-(1/(r**(ga-1)))#................#Air standard efficiency\n", + "gah=cph/cvh#................#Ratio of specific heats for Helium\n", + "rh=(t3/t1)**(1/((gah-1)*2))#..............#Compression ratio when Helium is used\n", + "etaottoh=1-(1/(rh**(gah-1)))#................#Air standard efficiency when Helium is used\n", + "print \"Air standard efficiency of the engine = %0.2f %%\"%(etaotto*100)\n", + "if ((round (etaotto)- round (etaottoh)) == 0):\n", + " print \"There is no change in efficiency when Helium is used as working fluid instead of air\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.15 PAGE 108" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power developed = 1.88 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "# Initialisation of Variables\n", + "t1=310#.........#Minimum temperature in K\n", + "t3=1450#............#maximum temperature in K\n", + "m=0.38#...........#Mass of working fluid in kg\n", + "cv=0.71#...........#Specific heat at constant volume in kJ/kg\n", + "#Calculations\n", + "t4=sqrt(t1*t3)#............#Temperature at the end of adiabatic expansion in K\n", + "t2=t4#\n", + "W=cv*(t3-t2-t4+t1)#..................#Work done in kJ/kg\n", + "P=W*(m/60)#.................#Power developed in kW\n", + "print \"Power developed = %0.2f kW\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.17 PAGE 113" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency of diesel engine = 61.19 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "r=15#...................#Compression ratio\n", + "ga=1.4#..............#Ratio os fpecific heats for air\n", + "perQ=6#................#Heat addition at constant pressure takes place at 6% of stroke\n", + "#Calculations\n", + "rho=1+((perQ/100)*(r-1))#.............#Cut off ratio\n", + "etad=1-((((rho**ga)-1)/(rho-1))*(1/(ga*(r**(ga-1)))))#..................#Efficiency of diesel engine\n", + "print \"Efficiency of diesel engine = %0.2f %%\"%(etad*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.18 PAGE 114" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency of diesel engine = 59.34 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "# Initialisation of Variables\n", + "L=0.25#...............#Engine stroke in m\n", + "D=0.15#..................#Engine bore in m\n", + "v2=0.0004#...............#Clearance volume in m**3\n", + "pers=5#...............#Percentage of stroke when fuel injection occurs\n", + "ga=1.4#..............#Ratio of specific heats\n", + "#Calculations\n", + "Vs=(pi/4)*D*D*L#..............#Swept volume in m**3\n", + "Vt=Vs+v2#....................#Total cylinder volume in m**3\n", + "v3=v2+((pers/100)*Vs)#..............#Volume at point of cut off\n", + "rho=v3/v2#............#Cut off ratio\n", + "r=1+(Vs/v2)#.............#Compression ratio\n", + "etad=1-((((rho**ga)-1)/(rho-1))*(1/(ga*(r**(ga-1)))))#..................#Efficiency of diesel engine\n", + "print \"Efficiency of diesel engine = %0.2f %%\"%(etad*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.19 PAGE 114" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Percentage loss in efficiency due to delay in cut off : 2.10\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "r=14#....................#Compression ratio\n", + "pers1=5#...............#Percentage of stroke when fuel cut off occurs\n", + "pers2=8#...............#Percentage of stroke when delayed fuel cut off occurs\n", + "v2=1#.....................#Clearance volume in m**3\n", + "ga=1.4#..................#Ratio of specific heats\n", + "#Calculations\n", + "#When the fuel is cut off at 5 %\n", + "rho1=((pers1/100)*(r-1))+1#.............#Cut off ratio\n", + "etad1=1-((((rho1**ga)-1)/(rho1-1))*(1/(ga*(r**(ga-1)))))#..................#Efficiency of diesel engine\n", + "#When the fuel is cut off at 8 %\n", + "rho2=((pers2/100)*(r-1))+1#.............# Delayed Cut off ratio\n", + "etad2=1-((((rho2**ga)-1)/(rho2-1))*(1/(ga*(r**(ga-1)))))#..................#Efficiency of diesel engine when cut off ratio is deyaled\n", + "print \"Percentage loss in efficiency due to delay in cut off : %0.2f\"%((etad1-etad2)*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.20 PAGE 115" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cut off Percentage : 11.17\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from scipy.optimize import fsolve\n", + "# Initialisation of Variables\n", + "pm=7.5#.................#Mean effective pressure in bar\n", + "r=12.5#..................#Compression ratio\n", + "p1=1#....................#Initial pressure in bar\n", + "ga=1.4#.................#Ratio of specific heats\n", + "#Calculations\n", + "k=(pm*(ga-1)*(r-1))/(p1*(r**ga))#\n", + "c1=(r**(1-ga))/k#\n", + "c2=(-ga)/k#\n", + "c=1+(ga/k)-((r**(1-ga))/k)#\n", + "def F(rho):\n", + " f=c1*(rho**ga)+c2*rho+c#\n", + " return f\n", + "#Initial guess\n", + "rho=2#\n", + "#Derivative\n", + "def D(rho):\n", + " z=c1*ga*(rho**(ga-1))+c2#\n", + " return z\n", + "y=fsolve(F,rho)\n", + "perc=((y-1)/(r-1))*100#..................#Percentage of cutoff\n", + "print \"Cut off Percentage : %0.2f\"%perc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.21 PAGE 115" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature after adiabatic compression: 886.253082 K\n", + "\n", + "Pressure after adiabatic compression: 44.312654 bar\n", + "\n", + "Volume after adiabatic compression: 0.000673 m**3\n", + "\n", + "Temperature after isobaric compression: 1878.856533 K\n", + "\n", + "Pressure after isobaric compression: 44.312654 bar\n", + "\n", + "Volume after isobaric compression: 0.001427 m**3\n", + "\n", + "Temperature after adiabatic expansion: 858.996630 K\n", + "\n", + "Pressure after adiabatic expansion: 2.863322 bar\n", + "\n", + "Volume after adiabatic expansion: 0.010098 m**3\n", + "\n", + "Efficiency of diesel engine = 59.77 %\n", + "Mean effective pressure : 7.42\n", + "Power developed = 44.27 kW \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "D=0.2#.................#Engine bore in m\n", + "L=0.3#.............#Engine stroke in m\n", + "p1=1#................#Initial pressure in bar\n", + "N=380#.................#No of working cycles per min\n", + "t1=300#..............#Initial temperature in K\n", + "co=8#................#Cut off percentage\n", + "r=15#..................#Compression ratio\n", + "R=287#.................#gas constant in J/kg\n", + "ga=1.4#................#Ratio of specific heats\n", + "#Calculations\n", + "Vs=(pi/4)*D*D*L#.............#Stroke volume in m\n", + "v1=(r/(r-1))*Vs#................#Volume at the end of isochoric compression in m**3\n", + "m=(p1*v1*10**5)/(R*t1)#................#Mass of air in cylinder in kg/cycle\n", + "p2=p1*(r**ga)#.......................#Pressure at the end of isentropic compression in bar\n", + "t2=t1*(r**(ga-1))#....................#Temperature at the end of isentropic compression in K\n", + "v2=Vs/(r-1)#..................#Volume at the end of isentropic compressionin m**3\n", + "p3=p2#\n", + "rho=((r-1)*(co/100))+1#................#Cut off ratio\n", + "v3=rho*v2#.......................#Volume at the end of isobaric expansion in m**3\n", + "t3=t2*(v3/v2)#..................#Temperature at the end of isobaric expansion in K\n", + "p4=((rho/r)**ga)*p3#..............#Pressure at the end of adiabatic expansion in bar\n", + "t4=((rho/r)**(ga-1))*t3#..............#Temperature at the end of adiabatic expansion in K\n", + "v4=v1#\n", + "print \"Temperature after adiabatic compression: %f K\\n\"%(t2)\n", + "print \"Pressure after adiabatic compression: %f bar\\n\"%(p2)\n", + "print \"Volume after adiabatic compression: %f m**3\\n\"%(v2)\n", + "print \"Temperature after isobaric compression: %f K\\n\"%(t3)\n", + "print \"Pressure after isobaric compression: %f bar\\n\"%(p3)\n", + "print \"Volume after isobaric compression: %f m**3\\n\"%(v3)\n", + "print \"Temperature after adiabatic expansion: %f K\\n\"%(t4)\n", + "print \"Pressure after adiabatic expansion: %f bar\\n\"%(p4)\n", + "print \"Volume after adiabatic expansion: %f m**3\\n\"%(v4)\n", + "etad=1-((((rho**ga)-1)/(rho-1))*(1/(ga*(r**(ga-1)))))#..................#Efficiency of diesel engine\n", + "print \"Efficiency of diesel engine = %0.2f %%\"%(etad*100)\n", + "pm=p1*(r**ga)*(ga*(rho-1)-((r**(1-ga))*((rho**ga)-1)))*(1/(ga-1))*1/(r-1)#.......#Mean effective pressure \n", + "print \"Mean effective pressure : %0.2f\"%pm\n", + "Wdc=(pm*Vs*10**5)/1000#..................#Work done per cycle in kJ/cycle\n", + "P=(Wdc*N)/60#...........................#Power developed in kW\n", + "print \"Power developed = %0.2f kW \"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.22 PAGE 118" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mean effective pressure = 7.02 bar \n", + "Ratio of maximum pressure to mean effective pressure 6.49\n", + "Cycle efficiency = 60.48 %\n", + "Fuel consumption = 0.35 kg/kWh\n" + ] + } + ], + "source": [ + "# Initialisation of Variables\n", + "rc=15.3#....................#Compression ratio\n", + "re=7.5#...................#Expansion ratio\n", + "cp=1.005#.................#Specific heat at constant pressure in kJ/kg K\n", + "cv=0.718#..................#Specific heat at constant volume in kJ/kgK\n", + "ga=1.4#....................#Ratio of specific heats\n", + "p1=1#....................#Initial pressure in bar\n", + "t1=300#..................#Initial temperature in K\n", + "etamech=0.8#..................#Mechanical efficiency\n", + "C=42000#...........................#Calorific value of fuel in kJ/kg\n", + "rita=0.5#.........................#Ratio of indicated thermal efficiency to air standard efficiency\n", + "R=287#..........................#Gas constant in kJ/kgK\n", + "#Calculations\n", + "t2=t1*(rc**(ga-1))#.................#Temperature at the end of adiabatic compression in K\n", + "p2=p1*(rc**ga)#...................#Pressure at the end of adiabatic compression in bar\n", + "t3=(rc*t2)/re#....................#Temperature at the end of constant pressure process in K\n", + "v2=1#..................#Volume at the end of adiabatic process in m**3\n", + "m=(p2*v2*10**5)/(R*t2)#..................#Mass of working fluid in kg\n", + "t4=t3*((1/re)**(ga-1))#...................#Temperature at the end of adiabatic expansion in K\n", + "W=(m*(cp*(t3-t2)))-(m*(cv*(t4-t1)))#........#Work done in kJ\n", + "pm=W/(rc-1)#..............................#Mean effective pressure in kN/m**2\n", + "print \"Mean effective pressure = %0.2f bar \"%(pm/100)\n", + "print \"Ratio of maximum pressure to mean effective pressure %0.2f\"%((p2*100)/(pm))\n", + "etacy=W/(m*cp*(t3-t2))#...............#Cycle efficiency\n", + "print \"Cycle efficiency = %0.2f %%\"%(etacy*100)\n", + "etaith=rita*etacy#..................#Indicated thermal efficiency\n", + "etabth=etaith*etamech#...............#Brake thermal efficiency\n", + "mf=3600/(etabth*C)#................#Fuel consumption per kWh\n", + "print \"Fuel consumption = %0.2f kg/kWh\"%mf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.23 PAGE 123" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency of dual cycle : 63.76\n" + ] + } + ], + "source": [ + "# Initialisation of Variables\n", + "Vs=0.0053#................#Swept volume in m**3\n", + "Vc=0.00035#...............#Clearance volume in m**3\n", + "v3=Vc#\n", + "v2=Vc#\n", + "p3=65#..................#Max pressure in bar\n", + "co=5#...................#Cut off percentage\n", + "p4=p3#ga=1.4#...............#Ratio of specific heats\n", + "t1=353#....................#Temperature at the start of compression in K\n", + "p1=0.9#...................#Pressure at the start of compression in bar\n", + "#Calculations\n", + "r=1+(Vs/Vc)#...................#Compression ratio\n", + "rho=(((co/100)*Vs)/Vc)+1#...................#Cut off ratio\n", + "p2=p1*(r**ga)#\n", + "Beta=p3/p2#.............................#Explosion ratio\n", + "etadual=1-((1/(r**(ga-1)))*((Beta*(rho**ga))-1)*(1/((Beta-1)+(Beta*ga*(rho-1)))))#............#Efficiency of dual cycle\n", + "print \"Efficiency of dual cycle : %0.2f\"%(etadual*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.24 PAGE 124" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency of dual cycle : 61.42\n" + ] + } + ], + "source": [ + "# Initialisation of Variables\n", + "r=14#......................#Compression ratio\n", + "Beta=1.4#................#Explosion ratio\n", + "co=6#..................#Cut off percentage\n", + "ga=1.4#.................#Ratio of specific heats\n", + "#Calculation\n", + "rho=((co/100)*(r-1))+1#...............#Cut off ratio\n", + "etadual=1-((1/(r**(ga-1)))*((Beta*(rho**ga))-1)*(1/((Beta-1)+(Beta*ga*(rho-1)))))#............#Efficiency of dual cycle\n", + "print \"Efficiency of dual cycle : %0.2f\"%(etadual*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.25 PAGE 124" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air standard efficiency = 57.55 %\n", + "Power developed = 51.39 kW\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Initialisation of Variables\n", + "D=0.25#.................#Engine bore in m\n", + "L=0.3#.............#Engine stroke in m\n", + "p1=1#................#Initial pressure in bar\n", + "N=3#...............#No of cycles per second\n", + "p3=60#................#Maximum pressure in bar\n", + "t1=303#..............#Initial temperature in K\n", + "co=4#................#Cut off percentage\n", + "r=9#..................#Compression ratio\n", + "R=287#.................#gas constant in J/kg\n", + "cv=0.71#...............#Specific heat at constant volume in kJ/kgK\n", + "cp=1.0#.................#Specific heat at constant pressure in kJ/kgK\n", + "ga=1.4#...............#Ratio of specific heats\n", + "#Calculations\n", + "p4=p3#\n", + "Vs=(pi/4)*D*D*L#.............#Stroke volume in m**3\n", + "Vc=Vs/(r-1)#..................#Clearance volume in m**3\n", + "rho=((r-1)*(co/100))+1#................#Cut off ratio\n", + "v1=Vc+Vs#.................#Volume after isochoric compression in m**3\n", + "p2=p1*(r**ga)#................#Pressure after adiabatic compression in bar\n", + "t2=t1*(r**(ga-1))#..............#Temperature after adiabatic expansion in K\n", + "t3=(p3*t2)/p2#..............#Temperature after isochoric compression in K\n", + "t4=t3*rho#.....................#Temperature after isobaric expansion in K\n", + "t5=t4*((rho/r)**(ga-1))#.........#Temperature after adiabatic expansion in K\n", + "p5=p4*(rho/r)**ga#...............#Pressure after adiabatic expansion in bar\n", + "Qs=(cv*(t3-t2)+cp*(t4-t3))#.....#Heat supplied in kJ/kg\n", + "Qr=cv*(t5-t1)#...................#Heat rejected in kJ/kg\n", + "etast=1-(Qr/Qs)#.................#Air standard efficiency\n", + "print \"Air standard efficiency = %0.2f %%\"%(etast*100)\n", + "m=(p1*v1*10**5)/(R*t1)#...............#Mass of air in cycle\n", + "W=m*(Qs-Qr)#....................#Work done per cycle in kJ\n", + "P=W*N#............................#Power developed in kW\n", + "print \"Power developed = %0.2f kW\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.26 PAGE 127" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature after adiabatic compression: 874.19 K\n", + "\n", + "Pressure after adiabatic compression: 21.67 bar\n", + "\n", + "Temperature after isochoric compression: 2742.67 K\n", + "\n", + "Pressure after isochoric compression: 68.00 bar\n", + "\n", + "Temperature after isobaric expansion: 3166.05 K\n", + "\n", + "Pressure after isobaric expansion: 68.00 bar\n", + "\n", + "Temperature after adiabatic expansion: 1392.38 K\n", + "\n", + "Pressure after adiabatic expansion: 3.84 bar\n", + "\n", + "Air standard efficiency = 58.24 %\n", + "Mean effective pressure = 11.09 bar\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#................#Initial pressure in bar\n", + "t1=363#.............#Initial temperature in K\n", + "r=9#.................#Compression ratio\n", + "p3=68#...............#Max pressure\n", + "p4=p3#\n", + "Qs=1750#..............#Total heat supplied\n", + "ga=1.4#...............#Ratio of specific heats\n", + "R=287#................#Gas constant in kJ/kgK\n", + "cv=0.71#..............#Specific heat at constant volume in kJ/kgK\n", + "cp=1#................#Specific heat at constant pressure in kJ/kgK\n", + "#Calculations\n", + "p2=p1*((r)**ga)#............#Pressure at the end of adiabatic compression in bar\n", + "t2=t1*((r)**(ga-1))#..........#Temperature at the end of adiabatic compression in K\n", + "t3=t2*(p3/p2)#............#Temperature at the end of isochoric compression in K\n", + "Qv=cv*(t3-t2)#.............#Heat added at constant volume in kJ/kg\n", + "Qp=Qs-Qv#.....................#Heat added at constant pressure in kJ/kg\n", + "t4=(Qp/cp)+t3#................#Temperature at the end of isobaric expansion in kJ/kg\n", + "rho=t4/t3#.....................#Cut off ratio\n", + "p5=p4*((rho/r)**ga)#................#Pressure at the end of adiabatic expansion in kJ/kg\n", + "t5=t4*((rho/r)**(ga-1))#...........#Temperature at the end of adiabatic expansion in kJ/kg\n", + "print \"Temperature after adiabatic compression: %0.2f K\\n\"%(t2)\n", + "print \"Pressure after adiabatic compression: %0.2f bar\\n\"%(p2)\n", + "print \"Temperature after isochoric compression: %0.2f K\\n\"%(t3)\n", + "print \"Pressure after isochoric compression: %0.2f bar\\n\"%(p3)\n", + "print \"Temperature after isobaric expansion: %0.2f K\\n\"%(t4)\n", + "print \"Pressure after isobaric expansion: %0.2f bar\\n\"%(p4)\n", + "print \"Temperature after adiabatic expansion: %0.2f K\\n\"%(t5)\n", + "print \"Pressure after adiabatic expansion: %0.2f bar\\n\"%(p5)\n", + "Qr=cv*(t5-t1)#....................#Heat rejected in kJ\n", + "etast=1-(Qr/Qs)#.................#Air standard efficiency\n", + "print \"Air standard efficiency = %0.2f %%\"%(etast*100)\n", + "pm=(1/(r-1))*((68*(rho-1))+(((p4*rho)-(p5*r))/(ga-1))-((p2-r)/(ga-1)))#................#Mean effective pressure in bar\n", + "print \"Mean effective pressure = %0.2f bar\"%pm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.27 PAGE 129" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air standard efficiency = 65.30 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=300#...............#Initial temperature\n", + "rmami=70#....................#Ratio of max pressure and min pressure\n", + "r=15#....................#Compression ratio\n", + "ga=1.4#.................#Ratio of specific heats\n", + "R=287#....................#Gas constant in kJ/kgK\n", + "t2=t1*(r**(ga-1))#.................#Temperature at the end of adiabatic compression in K\n", + "t3=t2*(rmami/(r**ga))#............#Temperature at the end of isochoric compression in K\n", + "t4=t3+((t3-t2)/ga)#..............#Temperature at the end of isobaric process in K\n", + "t5=t4/((1/(t4/(t3*r)))**(ga-1))#..........#Temperature at the end of adiabatic expansion in K\n", + "etast=1-((t5-t1)/((t3-t2)+ga*(t4-t3)))#..............#Air standard efficiency\n", + "print \"Air standard efficiency = %0.2f %%\"%(etast*100,)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.28 PAGE 131" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal efficiency = 61.80 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=373#.............#Initial temperature in K\n", + "p1=1#...............#Initial pressure in bar\n", + "p3=65#..............#Maximum pressure in bar\n", + "R=287#.................#Gas constant in kJ/kg\n", + "p4=p3#\n", + "ga=1.41#.................#Ratio of specific heats\n", + "Vs=0.0085#............#Swept volume in m**3\n", + "afr=21#...............#Air fuel ratio\n", + "r=15#.................#Compression ratio\n", + "C=43890#..............#Calorific value of fuel in kJ/kg\n", + "cp=1#................#Specific heat at constant pressure in kJ/kgK\n", + "cv=0.71#..............#Specific heat at constant volume in kJ/kgK\n", + "#Calculations\n", + "Vc=Vs/(r-1)#...............#Clearance volume in m**3\n", + "v2=Vc\n", + "v1=Vs+v2#\n", + "v3=Vc\n", + "v5=v1#\n", + "p2=p1*(r**ga)#.....................#Pressure at the end of adiabatic compression in bar\n", + "t2=t1*(r**(ga-1))#................#Temperature at the end of adiabatic compression in K\n", + "t3=(t2*p3)/p2#...................#Temperature at the end of isochoric compression in K\n", + "m=(p1*v1*10**5)/(R*t1)#............#Mass of air in the cycle in kg\n", + "Qv=m*cv*(t3-t2)#.....................#Heat added during constant volume process in kJ\n", + "fv=Qv/C#.............................#Fuel added during constant volume process in kg\n", + "mf=m/afr#..................#Total amount of fuel added in kg\n", + "mfib=mf-fv#....................#Total amount of fuel added in isobaric process in kg\n", + "Qib=mfib*C#....................#Total amount of heat added in isobaric process in kJ\n", + "t4=(Qib/((m+mf)*cp))+t3#........#Temperature at the end of isobaric process in K\n", + "v4=(v3*t4)/t3#..................#Volume at the end of isobaric process in m**3\n", + "t5=t4/((v5/v4)**(ga-1))#.........#Temperature at the end of isochoric expansion in K\n", + "Qrv=(m+mf)*cv*(t5-t1)#...............#Heat rejected during constant volume process in kJ\n", + "W=(Qib+Qv)-Qrv#................#Work done in kJ\n", + "etath=W/(Qib+Qv)#..................#Thermal efficiency\n", + "print \"Thermal efficiency = %0.2f %%\"%(etath*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.29 PAGE 133" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature after adiabatic compression: 524.81 K\n", + "\n", + "Pressure after adiabatic compression: 15.59 bar\n", + "\n", + "Temperature after isochoric compression: 1201.99 K\n", + "\n", + "Pressure after isochoric compression: 35.70 bar\n", + "\n", + "Temperature after isobaric expansion: 3283.19 K\n", + "\n", + "Pressure after isobaric expansion: 35.70 bar\n", + "\n", + "Temperature after adiabatic expansion: 2195.60 K\n", + "\n", + "Pressure after adiabatic expansion: 4.78 bar\n", + "\n", + "Mean effective pressure = 10.92 bar\n", + "Engine efficiency = 32.95 %\n", + "Power of the engine = 171.53 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "D=0.25#.............#Engine bore in m\n", + "L=0.4#..............#Engine stroke in m\n", + "t1=303#.............#Initial temperature in K\n", + "R=287#...............#Gas constant in kJ/kgK\n", + "p1=1#...............#Initial pressure in bar\n", + "N=8#................#No of working cycles per sec\n", + "cv=0.71#.............#Specific heat at constant volume in kJ/kgK\n", + "cp=1#.................#Specific heat at constant pressure in kJ/kgK\n", + "n=1.25#.............#Adiabatic index\n", + "rc=9#...............#Compression ratio\n", + "re=5#...............#Expansion ratio\n", + "rqptqe=2#...........#Ratio of heat liberated at constant pressure to heat liberated at constant volume\n", + "#Calculations\n", + "p2=p1*(rc**n)#.......................#Pressure at the end of adiabatic compression in bar\n", + "t2=t1*(rc**(n-1))#...................#Temperature at the end of adiabatic compression in K\n", + "rho=rc/re#..........................#Cut off ratio\n", + "t3=(2*cv*t2)/((2*cv)-(cp*(rho-1)))#...............#Temperature at the end of isochoric compression in K\n", + "p3=p2*(t3/t2)#....................................#Pressure at the end of isochoric compression in bar\n", + "p4=p3#t4=rho*t3#.................................#Temperature and pressure at the end of isobaric process\n", + "p5=p4*(1/(re**n))#.................................#Pressure at the end of adiabatic expansion in bar\n", + "t5=t4*(1/(re**(n-1)))#.............................#Temperature at the end of adiabatic expansion in K\n", + "pm=(1/(rc-1))*((p3*(rho-1))+(((p4*rho)-(p5*rc))/(n-1))-((p2-(p1*rc))/(n-1)))#...............#Mean effective pressure \n", + "print \"Temperature after adiabatic compression: %0.2f K\\n\"%(t2)\n", + "print \"Pressure after adiabatic compression: %0.2f bar\\n\"%(p2)\n", + "print \"Temperature after isochoric compression: %0.2f K\\n\"%(t3)\n", + "print \"Pressure after isochoric compression: %0.2f bar\\n\"%(p3)\n", + "print \"Temperature after isobaric expansion: %0.2f K\\n\"%(t4)\n", + "print \"Pressure after isobaric expansion: %0.2f bar\\n\"%(p4)\n", + "print \"Temperature after adiabatic expansion: %0.2f K\\n\"%(t5)\n", + "print \"Pressure after adiabatic expansion: %0.2f bar\\n\"%(p5)\n", + "print \"Mean effective pressure = %0.2f bar\"%pm\n", + "Vs=(pi/4)*D*D*L#....................#Swept volume in m**3\n", + "W=(pm*(10**5)*Vs)/1000#.................#Work done per cycle in kJ\n", + "m=(p1*(10**5)*(rc/(rc-1))*Vs)/(R*t1)#.....................#Mass of air per cycle in kg\n", + "Qs=m*(cv*(t3-t2)+cp*(t4-t3))#.....................#Heat supplied per cycle in kJ\n", + "eta=W/Qs#....................#Engine efficiency\n", + "print \"Engine efficiency = %0.2f %%\"%(eta*100)\n", + "P=W*N#.................#Power of the engine in kW\n", + "print \"Power of the engine = %0.2f kW\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.31 PAGE 140" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work done = 282.90 kJ/kg\n", + "Efficiency of cycle = 32.64 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "cp=0.92#..................#Specific heat at constant pressure in kJ/kgK\n", + "cv=0.75#..................#Specific heat at constant volume in kJ/kgK\n", + "p1=1#...................#Pressure at the end of adiabatic expansion in bar\n", + "p2=p1#...................#Pressure at the end of isobaric compression in bar\n", + "p3=4#....................#Pressure at the end of isobaric compression in bar\n", + "p4=16#...................#Final pressure after heat addition in bar\n", + "t2=300#.....................#Temperature at the end of isobaric compression in K\n", + "ga=1.22#................#Ratio of specific heats\n", + "#Calculations\n", + "t3=t2*((p3/p2)**((ga-1)/ga))#............#Temperature at the end of isobaric compression in K\n", + "t4=(p4*t3)/p3#........................#Final temperature after heat addition in K\n", + "t1=t4/((p4/p1)**((ga-1)/ga))#...................#Temperature at the end of adiabatic compression in K\n", + "Qs=cv*(t4-t3)#.........................#Heat supplied in kJ/kg\n", + "Qr=cp*(t1-t2)#.........................#Heat rejected in kJ/kg\n", + "W=Qs-Qr#.......................#Work done per kg of gas in kJ\n", + "print \"Work done = %0.2f kJ/kg\"%W\n", + "eta=W/Qs#......................#Efficiency of cycle\n", + "print \"Efficiency of cycle = %0.2f %%\"%(eta*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.32 PAGE 145" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cycle efficiency = 40.07 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=101.325#....................#Pressure of intake air in kPa\n", + "t1=300#.......................#Temperature of intake air in kPa\n", + "rp=6#.........................#Pressure ratio in the cycle\n", + "ga=1.4#.........................#Ratio of specific heats\n", + "rtc=2.5#...........................#Ratio of turbine work and compressor work\n", + "#Calculations\n", + "t2=t1*(rp**((ga-1)/ga))#..................#Temperature at the end of isentropic expansion in K\n", + "t3=(rtc*(t2-t1))/(1-(1/(rp**((ga-1)/ga))))#........#Temperature at the end of isobaric expansion in K\n", + "t4=t3/(rp**((ga-1)/ga))#.......................#Temperature at the end of isentropic compression in K\n", + "eta=(t3-t4-t2+t1)/(t3-t2)#...................#Cycle efficiency\n", + "print \"Cycle efficiency = %0.2f %%\"%(eta*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.33 PAGE 147" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power developed per kg of gas per second = 362.01 kW \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#....................#Intake pressure in bar\n", + "p2=5#....................#Supply pressure in bar\n", + "t3=1000#..................#Supply temperature in Kelvin\n", + "cp=1.0425#................#Specific heat at constant pressure in kJ/kgK\n", + "cv=0.7662#.................#Specific heat at constant volume in kJ/kgK\n", + "ga=cp/cv#..................#Ratio of specific heats\n", + "#Calculations\n", + "t4=t3*((p1/p2)**((ga-1)/ga))#\n", + "P=cp*(t3-t4)#.....................#Power developed per kg of gas per second in kW\n", + "print \"Power developed per kg of gas per second = %0.2f kW \"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.34 PAGE 147" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at turbine inlet = 423.51 K\n", + "Power developed = 13.85 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "ma=0.1#...................#Air supplied in kg/s\n", + "p1=1#.....................#Supply pressure in bar\n", + "t4=285#.................#Temperature of air when supplied to cabin in K\n", + "p2=4#...................#Pressure at inlet to turbine in bar\n", + "cp=1.0#..................#Specific heat at constant pressure in kJ/kgK\n", + "ga=1.4#..................#Ratio of specific heats\n", + "#Calculations\n", + "t3=t4*((p2/p1)**((ga-1)/ga))#................#Temperature at turbine inlet in K\n", + "print \"Temperature at turbine inlet = %0.2f K\"%t3\n", + "P=ma*cp*(t3-t4)#...........................#Power developed in kW\n", + "print \"Power developed = %0.2f kW\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.35 PAGE 148" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency of the cycle : 30.09\n", + "Heat supplied to the air = 456.17 kJ/kg \n", + "Work available at the shaft = 137.25 kJ/kg \n", + "Heat rejected in the cooler = 318.92 kJ/kg \n", + "Temperature of air leaving the turbine = 610.33 K \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#......................#Pressure of air entering the compressor in bar\n", + "p2=3.5#.................#Pressure of air while leaving the compressor in bar\n", + "t1=293#..................#Temperature of air at the onlet of the compressor in K\n", + "t3=873#.................#Temperature of air at the turbine inlet in K\n", + "cp=1.005#...............#Specific heat at constant pressure in kJ/kgK\n", + "ga=1.4#...................#Ratio of specific heats\n", + "#Calculations\n", + "rp=p2/p1#....................#Pressure ratio of the cycle\n", + "eta=1-(1/(rp**((ga-1)/ga)))#..............#Efficiency of the cycle\n", + "print \"Efficiency of the cycle : %0.2f\"%(eta*100)\n", + "t2=t1*((rp**((ga-1)/ga)))#................#Temperature of air while leaving the compressor in K\n", + "q1=cp*(t3-t2)#................#Heat supplied to the air in kJ/kg\n", + "print \"Heat supplied to the air = %0.2f kJ/kg \"%(q1)\n", + "W=eta*q1#........................#Work available at the shaft in kJ/kg\n", + "print \"Work available at the shaft = %0.2f kJ/kg \"%W\n", + "q2=q1-W#................#Heat rejected in the cooler in kJ/kg\n", + "print \"Heat rejected in the cooler = %0.2f kJ/kg \"%q2\n", + "t4=t3/(rp**((ga-1)/ga))#.......................#Temperature of air leaving the turbine in K\n", + "print \"Temperature of air leaving the turbine = %0.2f K \"%t4" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.36 PAGE 149" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum temperature = 1251.38 K \n", + "Cycle efficiency = 40.07 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "p1=1#...................#Pressure of air entering the compressor in bar\n", + "t1=300#.................#Temperature of air entering the compressor in bar\n", + "rp=6#...................#Pressure ratio\n", + "rtc=2.5#.................#Ratio of turbine work to compressor work\n", + "ga=1.4#............#Ratio of specific heats\n", + "#calculations\n", + "t2=t1*(rp**((ga-1)/ga))#..................#Temperature at the end of isentropic expansion in K\n", + "t3=(rtc*(t2-t1))/(1-(1/(rp**((ga-1)/ga))))#........#Temperature at the end of isobaric expansion in K\n", + "t4=t3/(rp**((ga-1)/ga))#.......................#Temperature at the end of isentropic compression in K\n", + "eta=(t3-t4-t2+t1)/(t3-t2)#...................#Cycle efficiency\n", + "print \"Maximum temperature = %0.2f K \"%t3\n", + "print \"Cycle efficiency = %0.2f %%\"%(eta*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.37 PAGE 150" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of fuel consumption = 0.02 kg/s \n", + "Mass flow rate of air = 1.68 kg/s \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "# Initialisation of Variables\n", + "t1=303#........................#Min temperature in K\n", + "t3=1073#........................#Max temperature in K\n", + "C=45000#.....................#Calorific value of fuel in kJ/kg\n", + "cp=1#....................#Specific heat at constant pressure in kJ/kgK\n", + "ga=1.4#........................#Ratio os specific heats\n", + "diftc=100#..................#Difference between work done by turbine and compressor in kW\n", + "#Calculations\n", + "t2=sqrt(t1*t3)# t4 = t2#.....#Assumed\n", + "mf=diftc/(C*(1-((t4-t1)/(t3-t2))))#................#Fuel used in kg per second\n", + "print \"Rate of fuel consumption = %0.2f kg/s \"%mf\n", + "ma=(diftc-(mf*(t3-t4)))/((t3-t4-cp*(t2-t1)))#............#Rate of air consumption in kg/s\n", + "print \"Mass flow rate of air = %0.2f kg/s \"%(ma)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.38 PAGE 151" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Compressor work = 259.32 kJ/kg \n", + "Turbine work = 351.64 kJ/kg \n", + "Heat supplied = 517.54 kJ/kg \n", + "Cycle efficiency = 17.84 %\n", + "Actual exhaust temperature of turbine = 723.11 K\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "t1=300#.................#Inlet temperature in K\n", + "p1=1#....................#Inlet pressure in bar\n", + "ma=1#....................#Mass of air in kg\n", + "rp=6.25#.............#Pressure ratio\n", + "t3=1073#.....#Maximum temperature in K\n", + "etac=0.8#............#Efficiency of compressor\n", + "etat=0.8#.............#Efficiency of turbine\n", + "ga=1.4#.................#Ratio of specific heats\n", + "cp=1.005#.............#Specific heat at constant pressure in kJ/kgK\n", + "#Calculations\n", + "t2=t1*(rp**((ga-1)/ga))#...........#Ideal Temperature of air while leaviing the compressor in K\n", + "t21=((t2-t1)/etac)+t1#............#Actual Temperature of air while leaviing the compressor in K\n", + "Wcomp=ma*cp*(t21-t1)#.............#Compressor work in kJ/kg\n", + "t4=t3/(rp**((ga-1)/ga))#........#Ideal temperature of air while leaving the turbine in K\n", + "t41=t3-(etat*(t3-t4))#..........#Actual temperature of air while leaving the turbine in K\n", + "Wtur=ma*cp*(t3-t41)#..............#Turbine work in kJ/kg\n", + "Wnet=Wtur-Wcomp#.................#Net work produced in kJ/kg\n", + "Qs=ma*cp*(t3-t21)#.................#Heat supplied in kJ/kg\n", + "print \"Compressor work = %0.2f kJ/kg \"%(Wcomp)\n", + "print \"Turbine work = %0.2f kJ/kg \"%Wtur\n", + "print \"Heat supplied = %0.2f kJ/kg \"%Qs\n", + "print \"Cycle efficiency = %0.2f %%\"%((Wnet/Qs)*100)\n", + "print \"Actual exhaust temperature of turbine = %0.2f K\"%t41" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.39 PAGE 153" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air fuel ratio is 56:1\n" + ] + } + ], + "source": [ + "# Initialisation of Variables\n", + "etat=0.85#..............#Turbine efficiency\n", + "etac=0.8#...............#Compressor efficiency\n", + "t3=1148#................#Max temperature in K\n", + "t1=300#................#Temperature of working fluid when entering the compressor in Kelvin\n", + "cp=1#...................#specific heat at constant pressure in kJ/kgK\n", + "ga=1.4#................#ratio of specific heats\n", + "p1=1#...................#Pressure of working fluid while entering the compressor in bar\n", + "rp=4#...................#Pressure ratio\n", + "C=42000#...............#Calorific value of fuel used in kJ/kgK\n", + "perlcc=10#.............#Percentage loss of calorific value in combustion chamber\n", + "\n", + "#calculations\n", + "p2=p1*rp#.................#pressure of air while leaving the compressor in bar\n", + "etacc=1-(perlcc/100)#............#efficiency of combustion chamber\n", + "t2=t1*(rp**((ga-1)/ga))#...........#Ideal Temperature of air while leaviing the compressor in K\n", + "t21=((t2-t1)/etac)+t1#............#Actual Temperature of air while leaviing the compressor in K\n", + "afr=((C*etacc)/(cp*(t3-t21)))-1#...........#Air fuel ratio\n", + "print \"Air fuel ratio is %d:1\"%round(afr)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 3.40 PAGE 154" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work done in process 1-2 = 148.93 kJ/kg \n", + "Heat rejected in process 1-2 = 148.93 kJ/kg \n", + "Work done in process 2-3 = 0.00 kJ/kg \n", + "Heat rejected in process 2-3 = 0.00 kJ/kg \n", + "Work done in process 3-4 = 446.80 kJ/kg \n", + "Heat rejected in process 3-4 = 446.80 kJ/kg\n", + "Work done in process 4-1 = 297.87 kJ/kg \n", + "Heat rejected in process 4-1 = 297.87 kJ/kg \n", + "Thermal efficiency of the cycle = 66.67 %\n" + ] + } + ], + "source": [ + "from math import log\n", + "# Initialisation of Variables\n", + "p1=1#...........#pressure before isothermal compression in bar\n", + "t1=310#.........#temperature before isothermal compression in K\n", + "p3=16#.........#pressure before isothermal expansion in bar\n", + "t3=930#.........#temperature before isothermal expansion in K\n", + "R=287#.............#Gas constant in kJ/kgK\n", + "#Calculations\n", + "v1=(R*t1)/(p1*10**5)#...............#Volume before isothermal compression in m**3\n", + "v3=(R*t3)/(p3*10**5)#...............#Volume before isothermal expansion in m**3\n", + "v2=v3#v4=v1#.................#2-3 and 1-4 are isochoric processes\n", + "r=v1/v2#...................#Compression ratio\n", + "q12=R*t1*log(r)#...............#Work done and heat rejected in process 1-2\n", + "w12=q12#\n", + "print \"Work done in process 1-2 = %0.2f kJ/kg \"%(q12/1000)\n", + "print \"Heat rejected in process 1-2 = %0.2f kJ/kg \"%(w12/1000)\n", + "q23=0#w23=q23#..................#COnstant volume process and hence work done is zero\n", + "print \"Work done in process 2-3 = %0.2f kJ/kg \"%(q23/1000)\n", + "print \"Heat rejected in process 2-3 = %0.2f kJ/kg \"%(q23/1000)\n", + "q34=R*t3*log(r)#...............#Work done and heat rejected in process 1-2\n", + "w34=q34#\n", + "print \"Work done in process 3-4 = %0.2f kJ/kg \"%(q34/1000)\n", + "print \"Heat rejected in process 3-4 = %0.2f kJ/kg\"%(w34/1000)\n", + "q41=q34-q12\n", + "w41=q41#\n", + "print \"Work done in process 4-1 = %0.2f kJ/kg \"%(q41/1000)\n", + "print \"Heat rejected in process 4-1 = %0.2f kJ/kg \"%(w41/1000)\n", + "etath=w41/q34#.....................#Thermal efficiency\n", + "print \"Thermal efficiency of the cycle = %0.2f %%\"%(etath*100)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER4_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER4_3.ipynb new file mode 100644 index 00000000..82c4af35 --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER4_3.ipynb @@ -0,0 +1,448 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 4 - Fuel-Air & Actual Cycles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 4.2 PAGE 187" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage change in the efficiency of otto cycle = -0.71 %\n" + ] + } + ], + "source": [ + "from math import log\n", + "# Initialisation of Variables\n", + "r=8#..........................#Compression Ratio\n", + "ga=1.4#.......................#Degree of freedom for the gas\n", + "Cvinc=1.1#....................#Increase of specific heat at constant volume in percentage\n", + "#Calculations\n", + "eta=1-1/(r**(ga-1))#...........#efficiency of otto cycle\n", + "deta=(1-eta)*(ga-1)*log(r)*(Cvinc/100)#............#Change in efficiency\n", + "etach=-deta/eta#............................#Percentage change in efficiency of change in efficiency\n", + "print \"The percentage change in the efficiency of otto cycle = %0.2f %%\"%(etach*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 4.3 PAGE 187" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage change in the efficiency of otto cycle = -1.98 %\n" + ] + } + ], + "source": [ + "from math import log\n", + "from __future__ import division\n", + "# Initialisation of Variables\n", + "r=7#..........................#Compression Ratio\n", + "ga=1.4#.......................#Degree of freedom for the gas\n", + "Cvinc=3#....................#Increase of specific heat at constant volume in percentage\n", + "#Calculations\n", + "eta=1-1/(r**(ga-1))#...........#efficiency of otto cycle\n", + "deta=(1-eta)*(ga-1)*log(r)*(Cvinc/100)#............#Change in efficiency\n", + "etach=-deta/eta#............................#Percentage change in efficiency of change in efficiency\n", + "print \"The percentage change in the efficiency of otto cycle = %0.2f %%\"%(etach*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 4.4 PAGE 188" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Percentage change = -1.15 efficiency\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "# Initialisation of Variables\n", + "r=18#..........................#Compression Ratio\n", + "co=5#..........................#Cut off percent of stroke\n", + "cv=0.71#.......................#Mean specific heat for cycle in kJ/kg K\n", + "R=0.285#.......................#Charecteristic gas constant in kJ/kh K\n", + "cvinc=2#.......................#Percentage increase in mean specific heat of the cycle\n", + "#Calculation\n", + "rho=(co/100)*(r-1)+1#\n", + "ga=1+(R/cv)#\n", + "eta=1-(1/(ga*(r**(ga-1))))*((rho**ga)-1)/(rho-1)#.....................#Efficiency of diesel cycle \n", + "etach=-((1-eta)/eta)*(ga-1)*(log(r)-(((rho**ga)*log(rho))/((rho**ga)-1))+(1/ga))*(cvinc/100)#...#Variation in the air standard efficiency\n", + "print \"Percentage change = %0.2f efficiency\"%(etach*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 4.5 PAGE 189" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Max pressure in the cylinder = 65.52 bar \n", + "Max pressure for constant specific heat = 93.52 bar \n" + ] + } + ], + "source": [ + "from math import sqrt,log\n", + "from __future__ import division\n", + "# Initialisation of Variables\n", + "r=7#......................#Compression ratio\n", + "C=44000#..................#Calorific value of fuel used in kJ/kg\n", + "afr=15#...................#Air fuel ratio\n", + "t1=338#....................#Temperature of the charge at the end of the stroke in Kelvin\n", + "p1=1#......................#Pressure of the charge at the end of the stroke in bar\n", + "n=1.33#....................#Index of compression\n", + "cv=0.71#......#Specific heat constant at constant volume in kJ/kgK\n", + "k=20*10**(-5)#\n", + "#Calculations\n", + "p2=p1*(r)**n#\n", + "t2=(t1*p2)/(p1*r)#\n", + "ha=C/(afr+1)#......................#Heat added per kg of charge in kJ\n", + "t3=((-2*cv)+sqrt((4*cv*cv)+(4*k*((2*cv*t2)+(k*t2*t2)+(2*ha)))))/(2*k)#\n", + "p3=(p2*t3)/t2#.............................#Max pressure for constant volume process in bar\n", + "P3=p2*((ha/cv)+t2)/t2#.....................#Max pressure for constant specific heat in bar\n", + "print \"Max pressure in the cylinder = %0.2f bar \"%(p3)\n", + "print \"Max pressure for constant specific heat = %0.2f bar \"%P3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 4.6 PAGE 190" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Max pressure in the cylinder = 102.27 bar \n", + "Max pressure for constant specific heat = 150.64 bar \n" + ] + } + ], + "source": [ + "from math import sqrt,log\n", + "from __future__ import division\n", + "# Initialisation of Variables\n", + "r=10#......................#Compression ratio\n", + "C=48000#..................#Calorific value of fuel used in kJ/kg\n", + "afr=15#...................#Air fuel ratio\n", + "t1=330#....................#Temperature of the charge at the end of the stroke in Kelvin\n", + "p1=1#......................#Pressure of the charge at the end of the stroke in bar\n", + "n=1.36#....................#Index of compression\n", + "cv=0.7117#......#Specific heat constant at constant volume in kJ/kgK\n", + "k=2.1*10**(-4)#\n", + "#Calculations\n", + "p2=p1*(r)**n#\n", + "t2=t1*((p2/p1)**((n-1)/n))#\n", + "ha=C/(afr+1)#......................#Heat added per kg of charge in kJ\n", + "t3=((-2*cv)+sqrt((4*cv*cv)+(4*k*((2*cv*t2)+(k*t2*t2)+(2*ha)))))/(2*k)#\n", + "p3=(p2*t3)/t2#.............................#Max pressure for constant volume process in bar\n", + "P3=p2*((ha/cv)+t2)/t2#.....................#Max pressure for constant specific heat in bar\n", + "print \"Max pressure in the cylinder = %0.2f bar \"%p3\n", + "print \"Max pressure for constant specific heat = %0.2f bar \"%P3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 4.7 PAGE 191" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Percentage of the stroke when the combustion is completed is 9.77\n" + ] + } + ], + "source": [ + "from math import sqrt,log\n", + "from __future__ import division\n", + "# Initialisation of Variables\n", + "r=15#......................#Compression ratio\n", + "C=43000#..................#Calorific value of fuel used in kJ/kg\n", + "afr=27#...................#Air fuel ratio\n", + "t2=870#....................#Temperature of the charge at the end of the stroke in Kelvin\n", + "cv=0.71#......#Specific heat constant at constant volume in kJ/kgK\n", + "R=0.287#.........................#Gas constant in kJ/kgK\n", + "k=20*10**(-5)#\n", + "#Calculations\n", + "cp=cv+R#............................#Specific heat at constant pressure\n", + "ha=C/(afr+1)#......................#Heat added per kg of charge in kJ\n", + "t3=((-2*cp)+sqrt((4*cp*cp)+(4*k*((2*cp*t2)+(k*t2*t2)+(2*ha)))))/(2*k)#\n", + "co=((t3/t2)-1)/(r-1)#.............#combustion occupies this amt of stroke \n", + "print \"Percentage of the stroke when the combustion is completed is \",round(co*100,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 4.8 PAGE 192" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Max pressure in the cylinder = 73.23 bar \n", + "Percentage of the stroke when the combustion is completed is 2.42\n" + ] + } + ], + "source": [ + "from math import sqrt,log\n", + "from __future__ import division\n", + "# Initialisation of Variables\n", + "r=14#......................#Compression ratio\n", + "t1=87+273#....................#Temperature of the charge at the end of the stroke in Kelvin\n", + "p1=1#......................#Pressure of the charge at the end of the stroke in bar\n", + "hsupa=1700#............................#heat supplied per kg of air in kJ\n", + "cv=0.71#......#Specific heat constant at constant volume in kJ/kgK\n", + "k=20*10**(-5)#\n", + "ga=1.4#.....................#Degree of freedom \n", + "R=0.287#......................#Gas constant in kJ/kgK\n", + "#Calculations\n", + "p2=p1*(r)**ga#\n", + "t2=t1*(r**(ga-1))#\n", + "ha=hsupa/2#......................#Heat added per kg of charge in kJ\n", + "t3=((-2*cv)+sqrt((4*cv*cv)+(4*k*((2*cv*t2)+(k*t2*t2)+(2*ha)))))/(2*k)#\n", + "p3=(p2*t3)/t2#.............................#Max pressure for constant volume process in bar\n", + "P3=p2*((ha/cv)+t2)/t2#.....................#Max pressure for constant specific heat in bar\n", + "print \"Max pressure in the cylinder = %0.2f bar \"%p3\n", + "cp=cv+R#.................................#Heat capacity at constant pressure in kJ/kgK\n", + "t4=((-2*cp)+sqrt((4*cp*cp)+(4*k*((2*cp*t3)+(k*t3*t3)+(2*ha)))))/(2*k)#\n", + "co=((t4/t3)-1)/(r-1)#.............#combustion occupies this amt of stroke \n", + "print \"Percentage of the stroke when the combustion is completed is \",round(co*100,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 4.9 PAGE 193" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum temperature ignoring molecular expansion = 4862.39 Kelvin \n", + "Maximum pressure ignoring molecular expansion = 113.41 bar \n", + "Maximum temperature considering molecular expansion = 4862.39 Kelvin \n", + "Maximum pressure considering molecular expansion = 122.86 bar\n" + ] + } + ], + "source": [ + "from math import sqrt,log\n", + "from __future__ import division\n", + "# Initialisation of Variables\n", + "r=8#...........................#Compression ratio\n", + "C=44000#............#Calorific value of fuel in kJ/kg\n", + "afr=13.8#.....................#Air fuel ratio\n", + "t1=343#.............................#Temperature of the mixture at the beginning of the compression in Kelvin\n", + "p1=1#........................#Pressure of the mixture at the beginning of the compression in bar\n", + "cv=0.716#.....................#Specific heat at constant volume in kJ/kgK\n", + "In=1.35#.......................#Index of compression\n", + "nc=6#..........................#No of carbon elements in the given fuel\n", + "nh=14#.........................#No of hydrogen elements in the given fuel\n", + "mc=12#...........................#Atomic mass of carbon in amu\n", + "mh=2#.............................#atomic mass of hydrogen molecule in amu\n", + "mo=32#...........................#Atomic mass of oxygen molecule in amu\n", + "#Calculations\n", + "#The chemical equation is C6H14 + xO2 ==> yCO2 + zH2O \n", + "#x is the no of oxygen molecules required for complete combustion\n", + "#y is the no of carbon dioxide molecules produced in complete combustion\n", + "#z is the no of Water molecules produced in complete combustion\n", + "y=nc#............................#As no of CO2 molecules is equal to no of C atoms in the fuel\n", + "z=nh/2#..........................#No of H2O molecules is equal to half the no of H atoms in the fuel\n", + "x=(z/2)+y#...........................#No of oxygen molecules required for combustionis half the no of water molecules plus the no of oxygen molecules \n", + "gafr=((x*32)*(100/23))/((mc*y)+(mh*z))#.................#Gravimetric air fuel ratio\n", + "ms=(gafr/afr)*100#......................#Actual mixture strength\n", + "#Since the mixture strength is greater than 100 %\n", + "#The mixture is rich in fuel. The combustion is therefore incompplete and hence CO will be formed\n", + "d=ms/100#......................#No of fuel molecules required for combustion\n", + "#The chemical equation is d(C6H14) + 9.5(O2) ==> a(CO2) + b(CO) + c(H2O) \n", + "c=(d*nh)/2#...............................#No of H2O molecules is equal to half the no of H atoms in the fuel\n", + "a=(x*2)-(d*nc)-c#........................#Equating atoms of the same element on both sides of equation\n", + "b=(d*nc)-a#\n", + "#By adding nitrogen on both sides, we are adding the same molecular weight on both sides.\n", + "#Air is 79 % nitrogen and 21 % oxygen\n", + "#Both N2 and O2 are diatomic molecules\n", + "n=x*(79/21)#.............................#No of nitrogen molecules \n", + "mbc=d+x+n#.............................#Moles before combustion\n", + "mac=a+b+c+n#.............................#Moles after expansion\n", + "me=(mac-mbc)/mbc#........................#Molecular expansion\n", + "t2=(t1*(r**(In-1)))#\n", + "t3=(t2+(C/((afr+1)*cv)))#..................#Maximum temperature ignoring molecular expansion in Kelvin\n", + "p3=p1*r*(t3/t1)#...........................#Maximum pressure ignoring molecular expansion in bar\n", + "t3me=t3#...............................#Maximum temperature considering molecular expansion in Kelvin\n", + "p3me=p3*(mac/mbc)#....................#Maximum pressure considering molecular expansion in bar\n", + "print \"Maximum temperature ignoring molecular expansion = %0.2f Kelvin \"%t3\n", + "print \"Maximum pressure ignoring molecular expansion = %0.2f bar \"%p3\n", + "print \"Maximum temperature considering molecular expansion = %0.2f Kelvin \"%t3me\n", + "print \"Maximum pressure considering molecular expansion = %0.2f bar\"%p3me" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 4.10 PAGE 194" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The cycle work = 872.05 kJ/kg \n", + "The cycle efficiency = 44.99 %\n" + ] + } + ], + "source": [ + "from math import sqrt,log\n", + "from __future__ import division\n", + "# Initialisation of Variables\n", + "r=7#....................#Compression Ratio\n", + "t2=715#.................#Temperature at the end of isentropic compression in Kelvin\n", + "t4=1610#................#Temperature at the end of expansion in Kelvin\n", + "#Calculations\n", + "vr2=65.8#..................#From steam table\n", + "u2=524.2#..................#From steam table\n", + "vr4=5.69#..................#From steam table\n", + "u4=1307.63#..................#From steam table\n", + "vr1=r*vr2#\n", + "t1=338#..................#From steam table\n", + "u1=241.38#..................#From steam table\n", + "vr3=vr4/r#\n", + "t3=2800#..................#From steam table\n", + "u3=2462.5#..................#From steam table\n", + "W=(u3-u2)-(u4-u1)#..................#Work done\n", + "Qa=(u3-u2)#..........................#Heat added\n", + "eta=W/Qa#...........................#Cycle efficiency\n", + "print \"The cycle work = %0.2f kJ/kg \"%W\n", + "print \"The cycle efficiency = %0.2f %%\"%(eta*100)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER5_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER5_3.ipynb new file mode 100644 index 00000000..522e0194 --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER5_3.ipynb @@ -0,0 +1,119 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 5 - Combustion in S.I. Engines" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 5.1 PAGE 218" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time for one combustion process = 4.51e-03 seconds \n", + "The crank position is = 12.47 degree\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "d=10.2#............#Engine bore in cm\n", + "spo=0.6#............#Spark plug offset in cm\n", + "vf=15.8#............#Average flame speed in m/s\n", + "thetas=20#.............#The angle of the crank when spark plug is fired\n", + "theta=6.5#..........#Angle by which the Engine rotates for combustion to develop (degree)\n", + "N=1200#................#Engine rpm\n", + "#calculations\n", + "dmax=(0.5*d)+spo#.........#Max distance of flame travel in cm\n", + "tf=(dmax)/(vf*100)#................#Time of flame travel in seconds\n", + "degs=(N/60)*360#...................#Conversion of engine rpm into degree/second\n", + "ctheta=tf*degs#...............#Crank angle for flame travel in degree\n", + "tc=theta/degs#..................#time for combustion to develop in seconds\n", + "top=tf+tc#......................#Time for one combustion process in seconds\n", + "thetatot=theta+ctheta#................#Total crank rotation in degree\n", + "thetacp = thetatot-thetas#..........#Crank position\n", + "print \"Time for one combustion process = %0.2e seconds \"%top\n", + "print \"The crank position is = %0.2f degree\"%thetacp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 5.2 PAGE 219" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time of spark = 15.00 Degrees before TDC \n", + "Time of spark = 16.54 degrees before TDC\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Initialisation of Variables\n", + "dp=22#............#Delay period in degree\n", + "cp=17#...............#Combustion period in degree\n", + "dper=14#...............#Delay Percentage\n", + "#Calculations\n", + "thetad=dp/2#.........#Full throttle half speed will result in delay angle being reduced for the same time\n", + "#Thus ignition timing should be arranged so that the total of thetad+cp ends 13 degree after TDC\n", + "tsp=(thetad+cp)-13#............#Time of spark in degree\n", + "print \"The time of spark = %0.2f Degrees before TDC \"%tsp\n", + "#Half throttle half speed will result in an increase of 14% in delay time over that at full throttle half speed \n", + "theta=(dper*thetad)/100#\n", + "dtheta=thetad+theta#............#Delay angle\n", + "tp=dtheta+cp#................#Total period\n", + "tsp=tp-13#..............#\n", + "print \"Time of spark = %0.2f degrees before TDC\"%tsp" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER7_3.ipynb b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER7_3.ipynb new file mode 100644 index 00000000..79ef34ae --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/CHAPTER7_3.ipynb @@ -0,0 +1,612 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 7 - Air capacity of four stroke engines" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.1 PAGE 246" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Gas Inhaled per minute : 0.23 m**3/min \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Initialisation of Variables\n", + "D=20.3#................#Diameter in cm\n", + "L=30.5#.................#Length in cm\n", + "N=300#................#Engine rpm\n", + "eta=78#.................#Efficiency in percentage\n", + "afr=4/1#.................#Air Fuel Ratio\n", + "\n", + "#Calculations\n", + "StV = ((pi)/4)*((D/100)**2)*(L/100)#.......#Calculating the stroke volume\n", + "Vinh= (eta/100)*StV#...................#Volume Inhaled\n", + "Gainh= (Vinh/(4+1))#..............#Gas Inhaled\n", + "Gainhpm = Gainh*(N/2)#\n", + "print \"Gas Inhaled per minute : %0.2f m**3/min \"%(Gainhpm)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.2 PAGE 247" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volumetric efficiency of the engine is : 0.85\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Initializing the variables\n", + "\n", + "N=3600#................#engine rpm\n", + "T=15#................#Inlet temperature in degree Celsius\n", + "Tk = T+273#..............#Inlet temperature in Kelvin\n", + "p=760#................#Inlet pressure in mm of Hg i.e. 1.013 x 10**5 Pa\n", + "ppa=1.013*(10**5)#.........# Inlet pressure in Pascals\n", + "pdv=4066#..............#Total piston displacement volume in cm**3\n", + "pdvsi=pdv*(10**(-6))#.............#Total piston displacement volume in m**3\n", + "afr=14/1#...................#Air fuel ratio is 14:1\n", + "bsfc=0.38#..................# b.s.f.c in kg/kWh\n", + "BP=86#.............#power output in kW\n", + "R=287#................#Gas constant for air in J/kg.K\n", + "#Calculations\n", + "m = (BP*bsfc*afr)/60#...............#Air consumption\n", + "V = (m*R*Tk)/ppa#\n", + "DV= pdvsi*(N/2)#.........#Displacement Volume\n", + "VE=V/DV#...............#Volumetric Efficiency\n", + "print \"Volumetric efficiency of the engine is : %0.2f\"%(VE)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.3 PAGE 248" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volumetric efficiency based on the free air condition : 0.86\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "#Initializing the variables\n", + "n=4#..............#No of cylinders\n", + "d=5#.............#diameter of orifice in cm\n", + "dsi=d/100#..........# diameter in m\n", + "Cd=0.6#.............#Co-efficient of discharge\n", + "D=10#..............#Engine bore in cm\n", + "Dsi=d/100#............#Engine bore in m\n", + "L=12#................#Engine stroke in cm\n", + "Lsi=L/100#............#Engine stroke in m\n", + "N=1200#...............#Engine rpm\n", + "hw=0.046#............#Pressure drop across orifice in m of water\n", + "T = 17#..........#Ambient Temparature in Degree Celsius\n", + "Tk = T+273#..........# Ambient Temperature in Kelvin\n", + "Pbar = 1#.............# Ambient pressure in bar\n", + "Ppa = 1 * (10**5)#.......#Ambient pressure in Pascal\n", + "R = 287#.............# Gas constant in J/kg.K\n", + "rhow = 1000#............#Density of water in kg/m**3\n", + "g=9.81#...............#Acceleration due to gravity\n", + "#Calculations\n", + "\n", + "rhoa= Ppa/(R*Tk)#.........#Density of air\n", + "ha= (hw*rhow)/rhoa#\n", + "av= sqrt(2*g*ha)#.............#Air velocity\n", + "area = (pi/4)*(dsi**2)#\n", + "Vact = Cd*area*av#.............# V actual\n", + "Vswt = n*(pi/4)*(Dsi**2)*Lsi*(N/60*2)#\n", + "eff = Vact/Vswt#...............#Volumetric efficiency\n", + "print \"The volumetric efficiency based on the free air condition : %0.2f\"%(eff)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.4 PAGE 249" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The given fuel C7H16 requires 11(O2) for complete combustion\n", + "Hence, Mass of fuel is : 100.00\n", + "Mass of Oxygen required is : 352.00\n", + "The air fuel ratio is : 15.30\n", + "The volumetric efficiency of the engine is = 75.90 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Initializing the variables\n", + "n=1#...................#No of cylinders\n", + "k=0.5#\n", + "Vs=7000#............#displacement volume in cm**3\n", + "Vssi= Vs*(10**(-6))#........#displacement volume in m**3\n", + "OP=14.7#...................#Power developed in kW\n", + "N=450#..................#Engine rpm\n", + "sfc=0.272#................#Specific fuel consumption in kg/kWh\n", + "#Fuel used is C7H16\n", + "mC=12#.............#mass of carbon in amu\n", + "mH=1#.................#mass of hydrogen in amu\n", + "mO=16#.................#mass of oxygen in amu\n", + "pi=1.013 * (10**5)#................#initial pressure in pascal\n", + "T=30#...................#initial temperature in degree celsius\n", + "Tk=30+273#................#initial temperature in degree kelvin\n", + "R=287#..................#Gas constant for air in J/kg.K\n", + "#calculations\n", + "print \"The given fuel C7H16 requires 11(O2) for complete combustion\"\n", + "mf=(7*mC)+(16*mH)#\n", + "print \"Hence, Mass of fuel is : %0.2f\"%(mf)\n", + "MO=11* 2 * mO#\n", + "print \"Mass of Oxygen required is : %0.2f\"%(MO)\n", + "ma = MO/0.23#.......#mass of air\n", + "#Air contains 23% of oxygen by weight\n", + "afr = ma/mf#...............#air fuel ratio is the ratio of mass of air to mass of fuel\n", + "print \"The air fuel ratio is : %0.2f\"%(afr)\n", + "MF = sfc * OP#...........#actual fuel consumed in kg/h\n", + "MA = afr*MF#\n", + "AAS = MA * (1+0.3)#....................#actual air supplied in kg/h\n", + "M = AAS + MF#................#mass of charge in kg/h\n", + "VCS = ((M/60)*R*Tk)/pi#.............#Volume of charge sucked in m**3/min\n", + "DVM = Vssi * (N/2)#..............#Displacement volume/min\n", + "eta = VCS/DVM#\n", + "print \"The volumetric efficiency of the engine is = %0.2f %%\"%(eta*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.5 PAGE 250" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volumetric efficiency is = 87.43 %\n", + "The brake thermal efficiency is = 23.38 %\n", + "The brake torque = 0.01 Nm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Initializing the variables\n", + "n=6#................#No of cylinders\n", + "vsi=730*(10**(-6))#..........#Piston displacement per cylinder in m**3\n", + "BP=80#.............#Power produced per cylinder in kW\n", + "N=3100#...........#Engine rpm\n", + "C=44*(10**6)#...........#Calorific value of petrol in J/kg\n", + "Pc=28#........#Petrol consumed per hour in kg\n", + "afr = 13/1#.......#air fuel ratio\n", + "pi=0.88*(10**5)#..............#Intake pressure in pa\n", + "T=300#............#Intake temperature in Kelvin\n", + "R = 287#.........#gas constant in J/kg.K\n", + "#calculations\n", + "ma = (Pc*afr)/60#...........#air comsumed\n", + "rhoa = pi/(R*T)#.......#Density of air\n", + "etaV=ma/(rhoa*vsi*n*(N/2))#\n", + "print \"The volumetric efficiency is = %0.2f %%\"%(etaV*100)\n", + "mf = Pc/3600#...............#Fuel consumed per sec\n", + "etaBT = (BP*1000)/(mf*C)#\n", + "print \"The brake thermal efficiency is = %0.2f %%\"%(etaBT*100)\n", + "T=(BP*60*1000)/(2*(pi)*N)#\n", + "print \"The brake torque = %0.2f Nm \"%(T)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.6 PAGE 251" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volumetric efficiency is = 81.44 %\n", + "Percentage Reduction in Output = 4.61 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "#Initializing the variables\n", + "etaV1 = 0.8#.........#Volumetric efficiency\n", + "pi1 = 1.013#.......#Inlet pressure\n", + "pe1= 1.013\n", + "pi2= 1.013#\n", + "pe2 = 1.15#.........#Exhaust pressure\n", + "Tk1 = 298#...........#Temperature in Kelvin\n", + "Tk2 = 318#...........#Temperature in Kelvin\n", + "r = 7.5#........#compression ratio\n", + "ga=1.4#..........#degree of freedom for gas\n", + "#calculations\n", + "#For pressure change\n", + "eta_V2 = r - (pe2/pi2)**(1/ga)#\n", + "eta_V1 = r - (pe1/pi1)**(1/ga)#\n", + "x=eta_V2/eta_V1#\n", + "#For inlet temperature change\n", + "y = sqrt(Tk2/Tk1)#\n", + "#For volumetric efficiency, considering both pressure and temperature\n", + "etaV2 = etaV1*x*y#\n", + "print \"The volumetric efficiency is = %0.2f %%\"%(etaV2*100)\n", + "PO=((etaV1/Tk1)-(etaV2/Tk2))/(etaV1/Tk1)#\n", + "print \"Percentage Reduction in Output = %0.2f %%\"%(PO*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.7 PAGE 252" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage increase in volumetric efficiency is = 6.68 %\n", + "The percentage increase in power = 23.34 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "#Initializing the variables\n", + "pi1 = 1.013#.......#Inlet pressure\n", + "pe1= 1.013\n", + "pi2= 1.3#\n", + "pe2 = 1.013#.........#Exhaust pressure\n", + "Tk1 = 300#...........#Temperature in Kelvin\n", + "Tk2 = 333#...........#Temperature in Kelvin\n", + "r = 14#........#compression ratio\n", + "ga=1.4#..........#degree of freedom for gas\n", + "R=287#...........#gas constant in J/kg.K\n", + "#calculations\n", + "#For pressure change\n", + "eta_V2 = r - (pe2/pi2)**(1/ga)#\n", + "eta_V1 = r - (pe1/pi1)**(1/ga)#\n", + "x=eta_V2/eta_V1#\n", + "#For inlet temperature change\n", + "y = sqrt(Tk2/Tk1)#\n", + "#For volumetric efficiency, considering both pressure and temperature\n", + "pive = ((x*y)-1)#.........#percentage increase in volumetric efficiency\n", + "print \"The percentage increase in volumetric efficiency is = %0.2f %%\"%(pive*100)\n", + "rho1 = (pi1*10**5)/(R*Tk1)#\n", + "rho2 = (pi2*10**5)/(R*Tk2)#\n", + "z = (rho2/rho1)*x*y#\n", + "pip = (z-1)#\n", + "print \"The percentage increase in power = %0.2f %%\"%(pip*100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.8 PAGE 253" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "There is no change in volumetric efficiency due to inlet and exhaust pressure change\n", + "For inlet temperature change\n", + "Percentage decrease = 4.99 %\n", + "Brake power of the engine on the hill station = 24.81 kW \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "#Initializing the variables\n", + "IP1 = 32#...............#Indicated power output in kW\n", + "etamech=80#............#Mechanical efficiency at sea level\n", + "p1=1.013#.............#initial pressure at sea level in bar\n", + "tk1 = 308#...............#Initial temperature at sea level in Kelvin\n", + "tk2 = 278#..................#temperature atop the hill in Kelvin\n", + "rhoHg=13600#..............#Density of mercury in kg/m**3\n", + "h=2000#.....................#Hill altitude \n", + "g = 9.81#................#Acceleration due to gravity\n", + "delp = 10#..............#drop of mercury in mm Hg per every 100 m climb\n", + "#calculations\n", + "\n", + "print \"There is no change in volumetric efficiency due to inlet and exhaust pressure change\"\n", + "print \"For inlet temperature change\"\n", + "x = sqrt(tk2/tk1)#................#for inlet temperature change\n", + "#x is the ratio of the efficiencies at the beginning and on hill top\n", + "print \"Percentage decrease = %0.2f %%\"%((1-x)*100)\n", + "dp = rhoHg*g*((delp/1000)*(h/100))*(10**(-5))#........#Drop in pressure at hill station\n", + "p2=p1-dp#\n", + "IP_1 = p1/tk1#\n", + "IP_2 = (x*p2)/tk2#\n", + "k = IP_2/IP_1#..............#Ratio of indicative power output during initial and final conditions\n", + "IP2 = (IP1 * k)/(etamech/100)#\n", + "#Since the engine speed is the same at two places, the friction and hence mechanical efficiency remains unchanged\n", + "BP2 = IP2*(etamech/100)#\n", + "print \"Brake power of the engine on the hill station = %0.2f kW \"%(BP2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.9 PAGE 254" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volumetric efficiency of supercharged engine = 86.36 %\n", + "Indicated power of supercharged engine is = 102.11 kW \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "#Initializing the variables\n", + "etaV1 = 0.81#.........#Volumetric efficiency\n", + "pi1 = 1.01#.......#Inlet pressure before supercharger\n", + "pe1= 1.01#...........#Exhaust pressure before supercharger\n", + "pi2= 1.38#............#Inlet pressure after supercharger\n", + "pe2 = 1.01#.........#Exhaust pressure in bar after addition of super charger\n", + "Tk1 = 300#...........#Temperature in Kelvin\n", + "Tk2 = 321#...........#Temperature in Kelvin\n", + "r = 7.5#........#compression ratio\n", + "ga=1.4#..........#degree of freedom for gas\n", + "R=287#.............#Gas constant for air in J/kgK\n", + "IP1=75#...............#Indicated power output before addition of supercharger\n", + "#calculations\n", + "#For pressure change\n", + "eta_V2 = r - (pe2/pi2)**(1/ga)#\n", + "eta_V1 = r - (pe1/pi1)**(1/ga)#\n", + "x=eta_V2/eta_V1#\n", + "#For inlet temperature change\n", + "y = sqrt(Tk2/Tk1)#\n", + "#For volumetric efficiency, considering both pressure and temperature\n", + "etaV2 = etaV1*x*y#\n", + "print \"The volumetric efficiency of supercharged engine = %0.2f %%\"%(etaV2*100)\n", + "rho1 = (pi1*10**5)/(R*Tk1)#....#density of air before addition of supercharger\n", + "rho2 = (pi2*10**5)/(R*Tk2)#..#density of air after addition of supercharger\n", + "IP2 = IP1 * (etaV2*rho2)/(etaV1*rho1)#\n", + "print \"Indicated power of supercharged engine is = %0.2f kW \"%IP2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.10 PAGE 255" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volumetric efficiency of the engine = 72.66 %\n", + "The heating value of 1 m**3 of charge at 25 degree Celsius = 1513.16 kJ.m**3\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "#Initializing the variables\n", + "n=1#.....#No of cylinders\n", + "D=0.32#.....#Bore of the cylinder in m\n", + "L=0.38#......#Stroke of the cylinder in m\n", + "N = 280#....#Engine rpm\n", + "CV = 18600#....#calorific value of fues in kJ/m**3\n", + "Tk1 = 300#....#Initial temperature in Kelvin\n", + "p1 = 1.013#.....#Initial pressure in bar\n", + "ma = 3.36#.......#mass of air consumed per min\n", + "tgc = 0.25#......#test gas consumption in m**3/min\n", + "pw = 120#.........#pressure of water in mm during the test gas consumption\n", + "tgct = 300#.......#Temperature in Kelvin during test gas consumption\n", + "rhow = 1000#.....#density of water in kg/m**3\n", + "R=287#...........#Gas constant in J/kg.K\n", + "#calculations\n", + "V= (ma*R*Tk1)/(p1*(10**5))#...#Volume of air consumed at inlet condition\n", + "\n", + "gsp = p1 +(pw/rhow)/10.2#...................#Gas supply pressure\n", + "#1 bar = 10.2 m\n", + "gcic = tgc*(gsp/p1)#..........#Gas consumption at inlet condition\n", + "Vi = gcic+V#.....#Volume of mixture at inlet condition\n", + "Vswt = (pi/4)*(D**2)*L*(N/2)#......#Swept volume \n", + "etaV = Vi/Vswt#.....#Volumetric efficiency\n", + "print \"The volumetric efficiency of the engine = %0.2f %%\"%(etaV*100)\n", + "hv = (gcic/Vi)*CV#......#Heating value\n", + "print \"The heating value of 1 m**3 of charge at 25 degree Celsius = %0.2f kJ.m**3\"%hv" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 7.12 PAGE 256" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)Nominal diameter of the inlet valve = 0.04 m or 44.95 mm\n", + "(ii)When the engine is modified to develop max indicative power at 2800rpm, nominal diameter of the inlet valve = 0.05 m or 48.55 mm\n", + "(iii)The new mach index value is : 0.64\n", + "Hence the volumetric efficiency drops (There is a steady decrease in volumetric efficiency of an engine if there is an increase in the mach index beyond 0.55, Refer the FIG 7_12)\n", + "(iv)The mach index for the unmodified engine : 1.10\n", + "The volumetric efficiency is approximately 56% (from the FIG 7_12)\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "#Initializing the variables\n", + "Z=0.55#.............#Mach Index\n", + "Dcy=0.11#............#Engine Bore in m\n", + "L = 0.14#............#stroke length in m\n", + "N = 2400#.............#Engine rpm\n", + "N1 = 2800#............#Engine rpm after modification\n", + "N2=4800#.............#Max rpm for unmodified engine\n", + "p = 0.88#.........#pressure at intake valve in bar\n", + "t=340#...............#temperature at intake valve in Kelvin\n", + "ki = 0.33#...........#Inlet flow co-efficient\n", + "ga = 1.4#...............#degree of freedom of the gas\n", + "R = 287#.................#Gas constant for air in J/kg.K\n", + "#calculations\n", + "\n", + "Us = sqrt(ga*R*t)#......#sonic velocity of air-fuel mixture at the inlet valve\n", + "Up = (2*L*N)/60#...........#piston speed\n", + "Div = sqrt(((Dcy**2)*Up)/(Z*ki*Us))#...............#Nominal diameter of the inlet valve in m\n", + "print \"(i)Nominal diameter of the inlet valve = %0.2f m or %0.2f mm\"%(Div,Div*1000)\n", + "Div1 = sqrt(((Dcy**2)*2*L*N1)/(Z*ki*Us*60))#.......# Nominal diameter of inlet valve for the modified engine in m\n", + "print \"(ii)When the engine is modified to develop max indicative power at 2800rpm, nominal diameter of the inlet valve = %0.2f m or %0.2f mm\"%(Div1, Div1*1000,)\n", + "Up1=(2*L*N1)/60#............#New piston speed for modified engine\n", + "Z1 = ((Dcy/Div)**2)*(Up1/(ki*Us))#\n", + "print \"(iii)The new mach index value is : %0.2f\"%(Z1)\n", + "print \"Hence the volumetric efficiency drops (There is a steady decrease in volumetric efficiency of an engine if there is an increase in the mach index beyond 0.55, Refer the FIG 7_12)\"\n", + "Up2 = (2*L*N2)/60#..............#Piston speed at max rpm for unmodified engine\n", + "Z2 = ((Dcy/Div)**2)*(Up2/(ki*Us))#\n", + "print \"(iv)The mach index for the unmodified engine : %0.2f\"%Z2\n", + "print \"The volumetric efficiency is approximately 56% (from the FIG 7_12)\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/screenshots/HeatTransfer(3)_3.png b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/screenshots/HeatTransfer(3)_3.png Binary files differnew file mode 100644 index 00000000..52b8656a --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/screenshots/HeatTransfer(3)_3.png diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/screenshots/cycleWorkEff(4)_3.png b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/screenshots/cycleWorkEff(4)_3.png Binary files differnew file mode 100644 index 00000000..e55702fe --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/screenshots/cycleWorkEff(4)_3.png diff --git a/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/screenshots/volumetric_Eff(7)_3.png b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/screenshots/volumetric_Eff(7)_3.png Binary files differnew file mode 100644 index 00000000..facf3212 --- /dev/null +++ b/A_textbook_of_Internal_Combustion_Engines_by_R._K._Rajput/screenshots/volumetric_Eff(7)_3.png diff --git a/Advance_Semiconductor_Devices_by_K._C._Nandi/README.txt b/Advance_Semiconductor_Devices_by_K._C._Nandi/README.txt new file mode 100644 index 00000000..1ce60d31 --- /dev/null +++ b/Advance_Semiconductor_Devices_by_K._C._Nandi/README.txt @@ -0,0 +1,10 @@ +Contributed By: Mohd Asif +Course: btech +College/Institute/Organization: Jagen Technologies, New Delhi +Department/Designation: Software Developer +Book Title: Advance Semiconductor Devices +Author: K. C. Nandi +Publisher: Tech. Max Publications Pune +Year of publication: 2013 +Isbn: 978-81-8492-983-6 +Edition: 2
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_4.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_4.ipynb new file mode 100644 index 00000000..7cd2fd61 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_4.ipynb @@ -0,0 +1,280 @@ +{
+ "metadata": {
+ "name": "chapter no.10.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Theory of Failures"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.1,Page No.401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P_e=300 #N/mm**2 #Elastic Limit in tension\n",
+ "FOS=3 #Factor of safety\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "P=12*10**3 #N Pull \n",
+ "Q=6*10**3 #N #Shear force\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let d be the diameter of the shaft\n",
+ "\n",
+ "#Direct stress\n",
+ "#P_x=P*(pi*4**-1*d**3)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P_x=48*10**3\n",
+ "\n",
+ "#Now shear stress at the centre of bolt\n",
+ "#q=4*3**-1*q_av\n",
+ "#After substituting values and further simplifying we get\n",
+ "#q=32*10**3*(pi*d**2)**-1\n",
+ "\n",
+ "#Principal stresses are\n",
+ "#P1=P_x*2**-1+((P_x*2**-1)**2+q**2)**0.5\n",
+ "#After substituting values and further simplifying we get\n",
+ "#p1=20371.833*(d**2)**-1\n",
+ "\n",
+ "#P2=P_x*2**-1-((P_x*2**-1)**2+q**2)**0.5\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P2=-5092.984*(d**2)**-1\n",
+ "\n",
+ "#q_max=((P_x*2**-1)**2+q**2)**0.5\n",
+ "\n",
+ "#From Max Principal stress theory\n",
+ "#Permissible stress in Tension\n",
+ "P1=100 #N/mm**2 \n",
+ "d=(20371.833*P1**-1)**0.5\n",
+ "\n",
+ "#Max strain theory\n",
+ "#e_max=P1*E**-1-mu*P2*E**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#e_max=21899.728*(d**2*E)**-1\n",
+ "\n",
+ "#According to this theory,the design condition is\n",
+ "#e_max=P_e*(E*FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d2=(21899.728*3*300**-1)**0.5 #mm\n",
+ "\n",
+ "#Max shear stress theory\n",
+ "#e_max=shear stress at elastic*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d3=(12732.421*6*300**-1)**0.5 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of Bolt by:Max Principal stress theory\",round(d,2),\"mm\"\n",
+ "print\" :Max strain theory\",round(d2,2),\"mm\"\n",
+ "print\" :Max shear stress theory\",round(d3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of Bolt by:Max Principal stress theory 14.27 mm\n",
+ " :Max strain theory 14.8 mm\n",
+ " :Max shear stress theory 15.96 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.2.Page No.402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "M=40*10**6 #N-mm #Bending moment\n",
+ "T=10*10**6 #N-mm #TOrque\n",
+ "mu=0.25 #Poissoin's ratio\n",
+ "P_e=200 #N/mm**2 #Stress at Elastic Limit\n",
+ "FOS=2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let d be the diameter of the shaft\n",
+ "\n",
+ "#Principal stresses are given by\n",
+ "\n",
+ "#P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5)\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P1=4.13706*10**8*(d**3)**-1 ............................(1)\n",
+ "\n",
+ "#P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5)\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P2=-6269718*(pi*d**3)**-1 ..............................(2)\n",
+ "\n",
+ "#q_max=(P1-P2)*2**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#q_max=2.09988*10**8*(d**3)**-1\n",
+ "\n",
+ "#Max Principal stress theory\n",
+ "#P1=P_e*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d=(4.13706*10**8*2*200**-1)**0.33333 #mm \n",
+ "\n",
+ "#Max shear stress theory\n",
+ "#q_max=shear stress at elastic limit*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d2=(2.09988*10**8*4*200**-1)**0.33333\n",
+ "\n",
+ "#Max strain energy theory\n",
+ "#P_3=0\n",
+ "#P1**2+P2**2-2*mu*P1*P2=P_e**2*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d3=(8.62444*10**12)**0.166666\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of shaft according to:MAx Principal stress theory\",round(d,2),\"mm\"\n",
+ "print\" :Max shear stress theory\",round(d2,2),\"mm\"\n",
+ "print\" :Max strain energy theory\",round(d3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of shaft according to:MAx Principal stress theory 160.52 mm\n",
+ " :Max shear stress theory 161.33 mm\n",
+ " :Max strain energy theory 143.2 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.3,Page No.403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "f_x=40 #N/mm**2 #Internal Fliud Pressure\n",
+ "d1=200 #mm #Internal Diameter\n",
+ "r1=d1*2**-1 #mm #Radius\n",
+ "q=300 #N/mm**2 #Tensile stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have,\n",
+ "\n",
+ "#Hoop Stress\n",
+ "#f_x=b*(x**2)**-1+a ..........................(1)\n",
+ "\n",
+ "#Radial Pressure\n",
+ "#p_x=b*(x**2)**-1-a .........................(2)\n",
+ "\n",
+ "#the boundary conditions are\n",
+ "x=d1*2**-1 #mm \n",
+ "#After sub values in equation 1 and further simplifying we get\n",
+ "#40=b*100**-1-a ..........................(3)\n",
+ "\n",
+ "#Max Principal stress theory\n",
+ "#q*(FOS)**-1=b*100**2+a ..................(4)\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "a=80*2**-1\n",
+ "#Sub value of a in equation 3 we get\n",
+ "b=(f_x+a)*100**2\n",
+ "\n",
+ "#At outer edge where x=r_0 pressure is zero\n",
+ "r_0=(b*a**-1)**0.5 #mm\n",
+ "\n",
+ "#thickness\n",
+ "t=r_0-r1 #mm\n",
+ "\n",
+ "#Max shear stress theory\n",
+ "P1=b*(100**2)**-1+a #Max hoop stress\n",
+ "P2=-40 #pressure at int radius (since P2 is compressive)\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(P1-P2)*2**-1\n",
+ "\n",
+ "#According max shear theory the design condition is\n",
+ "#q_max=P_e*2**-1*(FOS)**-1\n",
+ "#After sub values in equation we get and further simplifying we get\n",
+ "#80=b*(100**2)**-1+a\n",
+ "#After sub values in equation 1 and 3 and further simplifying we get\n",
+ "b2=120*100**2*2**-1\n",
+ "\n",
+ "#from equation(3)\n",
+ "a2=120*2**-1-a\n",
+ "\n",
+ "#At outer radius r_0,radial pressure=0\n",
+ "r_02=(b2*a2**-1)**0.5\n",
+ "\n",
+ "#thickness\n",
+ "t2=r_02-r1\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of metal by:Max Principal stress theory\",round(t,2),\"mm\"\n",
+ "print\" :Max shear stress thoery\",round(t2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of metal by:Max Principal stress theory 41.42 mm\n",
+ " :Max shear stress thoery 73.21 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_4.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_4.ipynb new file mode 100644 index 00000000..89760142 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_4.ipynb @@ -0,0 +1,22 @@ +{
+ "metadata": {
+ "name": "chapter no.1.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.1:Introduction"
+ ]
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_4.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_4.ipynb new file mode 100644 index 00000000..a37ee3c9 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_4.ipynb @@ -0,0 +1,2742 @@ +{
+ "metadata": {
+ "name": "chapter no.2.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Simple Stresses And Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.1,Page No.14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "P=45*10**3 #N #Load\n",
+ "E=200*10**3 #N/mm**2 #Modulus of elasticity of rod\n",
+ "L=500 #mm #Length of rod\n",
+ "d=20 #mm #Diameter of rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*d**2*4**-1 #mm**2 #Area of circular rod\n",
+ "p=P*A**-1 #N/mm**2 #stress\n",
+ "e=p*E**-1 #strain\n",
+ "dell_l=(P*L)*(A*E)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"The stress in bar due to Load is\",round(p,5),\"N/mm\"\n",
+ "print\"The strain in bar due to Load is\",round(e,5),\"N/mm\"\n",
+ "print\"The Elongation in bar due to Load is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stress in bar due to Load is 143.23945 N/mm\n",
+ "The strain in bar due to Load is 0.00072 N/mm\n",
+ "The Elongation in bar due to Load is 0.36 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.2,Page No.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A=15*0.75 #mm**2 #area of steel tape\n",
+ "P=100 #N #Force apllied\n",
+ "L=30*10**3 #mm #Length of tape\n",
+ "E=200*10**3 #N/m**2 #Modulus of Elasticity of steel tape\n",
+ "AB=150 #m #Measurement of Line AB \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "dell_l=P*L*(A*E)**-1 #mm #Elongation\n",
+ "l=L+dell_l*10**-3 #mm #Actual Length \n",
+ "AB1=AB*l*L**-1 #m Actual Length of AB\n",
+ "\n",
+ "#Result\n",
+ "print\"The Actual Length of Line AB is\",round(AB1,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Actual Length of Line AB is 150.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.3,Page No.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Let y be the yield stress\n",
+ "\n",
+ "y=250 #N/mm**2 #yield stress\n",
+ "FOS=1.75 #Factor of safety\n",
+ "P=140*10**3 #N #compressive Load\n",
+ "D=101.6 #mm #External diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "p=y*(FOS)**-1 #N/mm**2 #Permissible stress\n",
+ "A=P*p**-1 #mm**2 #Area of hollow tube\n",
+ "\n",
+ "#Let d be the internal diameter of tube\n",
+ "d=-((A*4*(pi)**-1)-D**2)\n",
+ "X=d**0.5\n",
+ "t=(D-X)*2**-1 #mm #Thickness of steel tube\n",
+ "\n",
+ "#result\n",
+ "print\"The thickness of steel tube is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness of steel tube is 3.17 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.4,Page No.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #diameter of steel\n",
+ "L=200 #mm #length of stee\n",
+ "P=80*10**3 #KN #Load\n",
+ "P1=160*10**3 #N #Load at Elastic Limit\n",
+ "P2=180*10**3 #N #Max Load\n",
+ "L1=56 #mm #Total Extension\n",
+ "dell_l=0.16 #mm #Extension\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*d**2*4**-1 #Area of steel #mm**2\n",
+ "\n",
+ "p=P1*A**-1 #Stress at Elastic Limit #N/mm**2\n",
+ "E=P*L*(A*dell_l)**-1\n",
+ "\n",
+ "#result\n",
+ "print E"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "203718.327158\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.5,Page No.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of bar\n",
+ "d2=14.7 #mm #Diameter at neck\n",
+ "L=200 #mm #guage Length \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,10,20,30,40,50,60]\n",
+ "Y1=[0,32,64,95,127,160,190]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Extension in divisions\")\n",
+ "plt.ylabel(\"Load in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of Bar\n",
+ "A2=pi*4**-1*d2**2\n",
+ "\n",
+ "P=45 #KN #Load obtained from graph\n",
+ "dell=0.143 #mm #Divisions\n",
+ "\n",
+ "#Modulus of Elasticity\n",
+ "E=P*L*(dell*A)**-1 \n",
+ "\n",
+ "BL=93*10**3 #N #Breaking Load\n",
+ "\n",
+ "#Nominal stress at Breaking point\n",
+ "sigma=BL*A**-1 #KN/mm**2 \n",
+ "\n",
+ "#True stress at breaking Point\n",
+ "sigma1=BL*A2**-1\n",
+ "\n",
+ "#Percentage Elongation \n",
+ "dell_l=(A-A2)*A**-1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The Value of ELongation is\",round(E,2),\"N/mm**2\"\n",
+ "print\"The Nominal stress at the Breaking Point\",round(sigma,2),\"KN/mm**2\"\n",
+ "print\"The True stress at the Breaking Point\",round(sigma1,2),\"KN/mm**2\"\n",
+ "print\"The Percentage Reduction in Area is\",round(dell_l,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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gwAFOnDjBRx99xNatWys876vr+8tf/kKzZs2Ii4vDVHG/iq+u7bL09HT279/P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\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x4d1d370>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Value of ELongation is 200.33 N/mm**2\n",
+ "The Nominal stress at the Breaking Point 296.03 KN/mm**2\n",
+ "The True stress at the Breaking Point 547.97 KN/mm**2\n",
+ "The Percentage Reduction in Area is 45.98\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.6,Page No.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=40*10**3 #N #Load\n",
+ "L1=160 #mm #Length of Bar1\n",
+ "L2=240 #mm #Length of bar2\n",
+ "L3=160 #mm #Length of bar3\n",
+ "d1=25 #mm #Diameter of Bar1\n",
+ "d2=20 #mm #diameter of bar2\n",
+ "d3=25 #mm #diameter of bar3\n",
+ "dell_l=0.285 #mm #Total Extension of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "E=P*4*(dell_l*pi)**-1*(L1*(d1**2)**-1+L2*(d2**2)**-1+L3*(d3**2)**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Young's Modulus of the material\",round(E,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Young's Modulus of the material 198714.72 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.7,Page No.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E1=2*10**5 #N/mm**2 #modulus of Elasticity of material1\n",
+ "E2=1*10**5 #N/mm**2 #modulus of Elasticity of material2\n",
+ "P=25*10**3 #N #Load \n",
+ "t=20 #mm #thickness of material\n",
+ "b1=40 #mm #width of material1\n",
+ "b2=30 #mm #width of material2\n",
+ "L1=500 #mm #Length of material1\n",
+ "L2=750 #mm #Length of material2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A1=b1*t #mm**2 #Area of materila1\n",
+ "A2=b2*t #mm**2 #Area of material2\n",
+ "\n",
+ "dell_l1=P*L1*(A1*E1)**-1 #Extension of Portion1\n",
+ "dell_l2=P*L2*(A2*E2)**-1 #Extension of portion2\n",
+ "\n",
+ "#Total Extension of Bar is\n",
+ "dell_l=dell_l1+dell_l2\n",
+ "\n",
+ "#Result\n",
+ "print\"The Total Extension of the Bar is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Total Extension of the Bar is 0.39 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.8,Page No.20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of Bar\n",
+ "l=400 #mm #Length upto which bire is drilled \n",
+ "D=30 #mm #diameter of bar\n",
+ "d1=10 #mm #diameter of bore\n",
+ "P=25*10**3 #N #Load\n",
+ "dell_l=0.185 #mm #Extension of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "L1=L-l #Length of bar above the bore\n",
+ "L2=400 #mm #Length of bore\n",
+ "\n",
+ "A1=pi*4**-1*D**2 #Area of bar\n",
+ "A2=pi*4**-1*(D**2-d1**2) #Area of bore\n",
+ "\n",
+ "E=P*dell_l**-1*(L1*A1**-1+L2*A2**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Modulus of ELasticity is\",round(E,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Modulus of ELasticity is 200735.96 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.11,Page No.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=10 #mm #Thickness of steel\n",
+ "b1=60 #mm #width of plate1\n",
+ "b2=40 #mm #width of plate2\n",
+ "P=60*10**3 #Load\n",
+ "L=600 #mm #Length of plate\n",
+ "E=2*10**5 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Extension of taperong bar of rectangular section\n",
+ "dell_l=P*L*(t*E*(b1-b2))**-1*log(b1*b2**-1)\n",
+ "\n",
+ "A_av=(b1*t+b2*t)*2**-1 #Average Area #mm**2\n",
+ "dell_l2=P*L*(A_av*E)**-1 \n",
+ "\n",
+ "#PErcentage Error\n",
+ "e=(dell_l-dell_l2)*(dell_l)**-1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The Percentage Error is\",round(e,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Percentage Error is 1.35\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.12,Page No.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1.5 #m #Length of steel bar\n",
+ "L1=1000 #m0 #Length of steel bar 1\n",
+ "L2=500 #m #Length of steel bar 2\n",
+ "d1=40 #Diameter of steel bar 1\n",
+ "d2=20 #diameter of steel bar 2\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "P=160*10**3 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A1=pi*4**-1*d1**2 #Area of Portion 1\n",
+ "\n",
+ "#Extension of uniform Portion 1\n",
+ "dell_l1=P*L1*(A1*E)**-1 #mm\n",
+ "\n",
+ "#Extension of uniform Portion 2\n",
+ "dell_l2=4*P*L2*(pi*d1*d2*E)**-1 #mm\n",
+ "\n",
+ "#Total Extension of Bar\n",
+ "dell_l=dell_l1+dell_l2\n",
+ "\n",
+ "#Result\n",
+ "print\"The Elongation of the Bar is\",round(dell_l,2),\"mm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.14,Page No.25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Portion AB\n",
+ "L_AB=600 #mm #Length of AB\n",
+ "A_AB=40*40 #mm**2 #Cross-section Area of AB\n",
+ "\n",
+ "#Portion BC\n",
+ "L_BC=800 #mm #Length of BC\n",
+ "A_BC=30*30 #mm #Length of BC\n",
+ "\n",
+ "#Portion CD\n",
+ "L_CD=1000 #mm #Length of CD\n",
+ "A_CD=20*20 #mm #Area of CD\n",
+ "\n",
+ "P1=80*10**3 #N #Load1\n",
+ "P2=60*10**3 #N #Load2\n",
+ "P3=40*10**3 #N #Load3\n",
+ "\n",
+ "E=2*10**5 #Modulus of Elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "P4=P1-P2+P3 #Load4\n",
+ "\n",
+ "#Now Force in AB\n",
+ "F_AB=P1\n",
+ "\n",
+ "#Force in BC\n",
+ "F_BC=P1-P2\n",
+ "\n",
+ "#Force in CD\n",
+ "F_CD=P4\n",
+ "\n",
+ "#Extension of AB\n",
+ "dell_l_AB=F_AB*L_AB*(A_AB*E)**-1\n",
+ "\n",
+ "#Extension of BC\n",
+ "dell_l_BC=F_BC*L_BC*(A_BC*E)**-1\n",
+ "\n",
+ "#Extension of CD\n",
+ "dell_l_CD=F_CD*L_CD*(A_CD*E)**-1\n",
+ "\n",
+ "#Total Extension\n",
+ "dell_l=dell_l_AB+dell_l_BC+dell_l_CD\n",
+ "\n",
+ "#Result\n",
+ "print\"The Total Extension in Bar is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Total Extension in Bar is 0.99 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.15,Page No.26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=800 #mm #Length of bar\n",
+ "F1=30*10**3 #N #Force acting on the bar\n",
+ "F2=60*10**3 #N #force acting on the bar\n",
+ "L=800 #mm #Length of bar\n",
+ "d=25 #mm #diameter of bar\n",
+ "L_AC=275 #mm #Length of AC\n",
+ "L_CD=150 #mm #Length of CD\n",
+ "L_DB=375 #mm #Length of DB\n",
+ "E=2*10**5 #Pa #Modulus of elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P be the Reaction on tne Bar from support at A\n",
+ "\n",
+ "#Shortening of Portion AC\n",
+ "#dell_l_AC1=P*L_AC*(A*E)**-1\n",
+ "\n",
+ "#Shortening of Portion CD\n",
+ "#dell_l_CD1=(30+P)*L_CD*(A*E)**-1\n",
+ "\n",
+ "#Extension of Portion DB\n",
+ "#dell_l_DB1=(30-P)*L_DB*(A*E)**-1\n",
+ "\n",
+ "#Total Extensions=1*(A*E)**-1*(P*L_AC-(30+P)*L_CD+(30-P)*L_DB)\n",
+ "#As Supports are unyielding,Total Extensions=0\n",
+ "\n",
+ "#After substituting values in above equation and Further simplifying we get\n",
+ "P=(30*375-150*30)*800**-1\n",
+ "\n",
+ "#Reaction of support A\n",
+ "R_A=P\n",
+ "\n",
+ "#Reaction of support B\n",
+ "R_B=30-P\n",
+ "\n",
+ "#Cross-sectional Area\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Stress in Portion AC\n",
+ "sigma1=P*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in Portion CD\n",
+ "sigma2=(30+P)*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in Portion DB\n",
+ "sigma3=(30-P)*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Shortening of Portion AC\n",
+ "dell_l_AC2=P*10**3*L_AC*(A*E)**-1 #mm \n",
+ "\n",
+ "#Shortening of Portion CD\n",
+ "dell_l_CD2=(30+P)*10**3*L_CD*(A*E)**-1 #mm \n",
+ "\n",
+ "#Extension of Portion DB\n",
+ "dell_l_DB2=(30-P)*10**3*L_DB*(A*E)**-1 #mm \n",
+ "\n",
+ "#result\n",
+ "print\"The Reactios at two Ends are:R_A\",round(R_A,2),\"KN\"\n",
+ "print\" :R_B\",round(R_B,2),\"KN\"\n",
+ "print\"Stress in Portion AC\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\"Stress in Portion CD\",round(sigma2,2),\"N/mm**2\"\n",
+ "print\"Stress in Portion DB\",round(sigma3,2),\"N/mm**2\"\n",
+ "print\"Shortening of Portion AC\",round(dell_l_AC2,3),\"mm\"\n",
+ "print\"Shortening of Portion CD\",round(dell_l_CD2,3),\"mm\"\n",
+ "print\"Shortening of Portion DB\",round(dell_l_DB2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Reactios at two Ends are:R_A 8.44 KN\n",
+ " :R_B 21.56 KN\n",
+ "Stress in Portion AC 17.19 N/mm**2\n",
+ "Stress in Portion CD 78.3 N/mm**2\n",
+ "Stress in Portion DB 43.93 N/mm**2\n",
+ "Shortening of Portion AC 0.024 mm\n",
+ "Shortening of Portion CD 0.059 mm\n",
+ "Shortening of Portion DB 0.082 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.19,Page No.29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "h=4 #m #height of Pillars\n",
+ "P=20 #KN #Load at M\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_A,P_B,P_C,P_D be the forces introduced in the Pillars\n",
+ "#Sun of All Vertical Forces\n",
+ "#P_A+P_B+P_C+P_D=20 ....................(1)\n",
+ "\n",
+ "#Sum of moment about AB, we get\n",
+ "#P_D+P_C=12 ....................(2)\n",
+ "\n",
+ "#Sum of Moment about AD\n",
+ "#P_C+P_B=8 ....................(3)\n",
+ "\n",
+ "#Let dell_l_A,dell_l_B,dell_l-C,dell_l_D be the deformations of Pillars A,B,C,D respectively\n",
+ "#Diagonals AC and BD will remain straight Lines even after the Load is applied.\n",
+ "#Deflection of central Point is given by (dell_l_A+dell_l_C)*2**-1 & (dell_l_B+dell_l_D)*2**-1\n",
+ "\n",
+ "#dell_l_A+dell_l_C=dell_l_B+ell_l_D\n",
+ "#P_A*L*(A*E)**-1+P_C*L*(A*E)**-1=P_B*L*(A*E)**-1+P_D*L*(A*E)**-1\n",
+ "\n",
+ "#Since Pillars are identical in Length,cross-sectional area,material Property\n",
+ "#P_A+P_C=P_B+P_D ..............(4)\n",
+ "\n",
+ "#From Equations 1 and 4 we get\n",
+ "#P_B+P_D=10 ....................(5)\n",
+ " \n",
+ "#Substracting Equation 3 from Equation 2 we get\n",
+ "#P_D-P_B=4 ....................(6)\n",
+ "\n",
+ "#Adding Equation 5 and 6 we get\n",
+ "\n",
+ "P_D=14*2**-1\n",
+ "P_C=12-P_D\n",
+ "P_B=8-P_C\n",
+ "\n",
+ "#Now substituting values of P_B,P_C,P_D in equation1 we get\n",
+ "P_A=20-(P_B+P_C+P_D)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Forces Developed in the Pillars are:P_A\",round(P_A,2),\"KN\"\n",
+ "print\" :P_B\",round(P_B,2),\"KN\"\n",
+ "print\" :P_C\",round(P_C,2),\"KN\"\n",
+ "print\" :P_D\",round(P_D,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Forces Developed in the Pillars are:P_A 5.0 KN\n",
+ " :P_B 3.0 KN\n",
+ " :P_C 5.0 KN\n",
+ " :P_D 7.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.20,Page No.31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma=150 #N/mm**2 #Stress\n",
+ "P=40*10**3 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt P_A.P_B,P_C,P_D be the forces developed in wires A,B,C,D respectively\n",
+ "\n",
+ "#Let sum of all Vertical Forces=0\n",
+ "#P_A+P_B+P_C+P_D=40 ..........................(1)\n",
+ "\n",
+ "#Let x be the distance between each wires\n",
+ "#sum of all moments=0\n",
+ "#P_B*x+P_C*2*x+P_D*3*x=40*2*x\n",
+ "\n",
+ "#After further simplifying we get\n",
+ "#P_B+2*P_C+3*P_D=80 ..........................(2)\n",
+ "\n",
+ "#As the equations of statics ae not enough to find unknowns,Consider compatibilit Equations\n",
+ "\n",
+ "#Let dell_l be the increse in elongation of wire\n",
+ "\n",
+ "#dell_l_B=dell_l_A+dell_l\n",
+ "#dell_l_C=dell_l_A+2*dell_l\n",
+ "#dell_l_D=dell_l_A+3*dell_l\n",
+ "\n",
+ "#Let P1 be the force required for the Elongation of wires,then\n",
+ "#P_B=P_A+P1 ]\n",
+ "#P_C=P_A+2*P1 ]\n",
+ "#P_D=P_A+3*P1 ] ................................(3) \n",
+ "\n",
+ "#from Equation (3) and (1) we get\n",
+ "#2*P_A+3*P1=20 ................................(4)\n",
+ "\n",
+ "#from Equation (3) and (2) we get\n",
+ "#6*P_A+14*P1=80 \n",
+ "\n",
+ "#subtracting 3 times equation (4) from (3) we get\n",
+ "P1=20*5**-1\n",
+ "\n",
+ "#from Equation 4 we get\n",
+ "P_A=(80-14*P1)*6**-1\n",
+ "P_B=P_A+P1\n",
+ "P_C=P_A+2*P1\n",
+ "P_D=P_A+3*P1\n",
+ "\n",
+ "#Let d be the diameter required,then\n",
+ "d=(P_D*10**3*4*(pi*150)**-1)**0.5\n",
+ "\n",
+ "#result\n",
+ "print\"The Required Diameter is\",round(d,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Required Diameter is 11.65 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.21,Page No.32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=20*10**3 #N #Load\n",
+ "d=6 #mm #diameter of wire\n",
+ "E=2*10**5 #N/mm**2 \n",
+ "L_BO=4000 #mm #Length of BO\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let theta be the angle between OA and OB and also between OC and OB\n",
+ "theta=30\n",
+ "\n",
+ "#Let P_OA,P_OB,P_OC be the Forces introduced in wires OA,OB,OC respectively\n",
+ "#Due to symmetry P_OA=P_OC (same angles)\n",
+ "\n",
+ "#Sum of all Vertical Forces=0\n",
+ "#P_OA*cos(theta)+P_OB+P_OC*cos(theta)=P\n",
+ "\n",
+ "#After further simplifyinf we get\n",
+ "#2*P_OA*cos(theta)+P_OB=20 ...............(1)\n",
+ "\n",
+ "#Let oo1 be the extension of BO\n",
+ "#oo1=L_A1o1*(cos(theta))**-1\n",
+ "\n",
+ "#From relation we get\n",
+ "#P_OB*L_BO=P_OA*L_AO*(cos(theta))**-1\n",
+ "\n",
+ "#But L_AO=L_BO*(cos(theta))**-1\n",
+ "\n",
+ "#After substituting value of L_AO in above equation we get\n",
+ "#P_OB=0.75*P_OA .......................(2)\n",
+ "\n",
+ "#substituting in Equation 1 we get\n",
+ "#2*P_OA*cos(theta)+0.75*P_OA=20\n",
+ "\n",
+ "P_OA=20*(2*cos(theta*pi*180**-1)+0.75)**-1\n",
+ "\n",
+ "P_OB=0.75*P_OA\n",
+ "\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Vertical displacement of Load\n",
+ "dell_l_BO=P_OB*10**3*L_BO*(A*E)**-1\n",
+ " \n",
+ "#Result\n",
+ "print\"Forces in each wire is:P_OA\",round(P_OA,2),\"KN\"\n",
+ "print\" :P_OB\",round(P_OB,2),\"KN\"\n",
+ "print\"Vertical displacement of Loadis\",round(dell_l_BO,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Forces in each wire is:P_OA 8.06 KN\n",
+ " :P_OB 6.04 KN\n",
+ "Vertical displacement of Loadis 4.27 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.22,Page No.34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_s=L_a=L=500 #mm #Length of bar\n",
+ "A_a=50*20 #mm #Area of aluminium strip\n",
+ "A_s=50*15 #mm #Area of steel strip\n",
+ "P=50*10**3 #N #Load\n",
+ "E_a=1*10**5 #N/mm**2 #Modulus of aluminium \n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_a and P_s br the Load shared by aluminium and steel strip\n",
+ "#P_a+P_s=P ..................(1)\n",
+ "\n",
+ "#For compatibility condition,dell_l_a=dell_l_s\n",
+ "#P_a*L_a*(A_a*E_a)**-1=P_s*L_s*(A_s*E_s)**-1 .....(2)\n",
+ "\n",
+ "#As L_a=L_s we get\n",
+ "#P_s=1.5*P_a .................(3)\n",
+ "\n",
+ "#From Equation 1 and 2 we get\n",
+ "P_a=P*2.5**-1\n",
+ "\n",
+ "#Substituting in equation 1 we get\n",
+ "P_s=P-P_a\n",
+ "\n",
+ "#stress in aluminium strip \n",
+ "sigma_a=P_a*A_a**-1\n",
+ "\n",
+ "#stress in steel strip\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Now from the relation we get\n",
+ "dell_l_a=dell_l_s=P_s*L_s*(A_s*E_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Stress in Aluminium strip is\",round(sigma_a,2),\"N/mm**2\"\n",
+ "print\"Stress in steel strip is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"The Extension of the bar is\",round(dell_l_s,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in Aluminium strip is 20.0 N/mm**2\n",
+ "Stress in steel strip is 40.0 N/mm**2\n",
+ "The Extension of the bar is 0.1 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.23,Page No.35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D_s=20 #mm #Diameter of steel\n",
+ "D_Ci=20 #mm #Internal Diameter of Copper\n",
+ "t=5 #mm #THickness of copper bar\n",
+ "P=100*10**3 #N #Load\n",
+ "E_s=2*10**5 #N/mm**2 #modulus of elasticity of steel\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of Copper\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=pi*4**-1*D_s**2 #mm**2 #Area of steel\n",
+ "D_Ce=D_s+2*t #mm #External Diameterof Copper Tube\n",
+ "\n",
+ "A_c=pi*4**-1*(D_Ce**2-D_Ci**2) #mm**2 #Area of Copper\n",
+ "\n",
+ "#From static Equilibrium condition\n",
+ "#Let P_s and P_c be the Load shared by steel and copper in KN\n",
+ "#P_s+P_c=100 ....................................(1)\n",
+ "\n",
+ "#From compatibility Equation,dell_l_s=dell_l_c\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "\n",
+ "#Substituting values in above Equation we get\n",
+ "#P_s=1.3333*P_C \n",
+ "\n",
+ "#Now Substituting value of P_s in Equation (1),we get\n",
+ "P_c=100*2.3333**-1 #KN\n",
+ "P_s=100-P_c #KN\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in copper\n",
+ "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\" :sigma_c\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Two material are:sigma_s 181.89 N/mm**2\n",
+ " :sigma_c 109.14 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.24,Page No.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A_C=230*400 #mm #Area of column\n",
+ "D_s=12 #mm #Diameter of steel Bar\n",
+ "P=600*10**3 #N #Axial compression\n",
+ "#E_s*E_c=18.67\n",
+ "n=8 #number of steel Bars\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=pi*4**-1*D_s**2*n #Area of steel #mm**2 \n",
+ "A_c=A_C-A_s #mm**2 #Area of concrete\n",
+ "\n",
+ "#From static Equilibrium condition\n",
+ "#P_s+P_c=600 .........(1)\n",
+ "\n",
+ "#Now from compatibility Equation dell_l_s=dell_l_c we get,\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "\n",
+ "#Substituting values in above Equation we get\n",
+ "#P_s=0.1854*P_c\n",
+ "\n",
+ "#Now Substituting value of P_s in Equation (1),we get\n",
+ "P_c=600*1.1854**-1\n",
+ "P_s=600-P_c\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in copper\n",
+ "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\" :sigma_c\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Two material are:sigma_s 103.72 N/mm**2\n",
+ " :sigma_c 5.56 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.25,Page No.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=200*10**3 #N #Load\n",
+ "A_a=1000 #mm**2 #Area of Aluminium\n",
+ "A_s=800 #mm**2 #Area of steel\n",
+ "E_a=1*10**5 #N/mm**2 #Modulus of Elasticity of Aluminium\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel\n",
+ "sigma_a1=65 #N/mm**2 #stress in aluminium\n",
+ "sigma_s1=150 #N/mm**2 #Stress in steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_a and P_s be the force in aluminium and steel pillar respectively\n",
+ "\n",
+ "#Now,sum of forces in Vertical direction we get\n",
+ "#2*P_a+P_s=200 .........................................(1)\n",
+ "\n",
+ "#By compatibility Equation dell_l_s=dell_l_a we get\n",
+ "#P_s=1.28*P_a ..........................................(2)\n",
+ "\n",
+ "#Now substituting value of P_s in Equation 1 we get\n",
+ "P_a=200*3.28**-1 #KN\n",
+ "P_s=200-2*P_a #KN\n",
+ "\n",
+ "#Stress developed in aluminium\n",
+ "sigma_a=P_a*10**3*A_a**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress developed in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Let sigma_a1 and sigma_s1 be the stresses in Aluminium and steel due to Additional LOad\n",
+ "\n",
+ "P_a1=sigma_a1*A_a #Load carrying capacity of aluminium\n",
+ "P_s1=1.28*P_a1\n",
+ "\n",
+ "#Total Load carrying capacity \n",
+ "P1=2*P_a1+P_s1 #N \n",
+ "\n",
+ "P_s2=sigma_s1*A_s #Load carrying capacity of steel\n",
+ "P_a2=P_s2*1.28**-1\n",
+ "\n",
+ "#Total Load carrying capacity\n",
+ "P2=2*P_a2+P_s2\n",
+ "\n",
+ "#Additional Load\n",
+ "P3=P1-P\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Each Pillar is:sigma_a\",round(sigma_a,2),\"N/mm**2\"\n",
+ "print\" :sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"Additional Load taken by pillars is\",round(P3,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "78.0487804878\n",
+ "Stresses Developed in Each Pillar is:sigma_a 60.98 N/mm**2\n",
+ " :sigma_s 97.56 N/mm**2\n",
+ "Additional Load taken by pillars is 13200.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.26,Page No.37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=500 #mm #Length of assembly\n",
+ "D=16 #mm #Diameter of steel bolt\n",
+ "Di=20 #mm #internal Diameter of copper tube\n",
+ "Do=30 #mm #External Diameter of copper tube\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of copper\n",
+ "p=2 #mm #Pitch of nut\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the Force in bolt and P_c be the FOrce in copper tube\n",
+ "#P_s=-P_s\n",
+ "\n",
+ "dell=1*4**-1*2 #Quarter turn of nut total movement\n",
+ "\n",
+ "#dell=dell_s+dell_c\n",
+ "\n",
+ "#Area of steel\n",
+ "A_s=pi*4**-1*D**2\n",
+ "\n",
+ "#Area of copper\n",
+ "A_c=pi*4**-1*(Do**2-Di**2)\n",
+ "\n",
+ "#dell=P*L*(A_s*E_s)**-1+P*L*(A_c*E_c)**-1\n",
+ "P=dell*(1*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1*L**-1 #LOad\n",
+ "\n",
+ "P_s=P*A_s**-1\n",
+ "P_c=P*A_c**-1\n",
+ "\n",
+ "#result\n",
+ "print\"stress introduced in bolt is\",round(P_s,2),\"N/mm**2\"\n",
+ "print\"stress introduced in tube is\",round(P_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "stress introduced in bolt is 107.91 N/mm**2\n",
+ "stress introduced in tube is 55.25 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.27,Page No.39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=20 #mm #Diameter of Bolts\n",
+ "Di=25 #m #internal Diameter\n",
+ "t=10 #mm #Thickness of bolt\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "p=3 #mm #Pitch\n",
+ "theta=30 #degree\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the Force in each bolt and P_c be the FOrce in copper tube\n",
+ "#From Static Equilibrium condition\n",
+ "#P_c=2*P_s\n",
+ "\n",
+ "#As nut moves by 60 degree.If nut moves by 360 degree its Longitudinal movement is by 3 mm\n",
+ "dell=theta*360**-1*p\n",
+ "\n",
+ "#From Compatibility Equaton we get\n",
+ "#dell=dell_c+dell_s\n",
+ "\n",
+ "\n",
+ "A_s=pi*4**-1*"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.28,Page No.40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=9 #m #Length of rigid bar\n",
+ "L_b=3000 #Length of bar\n",
+ "A_b=1000 #mm**2 #Area of bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brasss bar\n",
+ "L_s=5000 #mm #Length of steel bar\n",
+ "A_s=445 #mm**2 #Area of steel bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of elasticity of steel bar\n",
+ "P=3000 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From static equilibrium Equation of the rod after appliying Load is\n",
+ "#P_b+P_s=P ......................(1)\n",
+ "\n",
+ "#P_b=1.8727*P_s ..................(2)\n",
+ "\n",
+ "#NOw substituting equation 2 in equation 1 we get\n",
+ "P_s=P*2.8727**-1\n",
+ "P_b=P-P_s\n",
+ "\n",
+ "d=P_s*L*P**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance at which Load applied even after which bar remains horizontal is\",round(d,2),\"m\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance at which Load applied even after which bar remains horizontal is 3.13 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.29,Page No.41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A_b=1000 #MM**2 #Area of brass bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass\n",
+ "A_s=600 #N/mm**2 #Area of steel rod\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of eLasticity of steel bar\n",
+ "P=10*10**2 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",
+ "#Now taking moment about A we get static Equilibrium condition as\n",
+ "#P_b+2*P_s=27500 ......................................(1)\n",
+ "\n",
+ "#Now from deformed shape we get\n",
+ "#dell_s=2*dell_b\n",
+ "\n",
+ "#P_s*L_s*(A_s*E_s)**-1=P_b*L_b*(A_b*E_b)**-1\n",
+ "#Further simplifying we get\n",
+ "#P_s=1.2*P_b .........................................(2)\n",
+ "\n",
+ "#Now substituting equation 1 in equation 2 we get\n",
+ "P_b=27500*3.4**-1\n",
+ "P_s=1.2*P_b\n",
+ "\n",
+ "#Tensile stress in brass bar \n",
+ "sigma_b=P_b*A_b**-1\n",
+ "\n",
+ "#compressive stress in steel bar\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Compressive Stress in Bar is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"tensile Stress in Bar is\",round(sigma_b,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Compressive Stress in Bar is 16.18 N/mm**2\n",
+ "tensile Stress in Bar is 8.09 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.30,Page No.44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=12.6 #m #Length of rail\n",
+ "t1=24 #Degree celsius\n",
+ "t2=44 #degree celsius\n",
+ "alpha=12*10**-6 #Per degree celsius\n",
+ "E=2*10**5 #N/mm**2 #Modulus of ELasticity\n",
+ "gamma=2 #mm #Gap provided for Expansion\n",
+ "sigma=20 #N/mm**2 #Stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "t=t2-t1 #Temperature Difference\n",
+ "\n",
+ "#Free Expansion of the rails\n",
+ "dell=alpha*t*L*1000 #mm \n",
+ "\n",
+ "#When no expansion joint is provided then\n",
+ "p=dell*E*(L*10**3)**-1\n",
+ "\n",
+ "#When a gap of 2 mm is provided,then free expansion prevented is\n",
+ "dell_1=dell-gamma\n",
+ "p2=dell_1*E*(L*10**3)**-1\n",
+ "\n",
+ "#When stress is developed,then gap left is\n",
+ "gamma2=-(sigma*L*10**3*E**-1-dell)\n",
+ "\n",
+ "#Result\n",
+ "print\"The minimum gap between the two rails is\",round(dell,2),\"mm\"\n",
+ "print\"Thermal Developed in the rials if:No expansionn joint is provided:p\",round(p,2),\"N/mm**2\"\n",
+ "print\" :If a gap of is provided then :p2\",round(p2,2),\"N/mm**2\"\n",
+ "print\"When stress is developed gap left between the rails is\",round(gamma2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum gap between the two rails is 3.02 mm\n",
+ "Thermal Developed in the rials if:No expansionn joint is provided:p 48.0 N/mm**2\n",
+ " :If a gap of is provided then :p2 16.25 N/mm**2\n",
+ "When stress is developed gap left between the rails is 1.76 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.31,Page No.45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=20 #degree celsius\n",
+ "E_a=70*10**9 #N/mm**2 #Modulus of Elasticicty of aluminium\n",
+ "alpha_a=11*10**-6 #per degree celsius #Temperature coeff of aluminium\n",
+ "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel\n",
+ "L_a=1000 #mm #Length of aluminium \n",
+ "L_s=3000 #mm #Length of steel\n",
+ "E_a=7*10**4 #N/mm**2 #Modulus of Elasticity of aluminium\n",
+ "E_s=2*10**5 #N/mm*2 #Modulus of Elasticity of steel\n",
+ "A_a=600 #mm**2 #Area of aluminium\n",
+ "A_s=300 #mm**2 #Area of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Free Expansion \n",
+ "dell=alpha_a*t*L_a+alpha_s*t*L_s\n",
+ "\n",
+ "#support Reaction\n",
+ "P=dell*(L_a*(A_a*E_a)**-1+L_s*(A_s*E_s)**-1)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Reaction at support is\",round(P,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reaction at support is 12735.48 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.33,Page No.48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=25 #mm #Diameter of Brass\n",
+ "De=50 #mm #External Diameter of steel tube\n",
+ "Di=25 #mm #Internal Diameter of steel tube\n",
+ "L=1.5 #m #Length of both bars\n",
+ "t1=30 #degree celsius #Initial Temperature\n",
+ "t2=100 #degree celsius #final Temperature\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass bar\n",
+ "alpha_s=11.6*10**-6 #Temperature Coeff of steel\n",
+ "alpha_b=18.7*10**-6 #Temperature coeff of brass bar\n",
+ "d=20 #mm #diameter of pins\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "t=t2-t1 #Temperature Difference\n",
+ "A_s=pi*4**-1*(De**2-Di**2) #mm**2 #Area of steel\n",
+ "A_b=pi*4**-1*D**2 #mm**2 #Area of brass\n",
+ "\n",
+ "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",
+ "#But from Equilibrium of Forces \n",
+ "#P_b=P_s=P\n",
+ "\n",
+ "#Let dell=dell_s+dell_b\n",
+ "dell=(alpha_b-alpha_s)*t*L*1000\n",
+ "\n",
+ "P=dell*(1*(A_s*E_s)**-1+1*(A_b*E_b)**-1)**-1*(L*1000)**-1\n",
+ "P_b=P_s=P\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Stress in Brass\n",
+ "sigma_b=P_b*A_b**-1\n",
+ "\n",
+ "#Area of Pins\n",
+ "A_p=pi*4**-1*d**2\n",
+ "\n",
+ "#Since,the force is resisted by two cross section of pins\n",
+ "tou=P*(2*A_p)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in steel bar is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"Stress in Brass bar is\",round(sigma_b,2),\"N/mm**2\"\n",
+ "print\"Shear Stresss induced in pins is\",round(tou,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in steel bar is 14.2 N/mm**2\n",
+ "Stress in Brass bar is 42.6 N/mm**2\n",
+ "Shear Stresss induced in pins is 33.28 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.34,Page No.49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b_s=60 #mm #width of steel Bar\n",
+ "t_s=10 #mm #thickness of steel Bar\n",
+ "b_c=40 #mm #width of copper bar\n",
+ "t_c=5 #mm #thickness of copper bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",
+ "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",
+ "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel bar\n",
+ "alpha_c=17*10**-6 #Per degree celsius #Temperature coeff of copper bar\n",
+ "L_s=L_c=L=1000 #mm #Length of bar\n",
+ "t=80 #degree celsius\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=b_s*t_s #Area of steel bar\n",
+ "A_c=b_c*t_c #Area of copper bar\n",
+ "\n",
+ "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",
+ "#The equilibrium of forces gives \n",
+ "#P_s=2*P_c\n",
+ "\n",
+ "#Let dell=dell_s+dell_b\n",
+ "dell=(alpha_c-alpha_s)*t\n",
+ "\n",
+ "P_c=dell*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1\n",
+ "P_s=2*P_c\n",
+ "\n",
+ "#Stress in copper \n",
+ "sigma_c=P_c*A_c**-1\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Change in Length of bar\n",
+ "dell_2=alpha_s*t*L+P_s*L_s*(A_s*E_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Stress in copper is\",round(sigma_c,2),\"N/mm**2\"\n",
+ "print\"Stress in steel is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"the change in Length is\",round(dell_2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in copper is 30.0 N/mm**2\n",
+ "Stress in steel is 20.0 N/mm**2\n",
+ "the change in Length is 1.06 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.35,Page No.50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=2*10**5 #N #Weight\n",
+ "L=1 #m #Length of each rod\n",
+ "A_c=A_s=A=500 #mm**2 #Area of each rod\n",
+ "t=40 #degree celsius #temperature\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel rod\n",
+ "E_c=1*10**5 #N/mm**2 #modulus of Elastictiy of copper rod\n",
+ "alpha_s=1.2*10**-5 #Per degree Celsius #temp coeff of steel rod\n",
+ "alpha_c=1.8*10**-5 #Per degree Celsius #Temp coeff of copper rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the force in each one of the copper rods and P_s be the force in steel rod\n",
+ "#2*P_c+P_s=P .....................(1)\n",
+ "\n",
+ "#Extension of copper bar=Extension of steel bar\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "#after simplifying above equation we get\n",
+ "#P_s=2*P_c ........................(2)\n",
+ "\n",
+ "#Now substituting value of P_s in Equation 1 we get\n",
+ "P_c=P*4**-1\n",
+ "P_s=2*P_c\n",
+ "\n",
+ "#Now EXtension due to copper Load\n",
+ "dell_1=P_c*L*1000*(A_c*E_c)**-1\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Due to rise of temperature of40 degree celsius\n",
+ "\n",
+ "#As bars are rigidly joined,let P_c1 be the compressive forccesdeveloped in copper bar and P_s1 be the tensile force in steel causing changes\n",
+ "#P_s1=2*P_c1\n",
+ "\n",
+ "#dell_s+dell_c=(alpha_c-alpha_s)*t*L .......................................(3)\n",
+ "#P_s1*L*(A_s*E_s)**-1+P_c1*L*(A_c*E_c)**-1=(alpha_c-alpha_s)*t*L ................(4)\n",
+ "#After substituting values in above equation and further simplifying we get,\n",
+ "P_c1=(alpha_c-alpha_s)*t*L*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1 #.................(5)\n",
+ "P_s1=2*P_c1\n",
+ "\n",
+ "#Extension of bar due to temperature rise\n",
+ "dell_2=alpha_s*t*L+P_s1*L*(A_s*E_s)**-1\n",
+ "\n",
+ "#Amount by which bar will descend\n",
+ "dell_3=dell_1+dell_2\n",
+ "\n",
+ "#Load carried by steel bar\n",
+ "P_S=P_s+P_s1\n",
+ "\n",
+ "#Load carried by copper bar\n",
+ "P_C=P_c-P_c1\n",
+ "\n",
+ "#Part-3\n",
+ "\n",
+ "#Let P_c1_1=P_c #For convenience\n",
+ "#Rise in temperature if Load is to be carried out by steel rod alone\n",
+ "P_c1_1=P_c\n",
+ "\n",
+ "#From equation 5 \n",
+ "t=P_c1_1*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)*(alpha_c-alpha_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Extension Due top copper Load\",round(dell_1,2),\"mm\"\n",
+ "print\"Load carried by each rod:P_s\",round(P_s,2),\"N\"\n",
+ "print\" :P_c\",round(P_c,2),\"N\"\n",
+ "print\"Rise in Temperature of steel rod should be\",round(t,2),\"degree Celsius\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extension Due top copper Load 1.0 mm\n",
+ "Load carried by each rod:P_s 100000.0 N\n",
+ " :P_c 50000.0 N\n",
+ "Rise in Temperature of steel rod should be 333.33 degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.36,Page No.53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=40 #degree celsius #temperature\n",
+ "A_s=400 #mm**2 #Area of steel bar\n",
+ "A_c=600 #mm**2 #Area of copper bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",
+ "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",
+ "alpha_s=12*10**-6 #degree celsius #Temperature coeff of steel bar\n",
+ "alpha_c=18*10**-6 #degree celsius #Temperature coeff of copper bar\n",
+ "L_c=800 #mm #Length of copper bar\n",
+ "L_s=600 #mm #Length of steel bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",
+ "#Static Equilibrium obtained by taking moment about A\n",
+ "#P_c=2*P_s\n",
+ "\n",
+ "#From property of similar triangles we get\n",
+ "#(alpha_c*Lc-dell_c)*1**-1=(alpha_s*L_s-dell_s)*2**-1\n",
+ "#After substituting values in above equations and further simplifying we get\n",
+ "P_s=(2*alpha_c*L_c-alpha_s*L_s)*t*(L_s*(A_s*E_s)**-1+4*L_c*(A_c*E_c)**-1)**-1\n",
+ "P_c=2*P_s\n",
+ "\n",
+ "#Stress in steel rod\n",
+ "sigma_s=P_s*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress in copper rod\n",
+ "sigma_c=P_c*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in steel rod is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"STress in copper rod is\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in steel rod is 35.51 N/mm**2\n",
+ "STress in copper rod is 47.34 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.37,Page No.61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of bar\n",
+ "P=37.7*10**3 #N #Load\n",
+ "L=200 #mm #Guage Length \n",
+ "dell=0.12 #mm #Extension\n",
+ "dell_d=0.0036 #mm #contraction in diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of bar\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Let s and dell_s be the Linear strain and Lateral strain\n",
+ "s=dell*L**-1\n",
+ "dell_s=dell_d*d**-1\n",
+ "mu=dell_s*s**-1 #Poissoin's ratio \n",
+ "\n",
+ "#dell=P*L*(A*E)**-1\n",
+ "E=P*L*(dell*A)**-1 #N/mm**2 #Modulus of Elasticity of bar\n",
+ "\n",
+ "#Modulus of Rigidity\n",
+ "G=E*(2*(1+mu))**-1 #N/mm**2\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "#result\n",
+ "print\"Poisson's ratio is\",round(mu,2)\n",
+ "print\"The Elastic constant are:E\",round(E,2)\n",
+ "print\" :G\",round(G,2)\n",
+ "print\" :K\",round(K,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Poisson's ratio is 0.3\n",
+ "The Elastic constant are:E 200004.71\n",
+ " :G 76924.89\n",
+ " :K 166670.59\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.38,Page No.62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=100 #mm #Diameter of circular rod\n",
+ "P=1*10**6 #N #Tensile Force\n",
+ "mu=0.3 #Poisson's ratio\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "L=500 #mm #Length of rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Modulus of Rigidity\n",
+ "G=E*(2*(1+mu))**-1 #N/mm**2\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of Circular rod\n",
+ "#Let sigma be the Longitudinal stress\n",
+ "sigma=P*A**-1 #N/mm**2 \n",
+ "\n",
+ "s=sigma*E**-1 #Linear strain\n",
+ "e_x=s\n",
+ "\n",
+ "#Volumetric strain\n",
+ "e_v=e_x*(1-2*mu)\n",
+ "\n",
+ "v=pi*4**-1*d**2*L\n",
+ "#Change in VOlume\n",
+ "dell_v=e_v*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Bulk Modulus is\",round(E,2),\"N/mm**2\"\n",
+ "print\"Modulus of Rigidity is\",round(G,2),\"N/mm**2\"\n",
+ "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bulk Modulus is 200000.0 N/mm**2\n",
+ "Modulus of Rigidity is 76923.08 N/mm**2\n",
+ "The change in Volume is 1000.0 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.39,Page No.62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=500 #mm #Length of rectangular cross section bar\n",
+ "A=20*40 #mm**2 #Area of rectangular cross section bar\n",
+ "P1=4*10**4 #N #Tensile Force on 20mm*40mm Faces\n",
+ "P2=2*10**5 #N #compressive force on 20mm*500mm Faces\n",
+ "P3=3*10**5 #N #Tensile Force on 40mm*500mm Faces\n",
+ "E=2*10**5 #N/mm**2 #young's Modulus \n",
+ "mu=0.3 #Poisson's Ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_x,P_y,P_z be the forces n x,y,z directions\n",
+ "\n",
+ "P_x=P1*A**-1\n",
+ "P_y=P2*A**-1\n",
+ "P_z=P3*A**-1\n",
+ "\n",
+ "#Let e_x,e_y,e_z be the strains in x,y,z directions\n",
+ "e_x=1*E**-1*(50+mu*20-15*mu)\n",
+ "e_y=1*E**-1*(-mu*50-20-mu*15)\n",
+ "e_z=1*E**-1*(-mu*50+mu*20+15)\n",
+ "\n",
+ "#Volumetric strain\n",
+ "e_v=e_x+e_y+e_z\n",
+ "\n",
+ "#Volume\n",
+ "V=20*40*500 #mm**3\n",
+ "#Change in Volume \n",
+ "dell_v=e_v*V #mm**3\n",
+ "\n",
+ "#Result\n",
+ "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in Volume is 36.0 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.41,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2.1*10**5 #N/mm**2 #Young's Modulus \n",
+ "G=0.78*10**5 #N/mm**2 #Modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Now using the relation\n",
+ "#E=2*G*(1+mu)\n",
+ "mu=E*(2*G)**-1-1 #Poisson's ratio\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The Poisson's Ratio is\",round(mu,2)\n",
+ "print\"The modulus of Rigidity\",round(K,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Poisson's Ratio is 0.35\n",
+ "The modulus of Rigidity 227500.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.42,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=0.4*10**5 #N/mm**2 #Modulus of rigidity\n",
+ "K=0.75*10**5 #N/mm**2 #Bulk Modulus \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Young's Modulus\n",
+ "E=9*G*K*(3*K+G)**-1\n",
+ "\n",
+ "#Now from the relation\n",
+ "#E=2*G(1+2*mu)\n",
+ "mu=E*(2*G)**-1-1 #POissoin's ratio \n",
+ "\n",
+ "#result\n",
+ "print\"Young's modulus is\",round(E,2),\"N/mm**2\"\n",
+ "print\"Poissoin's ratio is\",round(mu,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Young's modulus is 101886.79 N/mm**2\n",
+ "Poissoin's ratio is 0.27\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.43,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=60 #mm #width of bar\n",
+ "d=30 #mm #depth of bar\n",
+ "L=200 #mm #Length of bar\n",
+ "A=30*60 #mm**2 #Area of bar\n",
+ "A2=30*200 #mm**2 #Area of bar along which expansion is restrained\n",
+ "P=180*10**3 #N #Compressive force\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#The bar is restrained from expanding in Y direction\n",
+ "P_z=0\n",
+ "P_x=P*A**-1 #stress developed in x direction\n",
+ "\n",
+ "#Now taking compressive strain as positive\n",
+ "#e_x=P_x*E**-1-mu*P_y*E**-1 .......................(1)\n",
+ "#e_y=-mu*P_x*E**-1+P_y*E**-1 ....................(2)\n",
+ "#e_z=-mu*P_x*E**-1-mu*P_y*E**-1 ......................(3)\n",
+ "\n",
+ "#Part-1\n",
+ "#When it is fully restrained\n",
+ "e_y=0\n",
+ "P_y=30 #N/mm**2 \n",
+ "e_x=P_x*E**-1-mu*P_y*E**-1\n",
+ "e_z=-mu*P_x*E**-1-mu*P_y*E**-1\n",
+ "\n",
+ "#Change in Length \n",
+ "dell_l=e_x*L #mm\n",
+ "\n",
+ "#Change in width\n",
+ "dell_b=b*e_y\n",
+ "\n",
+ "#change in Depth\n",
+ "dell_d=d*e_z\n",
+ "\n",
+ "#Volume of bar\n",
+ "V=b*d*L #mm**3\n",
+ "#Change in Volume\n",
+ "e_v=(e_x+e_y+e_z)*V #mm**3\n",
+ "\n",
+ "#Part-2\n",
+ "#When 50% is restrained\n",
+ "\n",
+ "#Free strain in Y direction\n",
+ "e_y1=mu*P_x*E**-1\n",
+ "\n",
+ "#As 50% is restrained,so\n",
+ "e_y2=-50*100**-1*e_y1\n",
+ "\n",
+ "#But form Equation 2 we have e_y=-mu*P_x*E**-1+P_y*E**-1 \n",
+ "#After substituting values in above equation and furthe simplifying we get\n",
+ "P_y=e_y2*E+d\n",
+ "\n",
+ "e_x2=P_x*E**-1-mu*P_y*E**-1 \n",
+ "e_z2=-mu*P_x*E**-1-mu*P_y*E**-1\n",
+ "\n",
+ "#Change in Length \n",
+ "dell_l2=e_x2*L #mm\n",
+ "\n",
+ "#Change in width\n",
+ "dell_b2=b*e_y2\n",
+ "\n",
+ "#change in Depth\n",
+ "dell_d2=d*e_z2\n",
+ "\n",
+ "#Change in Volume\n",
+ "e_v2=(e_x2+e_y2+e_z2)*V #mm**3\n",
+ "\n",
+ "#REsult\n",
+ "print\"Change in Dimension of bar is:dell_l\",round(dell_l,2),\"mm\"\n",
+ "print\" :dell_b\",round(dell_b,4),\"mm\"\n",
+ "print\" :dell_d\",round(dell_d,2),\"mm\"\n",
+ "print\"Change in Volume is\",round(e_v,2),\"mm**3\"\n",
+ "print\"Changes in material when only 50% of expansion can be reatrained:dell_l2\",round(dell_l2,2),\"mm\"\n",
+ "print\" :dell_b2\",round(dell_b2,4),\"mm\"\n",
+ "print\" :dell_d2\",round(dell_d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in Dimension of bar is:dell_l 0.09 mm\n",
+ " :dell_b 0.0 mm\n",
+ " :dell_d -0.01 mm\n",
+ "Change in Volume is 93.6 mm**3\n",
+ "Changes in material when only 50% of expansion can be reatrained:dell_l2 0.1 mm\n",
+ " :dell_b2 -0.0045 mm\n",
+ " :dell_d2 -0.01 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.44,Page No.72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=10*10**3 #N #Load\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "d2=12 #mm #Diameter of bar1\n",
+ "d1=16 #mm #diameter of bar2\n",
+ "L1=200 #mm #Length of bar1\n",
+ "L2=500 #mm #Length of bar2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let A1 and A2 be the cross Area of Bar1 & bar2 respectively\n",
+ "A1=pi*4**-1*d1**2 #mm**2\n",
+ "A2=pi*4**-1*d2**2 #mm**2\n",
+ "\n",
+ "#Let p1 and p2 be the stress in Bar1 nad bar2 respectively\n",
+ "p1=P*A1**-1 #N/mm**2\n",
+ "p2=P*A2**-1 #N/mm**2\n",
+ "\n",
+ "#Let V1 nad V2 be the Volume of of Bar1 and Bar2\n",
+ "V1=A1*(L1+L1)\n",
+ "V2=A2*L2\n",
+ "\n",
+ "#Let E be the strain Energy stored in the bar\n",
+ "E=p1**2*(2*E)**-1*V1+p2**2*V2*(2*E)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"The Strain Energy stored in Bar is\",round(E,2),\"N-mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Strain Energy stored in Bar is 1602.6 N-mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.45,Page No.73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Bar-A\n",
+ "d1=30 #mm #Diameter of bar1\n",
+ "L=600 #mm #length of bar1\n",
+ "\n",
+ "#Bar-B\n",
+ "d2=30 #mm #Diameter of bar2\n",
+ "d3=20 #mm #Diameter of bar2\n",
+ "L2=600 #mm #length of bar2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of bar-A\n",
+ "A1=pi*4**-1*d1**2\n",
+ "\n",
+ "#Area of bar-B\n",
+ "A2=pi*4**-1*d2**2\n",
+ "A3=pi*4**-1*d3**2\n",
+ "\n",
+ "#let SE be the Strain Energy\n",
+ "#Strain Energy stored in Bar-A\n",
+ "#SE=p**2*(2*E)**-1*V\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE=P**2*E**-1*0.4244\n",
+ "\n",
+ "#Strain Energy stored in Bar-B\n",
+ "#SE2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE2=0.6897*P**2*E**-1\n",
+ "\n",
+ "#Let X be the ratio of SE in Bar-B and SE in Bar-A\n",
+ "X=0.6897*0.4244**-1\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#When Max stress is produced is same:Let p be the max stress produced\n",
+ "\n",
+ "#Stress in bar A is p throughout \n",
+ "#In bar B:stress in 20mm dia.portion=p2=p\n",
+ "\n",
+ "#Stress in 30 mm dia.portion\n",
+ "#p1=P*A2*A3**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#p1=4*9**-1*p\n",
+ "\n",
+ "#Strain Energy in bar A\n",
+ "#SE_1=p**2*(2*E)**-1*A1*L1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE_1=67500*p**2*pi*E**-1\n",
+ "\n",
+ "#Strain Energy in bar B\n",
+ "#SE_2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE_2=21666.67*pi*p**2*E**-1\n",
+ "\n",
+ "#Let Y be the Ratio of SE in bar B and SE in bar A\n",
+ "Y=21666.67*67500**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Gradually applied Load is\",round(X,2)\n",
+ "print\"Gradually applied Load is\",round(Y,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gradually applied Load is 1.63\n",
+ "Gradually applied Load is 0.32\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.46,Page No.74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables \n",
+ "\n",
+ "W=100 #N #Load\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "h=60 #mm #Height through Load falls down\n",
+ "L=400 #mm #Length of collar\n",
+ "d=30 #mm #diameter of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of bar\n",
+ "\n",
+ "#Instantaneous stress produced is\n",
+ "p=W*A**-1*(1+(1+(2*A*E*h*(W*L)**-1))**0.5)\n",
+ "\n",
+ "#Now the EXtension of the bar is neglected in calculating work doneby the Load,then\n",
+ "P=(2*E*h*W*(A*L)**-1)**0.5\n",
+ "\n",
+ "#Let percentage error be denoted by E1\n",
+ "#Percentage error in approximating is\n",
+ "E1=(p-P)*p**-1*100\n",
+ "\n",
+ "#Instantaneous Extension produced is\n",
+ "dell_l=round(P,3)*E**-1*L\n",
+ "\n",
+ "#Result\n",
+ "print\"The Instantaneous stress is\",round(p,2),\"N/mm\"\n",
+ "print\"Percentage Error is\",round(E1,2)\n",
+ "print\"The Instantaneous extension is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Instantaneous stress is 92.27 N/mm\n",
+ "Percentage Error is 0.15\n",
+ "The Instantaneous extension is 0.18 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.47,Page No.75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of steel bar\n",
+ "L=1000 #mm #Length of bar\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "p=300 #N/mm**2 #max Permissible stress\n",
+ "h=50 #mm #Height through which weight will fall\n",
+ "w=600 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#ARea of steel bar\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Instantaneous extension is\n",
+ "dell_l=p*L*E**-1 #mm \n",
+ "\n",
+ "#Work done by Load \n",
+ "#W=W1*(h+dell_l)\n",
+ "\n",
+ "#Volume of bar\n",
+ "V=round(A,2)*L\n",
+ "#Let E1 be the strain Energy\n",
+ "E1=p**2*(2*E)**-1*V\n",
+ "\n",
+ "#Answer in Book for Strain Energy is Incorrect \n",
+ "\n",
+ "#Now Equating Workdone by Load to strain Energy \n",
+ "W1=E1*51.5**-1\n",
+ "\n",
+ "#Now when w=600 N\n",
+ "#Let W2 be the Work done by the Load\n",
+ "#W2=w(h2*dell_l)\n",
+ "\n",
+ "h=E1*w**-1-dell_l\n",
+ "\n",
+ "#Result\n",
+ "print\"The Max Lodad which can Fall from a height of 50 mm on the collar is\",round(W1,2),\"N\"\n",
+ "print\"the Max Height from which a 600 N Load can fall on the collar is\",round(h,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Max Lodad which can Fall from a height of 50 mm on the collar is 1372.54 N\n",
+ "the Max Height from which a 600 N Load can fall on the collar is 116.31 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.48,Page No.76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D_s=30 #mm #Diameter of steel rod\n",
+ "d=30 #mm #Internal Diameter of copper tube\n",
+ "D=40#mm #External Diameter of copper tube\n",
+ "E_s=2*10**5 #N/mm**2 #Young's Modulus of Steel rod\n",
+ "E_c=1*10**5#N/mm**2 #Young's Modulus of copper tube\n",
+ "P=100 #N #Load\n",
+ "h=40 #mm #height from which Load falls\n",
+ "L=800 #mm #Length \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of steel rod\n",
+ "A_s=pi*4**-1*D_s**2\n",
+ "\n",
+ "#Area of copper tube\n",
+ "A_c=pi*4**-1*(D**2-d**2)\n",
+ "\n",
+ "#But Dell_s=dell_c=dell\n",
+ "#p_s*E_s**-1*L=p_c*L*E_c\n",
+ "#After simplifying furthe we get\n",
+ "#p_s=2*p_c\n",
+ "\n",
+ "#Now Equating internal Energy to Workdone we get\n",
+ "p_c=(2*P*h*L**-1*(4*A_s*E_s**-1+A_c*E_c**-1))**0.5\n",
+ "p_s=2*p_c\n",
+ "\n",
+ "#Result\n",
+ "print A_s\n",
+ "print\"STress produced in steel is\",round(p_s,2),\"N/mm**2\"\n",
+ "print\"STress produced in copper is\",round(p_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "706.858347058\n",
+ "STress produced in steel is 0.89 N/mm**2\n",
+ "STress produced in copper is 0.44 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.49,Page No.77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "dell=0.25 #mm #Instantaneous Extension\n",
+ "\n",
+ "#Bar-A\n",
+ "b1=25 #mm #width of bar\n",
+ "D1=500 #mm #Depth of bar\n",
+ "\n",
+ "#Bar-B\n",
+ "b2_1=25 #mm #width of upper bar\n",
+ "b2_2=15 #mm #Width of Lower Bar\n",
+ "L2=200 #mm #Length of upper bar\n",
+ "L1=300 #mm #Length of Lower bar\n",
+ "\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Strain\n",
+ "e=dell*D1**-1 \n",
+ "\n",
+ "#Load\n",
+ "p=e*E\n",
+ "\n",
+ "#Area of bar-A\n",
+ "A=pi*4**-1*25**2\n",
+ "\n",
+ "#Volume of bar-A\n",
+ "V=A*D1\n",
+ "\n",
+ "#Let E1 be the Energy of Blow\n",
+ "#Energy of Blow\n",
+ "E1=p**2*(E)**-1*V\n",
+ "\n",
+ "#Let p2 be the Max stress in bar B When this blow is applied.\n",
+ "#the max stress occurs in the 15mm dia. portion,Hence, the stress in 25 mm dia.portion is\n",
+ "#p2*pi*4**-1*b2_2**2*(pi*4**-1*b2_2**2=0.36*p\n",
+ "\n",
+ "#Strain Energy of bar B\n",
+ "#E2=p**2*(2*E)**-1*v1+1*(2*E)**-1*(0.36*p2)**2*v2\n",
+ "#After substituting values and Further substituting values we get\n",
+ "#E2=0.1643445*p2**2\n",
+ "\n",
+ "#Equating it to Energy of applied blow,we get\n",
+ "p2=(12271.846*0.1643445**-1)**0.5\n",
+ "\n",
+ "#Stress in top portion\n",
+ "sigma=0.36*p2\n",
+ "\n",
+ "#Extension in Bar-1\n",
+ "dell_1=p2*E**-1*L1\n",
+ "\n",
+ "#Extension in Bar-2\n",
+ "dell_2=0.36*p2*E**-1*L2\n",
+ "\n",
+ "#Extension of bar\n",
+ "dell_3=dell_1+dell_2\n",
+ "\n",
+ "#Result\n",
+ "print\"Instantaneous Max stress is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\"extension in Bar is\",round(dell_3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous Max stress is 98.37 N/mm**2\n",
+ "extension in Bar is 0.51 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_4.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_4.ipynb new file mode 100644 index 00000000..f136c4ad --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_4.ipynb @@ -0,0 +1,1588 @@ +{
+ "metadata": {
+ "name": "chapter no.3.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Shear Force And Bending Moment Diagrams in Statically Determinate Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.1,Page No.100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=L_CD=1 #m #Length of AC & CD\n",
+ "L_DB=1.5 #m #Lengh of DB\n",
+ "L=3.5 #m #Length of Beam\n",
+ "F_B=10 #KN #Force at pt B\n",
+ "F_C=F_D=20 #KN #Force at pt C & D\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R_A=F_C+F_D+F_B #KN #Force at support A \n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At pt B\n",
+ "V_B1=0 #KN \n",
+ "V_B2=F_B #KN\n",
+ "\n",
+ "#S.F At pt D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1+F_D #KN\n",
+ "\n",
+ "#S.F At pt C \n",
+ "V_C1=V_D2 #KN\n",
+ "V_C2=V_D2+F_C #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2 #KN\n",
+ "V_A2=V_C2-R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M AT Pt D\n",
+ "M_D=F_B*L_DB #KN.m\n",
+ "\n",
+ "#B.M At pt C\n",
+ "M_C=F_B*(L_DB+L_CD)+F_D*L_CD #KN.m\n",
+ "\n",
+ "#B.M At pt A\n",
+ "M_A=F_B*L+F_D*(L_CD+L_AC)+F_C*L_AC\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5dde2d0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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pqQavCQgIEADwi1/84he/rPgKCAiw6e+yYlJMDQ0NuPfee/H999+jZ8+eGDhw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\no9ggiIjIKDYIIiIy6v8BsxKV3Pt7cnUAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5df0ab0>"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.2,Page No.101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "w1=10 #KN/m #u.d.L\n",
+ "F_D=20 #KN #Force at pt D\n",
+ "F_C=30 #KN #Force at pt C\n",
+ "L_DB=4 #m #Length of DB\n",
+ "L_CD=L_AC=2 #m #Length of AC & CD\n",
+ "L=8 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A And R_B be the Reactions at pt A and B \n",
+ "#R_A+R_B=90 \n",
+ "#Now Taking moment at A,M_A we get\n",
+ "R_A=(w1*L_DB*(L_DB*2**-1)+F_D*L_DB+F_C*(L_CD+L_DB))*L**-1\n",
+ "R_B=90-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=0 #KN\n",
+ "V_B2=R_B #KN\n",
+ "\n",
+ "#S.F At pt D\n",
+ "V_D1=R_B-w1*L_DB #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F at Pt C\n",
+ "V_C1=V_D2 #KN\n",
+ "V_C2=V_C1-F_C \n",
+ "\n",
+ "#S.F at PT A\n",
+ "V_A1=V_C2 #KN\n",
+ "V_A2=V_C2+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D=-R_B*L_DB+w1*L_DB*L_DB*2**-1 #KN.m\n",
+ "\n",
+ "#B.M At PT C\n",
+ "M_C=-R_B*(L_DB+L_CD)+w1*L_DB*(L_DB*2**-1+L_CD)+F_D*L_CD #KN.m\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=-R_B*L+w1*L_DB*(L_DB*2**-1+L_CD+L_AC)+F_D*(L_CD+L_AC)+F_C*L_AC\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Pni0aGhr0vo4HrxERkcSmlo+IiMi8WApERCRhKRARkYSlQEREEpYCERFJWApE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\nICIiCUuBiIgk/w9P4ODmR+1IBgAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5be93d0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Gt27dTE9mRiwY9ovtQ+zHlSuAv780HCkgQO00tkGRgnHr1i2kp6ejpKQEdXV1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\nx6hRo1BfX48XX3wRgYGBasd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+ "text": [
+ "<matplotlib.figure.Figure at 0x5de54f0>"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.3,Page No.102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DB=L_CD=1.5 #m #Length of DB & CD\n",
+ "L_AC=3 #m #Length of AC\n",
+ "F_D=80 #KN #Force at Pt D\n",
+ "w=40 #KN/m #u.v.l\n",
+ "L=6 #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the Reactions at Pt A & B respectively\n",
+ "#R_A+R_B=140 \n",
+ "#Taking moment at B we get,M_B\n",
+ "R_A=(1*2**-1*L_AC*w*(1*3**-1*L_AC+(L_CD+L_DB))+F_D*L_DB)*L**-1\n",
+ "R_B=140-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F at B\n",
+ "V_B1=0 #KN\n",
+ "V_B2=R_B #KN\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F at C\n",
+ "V_C=V_D2 #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_C-1*2**-1*w*L_AC #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=-R_B*L_DB\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_D*L_CD-R_B*(L_DB+L_CD)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=F_D*(L_CD+L_AC)-R_B*L+1*2**-1*w*L_AC*(1*3**-1*L_AC)+R_A\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x58f0c10>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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CBUEIDRWEv/9d7kpIbVz53SnLDGPjxo0YOHAg9u3bh/T0dKSlpQEAjEYjsrKy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\ngg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x562f910>"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.4,Page No.104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "M_D=120 #KN.m #B.M at Pt D\n",
+ "F_C=40 #KN #Force at Pt C\n",
+ "w1=20 #KN.m\n",
+ "L_DB=1.5 #m #Length of DB\n",
+ "L_CD=1.5 #m #Length of CD\n",
+ "L_AC=3 #m #Length of AC\n",
+ "L=6 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A And R_B be the Reactions at pt A and B \n",
+ "#R_A+R_B=100\n",
+ "#Now Taking Moment At Pt B We get,M_B\n",
+ "R_A=-(M_D-F_C*(L_CD+L_DB)-w1*L_AC*(L_AC*2**-1+L_CD+L_DB))*L**-1\n",
+ "R_B=100-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=0\n",
+ "V_B2=R_B\n",
+ "\n",
+ "#S.F at Pt D\n",
+ "V_D=V_B2 #KN\n",
+ "\n",
+ "#S.F At Pt C\n",
+ "V_C1=V_D #KN\n",
+ "V_C2=V_C1-F_C\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2-w1*L_AC #KN\n",
+ "V_A2=V_A1+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=M_B-R_B*L_DB #KN.m\n",
+ "M_D2=M_B+M_D-R_B*L_DB\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=M_D-R_B*(L_CD+L_DB)\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=M_D-R_B*L+F_C*L_AC+w1*L_AC*L_AC*2**-1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB+L_CD,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D1,M_D2,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x55dcbf0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x55ca4b0>"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.5,Page No.105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_C=20 #KN #Force at Pt C\n",
+ "F_D=40 #KN #Force at pt D\n",
+ "w=20 #KN.m #u.d.l \n",
+ "L_AD=L_DB=2 #m #Length of AD & DB\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L=5 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_A and R_B be the reactions at A & B respectively\n",
+ "#R_A+R_B=100 \n",
+ "#Now Taking Moment at B,M_B we get\n",
+ "R_A=-(F_C*L_BC-F_D*L_DB-w*L_AD*(L_AD*2**-1+L_DB))*(L_AD+L_DB)**-1\n",
+ "R_B=100-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At pt C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-F_C #KN\n",
+ "\n",
+ "#S.F At PT B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN\n",
+ "\n",
+ "#S.F At Pt D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_D2-w*L_AD #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=0 \n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=F_C*L_BC\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D=F_C*(L_BC+L_DB)-R_B*L_DB\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=F_C*L-R_B*(L_DB+L_AD)+F_D*L_AD+w*L_AD*L_AD*2**-1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_BC+L_DB,L_BC+L_DB,L_BC+L_DB+L_AD,L_BC+L_DB+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_D1,V_D2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_C,M_B,M_D,M_A]\n",
+ "X2=[0,L_BC,L_BC+L_DB,L_BC+L_DB+L_AD]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d9c6f0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Zb7/9lh07drB8+XLy8/N58cUXmTVrlrt+NSKaPpLAUlFRwZdffsnI\nkSOd91VWVgLGm/WZE1a7du3K4cOHAfj8889JSUkBcPYoONvw4cMB6NGjBx9++KHzfldOkKntmufr\n0KED3bp1A6Bbt24kJiYCEB0dTUlJSZ3XEXGVkoIElNOnT3PFFVdQWFhY48+bNm3q/PrMm/r5dYPz\n3+wvueQSAJo0aUJVVVW9Y6rpmuc7cw0w+iWceU5QUFCDrilSG00fSUC5/PLL6dChAx988AFgvAn/\n61//uuhz+vTpw4oVK3A4HBw+fJi8vLw6r9O8eXPnUcXn0xmUYmVKCuLXTpw4Qbt27Zy3//3f/+Xd\nd99l0aJFxMbGEh0dfU4z97Pn+898PWLECMLCwoiKiuK+++6jR48etGjR4oJr2Ww253OGDh3KypUr\niYuLIz8/v9bH1XbNml67tu8D+UhocT8dnS3igp9//pnLLruM//znPyQkJPDFF1/QqlUrs8MScTvV\nFERcMGTIEMrLy6msrGTatGlKCOK3NFIQEREn1RRERMRJSUFERJyUFERExElJQUREnJQURETESUlB\nRESc/j+iaB8fTwtkoQAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5c1ead0>"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.6,Page No.107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=L_EB=L_AD=1 #m #Length of spans BC,ED,AD\n",
+ "L_ED=2 #m #Length of ED\n",
+ "w=60 #KNm #u.d.l\n",
+ "F_C=20 #KN Pt Load at C\n",
+ "L=5 #m #Span of beam \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A & R_B be the reactions at A & B respectively\n",
+ "#R_A+R_B=80 \n",
+ "#Taking Moment At A,we get M_A\n",
+ "R_B=(F_C*L+1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD))*(L_AD+L_ED+L_EB)**-1\n",
+ "R_A=80-R_B\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-F_C #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN \n",
+ "\n",
+ "#S.F aT E\n",
+ "V_E=V_B2 #KN\n",
+ "\n",
+ "#S.F AT D\n",
+ "V_D=V_B2-1*2**-1*L_ED*w #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_D #KN \n",
+ "V_A2=V_D+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at C\n",
+ "M_C=0 #KN.m\n",
+ "\n",
+ "#B.M at B\n",
+ "M_B=F_C*L_BC #KN.m\n",
+ "\n",
+ "#B.M at E\n",
+ "M_E=F_C*(L_EB+L_BC)-R_B*L_EB #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D=F_C*(L_ED+L_EB+L_BC)-R_B*(L_ED+L_EB)+1*2**-1*L_ED*w*1*3**-1*L_ED #KN.m\n",
+ "\n",
+ "#B.M at A\n",
+ "M_A=1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD)-R_B*(L_AD+L_ED+L_EB)+F_C*L\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_EB+L_BC,L_ED+L_EB+L_BC,L_AD+L_ED+L_EB+L_BC,L_ED+L_EB+L_BC+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_E,V_D,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_BC,L_BC+L_EB,L_EB+L_BC+L_ED,L_EB+L_BC+L_ED+L_AD]\n",
+ "Y2=[M_C,M_B,M_E,M_D,M_A]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Cr9fzLAHAgQMHUFpaig8++AAbNmzAvn37Ot1OUUUhJCSkwx1Vq6qqEBoaKjAR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\nAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58a6f70>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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UlD0DZoHZoDk7OxvPPfec/SqwAoNm98HQmcjAloBZINmcwq5du5rcTwEAHnnkEeuqkgCb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\nkeDZfAAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x567bcf0>"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.7,Page No.109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L_DB=2 #m #Length of DB\n",
+ "L_AD=4 #m #Length 0f AD\n",
+ "M_D=30 #KN.m #Moment at D\n",
+ "w=45 #KN/m #u.d.l\n",
+ "L=7 #m #Span of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_B & R_A be the Reactions at B & A respectively\n",
+ "#R_B+R_A=180+P ............(1)\n",
+ "\n",
+ "#Now Taking Moment about A,we get\n",
+ "#R_B=7*P+390 ...............(2)\n",
+ "\n",
+ "#Since R_A & R_B Are Equal\n",
+ "#2*R_B=180+P ...................(3)\n",
+ "\n",
+ "#From equation 1 and 3 we get\n",
+ "#3*(180+P)=7P+390\n",
+ "#After simplifying Further above equation we get\n",
+ "P=150*4**-1 #KN\n",
+ "R_A=R_B=(180+P)*2**-1\n",
+ "F_C=P\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-P #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN \n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=V_B2 #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_D-w*L_AD #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at C\n",
+ "M_C=0 #KN.m \n",
+ "\n",
+ "#B.M at B\n",
+ "M_B=F_C*L_BC #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D1=F_C*(L_BC+L_DB)-R_B*L_DB #KN.m\n",
+ "M_D2=M_D1+M_D\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=w*L_AD*L_AD*2**-1+M_D-R_B*(L_AD+L_DB)+P*L\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_DB+L_BC,L_DB+L_BC+L_AD,L_DB+L_BC+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_D,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_BC,L_DB+L_BC,L_DB+L_BC,L_AD+L_DB+L_BC]\n",
+ "Y2=[M_C,M_B,M_D1,M_D2,M_A]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d6f150>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d79470>"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.8,Page No.110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6 #m #Span Of beam\n",
+ "w=30 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Due to Symmetry\n",
+ "#Let R_B and R_C be the reactions at B & C Respectively\n",
+ "R_B=R_C=w*L*2**-1 #KN\n",
+ "\n",
+ "#Let a be the overhang.The Max -ve moment occurs at the support and max +ve moment at middle of the beam\n",
+ "#Now Equating these two equations we get\n",
+ "#30*a*a*2**-1=90*(3-a)-w*L*2**-1*L*4**-1\n",
+ "#After simplifying we get an equation as\n",
+ "#a**2+6*a-9=0\n",
+ "x=1\n",
+ "y=6\n",
+ "z=-9\n",
+ "\n",
+ "p=y**2-4*x*z\n",
+ "\n",
+ "a1=(-y+p**0.5)*2**-1\n",
+ "a2=(-y-p**0.5)*2**-1\n",
+ "\n",
+ "#Now Length cannot be negative,so taking a1 into Consideration\n",
+ "\n",
+ "L_CD=L_AB=a1\n",
+ "L_BC=L-2*a1\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=0\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=V_D-w*L_CD #KN\n",
+ "V_C2=V_C1+R_C #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=-w*(L_BC+L_CD)+R_C\n",
+ "V_B2=V_B1+R_B\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A=round(V_B2,2)-round(w*L_AB,2)\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=0\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=w*L_CD*L_CD*2**-1 #KN.m\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=w*(L_BC+L_CD)*(L_BC+L_CD)*2**-1-R_C*L_BC*L_BC*2**-1\n",
+ "\n",
+ "#B.M At A\n",
+ "X=w*L*L*2**-1\n",
+ "Y=-R_C*(L_AB+L_BC)-R_B*L_AB\n",
+ "M_A=X+Y\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,L_CD,L_CD,L_CD+L_BC,L_CD+L_BC,L_CD+L_BC+L_AB]\n",
+ "Y1=[V_D,V_C1,V_C2,V_B1,V_B2,V_A]\n",
+ "Z1=[0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_CD,L_BC+L_CD,L_AB+L_BC+L_CD]\n",
+ "Y2=[M_D,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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lXLhwAQBw6dIlHDhwwKleBejt7Q0/Pz+UlpYCAA4ePIiwsLAub6vKm9f6yt3d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\nAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5945590>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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LUVxcrNtAqS48PDxqdH5WVhby8vIQEBAAAJg4cSJ27NjBhEGkEDc3ID5e7BWe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\nMQcOHJASExMlb29vtUPRi6y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+ "text": [
+ "<matplotlib.figure.Figure at 0x58e1750>"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.9,Page No.112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_F=6 #KN #Force at F\n",
+ "w1=w2=w=3 #KN.m #u.d.l\n",
+ "M_D=24 #KN.m \n",
+ "L_AB=L_CD=L_DE=L_EF=4 #m #Length of AB,CD,DE,EF\n",
+ "L_BC=2 #m #Length of BC\n",
+ "L=18 #m #Span of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_B and R_E be the Reactions at B & E respectively\n",
+ "#R_B+R_E=42\n",
+ "\n",
+ "#Taking Moment At Pt B,M_B\n",
+ "R_E=(F_F*(L_BC+L_CD+L_DE+L_EF)+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)-w*L_AB*L_AB*2**-1-M_D)*(L_BC+L_CD+L_DE)**-1\n",
+ "R_B=42-R_E #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F aT F\n",
+ "V_F1=0 #KN \n",
+ "V_F2=-F_F #KN\n",
+ "\n",
+ "#S.F at E\n",
+ "V_E1=V_F2 #KN\n",
+ "V_E2=V_E1+R_E #KN\n",
+ "\n",
+ "#S.F aT C\n",
+ "V_C=V_E2-w*(L_CD+L_DE) #KN\n",
+ "\n",
+ "#S.F at B\n",
+ "V_B1=V_C #KN \n",
+ "V_B2=V_C+R_B #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A=V_B2-w*L_AB #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At F\n",
+ "M_F=0\n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=F_F*L_EF #KN.m\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D1=F_F*(L_DE+L_EF)-R_E*L_DE+w*L_DE*L_DE*2**-1 #KN.m\n",
+ "M_D2=M_D1-M_D\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_F*(L_CD+L_DE+L_EF)-R_E*(L_CD+L_DE)+w*(L_CD+L_DE)*(L_CD+L_DE)*2**-1-M_D\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_F*(L_BC+L_CD+L_DE+L_EF)-R_E*(L_BC+L_CD+L_DE)-M_D+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=w*L_AB*L_AB*2**-1-R_B*L_AB+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC+L_AB)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_F*L-M_D\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_EF,L_EF,L_EF+L_DE+L_CD,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y1=[V_F1,V_F2,V_E1,V_E2,V_C,V_B1,V_B2,V_A]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB]\n",
+ "Y2=[M_F,M_E,M_D1,M_D2,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d75ab0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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VxMfHIzs7G/369cOvv/6KcePG4bHHHnNknEROixVL5KzMziBqa2uRmJiIMWPG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\nfyRzHAAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5c44130>"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.10,Page No.114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DC=L_BA=2 #m #Length of BA & DC\n",
+ "L_CB=1 #m #Length of CB\n",
+ "F_A=10 #KN #Force at pt A\n",
+ "F_B=20 #KN #Force at pt B\n",
+ "w=4 #KN.m #u.d.l\n",
+ "L=5 #m #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_D be the reactions at Pt D\n",
+ "R_D=F_B+F_A+w*L_DC #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F at Pt A\n",
+ "V_A1=0 #KN\n",
+ "V_A2=F_A #KN\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=V_A2\n",
+ "V_B2=F_B+F_A\n",
+ "\n",
+ "#S.F at Pt C\n",
+ "V_C=F_B+F_A #KN \n",
+ "\n",
+ "#S.F At Pt D\n",
+ "V_D1=V_B2+w*L_DC\n",
+ "V_D2=F_B+F_A+w*L_DC-R_D\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=0\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=F_A*L_BA\n",
+ "\n",
+ "#B.M at Pt C\n",
+ "M_C=F_B*L_CB+F_A*(L_BA+L_CB) #KN\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1\n",
+ "M_D2=(F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1)-M_D1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BA,L_BA,L_BA+L_CB,L_BA+L_CB+L_DC,L_BA+L_CB+L_DC]\n",
+ "Y1=[V_A1,V_A2,V_B1,V_B2,V_C,V_D1,V_D2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_A,M_B,M_C,M_D1,M_D2]\n",
+ "X2=[0,L_BA,L_CB+L_BA,L_CB+L_BA+L_DC,L_CB+L_BA+L_DC]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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e3542bZr69++vhIQEzZ49WyNGjNAFF1wgSbrxxht15ZVXen+5e/LxJ0/9axiG\n8vLy9Oc//1kxMTFatWqV8vLyvMNNvo71tX3yMb6+tvMUxDh7XM4JWzp8+LAiIiL0ww8/aPTo0dq8\nebP69OljdSwgKBjjhy1NnjxZNTU1amho0L333kvpw1Y44wcAm2GMHwBshuIHAJuh+AHAZih+ALAZ\nih8AbIbiBwCb+T+gqVKkQGpm/QAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5738d10>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5788db0>"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.11,Page No.115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "w=20 #KN/m #u.v.l\n",
+ "F_C=40 #KN #Force at Pt C\n",
+ "M_D=40 #KN.m #Moment at pt D\n",
+ "L_AB=3 #m #Length of AB\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L_CD=L_DE=2 #m #Length of CD & DE\n",
+ "L=8 #8 #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A & R_E be the Reactions at A & E respectively\n",
+ "#R_A+R_E=70\n",
+ "\n",
+ "#Taking Moments At Pt A we get,M_A\n",
+ "R_E=(F_C*(L_AB+L_BC)+1*2**-1*L_AB*w*2+40)*L**-1\n",
+ "R_A=70-R_E\n",
+ "\n",
+ "#shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt E\n",
+ "V_E1=0\n",
+ "V_E2=R_E #KN\n",
+ "\n",
+ "#S.F aT pt D\n",
+ "V_D=V_E2\n",
+ "\n",
+ "#S.F At PT C\n",
+ "V_C1=V_D\n",
+ "V_C2=V_D-F_C #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2-(1*2**-1*w*L_AB)\n",
+ "V_A2=V_A1+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt E\n",
+ "M_E=0\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=M_E-R_E*L_DE\n",
+ "M_D2=M_D1+M_D\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=-R_E*(L_DE+L_CD)+M_D\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=-R_E*(L_DE+L_CD+L_BC)+M_D+F_C*L_BC\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=-R_E*L+M_D+(1*2**-1*L_AB*w*2)+F_C*(L_BC+L_AB)\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DE,L_CD+L_DE,L_CD+L_DE,L_CD+L_DE+L_AB,L_CD+L_DE+L_AB]\n",
+ "Y1=[V_E1,V_E2,V_D,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_E,M_D1,M_D2,M_C,M_B,M_A]\n",
+ "X2=[0,L_DE,L_DE,L_CD+L_DE,L_DE+L_CD+L_BC,L_AB+L_BC+L_CD+L_DE]\n",
+ "Z2=[0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5df5230>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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PZcuW8dRTT5GZmcnOnTtp1KgRI0aMsPs+cvTnxbp1U4fwLFyoO4nzTCaYMkWt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\nd/r0aduBO+VVUFCA2Wxm+/b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6ePR5C6Jyk56CqHT27dtHt27daNu2LXfddRdpaWkAPPbY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\nGhPHAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5dff130>"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.12,Page No.116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_G=10 #KN #Force at Pt G\n",
+ "F_B=F_E=15 #KN #Force at Pt B & E\n",
+ "w=20 #KN/m #U.d.L\n",
+ "L_FG=L_EF=L_DE=L_CD=L_BC=L_AB=1 #m #Lengths of FG,EF,DE,CD,BC,AB respectively\n",
+ "L=6 #m #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_F & R_A be the Reactions at E & A respectively\n",
+ "#R_F+R_A=60\n",
+ "\n",
+ "#Taking Moment At Pt A,M_A\n",
+ "R_F=(F_G*L+F_E*(L_AB+L_BC+L_CD+L_DE)+w*L_CD*(L_AB+L_BC+L_CD*2**-1)+F_B*L_AB)*(L_AB+L_BC+L_CD+L_DE+L_EF)**-1\n",
+ "R_A=60-R_F\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At G\n",
+ "V_G1=0 #KN \n",
+ "V_G2=F_G #KN\n",
+ "\n",
+ "#S.F At F\n",
+ "V_F1=V_G2 #KN\n",
+ "V_F2=V_F1-R_F\n",
+ "\n",
+ "#S.F At E\n",
+ "V_E1=V_F2 #KN\n",
+ "V_E2=V_F2+F_E\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=V_E2\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C=V_E2+w*L_CD\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C\n",
+ "V_B2=V_B1+F_B\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_B2\n",
+ "V_A2=V_B2-R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt G\n",
+ "M_G=0\n",
+ "\n",
+ "#B.M At F\n",
+ "M_F=F_G*L_FG \n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=F_G*(L_FG+L_EF)-R_F*L_EF\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=F_G*(L_FG+L_EF+L_DE)-R_F*(L_EF+L_DE)+F_E*L_DE\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_G*(L_FG+L_EF+L_DE+L_CD)-R_F*(L_EF+L_DE+L_CD)+F_E*(L_DE+L_CD)+w*L_CD*L_CD*2**-1\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_G*(L_FG+L_EF+L_DE+L_CD+L_BC)-R_F*(L_EF+L_DE+L_CD+L_BC)+F_E*(L_DE+L_CD+L_BC)+w*L_CD*(L_CD*2**-1+L_BC)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=F_G*L-R_F*(L_EF+L_DE+L_CD+L_BC+L_AB)+F_E*(L_DE+L_CD+L_BC+L_AB)+F_B*L_AB+w*L_CD*(L_CD*2**-1+L_BC+L_AB)\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_FG,L_FG,L_FG+L_EF,L_FG+L_EF,L_FG+L_EF+L_DE,L_FG+L_EF+L_DE+L_CD,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y1=[V_G1,V_G2,V_F1,V_F2,V_E1,V_E2,V_D,V_C,V_B1,V_B2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_FG,L_EF+L_FG,L_EF+L_FG+L_DE,L_EF+L_FG+L_DE+L_CD,L_EF+L_FG+L_DE+L_CD+L_BC,L_EF+L_FG+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y2=[M_G,M_F,M_E,M_D,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x57502f0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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9jlmzZjkcDuPv6NGjR8t9nl+NFHJycmjatCnR0dGEh4dz1113kZmZ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\nkcIBAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58b0ed0>"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.13,Page No.117"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "L_AB=L_BC=L_CD=L_DE=L_EF=1 #m #LEngth of AB,BC,CD,DE,EF respectively\n",
+ "M_A=50 #KN/m #Moment at A\n",
+ "w=5 #KN/m #u.v.l\n",
+ "F_D=10 #KN\n",
+ "w2=5 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_B & R_E be the Reactions at B and E respectively\n",
+ "#R_B+R_E=20\n",
+ "\n",
+ "#Taking Moment At Pt B,M_B\n",
+ "R_E=(w2*L_EF*(L_EF*2**-1+L_DE+L_CD+L_BC)+w*L_BC*2**-1*2*3**-1+50+F_D*(L_BC+L_CD))*3**-1\n",
+ "R_B=17.5-R_E #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At F\n",
+ "V_F=0\n",
+ "\n",
+ "#S.F aT E\n",
+ "V_E1=-w2*L_EF #KN\n",
+ "V_E2=V_E1+R_E\n",
+ "\n",
+ "#S.F at D\n",
+ "V_D1=R_E-w2*L_EF #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C=V_D2\n",
+ "\n",
+ "#S.F aT B\n",
+ "V_B1=-L_BC*w*2**-1-F_D+R_E-w2*L_EF\n",
+ "V_B2=V_B1+R_B\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at F\n",
+ "M_F=0 #KN.m\n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=w2*L_EF*L_EF*2**-1 #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D=-R_E*L_DE+w2*L_EF*(L_EF*2**-1+L_DE) #KN.m\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_D*L_CD*R_E*(L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_DE+L_CD) #KN.m\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_D*(L_CD+L_BC)-R_E*(L_BC+L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_BC+L_CD+L_DE)+1*2**-1*L_BC*w*2*3**-1\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A1=w*L_EF*(L_EF*2**-1+L_AB+L_BC+L_CD+L_DE)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_D*(L_AB+L_BC+L_CD)+1*2**-1*L_BC*w*(2*3**-1*L_BC+L_AB)-R_B*L_AB\n",
+ "M_A2=M_A1+M_A\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,L_EF,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC]\n",
+ "Y1=[V_F,V_E1,V_E2,V_D1,V_D2,V_C,V_B1,V_B2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB,L_CD+L_DE+L_EF+L_BC+L_AB]\n",
+ "Y2=[M_F,M_E,M_D,M_C,M_B,M_A1,M_A2]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5c2a7b0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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LlixREhMTLY8fMWKEsm/fvjuex4ZfSQi7euUVRXnxRa1TuIdr1xRFr1eUb77R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\ntnx2yhiGEEIIm0jBEEIIYRM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+ "text": [
+ "<matplotlib.figure.Figure at 0x1a108f0>"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_4.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_4.ipynb new file mode 100644 index 00000000..09aec3f9 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_4.ipynb @@ -0,0 +1,1628 @@ +{
+ "metadata": {
+ "name": "chapter no.4.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Stresses in Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.1,Page no.130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=5000 #mm #Length of Beam\n",
+ "a=2000 #mm #Length of start of beam to Pt Load\n",
+ "b=3000 #mm #Length of Pt load to end of beam\n",
+ "A=150*250 #m**2 #Area of beam\n",
+ "b=150 #mm #Width of beam\n",
+ "d=250 #mm #Depth of beam\n",
+ "sigma=10#N/mm**2 #stress\n",
+ "l=2000 #m #Load applied from one end\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*d**3 #m**4\n",
+ "\n",
+ "#Distance from N.A to end\n",
+ "y_max=d*2**-1 #m\n",
+ "\n",
+ "#Section Modulus\n",
+ "Z=1*6**-1*b*d**2 #mm**3\n",
+ "\n",
+ "#Moment Carrying Capacity\n",
+ "M=sigma*Z #N-mm\n",
+ "\n",
+ "#Let w be the Intensity of the Load in N/m,then Max moment\n",
+ "#M_max=w*L**2*8**-1 #N-mm\n",
+ "#After substituting values and further simplifying we get\n",
+ "#M_max=w*25*100*8**-1\n",
+ "\n",
+ "#EQuating it to moment carrying capacity,we get max intensity load\n",
+ "w=M*(25*1000)**-1*8*10**-3\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Let P be the concentrated load,then max moment occurs under the load and its value\n",
+ "#M1=P*a*b*L**-1 #N-mm\n",
+ "\n",
+ "#Equting it to moment carrying capacity we get\n",
+ "P=M*1200**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Intensity of u.d.l it can carry\",round(w,3),\"KN-m\"\n",
+ "print\"MAx concentrated Load P apllied at 2 m from one end is\",round(P,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Intensity of u.d.l it can carry 5.0 N-mm\n",
+ "MAx concentrated Load P apllied at 2 m from one end is 13.021 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.2,Page no.131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=70 #mm #External Diameter\n",
+ "t=8 #mm #Thickness of pipe\n",
+ "L=2500 #mm #span \n",
+ "sigma=150 #N/mm**2 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Internal Diameter \n",
+ "d=D-2*t #mm\n",
+ "\n",
+ "#M.I Of Pipe\n",
+ "I=pi*64**-1*(D**4-d**4) #mm**4\n",
+ "\n",
+ "y_max=D*2**-1 #mm\n",
+ "Z=I*(y_max)**-1 #mm**3\n",
+ "\n",
+ "#Moment Carrying capacity\n",
+ "M=sigma*Z #N*mm\n",
+ "\n",
+ "#Max moment int the beam occurs at the mid-span and is equal to\n",
+ "#m=P*L*4**-1\n",
+ "\n",
+ "#Equating Max moment to moment carrying capacity we get,\n",
+ "#M=P*2.5*L*4**-1\n",
+ "#After substituting and simplifying we get\n",
+ "P=4*M*(L)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Max concentrated load that can be applied at the centre of span is\",round(P,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max concentrated load that can be applied at the centre of span is 5.22 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.3,Page no.132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges Dimension\n",
+ "b1=180 #mm #Width\n",
+ "d1=10 #mm #Thickness\n",
+ "\n",
+ "D=500 #mm #Overall depth\n",
+ "t=8 #mm #Thickness of web\n",
+ "\n",
+ "#Plate Dimensions\n",
+ "b2=240 #mm #Width\n",
+ "t2=12 #mm #Thickness\n",
+ "\n",
+ "sigma=150 #N/mm**2 #Stress\n",
+ "L=3000 #mm #span\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of centroid from bottom fibre\n",
+ "y_bar=(b2*t2*(D+t2*2**-1)+b1*d1*(D-t1*2**-1)+(D-2*t1)*t*D*2**-1+(b1*t1*t1*2**-1))*(b2*t2+b1*d1+b1*d1+(D-2*d1)*t)**-1\n",
+ "\n",
+ "#M.I of section\n",
+ "I=(1*12**-1*b2*t2**3+b2*t2*(D+t2*2**-1-y_bar)**2+1*12**-1*b1*d1**3+b1*d1*(D-t1*2**-1-y_bar)**2+1*12**-1*b1*t1**3+b1*t1*(t1*2**-1-y_bar)**2+1*12**-1*t*(D-2*t1)**3+t*(D-2*t1)*(D*2**-1-y_bar)**2)\n",
+ "\n",
+ "#Section Modulus\n",
+ "Z=I*(y_bar)**-1 #mm**3\n",
+ "\n",
+ "#Moment or Resistance\n",
+ "M=sigma*Z\n",
+ "\n",
+ "#Let Load on Cantilever be w/m Length \n",
+ "#Max M.I produced\n",
+ "#M_max=w*L**2**-1 \n",
+ "\n",
+ "#Now Equating Moment of resistance to Max moment,we get Max load\n",
+ "#4.5*w=M\n",
+ "#After rearranging and further simplifying we get\n",
+ "w=M*4.5**-1*10**3*10**-9\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of Resistance is\",round(M,2),\"KN-mm\"\n",
+ "print\"Load the section can carry is\",round(w,3),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of Resistance is 198770121.83 KN-mm\n",
+ "Load the section can carry is 44.171 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.4,Page no.134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flange (Top)\n",
+ "b1=80 #mm #Width \n",
+ "t1=40 #mm #Thickness\n",
+ "\n",
+ "#Flange (Bottom)\n",
+ "b2=160 #mm #width\n",
+ "t2=40 #mm #Thickness\n",
+ "\n",
+ "#web\n",
+ "d=120 #mm #Depth\n",
+ "t3=20 #mm #Thickness\n",
+ "\n",
+ "D=200 #mm #Overall Depth\n",
+ "sigma1=30 #N/mm**2 #Tensile stress\n",
+ "sigma2=90 #N/mm**2 #Compressive stress\n",
+ "L=6000 #mm #Span\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of centroid from bottom fibre\n",
+ "y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-round(y_bar,2))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-round(y_bar,2))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-round(y_bar,2))**2\n",
+ "\n",
+ "#Extreme fibre distance of top and bottom fibres are y_t and y_c respectively\n",
+ "\n",
+ "y_t=y_bar #mm\n",
+ "y_c=D-y_bar #mm\n",
+ "\n",
+ "#Moment carrying capacity considering Tensile strength \n",
+ "M1=sigma1*I*y_t**-1*10**-6 #KN-m\n",
+ "\n",
+ "#Moment carrying capacity considering compressive strength \n",
+ "M2=sigma2*I*y_c**-1*10**-6 #KN-m\n",
+ "\n",
+ "#Max Bending moment in simply supported beam 6 m due to u.d.l\n",
+ "#M_max=w*L*10**-3*8**-1\n",
+ "#After simplifying further we get\n",
+ "#M_max=4.5*w\n",
+ "\n",
+ "#Now Equating it to Moment carrying capacity, we get load carrying capacity\n",
+ "w=M1*4.5**-1 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Uniformly Distributed Load is\",round(w,3),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Uniformly Distributed Load is 5.096 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.5,Page no.136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from scipy.integrate import quad\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges\n",
+ "b=200 #mm #Width\n",
+ "t=25 #mm #Thickness \n",
+ "\n",
+ "D1=500 #mm #Overall Depth\n",
+ "t2=20 #mm #Thickness of web\n",
+ "\n",
+ "d=450 #mm #Depth of web\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider,Element of Thickness \"y\" at Distance \"dy\" from N.A \n",
+ "#Let Bending stress \"sigma_max\"\n",
+ "\n",
+ "#Stress on the element \n",
+ "#sigma=y*(D*2**-1)*sigma_max ..............(1)\n",
+ "\n",
+ "#Area of Element\n",
+ "#A=b*dy .................................(2)\n",
+ "\n",
+ "#Force on Element \n",
+ "#F=y*250**-1*sigma_max*b*dy\n",
+ "\n",
+ "#Let M be the Moment of resistance\n",
+ "#M=y*250**-1*sigma_max*b*dy*y\n",
+ "\n",
+ "#Moment of Resistance of top flange be M1\n",
+ "def integrand(y, b, D):\n",
+ " return b*y**2*D**-1\n",
+ "b=200 \n",
+ "D=250\n",
+ "\n",
+ "X = quad(integrand, 225, 250, args=(b,D))\n",
+ "\n",
+ "Y=2*X[0]\n",
+ "\n",
+ "#M1=Y*sigma\n",
+ "\n",
+ "#Now Moment of Inertia I section is\n",
+ "X=b*D1**3\n",
+ "Y=(b-t2)*d**3\n",
+ "I=(X-Y)*12**-1*10**-8\n",
+ "\n",
+ "#Moment acting on the entire section\n",
+ "#since sigmais the value at y=250\n",
+ "y_max=250\n",
+ "Z=I*10**8*y_max**-1\n",
+ "#M=sigma*Z \n",
+ "#After Simplifying Further we get\n",
+ "#M2=Z*sigma\n",
+ "\n",
+ "#Percentage Moment resisted by Flanges\n",
+ "P1=2258333.3*(2865833.3)**-1*100\n",
+ "\n",
+ "#Percentage Moment resisted by web\n",
+ "P2=100-P1\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage Moment resisted by Flanges\",round(P1,2),\"%\"\n",
+ "print\"Percentage Moment resisted by web\",round(P2,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage Moment resisted by Flanges 78.8 %\n",
+ "Percentage Moment resisted by web 21.2 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.6,Page no.137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges\n",
+ "b1=200 #mm #Width\n",
+ "t1=10 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=380 #mm #Depth \n",
+ "t2=8 #mm #Thickness\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "sigma=150 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area\n",
+ "A=b1*t1+d*t2+b1*t1 #mm**2\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*(b1*D**3-(b1-t2)*d**3)\n",
+ "\n",
+ "#Bending Moment\n",
+ "M=sigma*I*(D*2**-1)**-1\n",
+ "\n",
+ "#Square Section\n",
+ "\n",
+ "#Let 'a' be the side\n",
+ "a=A**0.5\n",
+ "\n",
+ "#Moment of Resistance of this section\n",
+ "M1=1*6**-1*a*a**2*sigma\n",
+ "\n",
+ "X=M*M1**-1\n",
+ "\n",
+ "#Rectangular section\n",
+ "#Let 'a' be the side and depth be 2*a\n",
+ "\n",
+ "a=(A*2**-1)**0.5\n",
+ "\n",
+ "#Moment of Rectangular secction\n",
+ "M2=1*6**-1*a*(2*a)**2*sigma\n",
+ "\n",
+ "X2=M*M2**-1\n",
+ "\n",
+ "#Circular section\n",
+ "#A=pi*d1**2*4**-1\n",
+ "\n",
+ "d1=(A*4*pi**-1)**0.5\n",
+ "\n",
+ "#Moment of circular section\n",
+ "M3=pi*32**-1*d1**3*sigma\n",
+ "\n",
+ "X3=M*M3**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of resistance of beam section\",round(M,2),\"mm\"\n",
+ "print\"Moment of resistance of square section\",round(X,2),\"mm\"\n",
+ "print\"Moment of resistance of rectangular section\",round(X2,2),\"mm\"\n",
+ "print\"Moment of resistance of circular section\",round(X3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of resistance of beam section 141536000.0 mm\n",
+ "Moment of resistance of square section 9.58 mm\n",
+ "Moment of resistance of rectangular section 6.78 mm\n",
+ "Moment of resistance of circular section 11.33 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.7,Page no.139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=12 #KN #Force at End of beam\n",
+ "L=2 #m #span\n",
+ "\n",
+ "#Square section \n",
+ "b=d=200 #mm #Width and depth of beam\n",
+ "\n",
+ "#Rectangular section\n",
+ "b1=150 #mm #Width\n",
+ "d1=300 #mm #Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max bending Moment\n",
+ "M=F*L*10**6 #N-mm\n",
+ "\n",
+ "#M=sigma*b*d**2\n",
+ "sigma=M*6*(b*d**2)**-1 #N/mm**2\n",
+ "\n",
+ "#Let W be the central concentrated Load in simply supported beam of span L1=3 m\n",
+ "#MAx Moment\n",
+ "#M1=W*L1*4**-1\n",
+ "#After Further simplifying we get\n",
+ "#M1=0.75*10**6 #N-mm\n",
+ "\n",
+ "#The section has a moment of resistance\n",
+ "M1=sigma*1*6**-1*b1*d1**2\n",
+ "\n",
+ "#Equating it to moment of resistance we get max load W\n",
+ "#0.75*10**6*W=M1\n",
+ "#After Further simplifying we get\n",
+ "W=M1*(0.75*10**6)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum Concentrated Load required to brek the beam\",round(W,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum Concentrated Load required to brek the beam 54.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.8,Page no.140"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3 #m #span\n",
+ "sigma_t=35 #N/mm**2 #Permissible stress in tension\n",
+ "sigma_c=90 #N/mm**2 #Permissible stress in compression\n",
+ "\n",
+ "#Flanges\n",
+ "t=30 #mm #Thickness\n",
+ "d=250 #mm #Depth\n",
+ "\n",
+ "#Web\n",
+ "t2=25 #mm #Thickness\n",
+ "b=600 #mm #Width\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let y_bar be the Distance of N.A from Extreme Fibres\n",
+ "y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2\n",
+ "\n",
+ "#Part-1\n",
+ "\n",
+ "#If web is in Tension\n",
+ "y_t=y_bar #mm\n",
+ "y_c=d-y_bar #mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of tensile stress\n",
+ "M=sigma_t*I*(y_bar)**-1 #N-mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of compressive stress\n",
+ "M1=sigma_c*I*(y_c)**-1 #N-mm\n",
+ "\n",
+ "#If w KN/m is u.d.l in beam,Max bending moment\n",
+ "#M=wl**2*8**-1\n",
+ "#After further simplifyng we get\n",
+ "#M=1.125*w*10**6 N-mm\n",
+ "w=M*(1.125*10**6)**-1 #KN\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#If web is in compression\n",
+ "y_t2=178.299 #mm\n",
+ "y_c2=71.71 #mm \n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of tensile stress\n",
+ "M2=sigma_t*I*(y_t2)**-1 #N-mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of compressive stress\n",
+ "M3=sigma_c*I*(y_c2)**-1 #N-mm\n",
+ "\n",
+ "#Moment of resistance is M2\n",
+ "\n",
+ "#Equating it to bending moment we get\n",
+ "#M2=1.125*10**6*w2\n",
+ "#After further simplifyng we get\n",
+ "w2=M2*(1.125*10**6)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Uniformly Distributed Load carrying capacity if:web is in Tension\",round(w,2),\"KN\"\n",
+ "print\" :web is in compression\",round(w2,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly Distributed Load carrying capacity if:web is in Tension 73.21 KN\n",
+ " :web is in compression 29.446 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.9,Page no.141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b1=200 #mm #Width at base\n",
+ "b2=100 #mm #Width at top\n",
+ "\n",
+ "L=8 #m Length\n",
+ "P=500 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider a section at y metres from top\n",
+ "\n",
+ "#At this section diameter d is\n",
+ "#d=b2+y*L**-1*(b1-b2)\n",
+ "#After Further simplifying we get\n",
+ "#d=b2+12.5*y #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "#I=pi*64**-1*d**4\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=pi*32**-1*(b1+12.5*y)**3\n",
+ "\n",
+ "#Moment \n",
+ "#M=5*10**5*y #N-mm\n",
+ "\n",
+ "#Let sigma be the fibre stress at this section then\n",
+ "#M=sigma*Z\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "#sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",
+ "\n",
+ "#For sigma to be Max,d(sigma)*(dy)**-1=0\n",
+ "#16*10**6*pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)\n",
+ "#After Further simplifying we get\n",
+ "#b2+12.5*y=37.5*y\n",
+ "#After Further simplifying we get\n",
+ "y=b2*25**-1 #m\n",
+ "\n",
+ "#Stress at this section\n",
+ "sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress at Extreme Fibre is max\",round(y,2),\"m\"\n",
+ "print\"Max stress is\",round(sigma,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress at Extreme Fibre is max 4.0 m\n",
+ "Max stress is 6.04 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.10,Page no.143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "H=10 #mm #Height\n",
+ "A1=160*160 #mm**2 #area of square section at bottom\n",
+ "L1=160 #mm #Length of square section at bottom\n",
+ "b1=160 #mm #width of square section at bottom\n",
+ "A2=80*80 #mm**2 #area of square section at top\n",
+ "L2=80 #mm #Length of square section at top\n",
+ "b2=80 #mm #Width of square section at top\n",
+ "P=100 #N #Pull\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider a section at distance y from top.\n",
+ "#Let the side of square bar be 'a'\n",
+ "#a=L2+y*(H)**-1*(b1-b2)\n",
+ "#After further simplifying we get\n",
+ "#a=L2+8*y\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "#I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3\n",
+ "#After further simplifying we get\n",
+ "#I=a**4*12**-1\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=a**4*(12*a*(2)**0.5)**-1\n",
+ "#After further simplifying we get\n",
+ "#Z=2**0.5*a**3*(12)**-1 #mm**3\n",
+ "\n",
+ "#Bending moment at this section=100*y N-mm\n",
+ "#M=100*10**3*y #N-mm\n",
+ "\n",
+ "#But\n",
+ "#M=sigma*Z\n",
+ "#After sub values in above equation we get\n",
+ "#sigma=M*Z**-1\n",
+ "#After further simplifying we get\n",
+ "#sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1)\n",
+ "\n",
+ "#For Max stress df*(dy)**-1=0\n",
+ "#After taking Derivative of above equation we get\n",
+ "#df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8)\n",
+ "#After further simplifying we get\n",
+ "y=80*16**-1 #m\n",
+ "\n",
+ "#Max stress at this level is\n",
+ "sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Bending stress is Developed at\",round(y,3),\"m\"\n",
+ "print\"Value of Max Bending stress is\",round(sigma,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Bending stress is Developed at 5.0 m\n",
+ "Value of Max Bending stress is 2.455 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.12,Page no.147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=200 #mm #Width of timber \n",
+ "d=400 #mm #Depth of timber\n",
+ "t=6 #mm #Thickness\n",
+ "b2=200 #mm #width of steel plate\n",
+ "t2=20 #mm #Thickness of steel plate\n",
+ "M=40*10**6 #KN-mm #Moment\n",
+ "#Let E_s*E_t**-1=X\n",
+ "X=20 #Ratio of Modulus of steel to timber\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#let y_bar be the Distance of centroidfrom bottom most fibre\n",
+ "y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*d**3+b*d*(b+t-round(y_bar,3))**2+1*12**-1*t2*b2*t**3+b2*t2*t*(round(y_bar,3)-t*2**-1)**2\n",
+ "\n",
+ "#distance of the top fibre from N-A\n",
+ "y_1=d+t-y_bar #mm\n",
+ "\n",
+ "#Distance of the junction of timber and steel From N-A\n",
+ "y_2=y_bar-t #mm\n",
+ "\n",
+ "#Stress in Timber at the top\n",
+ "Y=M*I**-1*y_1 #N/mm**2\n",
+ "\n",
+ "#Stress in the Timber at the junction point\n",
+ "Z=M*I**-1*y_2\n",
+ "\n",
+ "#Coressponding stress in steel at the junction point\n",
+ "Z2=X*Z #N/mm**2 \n",
+ "\n",
+ "#The stress in Extreme steel fibre \n",
+ "Z3=X*M*I**-1*y_bar\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in Extreme steel Fibre\",round(Z3,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in Extreme steel Fibre 69.67 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.13,Page no.149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Timber size\n",
+ "b=150 #mm #Width\n",
+ "b2=120 #mm \n",
+ "d=300 #mm #Depth\n",
+ "\n",
+ "t=6 #mm #Thickness of steel plate\n",
+ "L=6 #m #Span\n",
+ "\n",
+ "#E_s*E_t**-1=20 \n",
+ "#X=E_s*E_t**-1\n",
+ "X=20 \n",
+ "sigma_timber=8 #N/mm**2 #Stress in timber\n",
+ "sigma_steel=150 #N/mm**2 #Stress in steel plate\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Y\n",
+ "\n",
+ "#Due to synnetry cenroid,the neutral axis is half the depth\n",
+ "I=1*12**-1*\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print Z"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "153.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.14,Page no.151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6000 #mm #Span of beam\n",
+ "W=20*10**3 #N #Load\n",
+ "sigma=8 #N/mm**2 #Stress\n",
+ "b=200 #mm #Width of section\n",
+ "d=300 #mm #Depth of section\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#let x be the distance from left side of beam\n",
+ "\n",
+ "#Bending moment\n",
+ "#M=W*2**-1*x #Nmm .......(1)\n",
+ "\n",
+ "#But M=sigma*Z ..........(2)\n",
+ "\n",
+ "#Equating equation 1 and 2 we get\n",
+ "#W*2**-1*x=sigma*Z ............(3)\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=1*6*b*d**2 ...............(4)\n",
+ "\n",
+ "#Equating equation 3 and 4 we get\n",
+ "#b*d**2=3*W*x*sigma**-1 .............(5)\n",
+ "\n",
+ "#Beam of uniform strength of constant depth\n",
+ "#b=3*W*x*(sigma*d**2) \n",
+ "\n",
+ "#When x=0\n",
+ "b=0\n",
+ "\n",
+ "#When x=L*2**-1\n",
+ "b2=3*W*L*(2*sigma*d**2)**-1 #mm\n",
+ "\n",
+ "#Beam with constant width of 200 mm\n",
+ "\n",
+ "#We have\n",
+ "#d=(3*W*x*(sigma*d)**-1)**0.5\n",
+ "#thus depth varies as (x)**0.5\n",
+ "\n",
+ "#when x=0\n",
+ "d1=0\n",
+ "\n",
+ "#when x=L*2**-1\n",
+ "d2=(2*W*L*(2*sigma*300)**-1)**0.5 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Cross section of rectangular beam is:\",round(b2,2),\"mm\"\n",
+ "print\" :\",round(d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Cross section of rectangular beam is: 250.0 mm\n",
+ " : 223.61 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.15,Page no.154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=800 #mm #Span\n",
+ "n=5 #number of leaves\n",
+ "b=60 #mm #Width\n",
+ "t=10 #mm #thickness\n",
+ "sigma=250 #N/mm**2 #Stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#section Modulus\n",
+ "Z=n*6**-1*b*t**2 #mm**3\n",
+ "\n",
+ "#from the relation\n",
+ "#sigma*Z=M ...................(1)\n",
+ "#M=P*L*4**-1\n",
+ "#sub values of M in equation 1 we get\n",
+ "P=sigma*Z*4*L**-1*10**-3 #KN #Load\n",
+ "\n",
+ "#Length of Leaves\n",
+ "L1=0.2*L #mm\n",
+ "L2=0.4*L #mm\n",
+ "L3=0.6*L #mm\n",
+ "L4=0.8*L #mm\n",
+ "L5=L #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Load it can take is\",round(P,2),\"KN\"\n",
+ "print\"Length of leaves:L1\",round(L1,2),\"mm\"\n",
+ "print\" :L2\",round(L2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Load it can take is 6.25 KN\n",
+ "Length of leaves:L1 160.0 mm\n",
+ " :L2 320.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.16,Page no.161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=20*10**3 #N #Shear Force\n",
+ "\n",
+ "#Tee section\n",
+ "\n",
+ "#Flange\n",
+ "b=100 #mm #Width\n",
+ "t=12 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=88 #mm #Depth\n",
+ "t2=12 #mm #Thicknes\n",
+ "\n",
+ "D=100 #mm #Overall Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of C.G from Top Fibre\n",
+ "y=(b*t*t*2**-1+t2*d*(d*2**-1+t))*(b*t+d*t2)**-1 #mm \n",
+ "\n",
+ "#Moment Of Inertia\n",
+ "I=1*12**-1*b*t**3+b*t*(y-t*2**-1)**2+1*12**-1*t2*d**3+t2*d*(t+d*2**-1-y)**2 #mm**4\n",
+ "\n",
+ "#shear stress at bottom Flange\n",
+ "\n",
+ "#Area above this level\n",
+ "A=b*t #mm**2\n",
+ "\n",
+ "#C.G of this area from N-A\n",
+ "y2=y-t*2**-1\n",
+ "\n",
+ "#Stress at bottom of flange\n",
+ "sigma=F*A*y2*(b*I)**-1 #N/mm**2 \n",
+ "\n",
+ "#sigma2 at same level but in web where width is 12 mm\n",
+ "sigma2=F*A*y2*(t2*I)**-1 #N/mm**2 \n",
+ "\n",
+ "#To find shear stress at N-A\n",
+ "X=t*b*(y-t*2**-1)+t2*(y-t2)*(y-t2)*2**-1 #mm**3\n",
+ "\n",
+ "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Shear stress at top and bottom fibre is zero\n",
+ "#sigma4 and sigma5 are top and bottom fibre shear stress\n",
+ "sigma4=sigma5=0\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,t,t,y,D]\n",
+ "Y1=[sigma4,sigma,sigma2,sigma3,sigma5]\n",
+ "Z1=[0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x56a6270>"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.17,Page no.163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=40*10**3 #N #shear Force\n",
+ "\n",
+ "#I-section\n",
+ "\n",
+ "#Flanges\n",
+ "b=80 #mm #Width of flange\n",
+ "t=20 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=200 #mm #Depth\n",
+ "t2=20 #mm #Thickness\n",
+ "\n",
+ "#Flange-2\n",
+ "b2=160 #mm #Width\n",
+ "t3=20 #mm #Thickness\n",
+ "\n",
+ "D=240 #mm #Overall Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of N-A from Top Fibre \n",
+ "y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 #mm**4\n",
+ "\n",
+ "#Shear stress bottom of flange\n",
+ "sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 #N/mm**2\n",
+ "\n",
+ "#At same Level but in web\n",
+ "sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#for shear stress at N.A\n",
+ "X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 #mm**3\n",
+ "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Shear stress at bottom of web\n",
+ "\n",
+ "X=b2*t3*((D-y)-t3*2**-1) #mm**3\n",
+ "\n",
+ "#Stress at bottom of web\n",
+ "sigma4=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Stress at Lower flange\n",
+ "sigma5=F*X*(b2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force Diagram is the result\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,20,20,140,220,220,240]\n",
+ "Y1=[0,sigma,sigma2,sigma3,sigma4,sigma5,0]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length in mm\")\n",
+ "plt.ylabel(\"Shear Force in N\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force Diagram is the result\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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OnYL/+i/TF3DLLWZPgXbt7K5KRHyRwsAGJ0+afoBXXjHNQOvXQ1iY3VWJiC8r1gj183MN\nwGyHmZqa6tKivFVWlpko9uc/m72FP/3UDBlVEIiI3YoMg4SEBJ5//nnmzp0LQE5ODqNGjXJ5Yd4k\nMxNmzIDGjc0Cclu3wrvvQmio3ZWJiBhFhsGHH37ImjVrqPrH2sf169fn5MmTLi/MG/z8s1k5tEkT\nOHYMvv4ali6FZs3srkxE5FJFhkHlypWpcNF6B6dOnXJpQd4gIwMee8x86Gdlwc6dZrRQSIjdlYmI\nXF2RYTBs2DDuvfdesrKyWLx4MT179mT8+PHuqK3cOXLE7CvcsiWcO2f6Bf76V7j5ZrsrExEpXJGj\niaZMmcKGDRuoVq0a33//PbNmzSI6OtodtZUbP/1kJoqtXAnx8bB/PwQF2V2ViEjxFRkGqampdO3a\nld69ewNw5swZ0tLSCA4OdnVtHu/QIbNkxIcfwoQJcOAA1K1rd1UiIteuyGai2NhY/C5aKL9ChQrE\nxsa6tChPd+AAjBkDHTrATTfBwYMmFBQEIlJeFXllkJeXR6VKlZzfV65cmXPnzrm0KE+1b59ZQXTj\nRnjwQTNMtEYNu6sSESm9Iq8M6tSpw5o1a5zfr1mzhjp16ri0KE+zezfExpqVQ8PC4Mcfze5iCgIR\n8RZFXhm89tprjBw5ksmTJwPQoEEDli1b5vLCPMGOHWbG8PbtZqjo0qXwx3QLERGvUmgY5OXl8dpr\nr/H11187J5pVq1bNLYXZads2EwLffgtTp8KKFWaHMRERb1VoGPj5+bF161Ysy/KJEPjsMxMCBw/C\nE0+YUUJ/bPssIuLVimwmCg8PZ9CgQQwbNowqVaoAZqezoUOHlvigWVlZjB8/nn379uFwOFiyZAkd\nO3Ys8euVhmWZBeNmzjSTxqZNg9Gjwd/flnJERGxRZBicPXuWWrVq8emnn17yeGnC4MEHH6Rfv36s\nWrWK3Nxc25a42LkTJk+GEydMh3BcHFTUot4i4oMclmVZ7jzgr7/+SkRERKF7KDscDtxR1qhR8Kc/\nmaahi6ZSiIhcITISFi82Xz1VaT47ixxaevjwYYYMGULdunWpW7cuMTExHDlypEQHAzOjuW7duowd\nO5a2bdtyzz33cPr06RK/Xmm1bKkgEBEpslFk7NixjBw5kvfffx+A5cuXM3bsWDZu3FiiA+bm5rJz\n504WLlzILbfcwkMPPURiYiIzZ8685HkJCQnO+1FRUURFRZXoeCIi3iolJYWUlJQyea0im4nCwsLY\ns2dPkY8VV0ZGBp06dXLulrZ161YSExP5+OOPLxTlxmai2283X0VECuPzzUS1a9dm2bJl5OXlkZub\nyzvvvFOqGchBQUE0bNiQ77//HoBNmzYRqi2/RERsVWQz0ZIlS7j//vt55JFHAOjcubNzP+SSWrBg\nASNHjiQnJ4eQkJBSv56IiJROgWHw1Vdf0bFjR4KDg/noo4/K9KBhYWFs3769TF9TRERKrsBmookT\nJzrvd+rUyS3FiIiIPYrsMwAz8UxERLxXgc1EeXl5ZGZmYlmW8/7FatWq5fLiRETEPQoMg99++43I\nP8ZQWZblvA9m+FJhM4hFRKR8KTAM0tLS3FiGiIjYqVh9BiIi4t0UBiIiojAQEZEiwiA3N5dmzZq5\nqxYREbFJoWFQsWJFmjdvzk8//eSuekRExAZFrk2UmZlJaGgo7du3p2rVqoAZWrp27VqXFyciIu5R\nZBjMmjXLHXWIiIiNigwDbSojIuL9ihxNtG3bNm655RYCAgLw9/enQoUKVK9e3R21iYiImxQZBpMn\nT+bdd9+lSZMmnD17ljfeeINJkya5ozYREXGTYs0zaNKkCXl5efj5+TF27FiSk5NdXZeIiLhRkX0G\nVatW5ffffycsLIypU6cSFBTklv2JRUTEfYq8Mnj77bfJz89n4cKFVKlShSNHjpCUlOSO2kRExE2K\nvDIIDg7m9OnTZGRkkJCQ4IaSRETE3Yq8Mli7di0RERH06dMHgF27djFw4ECXFyYiIu5TZBgkJCTw\n9ddfU7NmTQAiIiK0sY2IiJcpMgz8/f2pUaPGpT9UQYudioh4kyI/1UNDQ1m+fDm5ubkcPHiQ+++/\nn86dO7ujNhERcZMiw2DBggXs27ePypUrExcXR/Xq1Zk3b547ahMRETcp1jyDOXPmMGfOHHfUIyIi\nNigyDA4cOMCLL75IWloaubm5gFnC+tNPP3V5cSIi4h5FhsGwYcOYOHEi48ePx8/PDzBhICIi3qPI\nMPD392fixInuqEVERGxSYAdyZmYmJ06cYMCAASxatIj09HQyMzOdt9LKy8sjIiKCAQMGlPq1RESk\ndAq8Mmjbtu0lzUEvvvii877D4Sj1xLP58+fTsmVLTp48WarXERGR0iswDNLS0lx20CNHjrBu3Tqm\nTZvGyy+/7LLjiIhI8RTYTLR9+3bS09Od3y9dupSBAwfywAMPlLqZ6OGHH+aFF17QTGYREQ9R4JXB\nhAkT2Lx5MwCfffYZTzzxBAsXLmTXrl1MmDCBVatWleiAH3/8MTfeeCMRERGkpKQU+LyLV0iNiorS\nXswiIpdJSUkp9HP0WjisAnaqCQsLY8+ePQDcd9991K1b1/kBffG/XaunnnqKZcuWUbFiRc6ePctv\nv/1GTEwMb7/99oWiHA63bKAzahTcfrv5KiJSmMhIWLzYfPVUpfnsLLCdJi8vj3PnzgGwadMmunfv\n7vy385PPSmLOnDkcPnyY1NRUVqxYQY8ePS4JAhERcb8Cm4ni4uLo1q0bderUoUqVKnTt2hWAgwcP\nXrGKaWloApuIiP0KDINp06bRo0cPMjIy6N27t7Oz17IsFixYUCYH79atG926dSuT1xIRkZIrdAZy\np06drnisadOmLitGRETsobGdIiKiMBAREYWBiIigMBARERQGIiKCwkBERFAYiIgICgMREUFhICIi\nKAxERASFgYiIoDAQEREUBiIigsJARERQGIiICAoDERFBYSAiIigMREQEhYGIiKAwEBERFAYiIoLC\nQEREUBiIiAgKAxERQWEgIiIoDEREBIWBiIhgQxgcPnyY7t27ExoaSqtWrXj11VfdXYKIiFymorsP\n6O/vzyuvvEJ4eDjZ2dlERkYSHR1NixYt3F2KiIj8we1XBkFBQYSHhwMQEBBAixYtOHbsmLvLEBGR\ni9jaZ5CWlsauXbvo0KGDnWWIiPg828IgOzub2NhY5s+fT0BAgF1liIgINvQZAJw7d46YmBhGjRrF\n4MGDr/qchIQE5/2oqCiioqLcU5yISDmRkpJCSkpKmbyWw7Isq0xeqZgsy2LMmDHUrl2bV1555epF\nORy4o6xRo+D2281XEZHCREbC4sXmq6cqzWen25uJvvjiC9555x3+8Y9/EBERQUREBMnJye4uQ0RE\nLuL2ZqJbb72V/Px8dx9WREQKoRnIIiKiMBAREYWBiIigMBAREXw4DDIy4IsvoF49uysREbGfT4ZB\nZiZER8O4cdCzp93ViIjYz+fC4ORJ6NvXTDabNs3uakREPINPhcGZMzBwIISHw/PPg8Nhd0UiIp7B\nZ8Lg3DkYPhyCguA//1NBICJyMZ8Ig7w8uOsuc//tt8HPz956REQ8jS2rlrqTZcHEiXD8OHzyCfj7\n212RiIjn8eowsCyYMgX27oWNG+H66+2uSETEM3l1GDz7LGzYACkpUK2a3dWISHmXm2t3Ba7jtX0G\n8+eb/oENG6BWLburEZHyrl8/GDkSduywuxLX8MowePNNePll2LTJjB4SESmtWbNg7lwTCi+/DN62\nEr/bdzorjtLs1rNqFTzwAPzjH9CsWRkXJiI+LzUV7rwT6tSBt96CunXtruiCcrXTmSslJ8N998G6\ndQoCEXGNRo1g61Zo1QoiIkyfpDfwmiuDzz+HoUNhzRro3NlFhYmIXOTvf4e774YJE+Dpp6GizUNy\nSnNl4BVh8D//Y9Ybevdd6NXLhYWJiFwmPR1Gj4acHFi+HBo2tK8Wn24m2r8f+veHxYsVBCLifvXq\nmVGLfftCu3awdq3dFZVMub4y+PFH6NbN9PCPGuWGwkRECvHllzBiBAwaZBbDrFzZvcf3ySuDo0fN\nngRPPqkgEBHP0Lkz7NoFhw9Dp07w/fd2V1R85TIMfvnFBME998CkSXZXIyJyQc2akJQE48dDly6w\nbJndFRVPuWsm+vVXsztZ794wZ46bCxMRuQZ79sB//Ad06ACLFkFAgGuP5zPNRKdPw4AB0LEjzJ5t\ndzUiIoULCzOjHf38IDISdu+2u6KClZswyMmBmBgIDoZXX9XmNCJSPlStCkuWwIwZpnl74UKzorKn\nKRfNRLm5EBdnvn7wgf0TO0RESuKHH0yzUcOGJiDKehFNr24mys83s/uysmDFCgWBiJRfjRub4ad/\n/rNZymLrVrsrusCWMEhOTqZ58+Y0adKE5557rsDnWRY88ggcOACrV7t/zK6ISFmrXNmserpoEcTG\nmn1X8vLsrsqGMMjLy2Py5MkkJyezf/9+3nvvPf75z39e9bkJCbBli9musmpV99bpKVK8ZRWsMqBz\ncYHOxQXl9Vz07286lzdtMn0Jx47ZW4/bw+Cbb76hcePGBAcH4+/vz5133smaNWuueN5LL8HKlWYh\nqBo13F2l5yivb3RX0Lm4QOfigvJ8LurXh82bzUoKbdvC+vX21eL2MDh69CgNL1rJqUGDBhw9evSK\n5y1YYPYtvvFGd1YnIuJefn7wzDPmj98JE+Cxx8zoSXdzexg4ijkmdONGe1f/ExFxp27dzFIWBw7A\nrbeaeVVuZbnZtm3brD59+ji/nzNnjpWYmHjJc0JCQixAN9100023a7iFhISU+LPZ7fMMcnNzadas\nGZs3b+amm26iffv2vPfee7Ro0cKdZYiIyEXcPmq/YsWKLFy4kD59+pCXl8e4ceMUBCIiNvPIGcgi\nIuJeHjcDubgT0rxRcHAwbdq0ISIigvbt2wOQmZlJdHQ0TZs2pXfv3mRlZdlcpWvEx8cTGBhI69at\nnY8V9rvPnTuXJk2a0Lx5czZs2GBHyS5ztXORkJBAgwYNiIiIICIigvUXjUH05nNx+PBhunfvTmho\nKK1ateLVV18FfPO9UdC5KLP3Rol7G1wgNzfXCgkJsVJTU62cnBwrLCzM2r9/v91luU1wcLB14sSJ\nSx6bMmWK9dxzz1mWZVmJiYnW448/bkdpLvfZZ59ZO3futFq1auV8rKDffd++fVZYWJiVk5Njpaam\nWiEhIVZeXp4tdbvC1c5FQkKC9dJLL13xXG8/F+np6dauXbssy7KskydPWk2bNrX279/vk++Ngs5F\nWb03POrKoLgT0ryZdVmr3dq1axkzZgwAY8aMYfXq1XaU5XJdu3alZs2alzxW0O++Zs0a4uLi8Pf3\nJzg4mMaNG/PNN9+4vWZXudq5gCvfG+D95yIoKIjw8HAAAgICaNGiBUePHvXJ90ZB5wLK5r3hUWFQ\n3Alp3srhcNCrVy/atWvH66+/DsDx48cJDAwEIDAwkOPHj9tZolsV9LsfO3aMBg0aOJ/nK++TBQsW\nEBYWxrhx45zNIr50LtLS0ti1axcdOnTw+ffG+XPRsWNHoGzeGx4VBsWdkOatvvjiC3bt2sX69etZ\ntGgRn3/++SX/7nA4fPYcFfW7e/t5mThxIqmpqezevZt69erx6KOPFvhcbzwX2dnZxMTEMH/+fKpV\nq3bJv/naeyM7O5vY2Fjmz59PQEBAmb03PCoM6tevz+HDh53fHz58+JJk83b16tUDoG7dugwZMoRv\nvvmGwMBAMjIyAEhPT+dGH1qfo6Df/fL3yZEjR6hfv74tNbrLjTfe6PzQGz9+vPNy3xfOxblz54iJ\niWH06NEMHjwY8N33xvlzMWrUKOe5KKv3hkeFQbt27Th48CBpaWnk5OSwcuVKBg4caHdZbnH69GlO\nnjwJwKlTp9iwYQOtW7dm4MCBLF26FIClS5c63wC+oKDffeDAgaxYsYKcnBxSU1M5ePCgc/SVt0pP\nT3fe//DDD50jjbz9XFiWxbhx42jZsiUPPfSQ83FffG8UdC7K7L3hil7v0li3bp3VtGlTKyQkxJoz\nZ47d5bjNjz/+aIWFhVlhYWFWaGio83c/ceKE1bNnT6tJkyZWdHS09X//9382V+oad955p1WvXj3L\n39/fatBlqDKuAAAD90lEQVSggbVkyZJCf/fZs2dbISEhVrNmzazk5GQbKy97l5+LN954wxo9erTV\nunVrq02bNtagQYOsjIwM5/O9+Vx8/vnnlsPhsMLCwqzw8HArPDzcWr9+vU++N652LtatW1dm7w1N\nOhMREc9qJhIREXsoDERERGEgIiIKAxERQWEgIiIoDEREBIWBlCMBAQEuff158+Zx5syZazreRx99\n5HNLrYt30jwDKTeqVavmnKXtCo0aNWLHjh3Url3bLccT8SS6MpBy7dChQ/Tt25d27dpx2223ceDA\nAQDuvvtuHnzwQbp06UJISAhJSUkA5OfnM2nSJFq0aEHv3r254447SEpKYsGCBRw7dozu3bvTs2dP\n5+tPnz6d8PBwOnXqxP/+7/9ecfy33nqL+++/v9BjXiwtLY3mzZszduxYmjVrxsiRI9mwYQNdunSh\nadOmbN++HTAblowZM4bbbruN4OBg/va3v/HYY4/Rpk0b+vbtS25ubpmfS/Fxrpw+LVKWAgICrnis\nR48e1sGDBy3LsqyvvvrK6tGjh2VZljVmzBhr+PDhlmVZ1v79+63GjRtblmVZH3zwgdWvXz/Lsiwr\nIyPDqlmzppWUlGRZ1pWbCzkcDuvjjz+2LMuypk6daj377LNXHP+tt96yJk+eXOgxL5aammpVrFjR\n+u6776z8/HwrMjLSio+PtyzLstasWWMNHjzYsizLeuaZZ6yuXbtaubm51p49e6zrr7/euZzAkCFD\nrNWrVxf/xIkUQ0W7w0ikpLKzs9m2bRvDhg1zPpaTkwOYpXrPL17WokUL53r3W7duZfjw4YBZ+bJ7\n9+4Fvn6lSpW44447AIiMjGTjxo2F1lPQMS/XqFEjQkNDAQgNDaVXr14AtGrVirS0NOdr9e3bFz8/\nP1q1akV+fj59+vQBoHXr1s7niZQVhYGUW/n5+dSoUYNdu3Zd9d8rVarkvG/90TXmcDgu2RXKKqTL\nzN/f33m/QoUKxWqaudoxL1e5cuVLXvf8z1x+jIsfL0ktItdCfQZSblWvXp1GjRqxatUqwHz47t27\nt9Cf6dKlC0lJSViWxfHjx9myZYvz36pVq8Zvv/12TTUUFial4arXFSmIwkDKjdOnT9OwYUPnbd68\neSxfvpw33niD8PBwWrVqxdq1a53Pv3hXp/P3Y2JiaNCgAS1btmT06NG0bduWG264AYAJEyZw++23\nOzuQL//5q+0SdfnjBd2//GcK+v78/cJet7DXFikpDS0Vn3Pq1CmqVq3KiRMn6NChA19++aVP7SAn\ncjXqMxCf079/f7KyssjJyWHGjBkKAhF0ZSAiIqjPQEREUBiIiAgKAxERQWEgIiIoDEREBIWBiIgA\n/w/qZz1xEBCKMwAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5614110>"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.18,Page no.164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=30*10**3 #N #Shear Force\n",
+ "\n",
+ "#Channel Section\n",
+ "d=400 #mm #Depth of web \n",
+ "t=10 #mm #THickness of web\n",
+ "t2=15 #mm #Thickness of flange\n",
+ "b=100 #mm #Width of flange\n",
+ "\n",
+ "#Rectangular Welded section\n",
+ "b2=80 #mm #Width\n",
+ "d2=60 #mm #Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of Centroid From Top Fibre\n",
+ "y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 #mm\n",
+ "\n",
+ "#Moment Of Inertia of the section about N-A\n",
+ "I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2\n",
+ "\n",
+ "#Shear stress at level of weld\n",
+ "sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Max Shear Stress occurs at Neutral Axis\n",
+ "X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1\n",
+ "\n",
+ "sigma_max=F*X*((b+t)*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Shear stress in the weld is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\"Max shear stress is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear stress in the weld is 3.62 N/mm**2\n",
+ "Max shear stress is 4.48 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.19,Page no.165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Wooden Section\n",
+ "b=300 #mm #Width\n",
+ "d=300 #mm #Depth\n",
+ "\n",
+ "D=100 #mm #Diameter of Bore\n",
+ "F=10*10**3 #N #Shear Force\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment Of Inertia Of Section\n",
+ "I=1*12**-1*b*d**3-pi*64**-1*D**4\n",
+ "\n",
+ "#Shear stress at crown of circle\n",
+ "sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1\n",
+ "\n",
+ "#Let a*y_bar=X\n",
+ "X=b*d*2**-1*d*4**-1-pi*8**-1*D**2*4*D*2**-1*(3*pi)**-1 #mm**3\n",
+ "\n",
+ "#Shear Stress at Neutral Axis\n",
+ "sigma2=F*X*((b-D)*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shearing Stress at Crown of Bore\",round(sigma,3),\"N/mm**2\"\n",
+ "print\"Shear Stress at Neutral Axis\",round(sigma2,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shearing Stress at Crown of Bore 0.149 N/mm**2\n",
+ "Shear Stress at Neutral Axis 0.246 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.20,Page no.166"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#flanges\n",
+ "b=200 #mm #width\n",
+ "t1=25 #mm #Thickness\n",
+ "\n",
+ "#web\n",
+ "d=450 #mm #Depth \n",
+ "t2=20 #mm #thickness\n",
+ "\n",
+ "D=500 #mm #Total Depth of section\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment Of Inertia of the section about N-A\n",
+ "I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 #mm**4\n",
+ "\n",
+ "#Consider an element in the web at distance y from y from N-A\n",
+ "#Depth of web section=225-y\n",
+ "\n",
+ "#C.G From N-A\n",
+ "#y2=y+(((D*2**-1-t)-y)*2**-1)\n",
+ "\n",
+ "#ay_bar for section at y\n",
+ "#Let ay_bar be X\n",
+ "#X=X1 be of Flange + X2 be of web above y\n",
+ "#X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1\n",
+ "#After Sub values and Further simplifying we get\n",
+ "#X=1187500+10*(225**2-y**2)\n",
+ "\n",
+ "#Shear stress at y\n",
+ "#sigma_y=F*(X)*(t2*I)**-1\n",
+ "\n",
+ "#Shear Force resisted by the Element\n",
+ "#F1=F*X*t2*dy*(t2*I)**-1\n",
+ "\n",
+ "#Shear stress resisted by web \n",
+ "#sigma=2*F*I**-1*(X)*dy\n",
+ "\n",
+ "#After Integrating above equation and further simplifying we get\n",
+ "#sigma=0.9578*F\n",
+ "\n",
+ "sigma=0.9578*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Shear Resisted by web\",round(sigma,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear Resisted by web 95.78 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.21,Page no.167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Wooden Beam\n",
+ "\n",
+ "b=150 #mm #width\n",
+ "d=250 #mm #Depth\n",
+ "\n",
+ "L=5000 #mm #span\n",
+ "m=11.2 #N/mm**2 #Max Bending stress\n",
+ "sigma=0.7 #N/mm**2 #Max shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let 'a' be the distance from left support\n",
+ "#Max shear force\n",
+ "#F=R_A=W*(L-a)*L**-1 \n",
+ "\n",
+ "#Max Moment\n",
+ "#M=W*(L-a)*a*L**-1\n",
+ "\n",
+ "#But M=sigma*Z\n",
+ "#W*(L-a)*a*L**-1=m*1*6**-1*b*d**2 .....................(1)\n",
+ "\n",
+ "#In Rectangular Section MAx stress is 1.5 times Avg shear stress\n",
+ "F=sigma*b*d*1.5**-1\n",
+ "\n",
+ "#W*(L-a)*L**-1=F .....................(2)\n",
+ "\n",
+ "#Dividing Equation 1 nad 2 we get\n",
+ "a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1\n",
+ "\n",
+ "#Sub above value in equation 2 we get\n",
+ "W=(L-a)**-1*L*F*10**-3 #KN \n",
+ "\n",
+ "#Result\n",
+ "print\"Load is\",round(W,2),\"KN\"\n",
+ "print\"Distance from Left support is\",round(a,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Load is 21.87 KN\n",
+ "Distance from Left support is 1000.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.22,Page no.168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #span\n",
+ "\n",
+ "#Rectangular Section\n",
+ "\n",
+ "b=200 #mm #width\n",
+ "d=400 #mm #depth\n",
+ "\n",
+ "sigma=1.5 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let AB be the cantilever beam subjected to load W KN at free end\n",
+ "\n",
+ "#MAx shear Force\n",
+ "#F=W*10**3 #KN\n",
+ "\n",
+ "#Since Max shear stress in Rectangular section\n",
+ "#sigma_max=1.5*F*A**-1 \n",
+ "#After sub values and further simplifyng we get\n",
+ "W=1.5*b*d*(1.5*1000)**-1 #KN\n",
+ "\n",
+ "#Moment at fixwed end\n",
+ "M=W*1 #KN-m\n",
+ "y_max=d*2**-1 #mm\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*d**3 #mm**3\n",
+ "\n",
+ "#MAx Stress\n",
+ "sigma_max=M*10**6*I**-1*y_max\n",
+ "\n",
+ "#Result\n",
+ "print\"Concentrated Load is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentrated Load is 15.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.24,Page no.170"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=4000 #mm #span\n",
+ "\n",
+ "#Rectangular Cross-section\n",
+ "b=100 #mm #Width\n",
+ "d=200 #mm #Thickness\n",
+ "\n",
+ "F_per=10 #N/mm**2 #Max Bending stress\n",
+ "q_max=0.6 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#If the Load W is in KN/m\n",
+ "\n",
+ "#Max shear Force\n",
+ "#F=w*l*2**-1 #KN\n",
+ "#After substituting values and further simplifying we get\n",
+ "#M=2*w #KN-m\n",
+ "\n",
+ "#Max Load from Consideration of moment\n",
+ "#M=1*6**-1*b*d**2*F_per\n",
+ "#After substituting values and further simplifying we get\n",
+ "w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 #KN/m\n",
+ "\n",
+ "#Max Load from Consideration of shear stress\n",
+ "#q_max=1.5*F*(b*d)**-1 #N\n",
+ "#After substituting values and further simplifying we get\n",
+ "F=q_max*(1.5)*b*d #N\n",
+ "\n",
+ "#If w is Max Load in KN/m,then\n",
+ "#2*w*1000=8000\n",
+ "#After Rearranging and Further simplifying we get\n",
+ "w2=8000*(2*1000)**-1 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Uniformly Distributed Load Beam can carry is\",round(w,2),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly Distributed Load Beam can carry is 3.33 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_4.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_4.ipynb new file mode 100644 index 00000000..8a54f301 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_4.ipynb @@ -0,0 +1,1013 @@ +{
+ "metadata": {
+ "name": "chapter no.5.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Deflections Of Beams By Double Integration Methods"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.2,Page No.192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #span of beam\n",
+ "a=2000 #mm\n",
+ "W1=20*10**3 #N #Pt Load Acting on beam\n",
+ "W2=30*10**3 #N #Pt Load Acting on beam\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "I=2*10**8 #mm**4 #M.I\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Deflection at free End Due to W2\n",
+ "dell1=W2*L**3*(3*E*I)**-1 #mm\n",
+ "\n",
+ "#Deflection at free end Due to W1\n",
+ "dell2=W1*a**3*(3*E*I)**-1+(L-a)*W1*a**2*(2*E*I)**-1 #mm\n",
+ "\n",
+ "#Total Deflection at free end\n",
+ "dell=dell1+dell2 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection at Free End is\",round(dell,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at Free End is 9.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.4,Page No.193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "I=180*10**6 #mm**4 #M.I\n",
+ "W1=20 #N/m #u.d.l\n",
+ "W2=20*10**3 #N #Pt load\n",
+ "L=3000 #m #Span of beam\n",
+ "a=2000 #m #Span of u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Displacement of free End due to 20 KN Pt load at free end\n",
+ "dell1=W2*L**3*(3*E*I)**-1 #mm\n",
+ "\n",
+ "#Displacement of free end due to u.d.l\n",
+ "dell2=W1*a**4*(8*E*I)**-1+(L-a)*W1*a**3*(6*E*I)**-1\n",
+ "\n",
+ "#Deflection at free end\n",
+ "dell=dell1+dell2 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"The Displacement of Free End of cantilever beam is\",round(dell,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Displacement of Free End of cantilever beam is 6.85 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.10,Page No.201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200*10**6 #KN/m**2 #Young's Modulus\n",
+ "I=15*10**-6 #m**4 #M.I\n",
+ "a=4000 #m \n",
+ "L_AB=6 #m #Span of beam\n",
+ "L_CB=2 #m #Length of CB\n",
+ "F_C=18 #KN #force at C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the Reactions at A & B Respectively\n",
+ "#V_A+V_B=18\n",
+ "#Now taking moment at B,we get M_B\n",
+ "V_A=(F_C*L_CB)*L_AB**-1\n",
+ "V_B=18-V_A\n",
+ "\n",
+ "#Now Taking Moment at distance x\n",
+ "#M_x=6*x-18*(x-4)\n",
+ "#EI*d**2*y*(d*x**2)**-1=6*x-18*(x-4)\n",
+ "\n",
+ "#Now Integrating above equation,we get\n",
+ "#EI*dy*(dx)**-1=C1+3*x**2-9(x-4)**4\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+x**3-3*(x-4)**3\n",
+ "\n",
+ "#The Boundary conditions\n",
+ "x=0\n",
+ "y=0 #.....(a)\n",
+ "\n",
+ "x=6\n",
+ "y=0 #....(b)\n",
+ "\n",
+ "#From Boundary Condition(B.C) a we get\n",
+ "C2=0\n",
+ "\n",
+ "#From Boundary Condition(B.C) b we get\n",
+ "#6*C1+216-3*8\n",
+ "#After Further simplifying we get\n",
+ "C1=-(216-24)*6**-1\n",
+ "\n",
+ "#EI*y=-32*x+x**3-3*(x-4)**3\n",
+ "#EI*dy*(dx)**-1=-32+3*x**2-9(x-4)**4\n",
+ "\n",
+ "#For Max Deflection\n",
+ "#Assume it inthe Porion AC i.e x=4=a\n",
+ "#0=-32+3*x**2\n",
+ "x=(32*3**-1)**0.5\n",
+ "\n",
+ "#Value of Max deflection is\n",
+ "ymax=(-32*x+x**3)*(E*I)**-1 #mm\n",
+ "\n",
+ "#slope at mid-span\n",
+ "\n",
+ "#EI*(dy*(dx)**-1)_centre=-32+3*x**2\n",
+ "#at centre ,\n",
+ "x1=3 #m\n",
+ "\n",
+ "#Let (dy*(dx)**-1)_centre=X\n",
+ "X=-(-32+3*x1**2)*(E*I)**-1 #Radian\n",
+ "\n",
+ "#Deflection at Load Point\n",
+ "x2=4 #m\n",
+ "#EI*y_c=-32*x2+x2**3\n",
+ "\n",
+ "y_c=-(-32*x2+x2**3)*(E*I)**-1\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of Max Deflection\",round(ymax,4),\"mm\"\n",
+ "print\"SLope at mid-span\",round(X,4),\"radian\"\n",
+ "print\"Deflection at the Load Point is\",round(y_c,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of Max Deflection -0.0232 mm\n",
+ "SLope at mid-span 0.0017 radian\n",
+ "Deflection at the Load Point is 0.0213 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.11,Page No.203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_CB=2 #m #Length of CB\n",
+ "L_AC=4 #m #Length of AB\n",
+ "M_C=15 #KN.m #Moment At Pt C\n",
+ "F_C=30 #KN\n",
+ "L=6 #m Span of Beam\n",
+ "\n",
+ "#Let X=E*I\n",
+ "X=10000 #KN-m**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A and V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=30\n",
+ "\n",
+ "#Taking Moment a A,we get\n",
+ "V_B=(F_C*L_AC+M_C)*L**-1\n",
+ "V_A=30-V_B\n",
+ "\n",
+ "#Now Taking Moment at distacnce x from A\n",
+ "#M_x=7.5*x-30*(x-4)+15\n",
+ "\n",
+ "#By using Macaulay's Method\n",
+ "#EI*(d**2*x/dx**2)=M_x=7.5*x-30*(x-4)+15\n",
+ "\n",
+ "#Now Integrating above Equation we get\n",
+ "#EI*(dy/dx)=C1+7.5*x**2*2**-1-15*(x-4)**2+15*(x-4) ............(1)\n",
+ "\n",
+ "#Again Integrating above Equation we get\n",
+ "#EIy=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1..........(2)\n",
+ "\n",
+ "#Boundary Cinditions\n",
+ "x=0\n",
+ "y=0\n",
+ "\n",
+ "#Substituting above equations we get \n",
+ "C2=0\n",
+ "\n",
+ "x=6 #m\n",
+ "y=0\n",
+ "\n",
+ "C1=-(7.5*6**3*6**-1-5*2**3+15*2**2*2**-1)*6**-1\n",
+ "\n",
+ "#EIy_c=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1\n",
+ "#Sub values in Above equation we get\n",
+ "y_c=(93.3333*(X)**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Deflection at C\",round(y_c,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Deflection at C 0.0093 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.12,Page No.204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=L_CD=L_DB=2 #m #Length of AC,CD,DB\n",
+ "F_C=40 #KN #Force at C\n",
+ "w=20 #KN/m #u.d.l\n",
+ "L=6 #m #span of beam\n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=15000 #KN-m**2\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=80\n",
+ "\n",
+ "#Taking Moment B,M_B\n",
+ "V_A=(F_C*(L_CD+L_DB)+w*L_DB*L_DB*2**-1)*L**-1 #KN\n",
+ "V_B=80-V_A #KN\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=33.333*x-40*(x-2)-20*(x-4)**2*2**-1\n",
+ "#EI*(d**2/dx**2)=33.333*x-40*(x-2)-10*(x-4)**2\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+33.333*x**2*2**-1-20*(x-2)**2-10*3**-1*(x-4)**3\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At\n",
+ "x=6\n",
+ "y=0\n",
+ "C1=-760*6**-1\n",
+ "\n",
+ "#Assuming Deflection to be max in portion CD and sustituting value of C1 in equation of slope we get\n",
+ "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",
+ "#0=-126.667+33.333*x**2**-1-20*(x-2)**2\n",
+ "\n",
+ "#After rearranging and simplifying further we get\n",
+ "\n",
+ "#x**2-24*x+62=0\n",
+ "#From above equations\n",
+ "a=1\n",
+ "b=-24\n",
+ "c=62\n",
+ "\n",
+ "y=(b**2-4*a*c)**0.5\n",
+ "\n",
+ "x1=(-b+y)*(2*a)**-1\n",
+ "x2=(-b-y)*(2*a)**-1\n",
+ "\n",
+ "#Taking x2 into account\n",
+ "x=2.945 #m\n",
+ "C1=-126.667\n",
+ "C2=0\n",
+ "\n",
+ "y_max=(C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3)*X**-1 #mm\n",
+ "\n",
+ "#Max slope occurs at the ends\n",
+ "#At A,\n",
+ "#EI*(dy/dx)_A=-126.667\n",
+ "#At B\n",
+ "#EI*(dy/dx)_B=126.667+33.333*6**2*2**-1-20*4**2-10*2**3\n",
+ "#After simplifying Further we get\n",
+ "#EI*(dy/dx)_B=73.3273\n",
+ "\n",
+ "#Now Max slope is EI(dy/dx)_A=-126.667\n",
+ "#15000*(dy/dx)_=-126.667\n",
+ "\n",
+ "#Let Y=dy/dx\n",
+ "Y=-126.667*X**-1 #Radians\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum Deflection for Beam is\",round(y_max,4),\"mm\"\n",
+ "print\"Maximum Slope for beam is\",round(Y,4),\"radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum Deflection for Beam is -0.0158 mm\n",
+ "Maximum Slope for beam is -0.0084 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.13,Page No.206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2*10**8 #KN/m**2\n",
+ "I=450*10**-6 #m**4\n",
+ "L_AC=1 #m #Length of AC\n",
+ "L_CD=3 #m #Length of CD\n",
+ "L_DB=2 #m #Length of DB\n",
+ "w=10 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=30\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",
+ "#EI*(d**2/dx**2)=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",
+ "\n",
+ "#Now Integrating Above equation we get\n",
+ "#EI(dy/dx)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**2+5*3**-1*(x-4)**3\n",
+ "\n",
+ "#Again Integrating Above equation we get\n",
+ "#EI*y=C2+C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4+5*12**-1*(x-4)**4\n",
+ "\n",
+ "#At \n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At \n",
+ "x=6 \n",
+ "y=0\n",
+ "C1=(-17.5*x**3*6**-1+5*12**-1*(x-1)**4-5*12**-1*(x-4)**4)*x**-1\n",
+ "\n",
+ "# 1)Slope at A .i.e at x=0\n",
+ "#EI*(dy/dx)_A=C1=-62.708 #KN-m**2\n",
+ "#let (dy/dx)=X\n",
+ "X=C1*(E*I)**-1 #radiams\n",
+ "\n",
+ "#Deflection at mid-span\n",
+ "x=3 #m\n",
+ "#EI*y_centre=C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**2\n",
+ "y_centre=-(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",
+ "\n",
+ "#Maximum Deflection\n",
+ "\n",
+ "#At point of Max deflection (dy/dx)=0\n",
+ "#Assuming it in portion CD\n",
+ "\n",
+ "#0=C1*x+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",
+ "\n",
+ "#Now Let\n",
+ "#F(x)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",
+ "\n",
+ "#Let F(x)=Y\n",
+ "#At \n",
+ "x=2.5\n",
+ "Y1=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#AT\n",
+ "x=3\n",
+ "Y2=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#At\n",
+ "x=2.9 #m\n",
+ "Y3=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#A curve may be plotted for (F(x) and the value for which F(x)=0 may be found\n",
+ "#For F(x)=0 for x=2.92 m\n",
+ "#Therefore y_max occur at x=2.92\n",
+ "\n",
+ "x=2.92 #m\n",
+ "y_max=(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Slope at A\",round(X,6),\"mm\"\n",
+ "print\"Deflection at mid-span\",round(y_centre,6),\"mm\"\n",
+ "print\"Maxmimum Deflection is\",round(y_max,5),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at A -0.000697 mm\n",
+ "Deflection at mid-span 0.001289 mm\n",
+ "Maxmimum Deflection is -0.00129 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.14,Page No.208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=LDE=L_EB=1 #m #Length of AC\n",
+ "L_CD=2 #m #Length of CD\n",
+ "E=200 #KN/mm**2\n",
+ "I=60*10**6 #mm**4 #M.I\n",
+ "F_C=20 #KN #Force at C\n",
+ "F_E=30 #KN #Force at E\n",
+ "w=10 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "X=E*I*10**-6 #KN-m**2\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=70\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",
+ "#EI*(d**2y/dx**2)=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",
+ "\n",
+ "#Now Integrating Above equation,we get\n",
+ "#EI*(dy/dx)=C1+17*x**2-10*(x-1)**2-5*3**-1*(x-1)**3+5*3**-1*(x-3)**3-15*(x-4)**2\n",
+ "\n",
+ "#Again Integrating Above equation,we get\n",
+ "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At \n",
+ "x=5 #m\n",
+ "y=0\n",
+ "C1=(-17*3**-1*x**3+10*3**-1*(x-1)**3+5*12**-1*(x-1)**4-5*12**-1*(x-3)**4+5*(x-4)**3)*5**-1\n",
+ "\n",
+ "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",
+ "C2=0\n",
+ "C1=-78\n",
+ "x=1\n",
+ "y_c=(-78*x+17*3**-1*x)*(X)**-1\n",
+ "\n",
+ "#EI*y_D=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4\n",
+ "x=3\n",
+ "C1-78\n",
+ "C2=0\n",
+ "y_D=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4)*(X**-1)\n",
+ "\n",
+ "#EI*y_E=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4\n",
+ "x=4\n",
+ "C1-78\n",
+ "C2=0\n",
+ "y_E=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4)*X**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflections at C\",round(y_c,5),\"mm\"\n",
+ "print\"Deflections at D\",round(y_D,5),\"mm\"\n",
+ "print\"Deflections at E\",round(y_E,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflections at C -0.00603 mm\n",
+ "Deflections at D -0.00953 mm\n",
+ "Deflections at E -0.0061 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.15,Page No.209"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200 #KN/mm**2 #Modulus of Elasticity\n",
+ "I=300*10**6 #mm\n",
+ "L_AB=L_BC=L_CD=L_DE=1 #m #Length of AB,BC,CD,DE respectively\n",
+ "F_A=20 #KN #Force at A\n",
+ "F_C=10 #KN #Force at C\n",
+ "w=30 #KN/m #u.d.l\n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=E*I*10**-6 #KN-2**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_E be the reactions at E\n",
+ "V_E=F_A+F_C+w*(L_BC+L_CD) #KN \n",
+ "\n",
+ "#Taking Moment at distance x\n",
+ "#EI*(d**2x/dy**2)=M=-20*x-30*(x-1)**2*2**-1-10*(x-2)+30*(x-3)**2*2**-1\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1-10*x**2-5*(x-1)**3-5*(x-2)**2+5*(x-3)**3\n",
+ "\n",
+ "#Again Integrating above equation\n",
+ "#EI*y=C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*(x-3)**4*4**-1-5*3*(x-2)**3\n",
+ "\n",
+ "#At\n",
+ "#dy/dx=0\n",
+ "x=4 #m\n",
+ "C1=10*x**2+5*(x-1)**3+5*(x-2)**2-5*(x-3)**3\n",
+ "\n",
+ "#AT\n",
+ "x=4\n",
+ "y=0\n",
+ "C2=-C1*4+10*x**3*3**-1+5*(x-1)**4*4**-1-5*(x-3)**4*4**-1+5*3**-1*(x-2)**3\n",
+ "\n",
+ "#Max Deflection and Max slopes occurs at Free end in case of cantilever\n",
+ "y_max=y_A=C2*X**-1\n",
+ "\n",
+ "#EI*(dy/dx)_max=C1\n",
+ "#Let (dy/dx)=Y\n",
+ "Y=C1*X**-1 #radian\n",
+ "\n",
+ "#Now deflection at x=1 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=1\n",
+ "y_B=(C2+C1*x-10*x**3*3**-1)*X**-1\n",
+ "\n",
+ "#Now Deflection at x=2 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=2 #m\n",
+ "y_C=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1)*X**-1\n",
+ "\n",
+ "#Now Deflection at x=3 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=3 #m\n",
+ "y_D=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*3**-1*(x-2)**3)*X**-1\n",
+ "\n",
+ "y_E=0\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Deflection for Beam\",round(y_A,4),\"mm\"\n",
+ "print\"Max Slope for beam\",round(Y,5),\"radians\"\n",
+ "\n",
+ "#Plotting the ELastic Curve\n",
+ "\n",
+ "Y2=[y_E,y_D,y_C,y_B,y_A]\n",
+ "X2=[L_AB+L_BC+L_CD+L_DE,L_AB+L_BC+L_CD,L_AB+L_BC,L_AB,0]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in mm\")\n",
+ "plt.ylabel(\"Deflection in mm\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Deflection for Beam -0.0152 mm\n",
+ "Max Slope for beam 0.00517 radians\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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8yjGZmZkkJSWRmZlJSkoKM2bMoLKy8pxilcZbsQJGj4Y//QkWL1aSEWnvaq3R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\nL+mlAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5a29310>"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.16,Page No.211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BD=L_CB=L_AC=2 #m #Length of BD,CB,AC\n",
+ "F_C=40 #KN #Force at C\n",
+ "F_D=10 #KN Force at D\n",
+ "L=6 #m spna of beam\n",
+ "\n",
+ "#EI is constant in this problem\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B Respectively\n",
+ "#V_A+V_B=50\n",
+ "\n",
+ "#Taking Moment at Pt A\n",
+ "V_B=(F_D*L+F_C*L_AC)*(L_AC+L_CB)**-1\n",
+ "V_A=50-V_B\n",
+ "\n",
+ "#Now Taking Moment at distance x from A,M_x\n",
+ "#M_x=15*x-40*(x-2)+35*(x-4)\n",
+ "#EI*(d**2*y/dx**2)=15*x-40*(x-2)+35*(x-4)\n",
+ "\n",
+ "#Now Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+7.5*x**2-20*(x-2)**2+17.5(x-4)**2\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+2.5*x**2-20*3**-1*(x-2)**3+17.5*(x-4)**3*3**-1\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "#we get\n",
+ "C2=0\n",
+ "\n",
+ "#At\n",
+ "x=4 \n",
+ "y=0\n",
+ "#we get\n",
+ "C1=(2.5*4**3-20*3**-1*2**3)*4**-1\n",
+ "\n",
+ "#Now Deflection at C\n",
+ "x=2\n",
+ "C1=-26.667\n",
+ "C2=0\n",
+ "y_C=C2+C1*x+2.5*x**3\n",
+ "\n",
+ "#Now Deflection at D\n",
+ "C1=-21.667\n",
+ "C2=0\n",
+ "y_D=-26.667*6+2.5*6**3-20*3**-1*4**3+17.5*2**3*3**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflections Under Loads are:y_D\",round(y_D,4)\n",
+ "print\" :y_C\",round(y_C,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflections Under Loads are:y_D -0.002\n",
+ " :y_C -33.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.17,Page No.212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=L_EB=2 #m #Length of BC & EB\n",
+ "E=200*10**6 #KN/m**2 #Modulus of eLasticity\n",
+ "I=45*10**-6 #mm**4 #M.I\n",
+ "L_DE=3 #m #Length of DE\n",
+ "L_AD=1 #m #Length of AD\n",
+ "w=20 #KN/m #u.d.l\n",
+ "L=8 #m #span of beam\n",
+ "F_C=30 #KN #Force at C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=90\n",
+ "\n",
+ "#Taking Moment at A,M_A\n",
+ "V_B=(w*L_DE*(L_DE*2**-1+L_AD)+F_C*L)*(L_AD+L_DE+L_EB)**-1\n",
+ "V_A=90-V_B\n",
+ "\n",
+ "#Taking Moment at distance x\n",
+ "#M_x=25*x-20*(x-1)**2*2**-1+20*(x-4)**2*2**-1+65*(x-6)\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(d**2*y/dx**2)=25*x-10*(x-1)**2+10*(x-4)**2+65*(x-6)\n",
+ "\n",
+ "#again Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+25*x**2*2**-1-10*3**-1*(x-1)**3+10*3**-1*(x-4)**2+65*2**-1*(x-6)**2\n",
+ "\n",
+ "#again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3\n",
+ "\n",
+ "x=0\n",
+ "y=0\n",
+ "#Sub these values in above equation,we get\n",
+ "C2=0\n",
+ "\n",
+ "x=6 #m\n",
+ "y=0\n",
+ "C1=-(25*6**-1*6**3-10*12**-1*5**4+10*12**-1*2**4)*6**-1\n",
+ "\n",
+ "#deflection at C is given by\n",
+ "x=8\n",
+ "y_c=(C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Assuming y is max in the portion DE,then\n",
+ "#(dy/dx)=0 for that point\n",
+ "\n",
+ "#0=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",
+ "\n",
+ "#Let F(x)=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",
+ "#Let z=F(x)\n",
+ "\n",
+ "#AT \n",
+ "x=3\n",
+ "z=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "x=2.5\n",
+ "z1=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "x=2.4\n",
+ "z2=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "#The assumption is max in portion DE\n",
+ "x=2.46\n",
+ "y_max=(-65.417*x+25*6**-1*x**3-10*12**-1*1.46**4)*(E*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection at free end C\",round(y_c,4),\"mm\"\n",
+ "print\"Max Deflection between A and B\",round(y_max,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at free end C -0.0101 mm\n",
+ "Max Deflection between A and B -0.0114 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.18,Page No.213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DB=L_AC=L_ED=2 #m #Length of DB & AC\n",
+ "L_CD=4 #m #Length of CD\n",
+ "L_CE=2 #m #Length of CE\n",
+ "F_A=40 #KN #Force at C\n",
+ "F_B=20 #KN #Force at A\n",
+ "E=200*10**6 #KN/mm**2 #Modulus of Elasticity\n",
+ "I=50*10**-6 #m**4 #M.I\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt V_C & V_D be the reactions at C & D respectively\n",
+ "#V_C+V_D=60\n",
+ "\n",
+ "#Taking Moment At D,M_D\n",
+ "V_C=-(-F_A*(L_AC+L_CE+L_ED)+F_B*L_DB)*L_CD**-1\n",
+ "V_D=60-V_C\n",
+ "\n",
+ "#Now Taking Moment at Distance x from A,\n",
+ "#M_x=-40*x+50*(x-2)+10*(x-6)\n",
+ "\n",
+ "#EI*(d**2*y/dx**2)=-40*x+50*(x-2)+10*(x-6)\n",
+ "\n",
+ "#Now Integrating above Equation we get\n",
+ "#EI*(dy/dx)=C1+20*x**2-25*(x-2)+5*(x-6)**2\n",
+ "\n",
+ "#Again Integrating above Equation we get\n",
+ "#EI*y=C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "#C2+2*C1=-53.33 ...............(1)\n",
+ "\n",
+ "#At \n",
+ "x=6\n",
+ "y=0\n",
+ "#C2+6*C1=906.667 ...............(2)\n",
+ "\n",
+ "#Subtracting Equation 1 from 2 we get\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=0\n",
+ "y_A=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Answer For y_A is incorrect in textbook\n",
+ "\n",
+ "#At Mid-span\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=4\n",
+ "y_E=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Answer For y_E is incorrect in textbook\n",
+ "\n",
+ "#At B\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=8\n",
+ "y_B=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection relative to the level of the supports:at End A\",round(y_A,4),\"mm\"\n",
+ "print\" :at End B\",round(y_B,4),\"mm\"\n",
+ "print\" :at Centre of CD\",round(y_E,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection relative to the level of the supports:at End A -0.08 mm\n",
+ " :at End B -0.0267 mm\n",
+ " :at Centre of CD 0.0107 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_4.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_4.ipynb new file mode 100644 index 00000000..e3bdfbe1 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_4.ipynb @@ -0,0 +1,1518 @@ +{
+ "metadata": {
+ "name": "chapter no.6.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.6:Torsion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.1,Page No.225"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=10000 #mm #Length of solid shaft\n",
+ "d=100 #mm #Diameter of shaft\n",
+ "n=150 #rpm\n",
+ "P=112.5*10**6 #N-mm/sec #Power Transmitted\n",
+ "G=82*10**3 #N/mm**2 #modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*d**4*(32)**-1 #mm**3 #Polar Modulus\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "r=50 #mm #Radius\n",
+ "\n",
+ "q_s=T*r*J**-1 #N/mm**2 #Max shear stress intensity\n",
+ "Theta=T*L*(G*J)**-1 #angle of twist\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress intensity\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist\",round(Theta,3),\"radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress intensity 36.48 N/mm**2\n",
+ "Angle of Twist 0.089 radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.2,Page No.226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=440*10**6 #N-m/sec #Power transmitted\n",
+ "n=280 #rpm\n",
+ "theta=pi*180**-1 #radian #angle of twist\n",
+ "L=1000 #mm #Length of solid shaft\n",
+ "q_s=40 #N/mm**2 #Max torsional shear stress\n",
+ "G=84*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#P=2*pi*n*T*(60)**-1 #Equation of Power transmitted\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #torsional moment\n",
+ "\n",
+ "#From Consideration of shear stress\n",
+ "d1=(T*16*(pi*40)**-1)**0.333333 \n",
+ "\n",
+ "#From Consideration of angle of twist\n",
+ "d2=(T*L*32*180*(pi*84*10**3*pi)**-1)**0.25\n",
+ "\n",
+ "#result\n",
+ "print\"Diameter of solid shaft is\",round(d1,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of solid shaft is 124.09 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.3,Page No.227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "q_s=80 #N/mm**2 #Max sheare stress\n",
+ "P=736*10**6 #N-mm/sec #Power transmitted\n",
+ "n=200\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now From consideration of angle of twist\n",
+ "theta=pi*180**-1\n",
+ "#L=15*d\n",
+ "\n",
+ "d=(T*32*180*15*(pi**2*G)**-1)**0.33333\n",
+ "\n",
+ "#Now corresponding stress at the surface is\n",
+ "q_s2=T*32*d*(pi*2*d**4)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Max diameter required is\",round(d,2),\"mm\"\n",
+ "print\"Corresponding shear stress is\",round(q_s2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max diameter required is 156.66 mm\n",
+ "Corresponding shear stress is 46.55 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.4,Page No.228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #Diameter of steel bar\n",
+ "p=50*10**3 #N #Pull\n",
+ "dell_1=0.095 #mm #Extension of bar\n",
+ "l=200 #mm #Guage Length\n",
+ "T=200*10**3 #N-mm #Torsional moment\n",
+ "theta=0.9*pi*180**-1 #angle of twist\n",
+ "L=250 #mm Length of steel bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*4**-1*d**2 #Area of steel bar #mm**2\n",
+ "E=p*l*(dell_1*A)**-1 #N/mm**2 #Modulus of elasticity \n",
+ "\n",
+ "J=pi*32**-1*d**4 #mm**4 #Polar modulus\n",
+ "\n",
+ "G=T*L*(theta*J)**-1 #Modulus of rigidity #N/mm**2\n",
+ "\n",
+ "#Now from the relation of Elastic constants\n",
+ "mu=E*(2*G)**-1-1\n",
+ "\n",
+ "#result\n",
+ "print\"The Poissoin's ratio is\",round(mu,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Poissoin's ratio is 0.292\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.5,Page No.229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6000 #mm #Length of circular shaft\n",
+ "d1=100 #mm #Outer Diameter\n",
+ "d2=75 #mm #Inner Diameter\n",
+ "R=100*2**-1 #Radius of shaft\n",
+ "T=10*10**6 #N-mm #Torsional moment\n",
+ "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
+ "\n",
+ "#Max Shear stress produced\n",
+ "q_s=T*R*J**-1 #N/mm**2\n",
+ "\n",
+ "#Angle of twist\n",
+ "theta=T*L*(G*J)**-1 #Radian\n",
+ "\n",
+ "#Result\n",
+ "print\"MAx shear stress produced is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist is\",round(theta,2),\"Radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "MAx shear stress produced is 74.5 N/mm**2\n",
+ "Angle of Twist is 0.11 Radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.6,Page No.229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=200 #mm #External Diameter of shaft\n",
+ "t=25 #mm #Thickness of shaft\n",
+ "n=200 #rpm\n",
+ "theta=0.5*pi*180**-1 #Radian #angle of twist\n",
+ "L=2000 #mm #Length of shaft\n",
+ "G=84*10**3 #N/mm**2\n",
+ "d2=d1-2*t #mm #Internal Diameter of shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
+ "\n",
+ "#Torsional moment\n",
+ "T=G*J*theta*L**-1 #N/mm**2 \n",
+ "\n",
+ "#Power Transmitted\n",
+ "P=2*pi*n*T*60**-1*10**-6 #N-mm\n",
+ "\n",
+ "#Max shear stress transmitted\n",
+ "q_s=G*theta*(d1*2**-1)*L**-1 #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Power Transmitted is\",round(P,2),\"N-mm\"\n",
+ "print\"Max Shear stress produced is\",round(q_s,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power Transmitted is 824.28 N-mm\n",
+ "Max Shear stress produced is 36.65 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.7,Page No.230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=3750*10**6 #N-mm/sec\n",
+ "n=240 #Rpm\n",
+ "q_s=160 #N/mm**2 #Max shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#d2=0.8*d2 #mm #Internal Diameter of shaft\n",
+ "\n",
+ "#J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar modulus\n",
+ "#After substituting value in above Equation we get\n",
+ "#J=0.05796*d1**4\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now from Torsion Formula\n",
+ "#T*J**-1=q_s*R**-1 ......................................(1)\n",
+ "\n",
+ "#But R=d1*2**-1 \n",
+ "\n",
+ "#Now substituting value of R and J in Equation (1) we get\n",
+ "d1=(T*(0.05796*q_s*2)**-1)**0.33333\n",
+ "\n",
+ "d2=d1*0.8\n",
+ "\n",
+ "#Result\n",
+ "print\"The size of the Shaft is:d1\",round(d1,3),\"mm\"\n",
+ "print\" :d2\",round(d2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The size of the Shaft is:d1 200.362 mm\n",
+ " :d2 160.289 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.8,Page No.231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=245*10**6 #N-mm/sec #Power transmitted\n",
+ "n=240 #rpm\n",
+ "q_s=40 #N/mm**2 #Shear stress\n",
+ "theta=pi*180**-1 #radian #Angle of twist\n",
+ "L=1000 #mm #Length of shaft\n",
+ "G=80*10**3 #N/mm**2\n",
+ "\n",
+ "#Tmax=1.5*T\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional Moment\n",
+ "Tmax=1.5*T\n",
+ "\n",
+ "#Now For Solid shaft\n",
+ "#J=pi*32*d**4\n",
+ "\n",
+ "#Now from the consideration of shear stress we get\n",
+ "#T*J**-1=q_s*(d*2**-1)**-1\n",
+ "#After substituting value in above Equation we get\n",
+ "#T=pi*16**-1*d**3*q_s\n",
+ "\n",
+ "#Designing For max Torque\n",
+ "d=(Tmax*16*(pi*40)**-1)**0.33333 #mm #Diameter of shaft\n",
+ "\n",
+ "#For max Angle of Twist\n",
+ "#Tmax*J**-1=G*theta*L**-1 \n",
+ "#After substituting value in above Equation we get\n",
+ "d2=(Tmax*32*180*L*(pi**2*G)**-1)**0.25\n",
+ "\n",
+ "#For Hollow Shaft\n",
+ "\n",
+ "#d1_2=Outer Diameter\n",
+ "#d2_2=Inner Diameter\n",
+ "\n",
+ "#d2_2=0.5*d1_2\n",
+ "\n",
+ "# Polar modulus\n",
+ "#J=pi*32**-1*(d1_2**4-d2_2**4)\n",
+ "#After substituting values we get\n",
+ "#J=0.092038*d1_2**4\n",
+ "\n",
+ "#Now from the consideration of stress\n",
+ "#Tmax*J**-1=q_s*(d1_2*2**-1)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d1_2=(Tmax*(0.092038*2*q_s)**-1)**0.33333\n",
+ "\n",
+ "#Now from the consideration of angle of twist\n",
+ "#Tmax*J**-1=G*theta*L**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d1_3=(Tmax*180*L*(0.092038*G*pi)**-1)**0.25\n",
+ "\n",
+ "d2_2=0.5*d1_2\n",
+ "\n",
+ "#result\n",
+ "print\"Diameter of shaft is:For solid shaft:d\",round(d,2),\"mm\"\n",
+ "print\" :For Hollow shaft:d1_2\",round(d1_2,3),\"mm\"\n",
+ "print\" : :d2_2\",round(d2_2,3),\"mm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of shaft is:For solid shaft:d 123.01 mm\n",
+ " :For Hollow shaft:d1_2 125.69 mm\n",
+ " : :d2_2 62.845 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.11,Page No.235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=250*10**6 #N-mm/sec #Power transmitted\n",
+ "n=100 #rpm\n",
+ "q_s=75 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Equation of Power we have\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now from torsional moment equation we have\n",
+ "#T=j*q_s*(d/2**-1)**-1\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T=pi*16**-1**d**3*q_s\n",
+ "d=(T*16*(pi*q_s)**-1)**0.3333 #mm #Diameter of solid shaft\n",
+ "\n",
+ "#PArt-2\n",
+ "\n",
+ "#Let d1 and d2 be the outer and inner diameter of hollow shaft\n",
+ "#d2=0.6*d1\n",
+ "\n",
+ "#Again from torsional moment equation we have\n",
+ "#T=pi*32**-1*(d1**4-d2**4)*q_s*(d1/2)**-1\n",
+ "d1=(T*16*(pi*(1-0.6**4)*q_s)**-1)**0.33333\n",
+ "d2=0.6*d1\n",
+ "\n",
+ "#Cross sectional area of solid shaft\n",
+ "A1=pi*4**-1*d**2 #mm**2\n",
+ "\n",
+ "#cross sectional area of hollow shaft\n",
+ "A2=pi*4**-1*(d1**2-d2**2)\n",
+ "\n",
+ "#Now percentage saving in weight\n",
+ "#Let W be the percentage saving in weight\n",
+ "W=(A1-A2)*100*A1**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage saving in Weight is\",round(W,3),\"%\"\n",
+ "print\"Size of shaft is:solid shaft:d\",round(d,3),\"mm\"\n",
+ "print\" :Hollow shaft:d1\",round(d1,3),\"mm\"\n",
+ "print\" : :d2\",round(d2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage saving in Weight is 29.735 %\n",
+ "Size of shaft is:solid shaft:d 117.418 mm\n",
+ " :Hollow shaft:d1 123.031 mm\n",
+ " : :d2 73.818 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.12,Page No.237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "d=100 #mm #Diameter of solid shaft\n",
+ "d1=100 #mm #Outer Diameter of hollow shaft\n",
+ "d2=50 #mm #Inner Diameter of hollow shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Torsional moment of solid shaft\n",
+ "#T_s=J*q_s*(d*2**-1)**-1 \n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T_s=pi*16*d**3*q_s ...............(1)\n",
+ "\n",
+ "#torsional moment for hollow shaft is\n",
+ "#T_h=J*q_s*(d1**4-d2**4)**-1*(d1*2**-1)\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T_h=pi*32**-1*2*d1**-1*(d1**4-d2**4)*q_s ...........(2)\n",
+ "\n",
+ "#Dividing Equation 2 by 1 we get\n",
+ "#Let the ratio of T_h*T_s**-1 Be X\n",
+ "X=1-0.5**4\n",
+ "\n",
+ "#Loss in strength \n",
+ "#Let s be the loss in strength\n",
+ "#s=T_s*T_h*100*T_s**-1\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "s=(1-0.9375)*100\n",
+ "\n",
+ "#Weight Ratio \n",
+ "#Let w be the Weight ratio\n",
+ "#w=W_h*W_s**-1\n",
+ "\n",
+ "A_h=pi*32**-1*(d1**2-d2**2) #mm**2 #Area of Hollow shaft\n",
+ "A_s=pi*32**-1*d**2 #mm**2 #Area of solid shaft\n",
+ "\n",
+ "w=A_h*A_s**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"Loss in strength is\",round(s,2)\n",
+ "print\"Weight ratio is\",round(w,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Loss in strength is 6.25\n",
+ "Weight ratio is 0.75\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.13,Page No.239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "T=8 #KN-m #Torque \n",
+ "d=100 #mm #Diameter of portion AB\n",
+ "d1=100 #mm #External Diameter of Portion BC\n",
+ "d2=75 #mm #Internal Diameter of Portion BC\n",
+ "G=80 #KN/mm**2 #Modulus of Rigidity\n",
+ "L1=1500 #mm #Radial Distance of Portion AB\n",
+ "L2=2500 #mm #Radial Distance ofPortion BC\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R=d*2**-1 #mm #Radius of shaft\n",
+ "\n",
+ "#For Portion AB,Polar Modulus\n",
+ "J1=pi*32**-1*d**4 #mm**4 \n",
+ "\n",
+ "#For Portion BC,Polar modulus \n",
+ "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Now Max stress occurs in portion BC since max radial Distance is sme in both cases\n",
+ "q_max=T*J2**-1*R*10**6 #N/mm**2 \n",
+ "\n",
+ "#Let theta1 be the rotation in Portion AB and theta2 be the rotation in portion BC\n",
+ "theta1=T*L1*(G*J1)**-1 #Radians\n",
+ "theta2=T*L2*(G*J2)**-1 #Radians\n",
+ "\n",
+ "#Total Rotational at end C\n",
+ "theta=(theta1+theta2)*10**3 #Radians\n",
+ "\n",
+ "#Result\n",
+ "print\"Max stress induced is\",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist is\",round(theta,3),\"radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max stress induced is 59.6 N/mm**2\n",
+ "Angle of Twist is 0.053 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.14,Page No.240"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "q_b=80 #N/mm**2 #Shear stress in Brass\n",
+ "q_s=100 #N/mm**2 #Shear stress in Steel\n",
+ "G_b=40*10**3 #N/mm**2 \n",
+ "G_s=80*10**3 \n",
+ "L_b=1000 #mm #Length of brass shaft\n",
+ "L_s=1200 #mm #Length of steel shaft\n",
+ "d1=80 #mm #Diameter of brass shaft\n",
+ "d2=60 #mm #Diameter of steel shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar modulus of brass rod\n",
+ "J_b=pi*32**-1*d1**4 #mm**4 \n",
+ "\n",
+ "#Polar modulus of steel rod\n",
+ "J_s=pi*32**-1*d2**4 #mm**4\n",
+ "\n",
+ "#Considering bras Rod:AB\n",
+ "T1=J_b*q_b*(d1*2**-1)**-1 #N-mm \n",
+ "\n",
+ "#Considering Steel Rod:BC\n",
+ "T2=J_s*q_s*(d2*2**-1)**-1 #N-mm\n",
+ "\n",
+ "#Max Torque that can be applied\n",
+ "T2\n",
+ "\n",
+ "#Let theta_b and theta_s be the rotations in Brass and steel respectively\n",
+ "theta_b=T2*L_b*(G_b*J_b)**-1 #Radians\n",
+ "theta_s=T2*L_s*(G_s*J_s)**-1 #Radians\n",
+ "\n",
+ "theta=theta_b+theta_s #Radians #Rotation of free end\n",
+ "\n",
+ "#Result\n",
+ "print\"Total of free end is\",round(theta,3),\"Radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total of free end is 0.076 Radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.15,Page No.241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
+ "d1=100 #mm #Outer diameter of hollow shft\n",
+ "d2=80 #mm #Inner diameter of hollow shaft\n",
+ "d=80 #mm #diameter of Solid shaft\n",
+ "d3=60 #mm #diameter of Solid shaft having L=0.5m\n",
+ "L1=300 #mm #Length of Hollow shaft\n",
+ "L2=400 #mm #Length of solid shaft\n",
+ "L3=500 #mm #LEngth of solid shaft of diameter 60mm\n",
+ "T1=2*10**6 #N-mm #Torsion in Shaft AB\n",
+ "T2=1*10**6 #N-mm #Torsion in shaft BC\n",
+ "T3=1*10**6 #N-mm #Torsion in shaft CD\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Now Polar modulus of section AB\n",
+ "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
+ "\n",
+ "#Polar modulus of section BC\n",
+ "J2=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Polar modulus of section CD\n",
+ "J3=pi*32**-1*d3**4 #mm**4\n",
+ "\n",
+ "#Now angle of twist of AB\n",
+ "theta1=T1*L1*(G*J1)**-1 #radians\n",
+ "\n",
+ "#Angle of twist of BC\n",
+ "theta2=T2*L2*(G*J2)**-1 #radians\n",
+ "\n",
+ "#Angle of twist of CD\n",
+ "theta3=T3*L3*(G*J3)**-1 #radians\n",
+ "\n",
+ "#Angle of twist\n",
+ "theta=theta1-theta2+theta3 #Radians\n",
+ "\n",
+ "#Shear stress in AB From Torsion Equation\n",
+ "q_s1=T1*(d1*2**-1)*J1**-1 #N/mm**2 \n",
+ "\n",
+ "#Shear stress in BC\n",
+ "q_s2=T2*(d*2**-1)*J2**-1 #N/mm**2 \n",
+ "\n",
+ "#Shear stress in CD\n",
+ "q_s3=T3*(d3*2**-1)*J3**-1 #N-mm**2\n",
+ "\n",
+ "#As max shear stress occurs in portion CD,so consider CD\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of twist at free end is\",round(theta,5),\"Radian\"\n",
+ "print\"Max Shear stress\",round(q_s3,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of twist at free end is 0.00496 Radian\n",
+ "Max Shear stress 23.58 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.16,Page No.242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of bar\n",
+ "L1=600 #mm #Length of Bar AB\n",
+ "L2=400 #mm #Length of Bar BC\n",
+ "d1=60 #mm #Outer Diameter of bar BC\n",
+ "d2=30 #mm #Inner Diameter of bar BC\n",
+ "d=60 #mm #Diameter of bar AB\n",
+ "T=2*10**6 #N-mm #Total Torque\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar Modulus of Portion AB\n",
+ "J1=pi*32**-1*d**4 #mm*4\n",
+ "\n",
+ "#Polar Modulus of Portion BC\n",
+ "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Let T1 be the torque resisted by bar AB and T2 be torque resisted by Bar BC\n",
+ "#Let theta1 and theta2 be the rotation of shaft in portion AB & BC\n",
+ "\n",
+ "#theta1=T1*L1*(G*J1)**-1 #radians\n",
+ "#After substituting values and further simplifying we get \n",
+ "#theta1=32*600*T1*(pi*60**4*G)**-1\n",
+ "\n",
+ "#theta2=T2*L*(J2*G)**-1 #Radians\n",
+ "#After substituting values and further simplifying we get \n",
+ "#theta2=32*400*T2*(pi*60**4*(1-0.5**4)*G)**-1 \n",
+ "\n",
+ "#Now For consistency of Deformation,theta1=theta2\n",
+ "#After substituting values and further simplifying we get \n",
+ "#T1=0.7111*T2 ..................................................(1)\n",
+ "\n",
+ "#But T1+T2=T=2*10**6 ...........................................(2)\n",
+ "#Substituting value of T1 in above equation\n",
+ "\n",
+ "T2=T*(0.7111+1)**-1\n",
+ "T1=0.71111*T2\n",
+ "\n",
+ "#Max stress in Portion AB\n",
+ "q_s1=T1*(d*2**-1)*(J1)**-1 #N/mm**2\n",
+ "\n",
+ "#Max stress in Portion BC\n",
+ "q_s2=T2*(d1*2**-1)*J2**-1 #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Portion:AB\",round(q_s1,2),\"N/mm**2\"\n",
+ "print\" :BC\",round(q_s2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Portion:AB 19.6 N/mm**2\n",
+ " :BC 29.4 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.17,Page No.243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=80 #mm #External Diameter of Brass tube\n",
+ "d2=50 #mm #Internal Diameter of Brass tube\n",
+ "d=50 #mm #Diameter of steel Tube\n",
+ "G_b=40*10**3 #N/mm**2 #Modulus of Rigidity of brass tube\n",
+ "G_s=80*10**3 #N/mm**2 #Modulus of rigidity of steel tube\n",
+ "T=6*10**6 #N-mm #Torque\n",
+ "L=2000 #mm #Length of Tube\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar Modulus of brass tube\n",
+ "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
+ "\n",
+ "#Polar modulus of steel Tube\n",
+ "J2=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Let T_s & T_b be the torque resisted by steel and brass respectively\n",
+ "#Then, T_b+T_s=T ............................................(1)\n",
+ "\n",
+ "#Since the angle of twist will be the same\n",
+ "#Theta1=Theta2\n",
+ "#After substituting values and further simplifying we get \n",
+ "#Ts=0.360*Tb ...........................................(2)\n",
+ "\n",
+ "#After substituting value of Ts in eqn 1 and further simplifying we get \n",
+ "T_b=T*(0.36+1)**-1 #N-mm\n",
+ "T_s=0.360*T_b\n",
+ "\n",
+ "#Let q_s and q_b be the max stress in steel and brass respectively\n",
+ "q_b=T_b*(d1*2**-1)*J1**-1 #N/mm**2\n",
+ "q_s=T_s*(d2*2**-1)*J2**-1 #N/mm**2\n",
+ "\n",
+ "#Since angle of twist in brass=angle of twist in steel\n",
+ "theta_s=T_s*L*(J2*G_s)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Materials are:Brass\",round(q_b,2),\"N/mm**2\"\n",
+ "print\" :Steel\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist in 2m Length\",round(theta_s,3),\"Radians\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Materials are:Brass 51.79 N/mm**2\n",
+ " :Steel 64.71 N/mm**2\n",
+ "Angle of Twist in 2m Length 0.065 Radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.18,Page No.245"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=60 #mm #External Diameter of aluminium Tube\n",
+ "d2=40 #mm #Internal Diameter of aluminium Tube\n",
+ "d=40 #mm #Diameter of steel tube\n",
+ "q_a=60 #N/mm**2 #Permissible stress in aluminium\n",
+ "q_s=100 #N/mm**2 #Permissible stress in steel tube\n",
+ "G_a=27*10**3 #N/mm**2 \n",
+ "G_s=80*10**3 #N/mm**2 \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar modulus of aluminium Tube\n",
+ "J_a=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Polar Modulus of steel Tube\n",
+ "J_s=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Now the angle of twist of steel tube = angle of twist of aluminium tube\n",
+ "#T_s*L_s*(J_s*theta_s)**-1=T_a*L_a*(J_a*theta_a)**-1\n",
+ "#After substituting values in above Equation and Further simplifyin we get\n",
+ "#T_s=0.7293*T_a .....................(1)\n",
+ "\n",
+ "#If steel Governs the resisting capacity\n",
+ "T_s1=q_s*J_s*(d*2**-1)**-1 #N-mm\n",
+ "T_a1=T_s1*0.7293**-1 #N-mm\n",
+ "T1=(T_s1+T_a1)*10**-6 #KN-m #Total Torque in steel Tube\n",
+ "\n",
+ "#If aluminium Governs the resisting capacity \n",
+ "T_a2=q_a*J_a*(d1*2**-1) #N-mm\n",
+ "T_s2=T_a2*0.7293 #N-mm\n",
+ "T2=(T_s2+T_a2)*10**-6 #KN-m #Total Torque in aluminium tube\n",
+ "\n",
+ "#Result\n",
+ "print\"Steel Governs the torque carrying capacity\",round(T1,2),\"KN-m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Steel Governs the torque carrying capacity 2.98 KN-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.19,Page No.247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=225*10**6 #N-mm/sec #Power Trasmitted\n",
+ "q_b=80 #N/mm**2 #Shear stress\n",
+ "n=200 #Rpm\n",
+ "q_k=100 #N/mm**2 #PErmissible stress in Keys\n",
+ "D=300 #mm #Diameter of bolt circle\n",
+ "L=150 #mm #Length of shear key\n",
+ "d=16 #mm #Diameterr of bolt\n",
+ "\n",
+ "#Calculations\n",
+ "T=60*P*(2*pi*n)**-1 #N-mm #Torque\n",
+ "\n",
+ "#Now From Torsion Formula\n",
+ "#T*J**-1=q_s*R**-1\n",
+ "#After substituting values we get\n",
+ "#T=pi*16*d**3*n\n",
+ "#After further simplifying we get\n",
+ "d1=(T*16*(pi*q_s)**-1)**0.33333\n",
+ "\n",
+ "#Let b be the width of shear Key\n",
+ "#T=q_k*L*b*R\n",
+ "#After simplifying further we get\n",
+ "b=T*(q_k*L*(d1*2**-1))**-1 #mm\n",
+ "\n",
+ "#Let n2 be the no. of bolts required at bolt circle of radius\n",
+ "R_b=D*2**-1 #mm \n",
+ "\n",
+ "n2=T*4*(q_b*pi*d**2*R_b)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Minimum no. of Bolts Required are\",round(n2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum no. of Bolts Required are 4.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.20,Page No.250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "T=2*10**6 #N-mm #Torque transmitted\n",
+ "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "d1=40 #mm \n",
+ "d2=80 #mm\n",
+ "r1=20 #mm\n",
+ "r2=40 #mm\n",
+ "L=2000 #mm #Length of shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Angle of twist \n",
+ "theta=2*T*L*(r1**2+r1*r2+r2**2)*(3*pi*G*r2**3*r1**3)**-1 #radians\n",
+ "\n",
+ "#If the shaft is treated as shaft of average Diameter\n",
+ "d_avg=(d1+d2)*2**-1 #mm\n",
+ "\n",
+ "theta1=T*L*(G*pi*32**-1*d_avg**4)**-1 #Radians\n",
+ "\n",
+ "#Percentage Error\n",
+ "#Let Percentage Error be E\n",
+ "X=theta-theta1\n",
+ "E=(X*theta**-1)*100 \n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage Error is\",round(E,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage Error is 32.28 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.21,Page No.252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 \n",
+ "P=1*10**9 #N-mm/sec #Power\n",
+ "n=300 \n",
+ "d1=150 #mm #Outer Diameter\n",
+ "d2=120 #mm #Inner Diameter\n",
+ "L=2000 #mm #Length of circular shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm\n",
+ "\n",
+ "#Polar Modulus \n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "q_s=T*J**-1*(d1*2**-1) #N/mm**2 \n",
+ "\n",
+ "\n",
+ "#Strain ENergy\n",
+ "U=q_s**2*(4*G)**-1*pi*4**-1*(d1**2-d2**2)*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Strain Energy stored in the shaft is\",round(U,2),\"N-mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress is 81.36 N/mm**2\n",
+ "Strain Energy stored in the shaft is 263181.37 N-mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.22,Page No.254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=12 #mm #Diameter of helical spring\n",
+ "D=150 #mm #Mean Diameter\n",
+ "R=D*2**-1 #mm #Radius of helical spring\n",
+ "n=10 #no.of turns\n",
+ "G=80*10**3 #N/mm**2 \n",
+ "W=450 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max shear stress \n",
+ "q_s=16*W*R*(pi*d**3)**-1 #N/mm**2\n",
+ "\n",
+ "#Strain Energy stored\n",
+ "U=32*W**2*R**3*n*(G*d**4)**-1 #N-mm\n",
+ "\n",
+ "#Deflection Produced\n",
+ "dell=64*W*R**3*n*(G*d**4)**-1 #mm\n",
+ "\n",
+ "#Stiffness Spring\n",
+ "k=W*dell**-1 #N/mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Strain Energy stored is\",round(U,2),\"N-mm\"\n",
+ "print\"Deflection Produced is\",round(dell,2),\"mm\"\n",
+ "print\"Stiffness spring is\",round(k,2),\"N/mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress is 99.47 N/mm**2\n",
+ "Strain Energy stored is 16479.49 N-mm\n",
+ "Deflection Produced is 73.24 mm\n",
+ "Stiffness spring is 6.14 N/mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.23,Page No.255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "K=5 #N/mm #Stiffness\n",
+ "L=100 #mm #Solid Length\n",
+ "q_s=60 #N/mm**2 #Max shear stress\n",
+ "W=200 #N #Max Load\n",
+ "G=80*10**3 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#K=W*dell**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#d=0.004*R**3*n ........(1) #mm #Diameter of wire\n",
+ "#n=L*d**-1 ........(2)\n",
+ "\n",
+ "#From Shearing stress\n",
+ "#q_s=16*W*R*(pi*d**3)**-1 \n",
+ "#After substituting values and further simplifying we get\n",
+ "#d**4=0.004*R**3*n .................(4)\n",
+ "\n",
+ "#From Equation 1,2,3\n",
+ "#d**4=0.004*(0.0785*d**3)**3*100*d**-1\n",
+ "#after further simplifying we get\n",
+ "d=5168.101**0.25\n",
+ "n=100*d**-1\n",
+ "R=(d**4*(0.004*n)**-1)**0.3333\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of Wire is\",round(d,2),\"mm\"\n",
+ "print\"No.of turns is\",round(n,2)\n",
+ "print\"Mean Radius of spring is\",round(R,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of Wire is 8.48 mm\n",
+ "No.fo turns is 11.79\n",
+ "Mean Radius of spring is 47.83 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.24,Page No.255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "m=5*10**5 #Wagon Weighing\n",
+ "v=18*1000*36000**-1 \n",
+ "d=300 #mm #Diameter of Beffer springs\n",
+ "n=18 #no.of turns\n",
+ "G=80*10**3 #N/mm**2\n",
+ "dell=225\n",
+ "R=100 #mm #Mean Radius\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Energy of Wagon\n",
+ "E=m*v**2*(9.81*2)**-1 #N-mm\n",
+ "\n",
+ "#Load applied\n",
+ "W=dell*G*d**4*(64*R**3*n)**-1 #N \n",
+ "\n",
+ "#Energy each spring can absorb is\n",
+ "E2=W*dell*2**-1 #N-mm\n",
+ "\n",
+ "#No.of springs required to absorb energy of Wagon\n",
+ "n2=E*E2**-1 *10**7\n",
+ "\n",
+ "#Result\n",
+ "print\"No.of springs Required for Buffer is\",round(n2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No.of springs Required for Buffer is 4.47\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.25,Page No.259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=180 #mm #width of flange\n",
+ "d=10 #mm #Depth of flange\n",
+ "t=10 #mm #Thickness of flange\n",
+ "D=400 #mm #Overall Depth \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I_xx=1*12**-1*(b*D**3-(b-t)*(D-2*d)**3)\n",
+ "I_yy=1*12**-1*((D-2*d)*t**3+2*t*b**3)\n",
+ "\n",
+ "#If warping is neglected\n",
+ "J=I_xx+I_yy #mm**4\n",
+ "\n",
+ "#Since b/d>1.6,we get\n",
+ "J2=1*3**-1*d**3*b*(1-0.63*d*b**-1)*2+1*3**-1*t**3*(D-2*d)*(1-0.63*t*b**-1)\n",
+ "\n",
+ "#Over Estimation of torsional Rigidity would have been \n",
+ "T=J*J2**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Error in assessing torsional Rigidity if the warping is neglected is\",round(T,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Error in assessing torsional Rigidity if the warping is neglected is 808.28\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.26,Page No.261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=100 #mm #Outer Diameter\n",
+ "d2=95 #mm #Inner Diameter\n",
+ "T=2*10**6 #N-mm #Torque\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus\n",
+ "\n",
+ "#Shear stress\n",
+ "q_max=T*J**-1*d1*2**-1 #N/mm**2 \n",
+ "\n",
+ "#Now theta*L**-1=T*(G*J)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#Let theta*L**-1=X\n",
+ "X=T*J**-1\n",
+ "\n",
+ "#Now Treating it as very thin walled tube\n",
+ "d=(d1+d2)*2**-1 #mm\n",
+ "\n",
+ "r=d*2**-1 \n",
+ "t=(d1-d2)*2**-1\n",
+ "q_max2=T*(2*pi*r**2*t)**-1 #N/mm**2\n",
+ "\n",
+ "X2=T*(2*pi*r**3*t)**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"When it is treated as hollow shaft:Max shear stress\",round(q_max,2),\"N/mm**2\"\n",
+ "print\" :Angle of Twist per unit Length\",round(X,3)\n",
+ "print\"When it is very thin Walled Tube :Max shear stress\",round(q_max2,2),\"N/mm**2\"\n",
+ "print\" :Angle of twist per Unit Length\",round(X2,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "When it is treated as hollow shaft:Max shear stress 54.91 N/mm**2\n",
+ " :Angle of Twist per unit Length 1.098\n",
+ "When it is very thin Walled Tube :Max shear stress 53.57 N/mm**2\n",
+ " :Angle of twist per Unit Length 1.099\n"
+ ]
+ }
+ ],
+ "prompt_number": 72
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_4.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_4.ipynb new file mode 100644 index 00000000..b75d886a --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_4.ipynb @@ -0,0 +1,1328 @@ +{
+ "metadata": {
+ "name": "chapter no.7.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.7:Compound Stresses And Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1,Page No.269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma1=30 #N/mm**2 #Stress in tension\n",
+ "d=20 #mm #Diameter \n",
+ "sigma2=90 #N/mm**2 #Max compressive stress\n",
+ "sigma3=25 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#In TEnsion\n",
+ "\n",
+ "#Corresponding stress in shear\n",
+ "P=sigma1*2**-1 #N/mm**2\n",
+ "\n",
+ "#Tensile force\n",
+ "F=pi*4**-1*d**2*sigma1\n",
+ "\n",
+ "#In Compression\n",
+ "\n",
+ "#Correspong shear stress\n",
+ "P2=sigma2*2**-1 #N/mm**2\n",
+ "\n",
+ "#Correspong compressive(axial) stress\n",
+ "p=2*sigma3 #N/mm**2 \n",
+ "\n",
+ "#Corresponding Compressive force\n",
+ "P3=p*pi*4**-1*d**2 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Failure Loads are:\",round(F,2),\"N\"\n",
+ "print\" :\",round(P3,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Failure Loads are: 9424.78 N\n",
+ " : 15707.96 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.2,Page No.270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #Diameter of circular bar\n",
+ "F=20*10**3 #N #Axial Force\n",
+ "theta=30 #Degree #angle \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Axial stresses\n",
+ "p=F*(pi*4**-1*d**2)**-1 #N/mm**2\n",
+ "\n",
+ "#Normal Stress\n",
+ "p_n=p*(cos(30*pi*180**-1))**2\n",
+ "\n",
+ "#Tangential Stress\n",
+ "p_t=p*2**-1*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Max shear stress occurs on plane where theta2=45 \n",
+ "theta2=45\n",
+ "sigma_max=p*2**-1*sin(2*theta2*pi*180**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses developed on a plane making 30 degree is:\",round(p_n,2),\"N/mm**2\"\n",
+ "print\" :\",round(p_t,2),\"N/mm**2\"\n",
+ "print\"stress on max shear stress is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses developed on a plane making 30 degree is: 30.56 N/mm**2\n",
+ " : 17.64 N/mm**2\n",
+ "stress on max shear stress is 20.37 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.3,Page No.272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "theta=30 #degree\n",
+ "\n",
+ "#Stresses acting on material\n",
+ "p1=120 #N/mm**2\n",
+ "p2=80 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress\n",
+ "P_n=(p1+p2)*2**-1+(p1-p2)*2**-1*cos(2*theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Tangential stress\n",
+ "P_t=(p1-p2)*2**-1*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Resultant stress\n",
+ "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination to the plane\n",
+ "phi=arctan(P_n*P_t**-1)*(180*pi**-1)\n",
+ "\n",
+ "#Angle made by resultant with 120 #N/mm**2 stress\n",
+ "phi2=phi+theta #Degree\n",
+ "\n",
+ "#Result\n",
+ "print\"Normal Stress is\",round(P_n,2),\"N/mm**2\"\n",
+ "print\"Tangential Stress is\",round(P_t,2),\"N/mm**2\"\n",
+ "print\"Angle made by resultant\",round(phi2,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal Stress is 110.0 N/mm**2\n",
+ "Tangential Stress is 17.32 N/mm**2\n",
+ "Angle made by resultant 111.05 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.4,Page No.272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Direct Stresses\n",
+ "P1=60 #N/mm**2 \n",
+ "P2=100 #N/mm**2\n",
+ "\n",
+ "Theta=25 #Degree #Angle\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress\n",
+ "P_n=(P1-P2)*2**-1+(P1+P2)*2**-1*cos(2*Theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Tangential Stress\n",
+ "P_t=(P1+P2)*2**-1*sin(Theta*2*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Resultant stress\n",
+ "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "theta2=arctan(P_n*P_t**-1)*(180*pi**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses on the plane AC is:\",round(P_n,2),\"N/mm**2\"\n",
+ "print\" \",round(P_t,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses on the plane AC is: 31.42 N/mm**2\n",
+ " 61.28 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.6,Page No.278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Stresses acting on material\n",
+ "p_x=180 #N/mm**2 \n",
+ "p_y=120 #N/mm**2\n",
+ "\n",
+ "q=80 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1) #degrees\n",
+ "theta2=theta*2**-1 #Degrees\n",
+ "theta3=theta+180 #Degrees\n",
+ "theta4=theta3*2**-1 #Degrees\n",
+ "\n",
+ "#Stresses\n",
+ "p_1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p_2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of Principal stress is:\",round(p_1,2),\"N/mm**2\"\n",
+ "print\" \",round(p_2,2),\"N/mm**2\"\n",
+ "print\"Magnitude of max shear stress is\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of Principal stress is: 235.44 N/mm**2\n",
+ " 64.56 N/mm**2\n",
+ "Magnitude of max shear stress is 85.44 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.7,Page No.279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=60 #N/mm**2\n",
+ "p_y=-40 #N/mm**2\n",
+ "\n",
+ "q=10 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Principal Stresses\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" :\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: 60.99 N/mm**2\n",
+ " : -40.99 N/mm**2\n",
+ "Max shear stresses 50.99 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.8,Page No.280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=-120 #N/mm**2\n",
+ "p_y=-80 #N/mm**2\n",
+ "\n",
+ "q=-60 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Principal Stresses\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" :\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: -36.75 N/mm**2\n",
+ " : -163.25 N/mm**2\n",
+ "Max shear stresses 63.25 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.9,Page No.282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=-40 #N/mm**2\n",
+ "p_y=80 #N/mm**2\n",
+ "\n",
+ "q=48 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=((((p_x-p_y)*2**-1)**2)+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Normal Corresponding stress\n",
+ "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*(theta2+45)*pi*180**-1)+q*sin(2*(theta2+45)*pi*180**-1) #Degrees\n",
+ "\n",
+ "#Resultant stress\n",
+ "p=((p_n**2+q_max**2)**0.5) #N/mm**2\n",
+ "\n",
+ "phi=arctan(p_n*q_max**-1)*(180*pi**-1) #Degrees\n",
+ "\n",
+ "#Inclination to the plane\n",
+ "alpha=round((theta2+45),2)+round(phi ,2)#Degree\n",
+ "\n",
+ "#Answer in book is incorrect of alpha ie41.25\n",
+ "\n",
+ "#Result\n",
+ "print\"Planes of max shear stress:\",round(p_n,2),\"N/mm**2\"\n",
+ "print\" \",round(q_max,2),\"N/mm*2\"\n",
+ "print\"Resultant Stress is\",round(p,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Planes of max shear stress: 20.0 N/mm**2\n",
+ " 76.84 N/mm*2\n",
+ "Resultant Stress is 79.4 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.10,Page No.283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Stresses\n",
+ "p_x=50*cos(35*pi*180**-1)\n",
+ "q=50*sin(35*pi*180**-1)\n",
+ "p_y=0\n",
+ "\n",
+ "theta=40 #Degrees #Plane AB inclined to vertical\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress on AB\n",
+ "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*theta*pi*180**-1)+q*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Tangential Stress on AB\n",
+ "p_t=(p_x-p_y)*2**-1*sin(2*theta*pi*180**-1)-q*cos(2*theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Resultant stress\n",
+ "p=(p_n**2+p_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Angle of resultant\n",
+ "phi=arctan(p_n*p_t**-1)*(180*pi**-1) #degrees\n",
+ "phi2=phi+theta #Degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of resultant stress is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Direction of Resultant stress is\",round(phi2,2),\"Degrees\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of resultant stress is 54.44 N/mm**2\n",
+ "Direction of Resultant stress is 113.8 Degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.12,Page No.285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Direct stresses\n",
+ "p_x=120 #N/mm**2 #Tensile stress\n",
+ "p_y=-100 #N/mm**2 #Compressive stress\n",
+ "p1=160 #N/mm**2 #Major principal stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let q be the shearing stress\n",
+ "\n",
+ "#p1=(p_x+p_y)*2**-1+((((p_x+p_y)*2**-1)**2)+q**2)**0.5\n",
+ "#After further simplifying we get\n",
+ "q=(p1-((p_x+p_y)*2**-1))**2-((p_x-p_y)*2**-1)**2 #N/mm**2\n",
+ "q2=(q)**0.5 #N/mm**2\n",
+ "\n",
+ "#Minimum Principal stress\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shearing stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shearing stress of material\",round(q,2),\"N/mm**2\"\n",
+ "print\"Min Principal stress\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shearing stress\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shearing stress of material 10400.0 N/mm**2\n",
+ "Min Principal stress -140.0 N/mm**2\n",
+ "Max shearing stress 150.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.14,Page No.291"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=40*10**3 #N #Shear Force\n",
+ "M=20*10**6 #Bending Moment\n",
+ "\n",
+ "#Rectangular section\n",
+ "b=100 #mm #Width\n",
+ "d=200 #mm #Depth\n",
+ "\n",
+ "x=20 #mm #Distance from Top surface upto point\n",
+ "y=80 #mm #Distance from point to Bottom\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I=1*12**-1*b*d**3 #mm**4 #M.I\n",
+ "\n",
+ "#At 20 mm Below top Fibre\n",
+ "f_x=M*I**-1*y #N/mm**2 #Stress\n",
+ "\n",
+ "#Assuming sagging moment ,f_x is compressive p_x=f_x=-24 #N/mm**2\n",
+ "p_x=f_x=-24 #N/mm**2\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=F*(b*I)**-1*(b*x*(b-x*2**-1)) #N/mm**2\n",
+ "\n",
+ "#Direct stresses\n",
+ "\n",
+ "p_y=0 #N/mm**2\n",
+ "\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Directions of principal stresses at a point below 20mm is:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" \",round(p2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Directions of principal stresses at a point below 20mm is: 0.05 N/mm**2\n",
+ " -24.05 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.15,Page No.292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=4000 #mm #Span\n",
+ "W1=W2=W3=2*10**3 #N #Load\n",
+ "\n",
+ "#SEction of beam\n",
+ "b=100 #mm #Width\n",
+ "d=240 #mm #Dept\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the reactions\n",
+ "R_A=R_B=(W1+W2+W3)*2**-1 #KN\n",
+ "\n",
+ "#Now at the section 1.5m from left support A\n",
+ "#Shear Force\n",
+ "F=R_A-W1 #KN\n",
+ "\n",
+ "#B.M\n",
+ "M=R_A*1.5-W1*0.5 #KN-m\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*d**3 #mm**4\n",
+ "\n",
+ "#Bending stress\n",
+ "#f=M*I**-1*y\n",
+ "#After Sub values and further simplifying we get\n",
+ "#f=3.04*10**-2*y\n",
+ "\n",
+ "#As it varies Linearly\n",
+ "\n",
+ "#at distance 0 From NA \n",
+ "f1=0\n",
+ "#at distance 60 mm from NA\n",
+ "f2=1.823 #N/mm**2\n",
+ "#at distance 120 mm from NA\n",
+ "f3=3.646 #N/mm**2\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=F*b*d*2**-1*d*4**-1*(b*I)**-1\n",
+ "\n",
+ "#At 60 mm above NA\n",
+ "q2=F*b*d*4**-1*(d*2**-1-d*8**-1)*(b*I)**-1\n",
+ "\n",
+ "#At 120 mm above NA\n",
+ "q3=0 \n",
+ "\n",
+ "#At NA element is under pure shear\n",
+ "p1=q #N/mm**2\n",
+ "p2=-q #N/mm**2 \n",
+ "\n",
+ "#Inclination of principal plane to vertical\n",
+ "#theta=2*q*0**-1\n",
+ "#Further simplifying we get\n",
+ "#theta=infinity\n",
+ "\n",
+ "#therefore\n",
+ "theta=90*2**-1 #degrees\n",
+ "theta2=270*2**-1 #degrees\n",
+ "\n",
+ "#At 60 mm From NA\n",
+ "p_x=-1.823 #N/mm**2 \n",
+ "p_y=0\n",
+ "q=0.0469 #N/mm**2\n",
+ "\n",
+ "#principal planes\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Principal planes inclination to hte plane of p_x is given by\n",
+ "theta3=(arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1))\n",
+ "theta4=theta3*2**-1#degrees\n",
+ "\n",
+ "theta5=theta3+180 #Degrees\n",
+ "\n",
+ "#At 120 mm From N-A\n",
+ "p_x2=3.646 #N/mm**2\n",
+ "p_y2=0 #N/mm**2\n",
+ "q2=0 #N/mm**2\n",
+ "\n",
+ "P3=p_x2 #N/mm**2\n",
+ "P4=0 #N/mm**2\n",
+ "\n",
+ "#Answer for P2 at 60 mm from NA is incorrect\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Planes at 60 mm from NA:\",round(p_x,2),\"N/mm**2\"\n",
+ "print\" \",round(p_y,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses at 60 mm From NA\",round(P1,4),\"N/mm**2\"\n",
+ "print\" \",round(P2,4),\"N/mm**2\"\n",
+ "print\"Principal Planes at 60 mm from NA:\",round(p_x2,4),\"N/mm**2\"\n",
+ "print\" \",round(p_y2,4),\"N/mm**2\"\n",
+ "print\"Principal Stresses at 60 mm From NA\",round(P3,4),\"N/mm**2\"\n",
+ "print\" \",round(P4,4),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Planes at 60 mm from NA: -1.82 N/mm**2\n",
+ " 0.0 N/mm**2\n",
+ "Principal Stresses at 60 mm From NA 0.0012 N/mm**2\n",
+ " -1.8242 N/mm**2\n",
+ "Principal Planes at 60 mm from NA: 3.646 N/mm**2\n",
+ " 0.0 N/mm**2\n",
+ "Principal Stresses at 60 mm From NA 3.646 N/mm**2\n",
+ " 0.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.16,Page No.295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=8000 #mm #Span of beam\n",
+ "w=40*10**6 #N/mm #udl\n",
+ "\n",
+ "#I-section\n",
+ "\n",
+ "#Flanges\n",
+ "b=100 #mm #Width\n",
+ "t=10 #mm #Thickness\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "t2=10 #mm #thickness of web\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the Reactions at A & B respectively\n",
+ "R_A=w*2**-1*L*10**-9 #KN\n",
+ "\n",
+ "#Shear force at 2m for left support\n",
+ "F=R_A-2*w*10**-6 #KN\n",
+ "\n",
+ "#Bending Moment\n",
+ "M=R_A*2-2*w*10**-6 #KN-m\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*D**3-1*12**-1*(b-t)*(D-2*t2)**3 #mm**4\n",
+ "\n",
+ "#Bending stress at 100 mm above N_A\n",
+ "f=M*10**6*I**-1*b\n",
+ "\n",
+ "#Shear stress \n",
+ "q=F*10**3*(t*I)**-1*(b*t*(D-t)*2**-1 +t2*(b-t2)*145) #N/mm**2\n",
+ "\n",
+ "p_x=-197.06 #N/mm**2 \n",
+ "p_y=0 #N/mm**2\n",
+ "q=21.38 #N/mm**2\n",
+ "\n",
+ "#Principal Stresses\n",
+ "\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(P1,2),\"N/mm**2\"\n",
+ "print\" \",round(P2,2),\"N/mm**2\"\n",
+ "print\"Max shear stress\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: 2.29 N/mm**2\n",
+ " -199.35 N/mm**2\n",
+ "Max shear stress 100.82 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.18,Page No.298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=100 #mm #Diameter of shaft\n",
+ "M=3*10**6 #N-mm #B.M\n",
+ "T=6*10**6 #N-mm #Twisting Moment\n",
+ "mu=0.3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max principal Stress\n",
+ "\n",
+ "P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5) #N/mm**2 \n",
+ "P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5) #N/mm**2 \n",
+ "\n",
+ "#Direct stress\n",
+ "P=round(P1,2)-mu*round(P2,2) #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Principal stresses are:\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Stress Producing the same strain is\",round(P,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal stresses are: 49.44 N/mm**2\n",
+ " : -18.89 N/mm**2\n",
+ "Stress Producing the same strain is 55.11 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.19,Page No.299"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=75 #mm #diameter \n",
+ "P=30*10**6 #W #Power transmitted\n",
+ "W=6 #N-mm/sec #Load\n",
+ "L=1000 #mm \n",
+ "N=300 #r.p.m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#B.M\n",
+ "M=W*L*4**-1 #N-mm\n",
+ "T=P*60*(2*pi*N)**-1 #Torque transmitted\n",
+ "\n",
+ "#M.I\n",
+ "I=pi*64**-1*d**4 #mm**4\n",
+ "\n",
+ "#Bending stress\n",
+ "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "#At A\n",
+ "p_x=f_A\n",
+ "p_y=0\n",
+ "\n",
+ "#Polar Modulus\n",
+ "J=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=T*J**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "#Principal Stresses\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Bending stress\n",
+ "p_x2=0\n",
+ "p_y2=0\n",
+ "\n",
+ "#Shearing stress\n",
+ "q2=T*J**-1*d*2**-1 #N/mm**2\n",
+ "\n",
+ "#Principal stresses\n",
+ "P3=(p_x2+p_y2)*2**-1+(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "P4=(p_x2+p_y2)*2**-1-(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max2=(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Answer for Principal Stresses P1,P2 and Max stress i.e q_max is incorrect in Book\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses at vertical Diameter:P1\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :P2\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Max stress at vertical Diameter : \",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses at Horizontal Diameter:P3\",round(P3,2),\"N/mm**2\"\n",
+ "print\" :P4\",round(P4,2),\"N/mm**2\"\n",
+ "print\"Max stress at Horizontal Diameter : \",round(q_max2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses at vertical Diameter:P1 11.55 N/mm**2\n",
+ " :P2 -11.51 N/mm**2\n",
+ "Max stress at vertical Diameter : 11.53 N/mm**2\n",
+ "Principal Stresses at Horizontal Diameter:P3 11.53 N/mm**2\n",
+ " :P4 -11.53 N/mm**2\n",
+ "Max stress at Horizontal Diameter : 11.53 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.20,Page No.302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=100 #mm #External Diameter\n",
+ "d2=50 #mm #Internal Diameter\n",
+ "N=500 #mm #r.p.m\n",
+ "P=60*10**6 #N-mm/sec #Power\n",
+ "p=100 #N/mm**2 #principal stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I\n",
+ "I=pi*(d1**4-d2**4)*64**-1 #mm**4\n",
+ "\n",
+ "#Bending Stress\n",
+ "#f=M*I*d1*2**-1 #N/mm**2\n",
+ "\n",
+ "#Principal Planes\n",
+ "#p_x=32*M*(pi*(d1**4-d2**4))*d1\n",
+ "#p_y=0\n",
+ "\n",
+ "#Shear stress\n",
+ "#q=T*J**-1*(d1*2**-1)\n",
+ "#After sub values and further simplifying we get\n",
+ "#q=16*T*d1*(pi*(d1**4-d2**4))*d1\n",
+ "\n",
+ "#Principal stresses\n",
+ "#P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "#After sub values and further simplifying we get\n",
+ "#P1=16*(pi*(d1**4-d2**4))*d1*(M+(M**2+t**2)**0.5) ...............(1)\n",
+ "\n",
+ "#P=2*pi*N*T*60**-1\n",
+ "#After sub values and further simplifying we get\n",
+ "T=P*60*(2*pi*N)**-1*10**-6 #N-mm\n",
+ "\n",
+ "#Again Sub values and further simplifying Equation 1 we get\n",
+ "M=(337.533)*(36.84)**-1 #KN-m\n",
+ "\n",
+ "#Min Principal stress\n",
+ "#P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "#Sub values and further simplifying we get\n",
+ "P2=16*(pi*(d1**4-d2**4))*d1*(M-(M**2+T**2)**0.5)*10**-11\n",
+ "\n",
+ "#Result\n",
+ "print\"Bending Moment safely applied to shaft is\",round(M,2),\"KN-m\"\n",
+ "print\"Min Principal Stress is\",round(P2,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bending Moment safely applied to shaft is 9.16 KN-m\n",
+ "Min Principal Stress is -0.336 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.21,Page No.303"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=150 #mm #Diameter\n",
+ "T=20*10**6 #N #Torque\n",
+ "M=12*10**6 #N-mm #B.M\n",
+ "F=200*10**3 #N #Axial Thrust\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I\n",
+ "I=(pi*64**-1*d**4)\n",
+ "\n",
+ "#Bending stress \n",
+ "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",
+ "f_B=-f_A #N/mm**2\n",
+ "\n",
+ "#Axial thrust due to thrust\n",
+ "sigma=F*(pi*4**-1*d**2)**-1\n",
+ "\n",
+ "#At A\n",
+ "p_x=f_A-sigma #N/mm**2\n",
+ "\n",
+ "#At B\n",
+ "p_x2=f_B-sigma #N/mm**2\n",
+ "\n",
+ "p_y=0 #At A and B\n",
+ "\n",
+ "#Polar Modulus\n",
+ "J=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Shearing stress at A and B\n",
+ "q=T*J**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Principal Stresses\n",
+ "#At A\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max1=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#At B\n",
+ "P1_2=(p_x2+p_y)*2**-1+(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2_2=(p_x2+p_y)*2**-1-(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max2=(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"MAx Principal Stresses:P1\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :P2\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Min Principal Stresses:P1_2\",round(P1_2,2),\"N/mm**2\"\n",
+ "print\" :P2_2\",round(P2_2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "MAx Principal Stresses:P1 45.1 N/mm**2\n",
+ " :P2 -20.2 N/mm**2\n",
+ "Min Principal Stresses:P1_2 14.65 N/mm**2\n",
+ " :P2_2 -62.18 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.22,Page No.311"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#strains\n",
+ "e_A=500 #microns\n",
+ "e_B=250 #microns\n",
+ "e_C=-150 #microns\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "theta=45 #Degrees\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "e_x=e_A=500\n",
+ "e_45=e_B=250\n",
+ "e_y=e_C=-150 \n",
+ "\n",
+ "#e_45=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "rho_x_y=(e_45-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2\n",
+ "\n",
+ "#Principal strains are given by\n",
+ "e1=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "e2=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "\n",
+ "#Principal Stresses\n",
+ "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Strains are:e1\",round(e1,2),\"N/mm**2\"\n",
+ "print\" :e2\",round(e2,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\" :sigma2\",round(sigma2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Strains are:e1 508.54 N/mm**2\n",
+ " :e2 -158.54 N/mm**2\n",
+ "Principal Stresses are:sigma1 101.31 N/mm**2\n",
+ " :sigma2 -1.31 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.23,Page No.313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Strains\n",
+ "e_A=600 #microns\n",
+ "e_B=-450 #microns\n",
+ "e_C=100 #micron\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "theta=240\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "e_x=e_A=600\n",
+ "\n",
+ "#e_A=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(theta)+rho_x_y*2**-1*sin(theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "#-450=(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(1)\n",
+ "\n",
+ "#e_C=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "#100=(e_x+e_y)*2**-1-0.5*(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(2)\n",
+ "\n",
+ "#Adding Equation 1 and 2 we get equations as\n",
+ "#-350=e_x+e_y-(e_x-e_y)*2**-1 ...............(3)\n",
+ "#Further simplifying we get\n",
+ "\n",
+ "e_y=(-700-e_x)*3**-1 #micron \n",
+ "\n",
+ "rho_x_y=(e_C-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2 #micron\n",
+ "\n",
+ "#Principal strains\n",
+ "e1=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "e2=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "\n",
+ "#Principal Stresses\n",
+ "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\" :sigma2\",round(sigma2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are:sigma1 -69.49 N/mm**2\n",
+ " :sigma2 117.11 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_4.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_4.ipynb new file mode 100644 index 00000000..08f57912 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_4.ipynb @@ -0,0 +1,1527 @@ +{
+ "metadata": {
+ "name": "chapter no.8.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.8:Thin And Thick Cyclinders And Spheres"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.1,Page No.322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #Length\n",
+ "d1=1000 #mm #Internal diameter\n",
+ "t=15 #mm #Thickness\n",
+ "P=1.5 #N/mm**2 #Fluid Pressure\n",
+ "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=P*d1*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Longitudinal Stress\n",
+ "f2=P*d1*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(f1-f2)*2**-1 #N/mm**2\n",
+ "\n",
+ "#Diametrical Strain\n",
+ "#Let e1=dell_d*d**-1 .....................(1)\n",
+ "e1=(f1-mu*f2)*E**-1 \n",
+ "\n",
+ "#Sub values in equation 1 and further simplifying we get\n",
+ "dell_d=e1*d1 #mm\n",
+ "\n",
+ "#Longitudinal strain\n",
+ "#e2=dell_L*L**-1 ......................(2)\n",
+ "e2=(f2-mu*f1)*E**-1 \n",
+ "\n",
+ "#Sub values in equation 2 and further simplifying we get\n",
+ "dell_L=e2*L #mm\n",
+ "\n",
+ "#Change in Volume \n",
+ "#Let Z=dell_V*V**-1 ................(3)\n",
+ "Z=2*e1+e2\n",
+ "\n",
+ "#Sub values in equation 3 and further simplifying we get\n",
+ "dell_V=Z*pi*4**-1*d1**2*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Intensity of shear stress\",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Change in the Dimensions of the shell is:dell_d\",round(dell_d,2),\"mm\"\n",
+ "print\" :dell_L\",round(dell_L,2),\"mm\"\n",
+ "print\" :dell_V\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Intensity of shear stress 12.5 N/mm**2\n",
+ "Change in the Dimensions of the shell is:dell_d 0.21 mm\n",
+ " :dell_L 0.15 mm\n",
+ " :dell_V 1119192.38 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.2,Page No.323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=2000 #mm #Length\n",
+ "d=200 #mm # diameter\n",
+ "t=10 #mm #Thickness\n",
+ "dell_V=25000 #mm**3 #Additional volume\n",
+ "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p be the pressure developed\n",
+ "\n",
+ "#Circumferential Stress\n",
+ "\n",
+ "#f1=p*d*(2*t)**-1 #N/mm**2\n",
+ "#After sub values and further simplifying\n",
+ "#f1=10*p\n",
+ "\n",
+ "#f1=p*d*(4*t)**-1 #N/mm**2\n",
+ "#After sub values and further simplifying\n",
+ "#f1=5*p\n",
+ "\n",
+ "#Diameterical strain = Circumferential stress\n",
+ "#Let X=dell_d*d**-1 ................................(1)\n",
+ "#X=e1=(f1-mu*f2)*E**-1 \n",
+ "#After sub values and further simplifying\n",
+ "#e1=8.5*p*E**-1\n",
+ "\n",
+ "#Longitudinal strain\n",
+ "#Let Y=dell_L*L**-1 ......................................(2)\n",
+ "#Y=e2=(f2-mu*f1)*E**-1 \n",
+ "#After sub values and further simplifying\n",
+ "#e2=2*p*E**-1\n",
+ "\n",
+ "#Volumetric strain\n",
+ "#Let X=dell_V*V**-1 \n",
+ "#X=2*e1+e2\n",
+ "#After sub values and further simplifying\n",
+ "#X=19*p*E**-1\n",
+ "#After further simplifying we get\n",
+ "p=dell_V*(pi*4**-1*d**2*L)**-1*E*19**-1 #N/mm**2\n",
+ "\n",
+ "#Hoop Stress\n",
+ "f1=p*d*(2*t)**-1\n",
+ "\n",
+ "X=e1=8.5*p*E**-1\n",
+ "#Sub value of X in equation 1 we get\n",
+ "dell_d=8.5*p*E**-1*d\n",
+ "\n",
+ "Y=e2=2*p*E**-1\n",
+ "#Sub value of Y in equation 2 we get\n",
+ "dell_L=2*p*E**-1*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Pressure Developed is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Hoop stress Developed is\",round(f1,2),\"N/mm**2\"\n",
+ "print\"Change in diameter is\",round(dell_d,2),\"mm\"\n",
+ "print\"Change in Length is\",round(dell_L,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pressure Developed is 4.19 N/mm**2\n",
+ "Hoop stress Developed is 41.88 N/mm**2\n",
+ "Change in diameter is 0.04 mm\n",
+ "Change in Length is 0.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.3,Page No.324"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=750 #mm #Diameter of water supply pipes\n",
+ "h=50*10**3 #mm #Water head\n",
+ "sigma=20 #N/mm**2 #Permissible stress\n",
+ "rho=9810*10**-9 #N/mm**3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Pressure of water\n",
+ "P=rho*h #N/mm**2\n",
+ "\n",
+ "#Stress\n",
+ "#sigma=p*d*(2*t)**-1\n",
+ "#After further simplifying\n",
+ "t=P*d*(2*sigma)**-1 #mm \n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of seamless pipe is\",round(t,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of seamless pipe is 9.197 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.4,Page No.326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=2500 #mm #Diameter of riveted boiler\n",
+ "P=1 #N/mm**2 #Pressure\n",
+ "rho1=0.7 #Percent efficiency\n",
+ "rho2=0.4 #Circumferential joints\n",
+ "sigma=150 #N/mm**2 #Permissible stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Equating Bursting force to longitudinal joint strength ,we get\n",
+ "#p*d*L=rho1*2*t*L*sigma\n",
+ "#After rearranging and further simplifying we get\n",
+ "t=P*d*(2*sigma*rho1)**-1 #mm\n",
+ "\n",
+ "#Considering Longitudinal force\n",
+ "#pi*d**2*4**-1*P=rho2*pi*d*t*sigma\n",
+ "#After rearranging and further simplifying we get\n",
+ "t2=P*d*(4*sigma*rho2)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of plate required is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of plate required is 11.9 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.5,Page No.326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Boiler Dimensions\n",
+ "t=16 #mm #Thickness\n",
+ "p=2 #N/mm**2 #internal pressure\n",
+ "f=150 #N/mm**2 #Permissible stress\n",
+ "rho1=0.75 #Longitudinal joints\n",
+ "rho2=0.45 #circumferential joints\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Equating Bursting force to longitudinal joint strength ,we get\n",
+ "d1=rho1*2*t*f*p**-1 #mm\n",
+ "\n",
+ "#Considering circumferential strength \n",
+ "d2=4*rho2*t*f*p**-1 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Largest diameter of Boiler is\",round(d1,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Largest diameter of Boiler is 1800.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.6,Page No.329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=250 #mm #Diameter iron pipe\n",
+ "t=10 #mm #Thickness\n",
+ "d2=6 #mm #Diameter of steel\n",
+ "p=80 #N/mm**2 #stress\n",
+ "P=3 #N/mm**2 #Pressure\n",
+ "E_c=1*10**5 #N/mm**2\n",
+ "mu=0.3 #poissoin's ratio\n",
+ "E_s=2*10**5 #N/mm**2\n",
+ "n=1 #No.of wires\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "L=6 #mm #Length of cyclinder\n",
+ "\n",
+ "#Force Exerted by steel wire at diameterical section\n",
+ "F=p*2*pi*d2**2*1*4**-1 #N\n",
+ "\n",
+ "#Initial stress in cyclinder\n",
+ "f_c=F*(2*t*d2)**-1 #N/mm**2\n",
+ "\n",
+ "#LEt due to fluid pressure alone stresses developed in steel wire be F_w and in cyclinder f1 and f2\n",
+ "f2=P*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Considering the equilibrium of half the cyclinder, 6mm long we get\n",
+ "#F_w*2*pi*4**-1*d2**2*n+f1*2*t*d2=P*d*d2\n",
+ "#After further simplifying we get\n",
+ "#F_w+2.122*f1=79.58 . ......................................(1)\n",
+ "\n",
+ "#Equating strain in wire to circumferential strain in cyclinder \n",
+ "#F_w=(f1-mu*f2)*E_s*E_c**-1 #N/mm**2\n",
+ "#After further simplifying we get\n",
+ "#F_w=2*f1-11.25 ....................................(2)\n",
+ "\n",
+ "#Sub in equation in1 we get\n",
+ "f1=(79.58+11.25)*(4.122)**-1 #N/mm**2\n",
+ "F_w=2*f1-11.25 #N/mm**2\n",
+ "\n",
+ "#Final stresses\n",
+ "#1) In steel Wire\n",
+ "sigma=F_w+p #N/mm**2\n",
+ "\n",
+ "#2) In Cyclinder\n",
+ "sigma2=f1-f_c\n",
+ "\n",
+ "#Result\n",
+ "print\"Final Stresses developed in:cyclinder is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\" :Steel is\",round(sigma2,2),\"N/mm**2\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Final Stresses developed in:cyclinder is 112.82 N/mm**2\n",
+ " :Steel is -15.66 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.7,Page No.332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=750 #mm #Diameter of shell\n",
+ "t=8 #mm #THickness\n",
+ "p=2.5 #N/mm**2\n",
+ "E=2*10**5 #N/mm**2\n",
+ "mu=0.25 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=f2=p*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Change in Diameter\n",
+ "dell_d=d*p*d*(1-mu)*(4*t*E)**-1 #mm\n",
+ "\n",
+ "#Change in Volume\n",
+ "dell_V=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",
+ "\n",
+ "#Answer for Change in diameter is incorrect in book\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress introduced is\",round(f1,2),\"N/mm**2\"\n",
+ "print\"Change in Diameter is\",round(dell_d,2),\"N/mm**2\"\n",
+ "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress introduced is 58.59 N/mm**2\n",
+ "Change in Diameter is 0.16 N/mm**2\n",
+ "Change in Volume is 145608.33 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.8,Page No.333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=600 #mm #Diameter of sherical shell\n",
+ "t=10 #mm #Thickness\n",
+ "f=80 #N/mm**2 #Permissible stress\n",
+ "rho=0.75 #Efficiency joint\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max Pressure\n",
+ "p=f*4*t*rho*d**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Pressure is\",round(p,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Pressure is 4.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.9,Page No.333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of shell\n",
+ "d=200 #mm #Diameter\n",
+ "t=6 #mm #Thickness\n",
+ "p=1.5 #N/mm**2 #Internal Pressure\n",
+ "E=2*10**5 #N/mm**2\n",
+ "mu=0.25 #Poissoin's Ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Change in Volume of sphere\n",
+ "dell_V_s=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=p*d*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Longitudinal stress\n",
+ "f2=p*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Principal strain\n",
+ "e1=(f1-mu*f2)*E**-1\n",
+ "e2=(f2-mu*f1)*E**-1\n",
+ "\n",
+ "V_c=1000 #mm**3\n",
+ "\n",
+ "#Change in Volume of cyclinder\n",
+ "dell_V_c=(2*e1+e2)*pi*4**-1*d**2*L\n",
+ "\n",
+ "#Total Change in Diameter\n",
+ "dell_V=dell_V_s+dell_V_c #mm**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in Volume is 8443.03 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.10,Page No.337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=400 #mm #Internal Diameter\n",
+ "t=100 #mm #Thickness\n",
+ "p=80 #N/mm**2 #Fluid pressure\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Internal Radius\n",
+ "r1=d1*2**-1 #mm\n",
+ "\n",
+ "#Outer Radius\n",
+ "r_o=r1+t #mm\n",
+ "\n",
+ "p1=80 #N/mm**2\n",
+ "p2=0\n",
+ "\n",
+ "#Now From Lame's Euation\n",
+ "#p_x=b*(x**2)**-1-a\n",
+ "#at x=200 #mm \n",
+ "p_x=80 #N/mm**2\n",
+ "#80=b*(200**2)**-1-a ..........................(1)\n",
+ "\n",
+ "#at x=300 #mm\n",
+ "#p_x2=0\n",
+ "#0=b*(300**2)**-1-a ...........................(2)\n",
+ "\n",
+ "#Sub equation 2 from 1\n",
+ "#80=b*(200**2)**-1-b*(300**2)**-1\n",
+ "#After Further simplifying we get\n",
+ "b=(50000)**-1*(200**2*300**2*80)\n",
+ "\n",
+ "#From equation 2 we get\n",
+ "a=b*(300**2)**-1\n",
+ "\n",
+ "#Variation of radial pressure p_x;\n",
+ "#p_x=b*(x**2)**-1-a\n",
+ "#After sub values and further simplifying we get\n",
+ "\n",
+ "#Radial pressure Variation\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "p_x=b*(x**2)**-1-a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=250 #mm\n",
+ "p_x2=b*(x2**2)**-1-a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x3=300 #mm\n",
+ "p_x3=b*(x3**2)**-1-a #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Hoop stress Distribution\n",
+ "#Variation of F_x\n",
+ "\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=250 #mm\n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x3=300 #mm\n",
+ "F_x3=b*(x3**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Hoop stress is\",round(F_x,2),\"N/mm**2\"\n",
+ "print\"Min Hoop stress is\",round(F_x3,2),\"N/mm**2\"\n",
+ "print\"Plot of Hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[p_x,p_x2,p_x3]\n",
+ "Y2=[-F_x,-F_x2,-F_x3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Y2,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Hoop stress is 208.0 N/mm**2\n",
+ "Min Hoop stress is 128.0 N/mm**2\n",
+ "Plot of Hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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skdWFW7ZsQffu3XH8+HGMGzcOY8aMAaA9wnfKlCkIDg7GmDFjEBcXJ7WK4uLi\nMGvWLPTq1QsBAQEcaCcishCN2iIlIiICycnJ5ojHZNwihYio6WTZImXVqlXSBxcWFuLDDz+UKlEo\nFJg3b55x0RIRUZuiN5GUlJRI3UqzZs1CSUmJ2YIiIqLWo1FdW60Ju7aIiJpO9oOtiIiI9GEiISIi\nkzCREBGRSRqdSBYsWIDExEQIIfDKK6/IGRMREbUijU4kkZGRWLlyJUJDQ3Hjxg05YyIiolZEbyL5\n+OOPcfHiRenxgw8+iNLSUri4uKB3795mCY6IiCyf3kTyr3/9Cz169AAAXLt2DSNHjkRQUBAOHz6M\nzZs3my1AIiKybHoTiUajQWlpKbKysjB06FBERUXhgw8+gJWVFSoqKswZIxERWTC9K9vnz58Pf39/\naDQa+Pv7w9nZGVlZWdiwYQO7toiISNLgynaNRiP9fe2117B3715ERETgo48+QpcuXcwWZFNwZTsR\nUdOZ8tvJLVKIiNqpalGNwpuFyC3Jhdpb3fy7/xIRUetVWVWJ/JJ85BTnILckV/u3OBc5JX/+Lc5B\nfmk+XDq4QOms/8TZxmCLhIiolSm5VaKbHGoniz//Xiu/Bk8nTyhdlPBx8YHS+a6/Lkp4O3vD3sYe\nALu2dDCREFFrVS2qcaXsSr3JoXaZplqjNznUPO7WsRusrawbXbesiaSiogKbNm1CVlaWNPiuUCjw\n1ltvGVWh3JhIiMgS3a66jbySvDpJofbf/JJ8ONk53UkKztq/dyeKTh06SedFNRdZTkis8fDDD8PV\n1RVqtRr29vZGVUJE1JaVVpbWTQ53jUcUlRfBw8mjTitC5aWSHns7e8PB1qGlv06TGWyR9OvXD7/9\n9pu54jEZWyRE1FyEENqupgbGI3KLc1FZVVmna+nuLiePjh5N6moyN1lbJIMHD0ZKSgpCQ0ONqoCI\nyBLdrrqN/NJ8neRwdysiryQPHe061kkOg7sP1ilztXdt9q6m1sRgiyQoKAhpaWnw8/NDhw4dtG9S\nKJCSkmKWAJuKLRIiull5s95B6tp/r5ZdRbeO3RpsRSidla2yq8kYsg62Z2VlSZUAkCry9fU1qkK5\nMZEQtV1CCFwtv2pwPOJW1S2Ds5o8nDxgY8WldDVkn/576tQpHD58GAqFAkOHDkVYWJhRldXYuHEj\nlixZgt9//x0nT56ESqUCoE1aQUFBCAwMBAAMGjQIcXFxAIDExEQ89dRTqKiowNixYxEbG1v/F2Ii\nIWqVNNUdY9fjAAAd4UlEQVQa5Jfk1zseUVOWV5IHBxuHOrOa7k4WbvZu7bqryRiyjpHExsbis88+\nw6RJkyCEwPTp0/Hss89izpw5RlUIACEhIdiyZQuee+65Os8FBAQgOTm5Tvns2bPxxRdfIDIyEmPH\njsXu3bsxevRoo2MgIvO5WXmzTjfT3a2IK2VX0LVjVykp1CSGMM+wO2UuSjjaOrb016G7GEwkn3/+\nOU6cOIGOHTsCAF599VUMHDjQpERS0+JorPz8fJSUlCAyMhIAEBMTg61btzKRELUwIQSKyosMrrKu\n0FRIiaAmKfRy74Vo32ipFeHp5MmuplaqUf+rWVlZ1XtfDpmZmYiIiECnTp3w3nvv4d5770Vubi58\nfHyk1yiVSuTm5soaB1F7p6nWoKC0QG9yyC3WdjnZ29jXGY+IUkZhUtAk6XFnh87samrDDCaSmTNn\nIioqSura2rp1K55++mmDHzxq1CgUFBTUKV+6dCnGjx9f73u8vb2RnZ0NNzc3JCUlYcKECThz5kwj\nvgYRNUXZ7TIpEegbjyi8WYgujl10ZjD5uPggpFuITllHu44t/XWohRlMJPPmzcOwYcNw5MgRKBQK\nrFu3DhEREQY/+Mcff2xyMHZ2drCzswMAqFQq+Pv748KFC1AqlcjJyZFel5OTA6VS/26VS5Yske5H\nR0cjOjq6ybEQtUZCCFyruKbTYqhvVlPZ7TLdrTeclQjoHIBhvsOkx55OnrC1tm3pr0QyiY+PR3x8\nfLN8lt5ZW8XFxXBxcUFRURGAO9N+a5qnnTt3Nrny4cOH44MPPoBarQYAXLlyBW5ubrC2tkZGRgb+\n8pe/4LfffoOrqyuioqKwevVqREZGYty4cZgzZ069YySctUVtVVV1lbarycAqaztrO4Ozmtwd3NnV\nRDpkmf47btw4/PDDD/D19a33P7jMzEyjKgSALVu2YM6cObhy5Qo6deqEiIgI7Nq1C5s2bcLixYth\na2sLKysrvPPOOxg3bhyAO9N/y8vLMXbsWKxevbr+L8REQq1Q+e3yBhfP5Rbn4vLNy3B3dK8zHlE7\nUShdlHCyc2rpr0OtELeRr4WJhCyNEAI5xTlILUxFTnFOvYniZuVNeDt7N7jK2svJi11NJBtZE8mI\nESOwf/9+g2WWgomEWpIQAlnXs5CUn4TE/EQk5SchKT8JVgorhHiEoLtL93pXWXdx7MKuJmpRsixI\nLC8vR1lZGQoLC6VxEkA7dsKpt0TapJF+LV2bNPISkVSgTRr2NvZQe6mh8lLh/wb8H9Teang5eTFR\nUJulN5F8+umniI2NRV5enjQYDgDOzs548cUXzRIckaWoFtW4cPWC1NJIzE9Ecn4yXDq4QO2thtpL\njbkD50LlpYKnk2dLh0tkVga7ttasWYOXXnrJXPGYjF1bZKqq6iqcu3pO28r4M3GcKjiFLo5doPJS\nSa0NlZcKXTt2belwiZqFrGMkX375Zb1N8piYGKMqlBsTCTWFplqDs4Vnta2MP7unfi34FV7OXlLS\nUHupEeEVgc4Opk95J7JUsm7aePLkSSmRlJeX48CBA1CpVBabSIj0qayqxJnLZ3QGwk9fPo3uLt2h\n9lZD5anC5ODJCPcMh6u9a0uHS9RqNHn67/Xr1/HYY49hz549csVkErZICABuaW7h9OXTOgPhZy6f\ngZ+bn9Q1pfZSI9wzHM4dnFs6XKIWJ2uL5G6Ojo4mLUYkam7lt8uRcilFamUk5ifi3JVz6OXeS0oY\nM8JnIMwjjPtCEcnAYCKpvcFidXU1UlNTMWXKFFmDItLnZuVN/HrpV6mVkZiXiLSiNAR2CZSSxrOq\nZxHqEdpujkglamkGu7ZqNvVSKBSwsbFBjx490L17d3PEZhR2bbUdJbdKkFyQrDOmkXktE3279dXp\nnurXrR862HRo6XCJWjXZt0jJz89HQkICrKysMGDAAHh6Wu48eSaS1ulGxQ1pFXhN0sguzkZIt5A7\nScNbjeCuwbCztmvpcInaHFkTyeeff4533nkHw4cPB6Btobz11lt45plnjKpQbkwklq+ovEhnEDwx\nLxEFpQUI8wyTptuqvFQI6hrEE/OIzETWRNK7d28cO3YM7u7uAICrV69i0KBBOH/+vFEVyo2JxLIU\n3izUaWUk5ifiatlVRHhFQOWpbWWovFTo494H1lbWLR0uUbsl66ytLl26wMnpzrbUTk5O6NKli1GV\nUdtWUFogtTRqEkfxrWJpFfjkoMl4/7730cu9F6wU8h7ZTETmozeRrFq1CgAQEBCAqKgoTJgwAQCw\nbds2hIaGmic6skhCCOSV5Om0MpLyk1ChqZAGwKeFTMOq+1fBz82PSYOojdObSEpKSqBQKODv74+e\nPXtKq9sffvhh7mLajgghkF2crbPvVFJ+EqpElTSe8VTYU1gzZg3u6XQP/9sgaod4sBVJhBDIvJ5Z\nZ1t0a4W1tMNtzUC4j4sPkwZRGyLLYPvLL7+M2NhYnQWJtSvcvn27URXKjYmkcapFNdKL0uscwORo\n6yjtO1UzEO7t7N3S4RKRzGRJJImJiVCr1Th06FCdD1coFBg2bJhRFcqNiaSuquoqXCi6oNM9lVyQ\nDFd7V52FfSovFTycPFo6XCJqAbJN/9VoNIiJicH69euNDs7c2nsi0VRrcO7KOZ2B8FMFp9CtY7c6\nZ2l0ceTsOyLSkm36r42NDS5evIhbt26hQwduQWFpblfdxtkrZ3Wm26ZcSoG3s7eUNMb3Hg+Vlwpu\nDm4tHS4RtVEG15H4+fnh3nvvxUMPPQRHR0cA2sw1b9482YOjOyqrKvHb5d90BsJ/u/wbenTqIbUy\nHg1+FOGe4ehk36mlwyWidsRgIvH394e/vz+qq6tRWlpqjpjavQpNBU5fOq1zPvjZwrPo6dZTGgh/\nIvQJhHuGw8nOyfAHEhHJyGAiCQ4OrrNt/IYNG0yqdMGCBdixYwfs7Ozg7++P//znP+jUSfuv6GXL\nlmHt2rWwtrbG6tWrcf/99wPQDv4/9dRTqKiowNixYxEbG2tSDJai7HaZ9iyNWgPh56+eRy/3XtJ0\n25nhMxHmGQZHW8eWDpeIqA6D60giIiKQnJxssKwpfvzxR4wYMQJWVlZ49dVXAQDLly9Hamoqpk2b\nhpMnTyI3NxcjR47EhQsXoFAoEBkZiX/+85+IjIzE2LFjMWfOHIwePbruF7LgwfbSylL8WvCrzkB4\nelE6groG6Uy3DfUIhb2NfUuHS0TtiCyD7bt27cLOnTuRm5uLOXPmSBWUlJTA1tbWuEj/NGrUKOl+\nVFQUNm3aBEC7/crUqVNha2sLX19fBAQE4MSJE7jnnntQUlKCyMhIAEBMTAy2bt1abyKxFMW3ipGc\nr3uWxh83/kDfrn2h8lJhSPchmBM1B3279uVZGkTUqulNJN7e3lCr1di2bRvUarWUSFxcXPCPf/yj\n2QJYu3Ytpk6dCgDIy8vDwIEDped8fHyQm5sLW1tb+Pj4SOVKpRK5ubnNFoOprldcr3OWRk5xDkI9\nQqH2UuM+v/uwYPACBHcNhq21aUmYiMjS6E0kYWFhCAsLwxNPPCG1QIqKipCTkwM3N8NTSUeNGoWC\ngoI65UuXLpVWy7///vuws7PDtGnTjI3f7K6WXa2zLfrlm5cR5qE9S2O0/2i8MfQNBHYJ5FkaRNQu\nGPylGzVqFLZv3w6NRgO1Wo2uXbtiyJAhBlslP/74Y4PPr1u3Djt37sT+/fulMqVSiezsbOlxTk4O\nfHx8oFQqkZOTo1OuVCr1fvaSJUuk+9HR0YiOjm4wFn0u37xc5wCmaxXXEOEZAZWXCg/3eRhvR7+N\n3u69eZYGEbUq8fHx0lHqpjI42B4eHo5Tp07h888/R3Z2Nt5++22EhITg9OnTRle6e/duzJ8/H4cO\nHdI526RmsD0hIUEabE9LS4NCoUBUVBRWr16NyMhIjBs3rtkH2/NL8nWm2yblJ6G0slS7CrzWQHhA\n5wBui05EbY6sB1tVVVUhPz8fGzZswHvvvSdVaIqXXnoJlZWV0qD7oEGDEBcXJ001Dg4Oho2NDeLi\n4qS64uLi8NRTT6G8vBxjx441eqBdCIHcktw626LfqrolTbedHjId/3jgH/Bz9eMOt0REBhhskWzc\nuBHvvvsuhgwZgo8//hjp6elYuHChNNPK0tTOqkIIXLxxUdvKqNU9BUDaFr1mK5EenXowaRBRuyXr\nme2tjUKhwKIfF0mzqOys7XQ2K1R7q6F0VjJpEBHVIkvX1ooVK7Bo0SK89NJLdSpQKBRYvXq1URWa\ng6OtI16OehkqLxW8nL1aOhwiojZNbyIJDg4GAKjV6jrPWfq/5t8a9lZLh0BE1G60ya6tNvaViIhk\nZ8pvZ4PzWNetWweVSgVHR0c4Ojqif//++PLLL42qiIiI2ia9XVtffvklYmNj8eGHHyIiIgJCCCQn\nJ2PBggVQKBSIiYkxZ5xERGSh9HZtRUVF4ZtvvoGfn59OeVZWFh577DGcOHHCLAE2Fbu2iIiaTpau\nrZKSkjpJBAB8fX1RUlJiVGVERNT26E0k9vb6z8No6DkiImpf9HZtOTg4ICAgoN43paeno6ysTNbA\njMWuLSKippNlQeLZs2eNDoiIiNoPriMhIiL51pEQEREZwkRCREQmaVIiKSoqQkpKilyxEBFRK2Qw\nkQwbNgzFxcUoKiqCWq3GrFmzMHfuXHPERkRErYDBRHLjxg24uLhg8+bNiImJQUJCAvbt22eO2IiI\nqBUwmEhqH7U7btw4AJa/jTwREZmPwUTy1ltv4YEHHoC/vz8iIyORnp6OXr16mSM2IiJqBbiOhIiI\n5F1HsnDhQhQXF+P27dsYMWIEunTpgv/9739GVUZERG2PwUSyZ88euLi4YMeOHfD19UV6ejr+/ve/\nmyM2IiJqBQwmEo1GAwDYsWMHHnnkEXTq1ImD7UREJDGYSMaPH4/AwEAkJiZixIgRuHz5ssnbyC9Y\nsABBQUEICwvDpEmTcOPGDQDaQ7McHBwQERGBiIgIvPDCC9J7EhMTERISgl69euHll182qX4iImpG\nohGuXr0qNBqNEEKI0tJSkZ+f35i36bV3715RVVUlhBBi0aJFYtGiRUIIITIzM0W/fv3qfc+AAQPE\niRMnhBBCjBkzRuzatave1zXyK7ULBw8ebOkQLAavxR28FnfwWtxhym+nwRbJzZs38a9//QvPP/88\nACAvLw+//PKLSclr1KhRsLLSVh0VFYWcnJwGX5+fn4+SkhJERkYCAGJiYrB161aTYmgP4uPjWzoE\ni8FrcQevxR28Fs3DYCKZOXMm7OzscPToUQCAt7c33njjjWYLYO3atRg7dqz0ODMzExEREYiOjsaR\nI0cAALm5ufDx8ZFeo1QqkZub22wxEBGR8fQebFUjPT0dGzZswDfffAMA6NixY6M+eNSoUSgoKKhT\nvnTpUowfPx4A8P7778POzg7Tpk0DoE1S2dnZcHNzQ1JSEiZMmIAzZ840+ssQEVELMNT3NWjQIFFW\nVibCw8OFEEKkpaWJAQMGGN2XVuM///mPGDx4sCgvL9f7mujoaJGYmCjy8vJEYGCgVL5+/Xrx3HPP\n1fsef39/AYA33njjjbcm3Pz9/Y3+PTfYIlmyZAlGjx6NnJwcTJs2DT///DPWrVtn6G0N2r17N/7+\n97/j0KFDOjPArly5Ajc3N1hbWyMjIwMXLlxAz5494erqChcXF5w4cQKRkZH43//+hzlz5tT72Wlp\naSbFRkRETdPgFinV1dXYuHEjRowYgePHjwPQDo537drVpEp79eqFyspKdO7cGQAwaNAgxMXFYdOm\nTVi8eDFsbW1hZWWFd955R9ooMjExEU899RTKy8sxduxYrF692qQYiIioeRjca0utViMxMdFc8RAR\nUStjcNbWqFGj8MEHHyA7OxtFRUXSrSVkZ2dj+PDh6Nu3L/r16ye1SoqKijBq1Cj07t0b999/P65f\nvy69Z9myZejVqxcCAwOxd+/eFolbDvquhb7FnkD7uxY1Vq1aBSsrK53/btvjtVizZg2CgoLQr18/\nLFq0SCpvb9ciISEBkZGRiIiIwIABA3Dy5EnpPW31WlRUVCAqKgrh4eEIDg7Ga6+9BqAZfzsNDaLc\nc889wtfXt86tJeTn54vk5GQhhBAlJSWid+/eIjU1VSxYsECsWLFCCCHE8uXLpQWOZ86cEWFhYaKy\nslJkZmYKf39/aSFka6fvWuhb7Nker4UQQly8eFE88MADwtfXV1y9elUI0T6vxYEDB8TIkSNFZWWl\nEEKIy5cvCyHa57UYNmyY2L17txBCiJ07d4ro6GghRNu+FkIIcfPmTSGEELdv3xZRUVHi8OHDzfbb\nabBF8vvvvyMzM1PndvbsWdPSo5E8PT0RHh4OAHByckJQUBByc3Oxfft2zJgxAwAwY8YMabHitm3b\nMHXqVNja2sLX1xcBAQFISEhokdibW33XIi8vT+9iz/Z4LQBg3rx5WLlypc7r29u1yM3NxSeffILX\nXnsNtra2ACCNc7bHa+Hl5SW11K9fvw6lUgmgbV8LAHB0dAQAVFZWoqqqCm5ubs3222kwkQwePLhR\nZeaWlZWF5ORkREVF4dKlS/Dw8AAAeHh44NKlSwC0q/BrL2T08fFpkwsZa1+L2mov9myP12Lbtm3w\n8fFBaGiozmva47U4f/48fvrpJwwcOBDR0dHS7hTt7VoMHDgQy5cvx/z589GjRw8sWLAAy5YtA9D2\nr0V1dTXCw8Ph4eEhdfk112+n3um/+fn5yMvLQ1lZGZKSkiCEgEKhQHFxMcrKyprruxmltLQUkydP\nRmxsLJydnXWeUygUDe5O3NZ2Li4tLcUjjzyC2NhYODk5SeV3L/asT1u+FlZWVli6dCl+/PFH6XnR\nwLyStnwtnJ2dodFocO3aNRw/fhwnT57ElClTkJGRUe972/K1cHJywoQJE7B69WpMnDgRGzduxNNP\nP63z30ltbelaWFlZ4dSpU7hx4wYeeOABHDx4UOd5U3479SaSPXv2YN26dcjNzcX8+fOlcmdnZyxd\nurQp8Ter27dvY/LkyXjyyScxYcIEANpMWlBQAE9PT+Tn56Nbt24AtFupZGdnS+/NycmRmrFtQc21\nmD59unQtAGDdunXYuXMn9u/fL5W1t2tx+vRpZGVlISwsDID2+6rVapw4caLdXQtA+y/KSZMmAQAG\nDBgAKysrXLlypV1ei4SEBOzbtw8A8Mgjj2DWrFkA2v7/R2p06tQJ48aNQ2JiYvP9dhoaoNm4caPp\nozzNpLq6Wjz55JPilVde0SlfsGCBWL58uRBCiGXLltUZMLp165bIyMgQPXv2FNXV1WaPWw76rsWu\nXbtEcHCwKCws1Clvj9eitvoG29vTtfjkk0/EW2+9JYQQ4ty5c6J79+5CiPZ5LSIiIkR8fLwQQoh9\n+/aJ/v37CyHa9rUoLCwU165dE0IIUVZWJoYOHSr27dvXbL+dehPJtm3bRGZmpvR4yZIlIiQkRIwf\nP15kZGQ0x3drssOHDwuFQiHCwsJEeHi4CA8PF7t27RJXr14VI0aMEL169RKjRo2SLpgQQrz//vvC\n399f9OnTR5qp0RbUdy127twpAgICRI8ePaSy2bNnS+9pb9eiNj8/PymRCNG+rsWuXbtEZWWlmD59\nuujXr59QqVQ626e3p2uxc+dOcfLkSREZGSnCwsLEwIEDRVJSkvSetnotUlJSREREhAgLCxMhISFi\n5cqVQgjRbL+dehckhoSE4MSJE3B0dMSOHTswd+5cfPPNN0hOTsbGjRuxZ8+e5m1vERFRq6R31paV\nlZU0XWzz5s145plnoFarMWvWLFy+fNlsARIRkWXTm0iEECgpKUF1dTX279+PESNGSM9VVFSYJTgi\nIrJ8emdtvfLKK4iIiICzszOCgoIwYMAAAEBSUhK8vb3NFiAREVm2BjdtzMnJweXLlxEeHi6tls7P\nz8ft27fRo0cPswVJRESWy+Duv0RERA0xuEUKERFRQ5hIqE2qvV2MHD766COUl5c3e33ff/89VqxY\n0SyfRWQueru2DJ05UnO6IZElcnZ2RklJiWyf7+fnh19++QXu7u5mqY/IkumdtaVSqRrcpCszM1OW\ngIjkkp6ejhdffBGFhYVwdHTEZ599hj59+uCpp55Cp06d8Msvv6CgoAArV67E5MmTUV1djRdffBEH\nDx5E9+7dYWtri6effhp5eXnIy8vD8OHD0bVrV2lPs7/97W/YsWMHHBwcsG3bNmnfohqvvPIK3N3d\n8eabb2LPnj1YunQpDh06pPOadevWITExEWvWrNEbV21ZWVkYPXo0Bg0ahKNHj6J///6YMWMG3n77\nbRQWFuKrr77CgAEDsGTJEukYiIsXL+LDDz/E0aNHsXfvXiiVSnz//fewsdH7c0DUMDmW4xO1NCcn\npzpl9913n7hw4YIQQojjx4+L++67TwghxIwZM8SUKVOEEEKkpqaKgIAAIYR2n7mxY8cKIYQoKCgQ\nbm5uYtOmTUII3b27hBBCoVCIHTt2CCG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+ "text": [
+ "<matplotlib.figure.Figure at 0x50e8310>"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.11,Page No.338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d2=200 #mm #Internal Diameter\n",
+ "p=14 #N/mm**2 #internal Fluid pressure\n",
+ "t=50 #mm #Thickness\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r2=100 #mm #Internal Diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation\n",
+ "#p_x=b*(x**2)**-1-a #N/mm**2 ...................(1)\n",
+ "#F_x=b*(x**2)**-1+a #N/mm**2 ...................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r2=100 #mm\n",
+ "p_x=14 #N/mm**2\n",
+ "\n",
+ "#Sub value of p_x in equation 1 we get\n",
+ "#14=(100)**-1*b-a ............................(3)\n",
+ "\n",
+ "#At\n",
+ "x2=r_o=150 #mm\n",
+ "p_x2=0 #N/mm**2\n",
+ "\n",
+ "#Sub value in equation 1 we get\n",
+ "#0=b*(150**2)**-1-a ......................(4)\n",
+ "\n",
+ "#From Equations 3 and 4 we get\n",
+ "#14=b*(100**2)**-1-b*(100**2)**-1\n",
+ "#After sub values and further simplifying we get\n",
+ "b=14*100**2*150**2*(150**2-100**2)**-1\n",
+ "\n",
+ "#From equation 4 we get\n",
+ "a=b*(150**2)**-1\n",
+ "\n",
+ "#Hoop Stress\n",
+ "#F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x=100 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=125 #mm\n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x3=150 #mm\n",
+ "F_x3=b*(x3**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#If thin Cyclindrical shell theory is used,hoop stress is uniform and is given by\n",
+ "F=p*d2*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Percentage error in estimating max hoop tension\n",
+ "E=(F_x-F)*F_x**-1*100 #%\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Hoop Stress Developed in the cross-section is\",round(F,2),\"N/mm**2\"\n",
+ "print\"Plot of Variation of hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[F_x,F_x2,F_x3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Hoop Stress Developed in the cross-section is 28.0 N/mm**2\n",
+ "Plot of Variation of hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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29vaQJAknTpx4bAP379/HiBEjMGzYMMycOdNwT41GA1dXV2RkZGDgwIEcMiIi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\nSUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58c2e30>"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.12,Page No.339"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d2=200 #mm #Internal Diameter\n",
+ "p=12 #N/mm**2 #internal Fluid pressure\n",
+ "F_max=16 #N/mm**2 #Tensile stress\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r2=100 #mm #Internal Diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p_o be the External Pressure applied.\n",
+ "#From LLame's theorem\n",
+ "#p_x=b*(x**2)**-1-a ..............(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Now At\n",
+ "x=100 #mm\n",
+ "p_x=12 #N/mm**2\n",
+ "#sub in equation 1 we get\n",
+ "#12=b*(100**2)**-1-a . ..................(3)\n",
+ "\n",
+ "#The Max Hoop stress occurs at least value of x where\n",
+ "x=r1=100 #mm\n",
+ "#16=b*(100**2)**-1+a .......................(4)\n",
+ "\n",
+ "#From Equations 1 and 2 we get\n",
+ "#28=b*(100**2)**-1+b*(100**2)**-1\n",
+ "#After furhter Simplifying we get\n",
+ "b=28*100**2*2**-1\n",
+ "\n",
+ "#sub in equation 1 we get\n",
+ "a=-(12-(b*(100**2)**-1))\n",
+ "\n",
+ "#Thus At\n",
+ "x2=150 #mm\n",
+ "p_o=b*(x2**2)**-1-a\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum External applied is\",round(p_o,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum External applied is 4.22 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.13,Page No.340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=160 #mm #Internal Diameter \n",
+ "r1=80 #mm #External Diameter\n",
+ "p1=40 #N/mm**2 #Internal Diameter\n",
+ "P_max=120 #N/mm**2 #Allowable stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=80 #N/mm**2 \n",
+ "#Sub in equation 1 we get\n",
+ "#120=b*(80**2)**-1+a ........................(3)\n",
+ "\n",
+ "#The hoop tension at inner edge is max stress\n",
+ "#Hence\n",
+ "#120=b*(80**2)**-1+a .............................(4)\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "b=160*80**2*2**-1 \n",
+ "\n",
+ "#Sub in equation 3 we get\n",
+ "a=-(40-(b*(80**2)**-1))\n",
+ "\n",
+ "#Let External radius be r_o.Since at External Surface is Zero,we get\n",
+ "#0=b*(r_o)**-1-a\n",
+ "#After Further simplifying we get\n",
+ "r_o=(b*a**-1)**0.5\n",
+ "\n",
+ "#Thickness of Cyclinder \n",
+ "t=r_o-r1 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness Required is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness Required is 33.14 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.14,Page No.341"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d1=180 #mm #Internal Diameter\n",
+ "p=12 #N/mm**2 #internal Fluid pressure\n",
+ "p_o=6 #N/mm**2 #External Pressure\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r=90 #mm #Internal Diameter\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=90 #N/mm**2 \n",
+ "p=42 #N/mm**2\n",
+ "#Sub in equation 1 we get\n",
+ "#42=b*(90**2)**-1-a ..............................(3)\n",
+ "\n",
+ "#At \n",
+ "x=r_o=150 #mm\n",
+ "p2=6 #N/mm**2\n",
+ "#sub in equation 1 we get\n",
+ "#6=b*(150**2)**-1-a ..............................(4)\n",
+ "\n",
+ "#From equations 3 and 4 weget\n",
+ "#36=b*(90**2)**-1-b2(150**2)**-1\n",
+ "#After further simplifying we get\n",
+ "b=36*90**2*150**2*(150**2-90**2)**-1\n",
+ "\n",
+ "#Sub value of b in equation 4 we get\n",
+ "a=b*(150**2)**-1-p_o\n",
+ "\n",
+ "#At \n",
+ "x=r1=90 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x2=r_o=150 #mm \n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#Now if External pressure is doubled i.e p_o2=12 #N/mm**2 We have\n",
+ "p_o2=12 #N/mm**2\n",
+ "#sub in equation 4 we get\n",
+ "#12=b2*(150**2)**-1-a2 ..........................(5)\n",
+ "\n",
+ "#Max Hoop stress is to be 70.5 #N/mm**2,which occurs at x=r1=90 #mm\n",
+ "#Sub in equation 4 we get\n",
+ "#70.5=b*(90**2)**-1+a2 ................................(6)\n",
+ "\n",
+ "#Adding equation 5 and 6\n",
+ "#82.5=b2*(150**2)**-1+b*(90**2)**-1\n",
+ "#After furhter simplifying we get\n",
+ "b2=82.5*150**2*90**2*(150**2+90**2)**-1\n",
+ "\n",
+ "#Sub in equation 5 we get\n",
+ "a2=b2*(150**2)**-1-12 \n",
+ "\n",
+ "#If p_i is the internal pressure required then from Lame's theorem\n",
+ "p_i=b2*(r1**2)**-1-a2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses int the material are:F_x\",round(F_x,2),\"N/mm**2\"\n",
+ "print\" :F_x2\",round(F_x2,2),\"N/mm**2\"\n",
+ "print\"Internal Pressure that can be maintained is\",round(p_i,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses int the material are:F_x 70.5 N/mm**2\n",
+ " :F_x2 34.5 N/mm**2\n",
+ "Internal Pressure that can be maintained is 50.82 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.15,Page No.344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "r1=200 #mm #Inner Radius\n",
+ "r2=250 #mm #Radius at common surface\n",
+ "r3=300 #mm #Outer radius\n",
+ "p=6 #N/mm**2 #Inital pressure\n",
+ "p2=80 #N/mm**2 #Pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Inner Cyclinder:\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=200 #mm\n",
+ "p_x=0\n",
+ "#0=b1*(250**2)**-1-a1 .................(3)\n",
+ "\n",
+ "#At x=r2=250 #mm\n",
+ "p_x2=6 #N/mm**2\n",
+ "#6=b1*(250**2)-a1 ...................(4)\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "b1=6*200**2*250**2*(200**2-250**2)**-1\n",
+ "\n",
+ "#From equation 3 we get\n",
+ "a1=b1*(200**2)**-1\n",
+ "\n",
+ "F_200=b1*(200**2)**-1+a1\n",
+ "F_250=b1*(250**2)**-1+a1\n",
+ "\n",
+ "#For outer cyclinder \n",
+ "#From Lame's Equation we have\n",
+ "#p_x2=b2*(x**2)**-1-a2 ..........................(5)\n",
+ "#F_x2=b2*(x**2)**-1+a2 ...........................(6)\n",
+ "\n",
+ "\n",
+ "#At \n",
+ "x2=r2=250 #mm\n",
+ "p_x2=6 #N/mm**2\n",
+ "#6=b2*(250**2)**-1-a2 ...........................(7) \n",
+ "\n",
+ "#At\n",
+ "x3=300 #mm\n",
+ "#p_x2=0\n",
+ "#0=b2**2*(300**2)**-1-a2 .................................(8)\n",
+ "\n",
+ "#from equation 7 and 8 we get\n",
+ "b2=6*250**2*300**2*(300**2-250**2)**-1\n",
+ "\n",
+ "#sub in equation 8 we get\n",
+ "a2=b2*(300**2)**-1\n",
+ "\n",
+ "F_250_2=b2*(250**2)**-1+a2\n",
+ "F_300_2=b2*(300**2)**-1+a2\n",
+ "\n",
+ "#When Fluid is admitted\n",
+ "#Let Lame's equation be\n",
+ "#p_x3=b3*(x**2)**-1-a3 ..........................(5)\n",
+ "#F_x3=b3*(x**2)**-1+a3 ...........................(6)\n",
+ "\n",
+ "\n",
+ "#At x=200\n",
+ "p_x3=80 #N/mm**2\n",
+ "#80=b3*(200**2)**-1-a3 ................................(7)\n",
+ "\n",
+ "#At x=300 #mm\n",
+ "#p_x=0\n",
+ "#0=b3*(300**2)**-1-a3 ..............................(8)\n",
+ "\n",
+ "#from Equation 7 and 8 we get\n",
+ "b3=80*200**2*300**2*(300**2-200**2)**-1\n",
+ "\n",
+ "#From Equation 8 we get\n",
+ "a3=b3*(300**2)**-1\n",
+ "\n",
+ "#Hoop stresses \n",
+ "F_200_3=b3*(200**2)**-1+a3 #N/mm**2\n",
+ "F_250_3=b3*(250**2)**-1+a3 #N/mm**2\n",
+ "F_300_3=b3*(300**2)**-1+a3 #N/mm**2\n",
+ "\n",
+ "#Pressure at common surface\n",
+ "p_250=b3*(250**2)**-1-a3 #N/mm**2\n",
+ "\n",
+ "#final stress\n",
+ "f_200=F_200+F_200_3 #N/mm**2\n",
+ "f_250=F_250+F_250_3 #N/mm**2\n",
+ "f_300=F_250_2+F_250_3 #N/mm**2\n",
+ "f_300_2=F_300_2+F_300_3 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"final Hoop stress are:f_200\",round(f_200,2),\"N/mm**2\"\n",
+ "print\" :f_250\",round(f_250,2),\"N/mm**2\"\n",
+ "print\" :f_300\",round(f_300,2),\"N/mm**2\"\n",
+ "print\" :f_300_2\",round(f_300_2,2),\"N/mm**2\"\n",
+ "print\"Variation of Hoop stress and Radial stress\"\n",
+ "\n",
+ "#Final stresses\n",
+ "#Variation of hoop stress \n",
+ " \n",
+ "X1=[x,x2,x3,x3]\n",
+ "Y1=[f_200,f_250,f_300,f_300_2]\n",
+ "Z1=[0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Due to Fluid\n",
+ "#Variation of hoop stress \n",
+ " \n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[F_200_3,F_250_3,F_300_3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "final Hoop stress are:f_200 174.67 N/mm**2\n",
+ " :f_250 128.83 N/mm**2\n",
+ " :f_300 189.43 N/mm**2\n",
+ " :f_300_2 155.27 N/mm**2\n",
+ "Variation of Hoop stress and Radial stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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MHz4cr7zyCmbPno1///vf1fIBnZeXp/l+8+bNmhVR0dHR+P7771FYWIjMzExcvHgRnTp1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\nVw+AnLO4d++eQY0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+ "text": [
+ "<matplotlib.figure.Figure at 0x5604510>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x56e5cb0>"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.16,Page No.348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "do=200 #mm #Inner Diameter\n",
+ "r_o=100 #mm #Inner radius\n",
+ "d1=300 #mm #outer diameter\n",
+ "r1=150 #mm #Outer radius\n",
+ "d2=250 #mm #Junction Diameter\n",
+ "r2=125 #mm #Junction radius\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "p=30 #N/mm**2 #radial pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#from Lame's Equation we get\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Then from Boundary condition \n",
+ "#p_x=0 at x=100 #mm\n",
+ "#0=b1*(100**2)**-1-a1 .....................(3)\n",
+ "\n",
+ "#p_x2=30 #N/mm**2 at x2=125 #mm\n",
+ "#30=b1*(125**2)**-1-a1 ................................(4)\n",
+ "\n",
+ "#From equation 3 and 4 we get\n",
+ "b1=30*125**2*100**2*(100**2-125**2)**-1\n",
+ "\n",
+ "#From Equation 3 we get\n",
+ "a1=b1*(100**2)**-1\n",
+ "\n",
+ "#therefore Hoop stress in inner cyclinder at junction\n",
+ "F_2_1=b1*(125**2)**-1+a1 #N/mm**2\n",
+ "\n",
+ "#Outer Cyclinder\n",
+ "#p_x=b*(x**2)**-1-a ..........................(5)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(6)\n",
+ "\n",
+ "#Now at x=125 #mm\n",
+ "#p_x3=30 #N/mm**2\n",
+ "#30=b2*(125**2)**-1-a2 ..................................(7)\n",
+ "\n",
+ "#At x=150 #mm\n",
+ "#p_x4=0\n",
+ "#0=b2*(150**2)**-1-a2 ...................................(8)\n",
+ "\n",
+ "#From equations 7 and 8\n",
+ "b2=30*150**2*125**2*(150**2-125**2)**-1\n",
+ "\n",
+ "#From eqauation 8 we get\n",
+ "a2=b2*(150**2)**-1\n",
+ "\n",
+ "#Hoop stress at junction \n",
+ "F_2_0=b2*(125**2)**-1+a2 #N/mm**2\n",
+ "\n",
+ "rho_r=(F_2_0-F_2_1)*E**-1*r2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shrinkage Allowance is\",round(rho_r,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shrinkage Allowance is 0.189 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.17,Page No.350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=500 #mm #Outer Diameter\n",
+ "r_o=250 #mm #Outer Radius\n",
+ "d1=300 #mm #Inner Diameter\n",
+ "r1=150 #mm #Inner Radius\n",
+ "d2=400 #mm #Junction Diameter\n",
+ "E=2*10**5 #N/mm**2 #Modulus ofElasticity\n",
+ "alpha=12*10**-6 #Per degree celsius\n",
+ "dell_d=0.2 #mm\n",
+ "dell_r=0.1 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p be the radial pressure developed at junction\n",
+ "#Let Lame's Equation for internal cyclinder be\n",
+ "#p_x=b*(x**2)**-1-a ................................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...............................(2)\n",
+ "\n",
+ "#At \n",
+ "x=150 #mm \n",
+ "p_x=0\n",
+ "#Sub in equation 1 we get\n",
+ "#0=b*(150**2)**-1-a .........................(3)\n",
+ "\n",
+ "#At \n",
+ "x2=200 #mm\n",
+ "#p_x2=p\n",
+ "#p=b*(200**2)**-1-a ......................(4)\n",
+ " \n",
+ "#From Equation 3 and 4\n",
+ "#p=b*(200**2)**-1-b(150**2)**-1\n",
+ "#after further simplifying we get\n",
+ "#b=-51428.571*p\n",
+ "\n",
+ "#sub in equation 3 we get\n",
+ "#a1=-2.2857*p\n",
+ "\n",
+ "#therefore hoop stress at junction is\n",
+ "#F_2_1=-21428.571*p*(200**2)**-1-2.2857*p\n",
+ "#after Further simplifying we geet\n",
+ "#F_2_1=3.5714*p\n",
+ "\n",
+ "#Let Lame's Equation for cyclinder be \n",
+ "#p_x=b*(x**2)**-1-a .........................5\n",
+ "#F_x=b*(x**2)**-1+a .............................6\n",
+ "\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "#p_x=p2\n",
+ "#p2=b2*(20**2)**-1-a2 ...................7\n",
+ "\n",
+ "#At\n",
+ "x2=200 #mm\n",
+ "p_x2=0\n",
+ "#0=b2*(250**2)**-1-a2 ....................8\n",
+ "\n",
+ "#from equation 7 and 8 we get\n",
+ "#p2=b2*(200**2)**-1-b2*(250**2)**-1\n",
+ "#After further simplifying we get\n",
+ "#p2=b2*(250**2-200**2)*(200**2*250**2)**-1\n",
+ "#b2=111111.11*p\n",
+ "\n",
+ "#from equation 7\n",
+ "#a2=b2*(250**2)**-1\n",
+ "#further simplifying we get\n",
+ "#a2=1.778*p\n",
+ "\n",
+ "#At the junctionhoop stress in outer cyclinder \n",
+ "#F_2_0=b2*(200**2)**-1+a2\n",
+ "#After further simplifying we get\n",
+ "#F_2_0=4.5556*p\n",
+ "\n",
+ "#Considering circumferential strain,the compatibility condition\n",
+ "#rho_r*r2**-1=1*E**-1*(F_2_1+F_2_0)\n",
+ "#where F_2_1 is compressive and F_2_0 is tensile\n",
+ "#furter simplifying we get\n",
+ "p=0.1*200**-1*2*10**5*(3.5714+4.5556)**-1\n",
+ "\n",
+ "#Let T be the rise in temperature required\n",
+ "#dell_d=d*alpha*T\n",
+ "#After sub values and further simplifying we get\n",
+ "d=250 #mm\n",
+ "T=dell_d*(d*alpha)**-1 #Per degree celsius\n",
+ "\n",
+ "#Result\n",
+ "print\"Radial Pressure Developed at junction\",round(p,2),\"N/mm**2\"\n",
+ "print\"Min Temperatureto outer cyclinder\",round(T,2),\"Per degree Celsius\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radial Pressure Developed at junction 12.3 N/mm**2\n",
+ "Min Temperatureto outer cyclinder 66.67 Per degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.18,Page No.355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=400 #mm #Outer Diameter\n",
+ "r_o=200 #mm #Outer radius\n",
+ "t=50 #mm #Thickness\n",
+ "r1=150 #mm #Internal Radius\n",
+ "p=50 #N/mm**2 #Internal Pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#The Radial Pressure and hoop stress at any radial distance x are given by\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Now at\n",
+ "x=150 #N/mm**2\n",
+ "p_x1=50 #N/mm**2\n",
+ "#Sub in equation 1 we get\n",
+ "#50=2*b*(150**3)**-1-a ...........................(3)\n",
+ "\n",
+ "#At x=200 #mm\n",
+ "p_x2=0\n",
+ "#0=2*b*(200**2)**-1-a ....................(4)\n",
+ "\n",
+ "#From equation 3 and 4 we get\n",
+ "#50=2*b*(150**3)**-1-2*b*(200**3)**-1\n",
+ "#After further simplifying we get\n",
+ "b=50*150**3*200**3*(200**3-150**3)**-1*2**-1\n",
+ "\n",
+ "#Sub in equation 3 we get\n",
+ "a=b*(200**3)**-1\n",
+ "\n",
+ "#Now At\n",
+ "x=150 #mm\n",
+ "F_x=b*(x**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x2=160 #mm\n",
+ "F_x2=b*(x2**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x3=170 #mm\n",
+ "F_x3=b*(x3**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x4=180 #mm\n",
+ "F_x4=b*(x4**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x5=190 #mm\n",
+ "F_x5=b*(x5**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x6=200 #mm\n",
+ "F_x6=b*(x6**3)**-1+a\n",
+ "\n",
+ "#Result\n",
+ "print\"Plot of Variation of hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3,x4,x5,x6]\n",
+ "Y1=[F_x,F_x2,F_x3,F_x4,F_x5,F_x6]\n",
+ "Z1=[0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Plot of Variation of hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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w3g2UlJRgzJgxCAsLw3vvvQcA8PT0hFqthpOTE7KyshAYGIhLly7VLIxzCERE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\nnp6uOVdYWBisrKzg4+OD8vJ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+ "text": [
+ "<matplotlib.figure.Figure at 0x56b4bf0>"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_4.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_4.ipynb new file mode 100644 index 00000000..cecacb12 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_4.ipynb @@ -0,0 +1,779 @@ +{
+ "metadata": {
+ "name": "chapter no.9.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9:Columns And Struts"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.1,Page No.377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "L=5000 #mm #Length of strut\n",
+ "dell=10 #mm #Deflection\n",
+ "W=10 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Central Deflection of a simply supported beam with central concentrated load is\n",
+ "#dell=W*L**3*(48*E*I)**-1 \n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=W*L**3*(48*dell)**-1 #mm\n",
+ "\n",
+ "#Euler's Load\n",
+ "#Let Euler's Load be P\n",
+ "P=pi**2*X*(L**2)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical Load of Bar is\",round(P,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical Load of Bar is 1028.08 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.2,Page No.377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=2000 #mm #Length of square column\n",
+ "E=12*10**3 #N/mm**2 #Modulus of Elasticity\n",
+ "sigma=12 #N/mm*2 #stress\n",
+ "W1=95*10**3 #N #Load1\n",
+ "W2=200*10**3 #N #Load2\n",
+ "FOS=3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Euler's Formula\n",
+ "#P=pi**2*E*I*(L**2)**-1 .........(1)\n",
+ "\n",
+ "#Working Load\n",
+ "#W=P*(FOS)**-1\n",
+ "\n",
+ "#Part-1\n",
+ "\n",
+ "#At W1=95*10**3 #N\n",
+ "#W1=P*(3*L**2)**-1\n",
+ "\n",
+ "#Let 'a' be the side of the square\n",
+ "#I=1*12**-1*a**4\n",
+ "\n",
+ "#sub value of I in Equation 1 and further rearranging we get\n",
+ "a=(W1*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",
+ "\n",
+ "#From Consideration of direct crushing\n",
+ "#sigma*a**2=W1\n",
+ "#After Reaaranging the above equation we get\n",
+ "a2=(W1*(sigma)**-1)**0.5 #mm\n",
+ "\n",
+ "#required size is 103.67*103.67 i.e a*a\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#At W2=200*10**3 #N\n",
+ "#W2=P*(3*L**2)**-1\n",
+ "#After substituting values and further Rearranging the above equation we get\n",
+ "a3=(W2*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",
+ "\n",
+ "#From consideration of direct compression,size required is\n",
+ "a4=(W2*sigma**-1)**0.5\n",
+ "\n",
+ "#required size is 129.10*129.10 i.e a4*a4\n",
+ "\n",
+ "#Result\n",
+ "print\"For W1 Load Required size is\",round(a*a,2),\"mm**2\"\n",
+ "print\"For W2 Load Required size is\",round(a4*a4,2),\"mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For W1 Load Required size is 10747.38 mm**2\n",
+ "For W1 Load Required size is 16666.67 mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.3,Page No.378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flange \n",
+ "b=100 #mm #Width\n",
+ "\n",
+ "D=80 #mm #Overall Depth\n",
+ "t=10 #mm #Thickness of web and flanges\n",
+ "L=3000 #mm #Length of strut\n",
+ "E=200*10**3 #N/mm**2 #Modulus of Elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let centroid be at depth y_bar from top fibre\n",
+ "y_bar=(b*t*t*2**-1+(D-t)*t*((D-t)*2**-1+t))*(b*t+(D-t)*t)**-1 #mm \n",
+ "\n",
+ "#M.I at x-x axis\n",
+ "I_x=1*12**-1*b*t**3+b*t*(y_bar-t*2**-1)**2+1*12**-1*t*((D-t))**3+t*((D-t))*((((D-t)*2**-1)+t)-y_bar)**2\n",
+ "\n",
+ "#M.I at y-y axis\n",
+ "I_y=1*12**-1*t*b**3+1*12**-1*(D-t)*t**3 #mm**3\n",
+ "\n",
+ "#Least M.I\n",
+ "I=I_y\n",
+ "\n",
+ "#Since both ends are hinged\n",
+ "#Feective Length=Actual Length\n",
+ "L=l=3000 #mm\n",
+ "\n",
+ "#Buckling Load \n",
+ "P=pi**2*E*I*(l**2)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"The Buckling Load for strut of tee section\",round(P,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Buckling Load for strut of tee section 184.05 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.4,Page No.379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "\n",
+ "#Flanges\n",
+ "b=300 #mm #Width\n",
+ "t=50 #mm #Thickness\n",
+ "\n",
+ "t2=30 #mm #Web Thickness\n",
+ "\n",
+ "dell=10 #mm #Deflection\n",
+ "w=40 #N/mm #Load\n",
+ "FOS=1.75 #Factor of safety\n",
+ "E=2*10**5 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I at x-x axis\n",
+ "I_x=1*12**-1*(b*D**3-(b-t2)*b**3) #mm**4\n",
+ "\n",
+ "#Central Deflection\n",
+ "#dell=5*w*L**4*(384*E*I)**-1\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "L=(dell*384*E*I_x*(5*w)**-1)**0.25\n",
+ "\n",
+ "#M.I aty-y axis\n",
+ "I=I_y=1*12**-1*t*b**3+1*12**-1*b*t2**3+1*12**-1*t*b**3 #mm**4\n",
+ "\n",
+ "#Both the Ends of column are hinged\n",
+ "\n",
+ "#Crippling Load\n",
+ "P=pi**2*E*I*(L**2)**-1 #N\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load if I-section is used as column with both Ends hhinged\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load if I-section is used as column with both Ends hhinged 4123.29 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.5,Page No.381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=200 #mm #External Diameter\n",
+ "t=20 #mm #hickness\n",
+ "d=200-2*t #mm #Internal Diameter\n",
+ "E=1*10**5 #N/mm**2\n",
+ "a=1*(1600)**-1 #Rankine's Constant\n",
+ "L=4.5 #m #Length\n",
+ "sigma=550 #N/mm**2 #Stress\n",
+ "FOS=2.5\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=pi*D**4*64**-1-pi*d**4*64**-1\n",
+ "\n",
+ "#Both Ends are fixed\n",
+ "\n",
+ "#Effective Length\n",
+ "l=1*2**-1*L*10**3 #mm\n",
+ "\n",
+ "#Euler's Critical Load\n",
+ "P_E=pi**2*E*I*(l**2)**-1\n",
+ "\n",
+ "A=pi*4**-1*(D**2-d**2) #mm*2\n",
+ "\n",
+ "k=(I*A**-1)**0.5\n",
+ "\n",
+ "#Rankine's Critical Load\n",
+ "P_R=sigma*A*(1+a*(l*k**-1)**2)**-1\n",
+ "\n",
+ "X=P_E*P_R**-1 \n",
+ "\n",
+ "#Safe Load using Rankine's Formula\n",
+ "S=P_R*(FOS)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load by Rankine's Formula is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load by Rankine's Formula is 1404.36 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.6,Page No.382"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #Length of column\n",
+ "W=800*10**3 #N #Load\n",
+ "a=1*1600**-1 #Rankine's constant\n",
+ "FOS=4 #Factor of safety\n",
+ "sigma=550 #N/mm**2 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Effective Length\n",
+ "l=L*2**-1 #mm \n",
+ "\n",
+ "#Let d1=outer diameter & d2=inner diameter\n",
+ "#d1=5*8**-1*d2\n",
+ "\n",
+ "#M.I\n",
+ "#I=pi*64**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Area of section\n",
+ "#A=pi4**-1*(d1**2-d2**2) #mm**2\n",
+ "\n",
+ "#k=(I*A**-1) \n",
+ "#substituting values in above equation \n",
+ "#k=1*16**-1*(d1**2-d2**2)\n",
+ "#after simplifying further we get\n",
+ "#k=0.2948119.d1\n",
+ "\n",
+ "#X=l*k**-1\n",
+ "#substituting values in above equation and after simplifying further we get\n",
+ "#X=5087.9898*d1**-1\n",
+ "\n",
+ "#Crtitcal Load\n",
+ "P=W*FOS #N\n",
+ "\n",
+ "#From Rankine's Load\n",
+ "#P2=sigma*A*(1+a*(X)**2)**-1\n",
+ "#substituting values in above equation and after simplifying further we get\n",
+ "#d1**4-12156618*d1**4-1.96691*10**8=0\n",
+ "#Solving Quadratic Equation we get\n",
+ "#d1**2-12156618*d1-196691000=0\n",
+ "a=1\n",
+ "b=-12156.618\n",
+ "c=-196691000\n",
+ "\n",
+ "Y=b**2-4*a*c\n",
+ "\n",
+ "d1_1=((-b+Y**0.5)*(2*a)**-1)**0.5 #mm\n",
+ "d1_2=((-b-Y**0.5)*(2*a)**-1) #mm\n",
+ "\n",
+ "d2=5*8**-1*d1_1\n",
+ "\n",
+ "#Result\n",
+ "print\"Section of cast iron hollow cylindrical column is:d1_1\",round(d1_1,2),\"mm\"\n",
+ "print\" :d2 \",round(d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Section of cast iron hollow cylindrical column is:d1_1 146.16 mm\n",
+ " :d2 91.35 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.7,Page No.383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Let X=(P*A**-1) #Average Stress at Failure \n",
+ "Lamda_1=70 #Slenderness Ratio\n",
+ "Lamda_2=170 #Slenderness Ratio\n",
+ "X1=200 #N/mm**2 \n",
+ "X2=69 #N/mm**2 \n",
+ "\n",
+ "#Rectangular section\n",
+ "b=60 #mm #width\n",
+ "t=20 #mm #Thickness\n",
+ "\n",
+ "L=1250 #mm #Length of strut\n",
+ "FOS=4 #Factor of safety\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Slenderness ratio\n",
+ "#Lamda=L*k**-1\n",
+ "\n",
+ "#The Rankine's Formula for strut\n",
+ "#P=sigma*A*(1+a*(L*k**-1)**-1\n",
+ "\n",
+ "#From test result 1,\n",
+ "#After sub values in above equation we get and further simplifying we get\n",
+ "#sigma_1=200+980000*a ...................(1)\n",
+ "\n",
+ "#From test result 2,\n",
+ "#After sub values in above equation we get and further simplifying we get\n",
+ "#sigma_2=69+1994100*a ...................(2)\n",
+ "\n",
+ "#Substituting it in equation (1) we get\n",
+ "a=131*1014100**-1 \n",
+ "\n",
+ "#Substituting a in equation 1\n",
+ "sigma_1=200+980000*a #N/mm**2\n",
+ "\n",
+ "#Effective Length \n",
+ "l=1*2**-1*L #mm\n",
+ "\n",
+ "#Least of M.I\n",
+ "I=1*12**-1*b*t**3 #mm**4\n",
+ "\n",
+ "#Area \n",
+ "A=b*t #mm**2 \n",
+ "\n",
+ "k=(I*A**-1)**0.5\n",
+ "\n",
+ "#Slenderness ratio\n",
+ "Lamda=l*k**-1\n",
+ "\n",
+ "#From Rankine's Ratio\n",
+ "P=sigma_1*A*(1+a*(Lamda)**2)**-1\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Constant in the Formula is:a \",round(a,6)\n",
+ "print\" :sigma_1\",round(sigma_1,2)\n",
+ "print\"Safe Load is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Constant in the Formula is:a 0.000129\n",
+ " :sigma_1 326.6\n",
+ "Safe Load is 38.98 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.8,Page No.385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=200 #mm #Depth\n",
+ "b=140 #mm #width\n",
+ "\n",
+ "#Plate\n",
+ "b2=160 #mm #Width\n",
+ "t2=10 #mm #Thickness\n",
+ "\n",
+ "L=l=4000 #mm #Length\n",
+ "FOS=4 #Factor of safety\n",
+ "sigma=315 #N/mm**2 #stress\n",
+ "a2=1*7500**-1 \n",
+ "I_xx=26.245*10**6 #mm**4 #M.I at x-x\n",
+ "I_yy=3.288*10**6 #mm**4 #M.I at y-y\n",
+ "a=3671 #mm**2 #Area\n",
+ "k_x=84.6#mm\n",
+ "k_y=29.9 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Total Area\n",
+ "A=a+2*t2*b2 #mm**2\n",
+ "\n",
+ "#M.I\n",
+ "I=I_yy+2*12**-1*t2*b2**3 #mm**4\n",
+ "\n",
+ "k=(I*A**-1)**0.5 #mm\n",
+ "\n",
+ "#Let X=L*k**-1\n",
+ "X=L*k**-1\n",
+ "\n",
+ "#Appliying Rankine's Formula\n",
+ "P=sigma*A*(1+a2*(X)**2)**-1 #N\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe axial Load is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe axial Load is 220.93 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.9,Page No.389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200*10**3 #N/mm**2 #Modulus of elasticity\n",
+ "sigma=330 #N/mm**2 #Stress\n",
+ "a=1*7500**-1 #Rankine's constant\n",
+ "A=5205 #mm**2 #area of column\n",
+ "I_xx=59.431*10**6 #mm**4 #M.I at x-x axis\n",
+ "I_yy=8.575*10**6 #mm**24#M.I at y-y axis\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Total M.I\n",
+ "I=I_xx+I_yy #mm**4\n",
+ "\n",
+ "#Area of compound Section \n",
+ "A2=2*A #mm**2\n",
+ "\n",
+ "k=(I*A2**-1)**0.5 #mm\n",
+ "\n",
+ "#Equating Euler's Load to Rankine's Load we get\n",
+ "#pi**2*E*I*(L**2)**-1=sigma*A*(1+a*(L*k)**2)**-1\n",
+ "#After Substitt=uting values and further simplifying we get\n",
+ "L=(39076198*(1-0.7975432)**-1)**0.5*10**-3 #m\n",
+ "\n",
+ "#Result\n",
+ "print\"Length of column for which Rankine's formula and Euler's Formula give the same result is\",round(L,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length of column for which Rankine's formula and Euler's Formula give the same result is 13.89 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.10,Page No.387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma=326 #N/mm**2 #stress\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "FOS=2 #Factor of safety\n",
+ "a=1*7500**-1 #Rankine's constant\n",
+ "D=350 #mm #Overall Depth \n",
+ "\n",
+ "#Cover plates\n",
+ "b1=500 #mm #width\n",
+ "t1=10 #mm #Thickness\n",
+ "\n",
+ "d=220 #mm #Distance between two channels\n",
+ "\n",
+ "L=6000 #mm #Length of column\n",
+ "\n",
+ "A=5366 #mm**2 #Area of Column section \n",
+ "I_xx=100.08*10**6 #mm**4 #M.I of x-x axis\n",
+ "I_yy=4.306*10**6 #mm**4 #M.I of y-y axis\n",
+ "C_yy=23.6 #mm #Centroid at y-y axis\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Symmetric axes are the centroidal axes is\n",
+ "\n",
+ "#M.I of Channel at x-x axis\n",
+ "I_xx_1=2*I_xx+2*(1*12**-1*b1*t1**3+b1*t1*(D*2**-1+t1*2**-1)**2)\n",
+ "\n",
+ "#M.I of Channel at y-y axis\n",
+ "I_yy_1=2*(I_yy+A*(d*2**-1+C_yy)**2)+2*12**-1*t1*b1**3\n",
+ "\n",
+ "#As I_yy<I_xx\n",
+ "#So\n",
+ "I=I_yy_1 #mm**4 \n",
+ "\n",
+ "A2=2*A+2*t1*b1 #Area of channel\n",
+ "\n",
+ "k=(I*A2**-1)**0.5 #mm\n",
+ "\n",
+ "#Critical Load\n",
+ "P=sigma*A2*(1+a*(L*k**-1)**2)**-1 \n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*2**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load carrying Capacity is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load carrying Capacity is 2717.35 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.11,Page No.390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "I=4.085*10**8 #mm**4 #M.I\n",
+ "A=20732.0 #mm**2 #area of column\n",
+ "f_y=250 #N/mm**2 \n",
+ "L=6000 #mm #Length of column\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "k=(I*A**-1)**0.5 #mm\n",
+ "lamda=L*k**-1 #Slenderness ratro\n",
+ "\n",
+ "#From Indian standard table\n",
+ "lamda_1=40 \n",
+ "sigma_a_c_1=139 #N/mm**2\n",
+ "lamda_2=50 \n",
+ "sigma_a_c_2=132 #N/mm**2 \n",
+ "\n",
+ "#Linearly interpolating between these values for lambda=42.744\n",
+ "\n",
+ "sigma_a_c_3=sigma_a_c_1-2.744*10**-1*(sigma_a_c_1-sigma_a_c_2)\n",
+ "\n",
+ "#Safe Load carrying capacity of column\n",
+ "P=sigma_a_c_3*A*10**-3\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load carrying capacity is\",round(P,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load carrying capacity is 2841.93 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_3.JPG b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_3.JPG Binary files differnew file mode 100644 index 00000000..9bbe723e --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_3.JPG diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1_3.jpg b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1_3.jpg Binary files differnew file mode 100644 index 00000000..0ccd99c0 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1_3.jpg diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_2_3.jpg b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_2_3.jpg Binary files differnew file mode 100644 index 00000000..3be9f649 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_2_3.jpg diff --git a/sample_notebooks/AshishKumar/Chapter2.ipynb b/sample_notebooks/AshishKumar/Chapter2.ipynb new file mode 100644 index 00000000..8d9af0c5 --- /dev/null +++ b/sample_notebooks/AshishKumar/Chapter2.ipynb @@ -0,0 +1,444 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 Switched communication systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2, page no 125" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum auxillary current is:10.00 mA\n", + "\n", + "MMF in the auxillary winding is:2.00AT \n", + "\n", + "MMF in main winding is:40.00 AT \n", + "\n", + "net MMF required in main winding is:44.00 AT \n", + "\n", + "operating current needed is:4.40 mA \n", + "\n", + "working voltage is:2.84 volts \n", + "\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#given\n", + "Io=4*10**-3 #rqueired operating current\n", + "N1=10000 #no of turns in the main winding\n", + "R1=645 #resistence of the main winding in ohms\n", + "N2=200 #no of turns in auxillary winding\n", + "B=2 #spacing bias\n", + "Iaux=B/N2 #maximum auxillary current\n", + "print \"maximum auxillary current is:%0.2f mA\\n\"%(Iaux*1e3)\n", + "MMFaux=N2*Iaux #MMF in the auxillary winding\n", + "print \"MMF in the auxillary winding is:%0.2fAT \\n\"%(MMFaux)\n", + "MMFop=Io*N1 #operating MFF in main winding\n", + "print \"MMF in main winding is:%0.2f AT \\n\"%(MMFop)\n", + "MMFnet=MMFop+(0.1*MMFop) #net MMF required in main winding\n", + "print \"net MMF required in main winding is:%0.2f AT \\n\"%(MMFnet)\n", + "Iop=MMFnet/N1 #operating current needed\n", + "print \"operating current needed is:%0.2f mA \\n\"%(Iop*1e3)\n", + "V=Iop*R1 #working voltage in volts\n", + "print \"working voltage is:%0.2f volts \\n\"%(V)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3,page no 125" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Busy hour calling rate is:1.20 \n", + "\n", + "Rate of traffic flow is 250.00 traffic unit \n" + ] + } + ], + "source": [ + "#given\n", + "C=6000#Tatol no of call in busy hour\n", + "SC=5000#no of subscribers\n", + "CR=C/SC#busy hour calling rate\n", + "print \"Busy hour calling rate is:%0.2f \\n\"%(CR)\n", + "T=2.5/60#avarage duration of calls in hours\n", + "\n", + "A=C*T#rate of traffic flow\n", + "print \"Rate of traffic flow is %0.2f traffic unit \"%(A)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4,page no 126" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maxixmum current is 33.33 mamps \n", + "\n", + "operate lag is 1.83 msec \n", + "\n", + "release lag is 2.85 msec \n", + "\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "#given\n", + "L=3#relay inductance in henry\n", + "R=1500#relay resistance in ohm\n", + "Io=20e-3#oparating current in amps\n", + "Ir=8e-3#release current in amps\n", + "\n", + "V=50#supply volatage in volts\n", + "Im=V/R#maxixmum current in amps\n", + "print \"maxixmum current is %0.2f mamps \\n\"%(Im*1e3)\n", + "to=(L/R)*log(1/(1-(Io/Im)))#operate lag in sec\n", + "print \"operate lag is %0.2f msec \\n\"%(to*1000)\n", + "tr=(L/R)*log(Im/Ir)#release lag in sec\n", + "print \"release lag is %0.2f msec \\n\"%(tr*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4.1,page no 126" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)\n", + "periods per character is:150.00 msec\n", + "\n", + "period per element is:20.00 msec\n", + "\n", + "speed is:50.00 bauds\n", + "\n", + "\n", + "(b)\n", + "periods per character is:100.00 msec\n", + "\n", + "period per element is:13.33 msec\n", + "\n", + "speed is 75.00 bauds\n", + "\n", + "\n", + "(c)\n", + "periods per character is:100.00 msec\n", + "\n", + "period per element is:10.00 msec\n", + "\n", + "speed is 100.00 bauds\n", + "\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#given\n", + "#a\n", + "C_S1=20/3#speed in characters per second\n", + "P_C1=1/C_S1#periods per character\n", + "print \"(a)\\nperiods per character is:%0.2f msec\\n\"%(P_C1*1e3)\n", + "E_C1=7.5#elements per character\n", + "P_E1=P_C1/E_C1#period per element\n", + "print \"period per element is:%0.2f msec\\n\"%(P_E1*1e3)\n", + "Sb1=1/P_E1#speed in bauds\n", + "print \"speed is:%0.2f bauds\\n\\n\"%(Sb1)\n", + "#b\n", + "C_S2=10#speed in characters per second\n", + "P_C2=1/C_S2#periods per character\n", + "print \"(b)\\nperiods per character is:%0.2f msec\\n\"%(P_C2*1e3)\n", + "E_C2=7.5#elements per character\n", + "P_E2=P_C2/E_C2#period per element\n", + "print \"period per element is:%0.2f msec\\n\"%(P_E2*1e3)\n", + "Sb2=1/P_E2#speed in bauds\n", + "print \"speed is %0.2f bauds\\n\\n\"%( Sb2)\n", + "#c\n", + "C_S3=10#speed in characters per second\n", + "P_C3=1/C_S3#periods per character\n", + "print \"(c)\\nperiods per character is:%0.2f msec\\n\"%(P_C3*1e3)\n", + "E_C3=10#elements per character\n", + "P_E3=P_C3/E_C3#period per element\n", + "print \"period per element is:%0.2f msec\\n\"%(P_E3*1e3)\n", + "Sb3=1/P_E3#speed in bauds\n", + "print \"speed is %0.2f bauds\\n\"%(Sb3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.5,page no 127" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "total inductance is 0.05 H \n", + "\n", + "maximum current is 10.00 mA \n", + "\n", + "operating current is 5.00 mA \n", + "\n", + "operate lag is 0.35 msec \n", + "\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#given\n", + "N=1000#no of turns\n", + "L1=5e-8#inductance per turn\n", + "L=N**2*L1#total inductance\n", + "print \"total inductance is %0.2f H \\n\"%(L)\n", + "R=100#resistance of winding in ohm\n", + "MMF=5#operating MMF in amp. turn\n", + "V=1#voltage of received signal in volts\n", + "Im=V/R#maximum current\n", + "print \"maximum current is %0.2f mA \\n\"%(Im*1e3)\n", + "Io=MMF/N#operating current\n", + "print \"operating current is %0.2f mA \\n\"%(Io*1e3)\n", + "to=(L/R)*log(1/(1-(Io/Im)))#operate lag\n", + "print \"operate lag is %0.2f msec \\n\"%(to*1e3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.6,page no 128" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Busy hour calling rate is:1.60 \n", + "\n", + "Rate of traffic flow is 693.33 traffic unit \n" + ] + } + ], + "source": [ + "#given\n", + "S=10000#no of subscribers\n", + "C=16000#Tatol no of call in busy hour\n", + "CR=C/S#busy hour calling rate\n", + "print \"Busy hour calling rate is:%0.2f \\n\"%(CR)\n", + "T=2.6#avarage duration of calls in min\n", + "\n", + "A=C*(T/60)#rate of traffic flow\n", + "print \"Rate of traffic flow is %0.2f traffic unit \"%(A)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7,page no 135" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duration of each element is:10.00 msec\n", + "\n", + "speed is 100.00 bauds\n", + "\n", + "total possible combinations are:128.00\n" + ] + } + ], + "source": [ + "#given\n", + "N=7#no of character elements\n", + "E_C=10#elements per character (1+7+1+1)\n", + "To=100e-3#duration of one character\n", + "Te=To/E_C#duration of each element\n", + "print \"duration of each element is:%0.2f msec\\n\"%(Te*1e3)\n", + "Sb=1/Te#speed in bauds\n", + "print \"speed is %0.2f bauds\\n\"%(Sb)\n", + "C=2**N#total possible combinations\n", + "print \"total possible combinations are:%0.2f\"%(C)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.8,page no 129" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total no of call in busy hour is:1500.00 calls per Hour\n", + "\n", + "Busy hour calling rate is:1.50 \n", + "\n", + "grade of service is: 0.02\n" + ] + } + ], + "source": [ + "#given\n", + "S=1000#no of subscribers\n", + "T=2.4/60#avarage duration of calls in hours\n", + "A=60#rate of traffic flow\n", + "C=A/T#Tatol no of call in busy hour\n", + "print \"Total no of call in busy hour is:%0.2f calls per Hour\\n\"%(C)\n", + "CR=C/S#busy hour calling rate\n", + "print \"Busy hour calling rate is:%0.2f \\n\"%(CR)\n", + "SCL=30#no of call lost per hour\n", + "\n", + "B=SCL/(C+SCL)#grade of service\n", + "print \"grade of service is: %0.2f\"%(B)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.9,page no 129" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "grade of service is: 2.00e-03\n", + "\n", + "traffic lost is: 1.80e-03\n" + ] + } + ], + "source": [ + "from math import factorial\n", + "#given\n", + "N=5#no of switches\n", + "A=0.9#traffic offered \n", + "#grade of service B=(A**N/N!)/(1+A+A**2/2!+A**3/3!+...+A**N/N!)\n", + "#here\n", + "B=(A**N/factorial(N))/(1+A+(A**2/factorial(2))+(A**3/factorial(3))+(A**4/factorial(4))+(A**5/factorial(5)))\n", + "print \"grade of service is: %0.2e\\n\"%(B)\n", + "Tl=A*B#traffic lost\n", + "print \"traffic lost is: %0.2e\"%(Tl)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/sample_notebooks/Mohd. Arif/Chapter4.ipynb b/sample_notebooks/Mohd. Arif/Chapter4.ipynb new file mode 100644 index 00000000..ac35f0f7 --- /dev/null +++ b/sample_notebooks/Mohd. Arif/Chapter4.ipynb @@ -0,0 +1,513 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 : Truncation Errors and the Taylor Series" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: 4.1 Page No:79" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of f at x=0 : 1.2\n", + "The value of f at x=1 due to zero order approximation : 1.2\n", + "Truncation error : -1.0\n", + "----------------------------------------------\n", + "The value of first derivative of f at x=0 : -0.4\n", + "The value of f at x=1 due to first order approximation : 0.8\n", + "Truncation error : -0.6\n", + "----------------------------------------------\n", + "The value of second derivative of f at x=0 : -1.8\n", + "The value of f at x=1 due to second order approximation : -0.1\n", + "Truncation error : 0.3\n", + "----------------------------------------------\n", + "The value of third derivative of f at x=0 : -0.9\n", + "The value of f at x=1 due to third order approximation : -0.25\n", + "Truncation error : 0.45\n", + "----------------------------------------------\n", + "The value of fourth derivative of f at x=0 : -2.4\n", + "The value of f at x=1 due to fourth order approximation : -0.35\n", + "Truncation error : 0.55\n", + "----------------------------------------------\n" + ] + } + ], + "source": [ + "from math import factorial\n", + "from scipy.misc import derivative\n", + "def f(x):\n", + " y=-0.1*x**4-0.15*x**3-0.5*x**2-0.25*x+1.2#\n", + " return y\n", + "xi=0#\n", + "xf=1#\n", + "h=xf-xi#\n", + "fi=f(xi)##function value at xi\n", + "ffa=f(xf)##actual function value at xf\n", + "\n", + "#for n=0, i.e, zero order approximation\n", + "ff=fi#\n", + "Et_1=ffa-ff##truncation error at x=1\n", + "print \"The value of f at x=0 :\",fi\n", + "print \"The value of f at x=1 due to zero order approximation :\",ff\n", + "print \"Truncation error :\",Et_1\n", + "print \"----------------------------------------------\"\n", + "\n", + "#for n=1, i.e, first order approximation\n", + "def f1(x):\n", + " y=derivative(f,x)\n", + " return y\n", + "f1i=f1(xi)##value of first derivative of function at xi\n", + "f1f=fi+f1i*h##value of first derivative of function at xf\n", + "Et_2=ffa-f1f##truncation error at x=1\n", + "print \"The value of first derivative of f at x=0 :\",f1i\n", + "print \"The value of f at x=1 due to first order approximation :\",f1f\n", + "print \"Truncation error :\",Et_2\n", + "print \"----------------------------------------------\"\n", + "\n", + "\n", + "#for n=2, i.e, second order approximation\n", + "def f2(x):\n", + " y=derivative(f1,x)\n", + " return y\n", + "f2i=f2(xi)##value of second derivative of function at xi\n", + "f2f=f1f+f2i*(h**2)/factorial(2)##value of second derivative of function at xf\n", + "Et_3=ffa-f2f##truncation error at x=1\n", + "print \"The value of second derivative of f at x=0 :\",f2i\n", + "print \"The value of f at x=1 due to second order approximation :\",f2f\n", + "print \"Truncation error :\",Et_3\n", + "print \"----------------------------------------------\"\n", + "\n", + "#for n=3, i.e, third order approximation\n", + "def f3(x):\n", + " y=derivative(f2,x)\n", + " return y\n", + "f3i=f3(xi)##value of third derivative of function at xi\n", + "f3f=f2f+f3i*(h**3)/factorial(3)##value of third derivative of function at xf\n", + "Et_4=ffa-f3f##truncation error at x=1\n", + "print \"The value of third derivative of f at x=0 :\",f3i\n", + "print \"The value of f at x=1 due to third order approximation :\",f3f\n", + "print \"Truncation error :\", Et_4\n", + "print \"----------------------------------------------\"\n", + "\n", + "#for n=4, i.e, fourth order approximation\n", + "def f4(x):\n", + " y=derivative(f3,x)\n", + " return y\n", + "f4i=f4(xi)##value of fourth derivative of function at xi\n", + "f4f=f3f+f4i*(h**4)/factorial(4)##value of fourth derivative of function at xf\n", + "Et_5=ffa-f4f##truncation error at x=1\n", + "print \"The value of fourth derivative of f at x=0 :\",f4i\n", + "print \"The value of f at x=1 due to fourth order approximation :\",f4f\n", + "print \"Truncation error :\",Et_5\n", + "print \"----------------------------------------------\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2: Page No:82" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of f at x=1 due to zero order approximation : 0.707106781187\n", + "% relative error : -41.4213562373\n", + "----------------------------------------------\n", + "The value of f at x=1 due to first order approximation : 0.551333569463\n", + "% relative error : -10.2667138927\n", + "----------------------------------------------\n", + "The value of f at x=1 due to second order approximation : 0.534175415889\n", + "% relative error : -6.83508317772\n", + "----------------------------------------------\n", + "The value of f at x=1 due to third order approximation : 0.535435376789\n", + "% relative error : -7.08707535775\n", + "----------------------------------------------\n", + "The value of f at x=1 due to fourth order approximation : 0.535504768061\n", + "% relative error : -7.10095361216\n", + "----------------------------------------------\n", + "The value of f at x=1 due to fifth order approximation : 0.535501917392\n", + "% relative error : -7.10038347839\n", + "----------------------------------------------\n", + "The value of f at x=1 due to sixth order approximation : 0.535501819651\n", + "% relative error : -7.10036393016\n", + "----------------------------------------------\n" + ] + } + ], + "source": [ + "from math import pi,cos,factorial\n", + "from scipy.misc import derivative\n", + "def f(x):\n", + " y=cos(x)\n", + " return y\n", + "xi=pi/4#\n", + "xf=pi/3#\n", + "h=xf-xi#\n", + "fi=f(xi)##function value at xi\n", + "ffa=f(xf)##actual function value at xf\n", + "\n", + "#for n=0, i.e, zero order approximation\n", + "ff=fi#\n", + "et1=(ffa-ff)*100/ffa##percent relative error at x=1\n", + "print \"The value of f at x=1 due to zero order approximation :\",ff\n", + "print \"% relative error :\",et1\n", + "print \"----------------------------------------------\"\n", + "\n", + "#for n=1, i.e, first order approximation\n", + "def f1(x):\n", + " y=derivative(f,x)\n", + " return y\n", + "f1i=f1(xi)##value of first derivative of function at xi\n", + "f1f=fi+f1i*h##value of first derivative of function at xf\n", + "et2=(ffa-f1f)*100/ffa##% relative error at x=1\n", + "print \"The value of f at x=1 due to first order approximation :\",f1f\n", + "print \"% relative error :\",et2\n", + "print \"----------------------------------------------\"\n", + "\n", + "\n", + "#for n=2, i.e, second order approximation\n", + "def f2(x):\n", + " y=derivative(f1,x)\n", + " return y\n", + "f2i=f2(xi)##value of second derivative of function at xi\n", + "f2f=f1f+f2i*(h**2)/factorial(2)##value of second derivative of function at xf\n", + "et3=(ffa-f2f)*100/ffa##% relative error at x=1\n", + "print \"The value of f at x=1 due to second order approximation :\",f2f\n", + "print \"% relative error :\",et3\n", + "print \"----------------------------------------------\"\n", + "\n", + "\n", + "#for n=3, i.e, third order approximation\n", + "def f3(x):\n", + " y=derivative(f2,x)\n", + " return y\n", + "f3i=f3(xi)##value of third derivative of function at xi\n", + "f3f=f2f+f3i*(h**3)/factorial(3)##value of third derivative of function at xf\n", + "et4=(ffa-f3f)*100/ffa##% relative error at x=1\n", + "print \"The value of f at x=1 due to third order approximation :\",f3f\n", + "print \"% relative error :\",et4\n", + "print \"----------------------------------------------\"\n", + "\n", + "\n", + "#for n=4, i.e, fourth order approximation\n", + "def f4(x):\n", + " y=derivative(f3,x)\n", + " return y\n", + "f4i=f4(xi)##value of fourth derivative of function at xi\n", + "f4f=f3f+f4i*(h**4)/factorial(4)##value of fourth derivative of function at xf\n", + "et5=(ffa-f4f)*100/ffa##% relative error at x=1\n", + "print \"The value of f at x=1 due to fourth order approximation :\",f4f\n", + "print \"% relative error :\",et5\n", + "print \"----------------------------------------------\"\n", + "\n", + "\n", + "#for n=5, i.e, fifth order approximation\n", + "f5i=(f4(1.1*xi)-f4(0.9*xi))/(2*0.1)##value of fifth derivative of function at xi (central difference method)\n", + "f5f=f4f+f5i*(h**5)/factorial(5)##value of fifth derivative of function at xf\n", + "et6=(ffa-f5f)*100/ffa##% relative error at x=1\n", + "print \"The value of f at x=1 due to fifth order approximation :\",f5f\n", + "print \"% relative error :\",et6\n", + "print \"----------------------------------------------\"\n", + "\n", + "\n", + "#for n=6, i.e, sixth order approximation\n", + "def f6(x):\n", + " y=derivative(f5,x)\n", + " return y\n", + "f6i=(f4(1.1*xi)-2*f4(xi)+f4(0.9*xi))/(0.1**2)##value of sixth derivative of function at xi (central difference method)\n", + "f6f=f5f+f6i*(h**6)/factorial(6)##value of sixth derivative of function at xf\n", + "et6=(ffa-f6f)*100/ffa##% relative error at x=1\n", + "print \"The value of f at x=1 due to sixth order approximation :\",f6f\n", + "print \"% relative error :\", et6\n", + "print \"----------------------------------------------\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3 : Page No:85" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Input value of m:4\n", + "Input value of h:5\n", + "\n", + "Remainder: 21 \n", + "The value by first order approximation: 1275\n", + "True Value at x2: 1296\n" + ] + } + ], + "source": [ + "from math import pi,cos,factorial\n", + "m=input(\"Input value of m:\")\n", + "h=input(\"Input value of h:\")\n", + "def f(x):\n", + " y=x**m\n", + " return y\n", + "x1=1#\n", + "x2=x1+h#\n", + "fx1=f(x1)#\n", + "fx2=fx1+m*(fx1**(m-1))*h#\n", + "if m==1:\n", + " R=0#\n", + "elif m==2 :\n", + " R=2*(h**2)/factorial(2)#\n", + " \n", + "elif m==3:\n", + " R=(6*(x1)*(h**2)/factorial(2))+(6*(h**3)/factorial(3))#\n", + " \n", + "elif m==4:\n", + " R=(12*(x1**2)*(h**2)/factorial(2))+(24*(x1)*(h**3)/factorial(3))+(24*(h**4)/factorial(4))\n", + " \n", + "print \"\\nRemainder:\",fx2,\"\\nThe value by first order approximation:\",R\n", + "print \"True Value at x2:\",f(x2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4: Page No:92" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Input h:1.232323\n", + "For h= 1.232323\n", + "and percent error= -2.70944264922 Derivative at x by forward difference method= 114.60931875\n", + "and percent error= -0.178591334206 Derivative at x by backward difference method= 85.854151746\n", + "and percent error= -1.44401699172 Derivative at x by central difference method= 14.3775835022\n" + ] + } + ], + "source": [ + "from scipy.misc import derivative\n", + "def f(x):\n", + " y=-0.1*(x**4)-0.15*(x**3)-0.5*(x**2)-0.25*(x)+1.2\n", + " return y\n", + "x=0.5#\n", + "h=input(\"Input h:\")\n", + "x1=x-h#\n", + "x2=x+h#\n", + "#forward difference method\n", + "fdx1=(f(x2)-f(x))/h##derivative at x\n", + "et1=abs((fdx1-derivative(f,x))/derivative(f,x))*100#\n", + "#backward difference method\n", + "fdx2=(f(x)-f(x1))/h##derivative at x\n", + "et2=abs((fdx2-derivative(f,x))/derivative(f,x))*100#\n", + "#central difference method\n", + "fdx3=(f(x2)-f(x1))/(2*h)##derivative at x\n", + "et3=abs((fdx3-derivative(f,x))/derivative(f,x))*100#\n", + "print \"For h=\",h\n", + "print \"and percent error=\",fdx1,\"Derivative at x by forward difference method=\",et1\n", + "print \"and percent error=\",fdx2,\"Derivative at x by backward difference method=\",et2\n", + "print \"and percent error=\",fdx3,\"Derivative at x by central difference method=\",et3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5: Page No: 95" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "true value is between : 15.4275 and 15.8225\n" + ] + } + ], + "source": [ + "from scipy.misc import derivative\n", + "def f(x):\n", + " y=x**3\n", + " return y\n", + "x=2.5#\n", + "delta=0.01#\n", + "deltafx=abs(derivative(f,x))*delta#\n", + "fx=f(x)#\n", + "print \"true value is between : \",fx-deltafx,\"and\",fx+deltafx" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.6: Page No: 96" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of y is between: 0.528721343471 and 0.596278656529\n", + "ymin is calculated at lower extremes of F, L, E, I values as = 0.524066539965\n", + "ymax is calculated at higher extremes of F, L, E, I values as = 0.602846335915\n" + ] + } + ], + "source": [ + "def f(F,L,E,I):\n", + " y=(F*(L**4))/(8*E*I)\n", + " return y\n", + "Fbar=50##lb/ft\n", + "Lbar=30##ft\n", + "Ebar=1.5*(10**8)##lb/ft**2\n", + "Ibar=0.06##ft**4\n", + "deltaF=2##lb/ft\n", + "deltaL=0.1##ft\n", + "deltaE=0.01*(10**8)##lb/ft**2\n", + "deltaI=0.0006##ft**4\n", + "ybar=(Fbar*(Lbar**4))/(8*Ebar*Ibar)#\n", + "def f1(F):\n", + " y=(F*(Lbar**4))/(8*Ebar*Ibar)\n", + " return y\n", + "def f2(L):\n", + " y=(Fbar*(L**4))/(8*Ebar*Ibar)\n", + " return y\n", + "def f3(E):\n", + " y=(Fbar*(Lbar**4))/(8*E*Ibar)\n", + " return y\n", + "def f4(I):\n", + " y=(Fbar*(Lbar**4))/(8*Ebar*I)\n", + " return y\n", + "\n", + "deltay=abs(derivative(f1,Fbar))*deltaF+abs(derivative(f2,Lbar))*deltaL+abs(derivative(f3,Ebar))*deltaE+abs(derivative(f4,Ibar))*deltaI#\n", + "\n", + "print \"The value of y is between:\",ybar-deltay,\"and\",ybar+deltay\n", + "ymin=((Fbar-deltaF)*((Lbar-deltaL)**4))/(8*(Ebar+deltaE)*(Ibar+deltaI))#\n", + "ymax=((Fbar+deltaF)*((Lbar+deltaL)**4))/(8*(Ebar-deltaE)*(Ibar-deltaI))#\n", + "print \"ymin is calculated at lower extremes of F, L, E, I values as =\",ymin\n", + "print \"ymax is calculated at higher extremes of F, L, E, I values as =\",ymax" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7 : Page No:98" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The condition number of function for x = 0.18201112073 is : 1.72787595947\n", + "The condition number of function for x = 0.0160083243793 is : 1.58650429006\n" + ] + } + ], + "source": [ + "from math import pi,tan\n", + "from scipy.misc import derivative\n", + "def f(x):\n", + " y=tan(x)\n", + " return y\n", + "x1bar=(pi/2)+0.1*(pi/2)#\n", + "x2bar=(pi/2)+0.01*(pi/2)#\n", + "#computing condition number for x1bar\n", + "condnum1=x1bar*derivative(f,x1bar)/f(x1bar)#\n", + "print \"The condition number of function for x =\",condnum1,\"is :\",x1bar\n", + "if abs(condnum1)>1:\n", + " print \"Function is ill-conditioned for x =\",x1bar\n", + "\n", + "#computing condition number for x2bar\n", + "condnum2=x2bar*derivative(f,x2bar)/f(x2bar)#\n", + "print \"The condition number of function for x =\",condnum2,\"is :\",x2bar\n", + "if abs(condnum2)>1:\n", + " print \"Function is ill-conditioned for x =\",x2bar" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/sample_notebooks/RavirajJadeja/Chapter2.ipynb b/sample_notebooks/RavirajJadeja/Chapter2.ipynb new file mode 100644 index 00000000..a3cca260 --- /dev/null +++ b/sample_notebooks/RavirajJadeja/Chapter2.ipynb @@ -0,0 +1,698 @@ +<!DOCTYPE html> +<html> +<head> + +<meta charset="utf-8" /> +<title>Chapter2_1</title> + +<script src="https://cdnjs.cloudflare.com/ajax/libs/require.js/2.1.10/require.min.js"></script> +<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script> +<script src="https://rawgit.com/fossee/custom/static/custom.js"></script> + +<style 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directory as the html file --> +<!--link rel="stylesheet" href="custom.css"--> +<link rel="stylesheet" href="https://rawgit.com/fossee/custom/static/custom.css"> + +<!-- Loading mathjax macro --> +<!-- Load mathjax --> + <script src="https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS_HTML"></script> + <!-- MathJax configuration --> + <script type="text/x-mathjax-config"> + MathJax.Hub.Config({ + tex2jax: { + inlineMath: [ ['$','$'], ["\\(","\\)"] ], + displayMath: [ ['$$','$$'], ["\\[","\\]"] ], + processEscapes: true, + processEnvironments: true + }, + // Center justify equations in code and markdown cells. Elsewhere + // we use CSS to left justify single line equations in code cells. + displayAlign: 'center', + "HTML-CSS": { + styles: {'.MathJax_Display': {"margin": 0}}, + linebreaks: { automatic: true } + } + }); + </script> + <!-- End of mathjax configuration --> + +</head> +<body> + <div tabindex="-1" id="notebook" class="border-box-sizing"> + <div class="container" id="notebook-container"> + +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h1 id="2:-Crystal-Structures">2: Crystal Structures<a class="anchor-link" href="#2:-Crystal-Structures">¶</a></h1> +</div> +</div> +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-number-2.1,-Page-number-2.23">Example number 2.1, Page number 2.23<a class="anchor-link" href="#Example-number-2.1,-Page-number-2.23">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [3]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#importing modules</span> +<span class="kn">import</span> <span class="nn">math</span> +<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> + +<span class="c">#Variable declaration</span> +<span class="n">M</span><span class="o">=</span><span class="mf">60.2</span><span class="p">;</span> <span class="c">#molecular weight</span> +<span class="n">Na</span><span class="o">=</span><span class="mf">6.023</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">26</span><span class="p">;</span> <span class="c">#avagadro number(kg/mole)</span> +<span class="n">n</span><span class="o">=</span><span class="mi">4</span><span class="p">;</span> +<span class="n">rho</span><span class="o">=</span><span class="mi">6250</span><span class="p">;</span> <span class="c">#density(kg/m**3)</span> + +<span class="c">#Calculation</span> +<span class="n">a</span><span class="o">=</span><span class="p">(</span><span class="n">n</span><span class="o">*</span><span class="n">M</span><span class="o">/</span><span class="p">(</span><span class="n">rho</span><span class="o">*</span><span class="n">Na</span><span class="p">))</span><span class="o">**</span><span class="p">(</span><span class="mi">1</span><span class="o">/</span><span class="mi">3</span><span class="p">);</span> <span class="c">#lattice constant(m)</span> + +<span class="c">#Result</span> +<span class="k">print</span> <span class="s">"lattice constant is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">a</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">10</span><span class="p">),</span><span class="s">"*10**-10 m"</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>lattice constant is 4.0 *10**-10 m +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-number-2.2,-Page-number-2.23">Example number 2.2, Page number 2.23<a class="anchor-link" href="#Example-number-2.2,-Page-number-2.23">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [7]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#importing modules</span> +<span class="kn">import</span> <span class="nn">math</span> +<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> + +<span class="c">#Variable declaration</span> +<span class="n">M</span><span class="o">=</span><span class="mf">63.5</span><span class="p">;</span> <span class="c">#molecular weight</span> +<span class="n">Na</span><span class="o">=</span><span class="mf">6.023</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">26</span><span class="p">;</span> <span class="c">#avagadro number(kg/mole)</span> +<span class="n">n</span><span class="o">=</span><span class="mi">4</span><span class="p">;</span> +<span class="n">r</span><span class="o">=</span><span class="mf">1.278</span><span class="o">*</span><span class="mi">10</span><span class="o">**-</span><span class="mi">8</span><span class="p">;</span> <span class="c">#atomic radius(cm)</span> + +<span class="c">#Calculation</span> +<span class="n">a</span><span class="o">=</span><span class="mi">2</span><span class="o">*</span><span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span><span class="o">*</span><span class="n">r</span><span class="p">;</span> <span class="c">#lattice constant(m)</span> +<span class="n">rho</span><span class="o">=</span><span class="n">n</span><span class="o">*</span><span class="n">M</span><span class="o">/</span><span class="p">(</span><span class="n">a</span><span class="o">**</span><span class="mi">3</span><span class="o">*</span><span class="n">Na</span><span class="p">);</span> <span class="c">#density(kg/cm**3)</span> + +<span class="c">#Result</span> +<span class="k">print</span> <span class="s">"density is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">rho</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">3</span><span class="p">,</span><span class="mi">2</span><span class="p">),</span><span class="s">"gm/cm**3"</span> +<span class="k">print</span> <span class="s">"answer in the book varies due to rounding off errors"</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>density is 8.93 gm/cm**3 +answer in the book varies due to rounding off errors +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-number-2.3,-Page-number-2.24">Example number 2.3, Page number 2.24<a class="anchor-link" href="#Example-number-2.3,-Page-number-2.24">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [10]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#importing modules</span> +<span class="kn">import</span> <span class="nn">math</span> +<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> + +<span class="c">#Variable declaration</span> +<span class="n">pf_BCC</span><span class="o">=</span><span class="n">math</span><span class="o">.</span><span class="n">pi</span><span class="o">*</span><span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span><span class="o">/</span><span class="mi">8</span><span class="p">;</span> <span class="c">#packing factor for BCC</span> +<span class="n">pf_FCC</span><span class="o">=</span><span class="n">math</span><span class="o">.</span><span class="n">pi</span><span class="o">/</span><span class="p">(</span><span class="mi">3</span><span class="o">*</span><span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="mi">2</span><span class="p">));</span> <span class="c">#packing factor of FCC</span> + +<span class="c">#Calculation</span> +<span class="n">r</span><span class="o">=</span><span class="n">pf_BCC</span><span class="o">/</span><span class="n">pf_FCC</span><span class="p">;</span> <span class="c">#ratio of densities</span> + +<span class="c">#Result</span> +<span class="k">print</span> <span class="s">"ratio of densities is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">r</span><span class="p">,</span><span class="mi">2</span><span class="p">)</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>ratio of densities is 0.92 +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-number-2.4,-Page-number-2.24">Example number 2.4, Page number 2.24<a class="anchor-link" href="#Example-number-2.4,-Page-number-2.24">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [14]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#importing modules</span> +<span class="kn">import</span> <span class="nn">math</span> +<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> + +<span class="c">#Variable declaration</span> +<span class="n">M</span><span class="o">=</span><span class="mf">55.85</span><span class="p">;</span> <span class="c">#molecular weight</span> +<span class="n">Na</span><span class="o">=</span><span class="mf">6.02</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">26</span><span class="p">;</span> <span class="c">#avagadro number(kg/mole)</span> +<span class="n">n</span><span class="o">=</span><span class="mi">2</span><span class="p">;</span> +<span class="n">rho</span><span class="o">=</span><span class="mi">7860</span><span class="p">;</span> <span class="c">#density(kg/m**3)</span> + +<span class="c">#Calculation</span> +<span class="n">a</span><span class="o">=</span><span class="p">(</span><span class="n">n</span><span class="o">*</span><span class="n">M</span><span class="o">/</span><span class="p">(</span><span class="n">rho</span><span class="o">*</span><span class="n">Na</span><span class="p">))</span><span class="o">**</span><span class="p">(</span><span class="mi">1</span><span class="o">/</span><span class="mi">3</span><span class="p">);</span> <span class="c">#lattice constant(m)</span> + +<span class="c">#Result</span> +<span class="k">print</span> <span class="s">"lattice constant is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">a</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">10</span><span class="p">,</span><span class="mi">4</span><span class="p">),</span><span class="s">"angstrom"</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>lattice constant is 2.8687 angstrom +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-number-2.5,-Page-number-2.24">Example number 2.5, Page number 2.24<a class="anchor-link" href="#Example-number-2.5,-Page-number-2.24">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [19]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#importing modules</span> +<span class="kn">import</span> <span class="nn">math</span> +<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> + +<span class="c">#Variable declaration</span> +<span class="n">M</span><span class="o">=</span><span class="mf">58.5</span><span class="p">;</span> <span class="c">#molecular weight</span> +<span class="n">Na</span><span class="o">=</span><span class="mf">6.02</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">26</span><span class="p">;</span> <span class="c">#avagadro number(kg/mole)</span> +<span class="n">n</span><span class="o">=</span><span class="mi">4</span><span class="p">;</span> +<span class="n">rho</span><span class="o">=</span><span class="mi">2189</span><span class="p">;</span> <span class="c">#density(kg/m**3)</span> + +<span class="c">#Calculation</span> +<span class="n">a</span><span class="o">=</span><span class="p">(</span><span class="n">n</span><span class="o">*</span><span class="n">M</span><span class="o">/</span><span class="p">(</span><span class="n">rho</span><span class="o">*</span><span class="n">Na</span><span class="p">))</span><span class="o">**</span><span class="p">(</span><span class="mi">1</span><span class="o">/</span><span class="mi">3</span><span class="p">);</span> <span class="c">#lattice constant(m)</span> + +<span class="c">#Result</span> +<span class="k">print</span> <span class="s">"lattice constant is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">a</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">10</span><span class="p">,</span><span class="mi">1</span><span class="p">),</span><span class="s">"angstrom"</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>lattice constant is 5.6 angstrom +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-number-2.6,-Page-number-2.25">Example number 2.6, Page number 2.25<a class="anchor-link" href="#Example-number-2.6,-Page-number-2.25">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [23]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#importing modules</span> +<span class="kn">import</span> <span class="nn">math</span> +<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> + +<span class="c">#Variable declaration</span> +<span class="n">M</span><span class="o">=</span><span class="mf">6.94</span><span class="p">;</span> <span class="c">#molecular weight</span> +<span class="n">Na</span><span class="o">=</span><span class="mf">6.02</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">26</span><span class="p">;</span> <span class="c">#avagadro number(kg/mole)</span> +<span class="n">n</span><span class="o">=</span><span class="mi">2</span><span class="p">;</span> +<span class="n">rho</span><span class="o">=</span><span class="mi">530</span><span class="p">;</span> <span class="c">#density(kg/m**3)</span> + +<span class="c">#Calculation</span> +<span class="n">a</span><span class="o">=</span><span class="p">(</span><span class="n">n</span><span class="o">*</span><span class="n">M</span><span class="o">/</span><span class="p">(</span><span class="n">rho</span><span class="o">*</span><span class="n">Na</span><span class="p">))</span><span class="o">**</span><span class="p">(</span><span class="mi">1</span><span class="o">/</span><span class="mi">3</span><span class="p">);</span> <span class="c">#lattice constant(m)</span> + +<span class="c">#Result</span> +<span class="k">print</span> <span class="s">"lattice constant is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">a</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">10</span><span class="p">,</span><span class="mi">3</span><span class="p">),</span><span class="s">"angstrom"</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>lattice constant is 3.517 angstrom +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-number-2.7,-Page-number-2.25">Example number 2.7, Page number 2.25<a class="anchor-link" href="#Example-number-2.7,-Page-number-2.25">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [29]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#importing modules</span> +<span class="kn">import</span> <span class="nn">math</span> +<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> + +<span class="c">#Variable declaration</span> +<span class="n">r1</span><span class="o">=</span><span class="mf">1.258</span><span class="o">*</span><span class="mi">10</span><span class="o">**-</span><span class="mi">10</span><span class="p">;</span> <span class="c">#radius(m)</span> +<span class="n">r2</span><span class="o">=</span><span class="mf">1.292</span><span class="o">*</span><span class="mi">10</span><span class="o">**-</span><span class="mi">10</span><span class="p">;</span> <span class="c">#radius(m)</span> + +<span class="c">#Calculation</span> +<span class="n">a_bcc</span><span class="o">=</span><span class="mi">4</span><span class="o">*</span><span class="n">r1</span><span class="o">/</span><span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="mi">3</span><span class="p">);</span> +<span class="n">v</span><span class="o">=</span><span class="n">a_bcc</span><span class="o">**</span><span class="mi">3</span><span class="p">;</span> +<span class="n">V1</span><span class="o">=</span><span class="n">v</span><span class="o">/</span><span class="mi">2</span><span class="p">;</span> +<span class="n">a_fcc</span><span class="o">=</span><span class="mi">2</span><span class="o">*</span><span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span><span class="o">*</span><span class="n">r2</span><span class="p">;</span> +<span class="n">V2</span><span class="o">=</span><span class="n">a_fcc</span><span class="o">**</span><span class="mi">3</span><span class="o">/</span><span class="mi">4</span><span class="p">;</span> +<span class="n">V</span><span class="o">=</span><span class="p">(</span><span class="n">V1</span><span class="o">-</span><span class="n">V2</span><span class="p">)</span><span class="o">*</span><span class="mi">100</span><span class="o">/</span><span class="n">V1</span><span class="p">;</span> <span class="c">#percent volume change is",V,"%"</span> + +<span class="c">#Result</span> +<span class="k">print</span> <span class="s">"percent volume change is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">V</span><span class="p">,</span><span class="mi">3</span><span class="p">),</span><span class="s">"%"</span> +<span class="k">print</span> <span class="s">"answer in the book varies due to rounding off errors"</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>percent volume change is 0.493 % +answer in the book varies due to rounding off errors +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-number-2.8,-Page-number-2.26">Example number 2.8, Page number 2.26<a class="anchor-link" href="#Example-number-2.8,-Page-number-2.26">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [31]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#importing modules</span> +<span class="kn">import</span> <span class="nn">math</span> +<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> + +<span class="c">#Variable declaration</span> +<span class="n">a</span><span class="o">=</span><span class="mf">0.356</span><span class="o">*</span><span class="mi">10</span><span class="o">**-</span><span class="mi">9</span><span class="p">;</span> <span class="c">#cube edge(m)</span> +<span class="n">w</span><span class="o">=</span><span class="mi">12</span><span class="p">;</span> <span class="c">#atomic weight</span> +<span class="n">Na</span><span class="o">=</span><span class="mf">6.02</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">26</span><span class="p">;</span> <span class="c">#avagadro number(kg/mole)</span> + +<span class="c">#Calculation</span> +<span class="n">n</span><span class="o">=</span><span class="mi">8</span><span class="o">/</span><span class="p">(</span><span class="n">a</span><span class="o">**</span><span class="mi">3</span><span class="p">);</span> <span class="c">#number of atoms/m**3</span> +<span class="n">m</span><span class="o">=</span><span class="n">w</span><span class="o">/</span><span class="n">Na</span><span class="p">;</span> <span class="c">#mass(kg)</span> +<span class="n">rho</span><span class="o">=</span><span class="n">m</span><span class="o">*</span><span class="n">n</span><span class="p">;</span> <span class="c">#density of diamond(kg/m**3)</span> + +<span class="c">#Result</span> +<span class="k">print</span> <span class="s">"number of atoms/m**3 is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">n</span><span class="o">/</span><span class="mi">10</span><span class="o">**</span><span class="mi">29</span><span class="p">,</span><span class="mi">2</span><span class="p">),</span><span class="s">"*10**29"</span> +<span class="k">print</span> <span class="s">"density of diamond is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">rho</span><span class="p">,</span><span class="mi">2</span><span class="p">),</span><span class="s">"kg/m**3"</span> +<span class="k">print</span> <span class="s">"answer in the book is wrong"</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>number of atoms/m**3 is 1.77 *10**29 +density of diamond is 3534.47 kg/m**3 +answer in the book is wrong +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-number-2.9,-Page-number-2.26">Example number 2.9, Page number 2.26<a class="anchor-link" href="#Example-number-2.9,-Page-number-2.26">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [1]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#importing modules</span> +<span class="kn">import</span> <span class="nn">math</span> +<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> +<span class="kn">from</span> <span class="nn">sympy</span> <span class="kn">import</span> <span class="n">Symbol</span> + +<span class="c">#Variable declaration</span> +<span class="n">r</span><span class="o">=</span><span class="n">Symbol</span><span class="p">(</span><span class="s">'r'</span><span class="p">)</span> + +<span class="c">#Calculation</span> +<span class="n">a</span><span class="o">=</span><span class="mi">4</span><span class="o">*</span><span class="n">r</span><span class="o">/</span><span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="mi">2</span><span class="p">);</span> +<span class="n">R</span><span class="o">=</span><span class="p">(</span><span class="mi">4</span><span class="o">*</span><span class="n">r</span><span class="o">/</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="mi">2</span><span class="p">)))</span><span class="o">-</span><span class="n">r</span><span class="p">;</span> <span class="c">#maximum radius of sphere</span> + +<span class="c">#Result</span> +<span class="k">print</span> <span class="s">"maximum radius of sphere is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">R</span><span class="o">/</span><span class="n">r</span><span class="p">,</span><span class="mi">3</span><span class="p">),</span><span class="s">"r"</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>maximum radius of sphere is 0.414 r +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-number-2.10,-Page-number-2.26">Example number 2.10, Page number 2.26<a class="anchor-link" href="#Example-number-2.10,-Page-number-2.26">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [2]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#importing modules</span> +<span class="kn">import</span> <span class="nn">math</span> +<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> +<span class="kn">from</span> <span class="nn">sympy</span> <span class="kn">import</span> <span class="n">Symbol</span> + +<span class="c">#Variable declaration</span> +<span class="n">r</span><span class="o">=</span><span class="n">Symbol</span><span class="p">(</span><span class="s">'r'</span><span class="p">)</span> + +<span class="c">#Calculation</span> +<span class="n">a</span><span class="o">=</span><span class="mi">4</span><span class="o">*</span><span class="n">r</span><span class="o">/</span><span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="mi">3</span><span class="p">);</span> +<span class="n">R</span><span class="o">=</span><span class="p">(</span><span class="n">a</span><span class="o">/</span><span class="mi">2</span><span class="p">)</span><span class="o">-</span><span class="n">r</span><span class="p">;</span> <span class="c">#radius of largest sphere</span> + +<span class="c">#Result</span> +<span class="k">print</span> <span class="s">"radius of largest sphere is"</span><span class="p">,</span><span class="nb">round</span><span class="p">(</span><span class="n">R</span><span class="o">/</span><span class="n">r</span><span class="p">,</span><span class="mi">3</span><span class="p">),</span><span class="s">"r"</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>radius of largest sphere is 0.155 r +</pre> +</div> +</div> + +</div> +</div> + +</div> + </div> + </div> +</body> +</html> diff --git a/sample_notebooks/Vaibhav Vajani/chapter1_2.ipynb b/sample_notebooks/Vaibhav Vajani/chapter1_2.ipynb new file mode 100644 index 00000000..46f134dd --- /dev/null +++ b/sample_notebooks/Vaibhav Vajani/chapter1_2.ipynb @@ -0,0 +1,582 @@ +<!DOCTYPE html> +<html> +<head> + +<meta charset="utf-8" /> +<title>chapter1_2</title> + +<script src="https://cdnjs.cloudflare.com/ajax/libs/require.js/2.1.10/require.min.js"></script> +<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script> +<script src="https://rawgit.com/fossee/custom/static/custom.js"></script> + +<style type="text/css"> + /*! +* +* Twitter Bootstrap +* +*//*! normalize.css v3.0.2 | MIT License | git.io/normalize 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directory as the html file --> +<!--link rel="stylesheet" href="custom.css"--> +<link rel="stylesheet" href="https://rawgit.com/fossee/custom/static/custom.css"> + +<!-- Loading mathjax macro --> +<!-- Load mathjax --> + <script src="https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS_HTML"></script> + <!-- MathJax configuration --> + <script type="text/x-mathjax-config"> + MathJax.Hub.Config({ + tex2jax: { + inlineMath: [ ['$','$'], ["\\(","\\)"] ], + displayMath: [ ['$$','$$'], ["\\[","\\]"] ], + processEscapes: true, + processEnvironments: true + }, + // Center justify equations in code and markdown cells. Elsewhere + // we use CSS to left justify single line equations in code cells. + displayAlign: 'center', + "HTML-CSS": { + styles: {'.MathJax_Display': {"margin": 0}}, + linebreaks: { automatic: true } + } + }); + </script> + <!-- End of mathjax configuration --> + +</head> +<body> + <div tabindex="-1" id="notebook" class="border-box-sizing"> + <div class="container" id="notebook-container"> + +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h1 id="Chapter-01:-The-Foundations:-Logic-and-Proofs">Chapter 01: The Foundations: Logic and Proofs<a class="anchor-link" href="#Chapter-01:-The-Foundations:-Logic-and-Proofs">¶</a></h1> +</div> +</div> +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-01:Page-02">Example 01:Page 02<a class="anchor-link" href="#Example-01:Page-02">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [1]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="k">print</span> <span class="s">"The following sentences are Propositions"</span> <span class="c">#Proposition should be a declarative sentence or should result in either a YES or a NO.</span> + +<span class="k">print</span> <span class="s">"1. Washington D.C is the capital of the United States of America</span><span class="se">\n</span><span class="s">2. Toronto is the capital of Canada</span><span class="se">\n</span><span class="s">3. 1+1=2.</span><span class="se">\n</span><span class="s">4. 2+2=3."</span> <span class="c">#Since these statements are declarative and they answer the question YES or NO they are called propositions.</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>The following sentences are Propositions +1. Washington D.C is the capital of the United States of America +2. Toronto is the capital of Canada +3. 1+1=2. +4. 2+2=3. +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-02:Page-02">Example 02:Page 02<a class="anchor-link" href="#Example-02:Page-02">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [2]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="k">print</span> <span class="s">"1. What time is it? </span><span class="se">\n</span><span class="s">2. Read this carefully. </span><span class="se">\n</span><span class="s">3. x+1=2.</span><span class="se">\n</span><span class="s">4. x+y=Z."</span> +<span class="k">print</span><span class="s">"Sentences 1 and 2 are not propositions since they are not declarative. Sentences 3 and 4 are neither true nor false and so they are not propositions."</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>1. What time is it? +2. Read this carefully. +3. x+1=2. +4. x+y=Z. +Sentences 1 and 2 are not propositions since they are not declarative. Sentences 3 and 4 are neither true nor false and so they are not propositions. +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-03:Page-03">Example 03:Page 03<a class="anchor-link" href="#Example-03:Page-03">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [9]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="k">print</span> <span class="s">"Propositon p=Michael's PC runs Linux."</span> +<span class="k">print</span> <span class="s">"</span><span class="se">\n</span><span class="s"> Negation of p is ~p : It is not the case that Michael's PC runs Linux."</span> +<span class="k">print</span> <span class="s">"</span><span class="se">\n</span><span class="s"> Negation of p is ~p : Michae's PC does not run."</span><span class="c">#Negation is opposite of the truth value of the proposition expressed with "it is not the case that" or with "not".</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Propositon p=Michael's PC runs Linux. + + Negation of p is ~p : It is not the case that Michael's PC runs Linux. + + Negation of p is ~p : Michae's PC does not run. +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-04:Page-03">Example 04:Page 03<a class="anchor-link" href="#Example-04:Page-03">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [10]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="k">print</span> <span class="s">"Let p=Vandana's smartphone has at least 32GB of memory."</span> +<span class="k">print</span> <span class="s">"The negation of p is ( ~p ) :It is not the case that Vandana's smartphone has at least 32GB of memory."</span> +<span class="k">print</span> <span class="s">"Or in simple English ( ~p ): Vandana's smartphone does not have at least 32GB of memory."</span> +<span class="k">print</span> <span class="s">"Or even more simple as ( ~p ): Vandana's smartphone has less than 32GB of memory."</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Let p=Vandana's smartphone has at least 32GB of memory. +The negation of p is ( ~p ) :It is not the case that Vandana's smartphone has at least 32GB of memory. +Or in simple English ( ~p ): Vandana's smartphone does not have at least 32GB of memory. +Or even more simple as ( ~p ): Vandana's smartphone has less than 32GB of memory. +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-05:Page-04">Example 05:Page 04<a class="anchor-link" href="#Example-05:Page-04">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [11]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="n">p</span><span class="o">=</span><span class="s">"Rebecca's PC has more than 16GB free hard disk space"</span> +<span class="n">q</span><span class="o">=</span><span class="s">"The processor in Rebecca's PC runs faster than 1GHz"</span> +<span class="k">print</span> <span class="s">"Let p,q be two propositions"</span> +<span class="k">print</span> <span class="s">"Let p="</span><span class="p">,</span><span class="n">p</span><span class="p">,</span><span class="s">"</span><span class="se">\n</span><span class="s">"</span><span class="p">,</span><span class="s">"Let q="</span><span class="p">,</span><span class="n">q</span> +<span class="k">print</span> <span class="s">"Conjunction of p^q is : "</span><span class="o">+</span><span class="n">p</span><span class="o">+</span><span class="s">" and "</span><span class="o">+</span><span class="n">q</span> <span class="c">#conjunction combines two propositons with "and"</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Let p,q be two propositions +Let p= Rebecca's PC has more than 16GB free hard disk space +Let q= The processor in Rebecca's PC runs faster than 1GHz +Conjunction of p^q is : Rebecca's PC has more than 16GB free hard disk space and The processor in Rebecca's PC runs faster than 1GHz +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-06:Page-05">Example 06:Page 05<a class="anchor-link" href="#Example-06:Page-05">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [12]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="n">p</span><span class="o">=</span><span class="s">"Rebecca's PC has more than 16GB free hard disk space"</span> +<span class="n">q</span><span class="o">=</span><span class="s">"The processor in Rebecca's PC runs faster than 1GHz"</span> +<span class="k">print</span> <span class="s">"Let p,q be two propositions"</span> +<span class="k">print</span> <span class="s">"Let p="</span><span class="p">,</span><span class="n">p</span><span class="p">,</span><span class="s">"</span><span class="se">\n</span><span class="s">"</span><span class="p">,</span><span class="s">"Let q="</span><span class="p">,</span><span class="n">q</span> +<span class="k">print</span> <span class="s">"Disjunction of p\/q is : "</span><span class="o">+</span><span class="n">p</span><span class="o">+</span><span class="s">" or "</span><span class="o">+</span><span class="n">q</span> <span class="c">#unavailability of cup symbol. So \/</span> +<span class="c">#Disjunction combines two propositons using OR </span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Let p,q be two propositions +Let p= Rebecca's PC has more than 16GB free hard disk space +Let q= The processor in Rebecca's PC runs faster than 1GHz +Disjunction of p\/q is : Rebecca's PC has more than 16GB free hard disk space or The processor in Rebecca's PC runs faster than 1GHz +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-07:Page-07">Example 07:Page 07<a class="anchor-link" href="#Example-07:Page-07">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [14]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="n">p</span><span class="o">=</span><span class="s">"Maria learns discrete mathematics"</span> +<span class="n">q</span><span class="o">=</span><span class="s">"Maria will find a good job"</span> +<span class="k">print</span><span class="s">"Let p="</span><span class="p">,</span><span class="n">p</span><span class="p">,</span><span class="s">"</span><span class="se">\n</span><span class="s">"</span><span class="p">,</span><span class="s">"Let q="</span><span class="p">,</span><span class="n">q</span> +<span class="k">print</span><span class="s">"p->q is : "</span><span class="o">+</span><span class="s">"If "</span><span class="o">+</span><span class="n">p</span><span class="o">+</span><span class="s">" then "</span><span class="o">+</span><span class="n">q</span> <span class="c">#p->q p implies q means If P then Q.</span> +<span class="k">print</span><span class="s">"p->q is also expressed as :"</span><span class="p">,</span><span class="n">q</span><span class="p">,</span><span class="s">" when "</span><span class="p">,</span><span class="n">p</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Let p= Maria learns discrete mathematics +Let q= Maria will find a good job +p->q is : If Maria learns discrete mathematics then Maria will find a good job +p->q is also expressed as : Maria will find a good job when Maria learns discrete mathematics +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-01:Page-37">Example 01:Page 37<a class="anchor-link" href="#Example-01:Page-37">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [15]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="k">def</span> <span class="nf">p</span><span class="p">(</span><span class="n">x</span><span class="p">):</span> <span class="c">#Function defined to check whether the given statements are true.</span> + <span class="k">if</span><span class="p">(</span><span class="n">x</span><span class="o">></span><span class="mi">3</span><span class="p">):</span> + <span class="k">print</span> <span class="s">"p("</span><span class="p">,</span><span class="n">x</span><span class="p">,</span><span class="s">") which is the statement"</span><span class="p">,</span><span class="n">x</span><span class="p">,</span><span class="s">">3, is true"</span> + <span class="k">else</span><span class="p">:</span> + <span class="k">print</span> <span class="s">"p("</span><span class="p">,</span><span class="n">x</span><span class="p">,</span><span class="s">") which is the statement"</span><span class="p">,</span><span class="n">x</span><span class="p">,</span><span class="s">">3, is false"</span> +<span class="n">p</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span><span class="c">#Fuction call </span> +<span class="n">p</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>p( 4 ) which is the statement 4 >3, is true +p( 2 ) which is the statement 2 >3, is false +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="Example-02:Page-38">Example 02:Page 38<a class="anchor-link" href="#Example-02:Page-38">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [17]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="n">x1</span><span class="o">=</span><span class="s">"CS1"</span> <span class="c">#Defining systems to check whether they are under attack through a function.</span> +<span class="n">x2</span><span class="o">=</span><span class="s">"CS2"</span> +<span class="n">x3</span><span class="o">=</span><span class="s">"MATH1"</span> +<span class="k">def</span> <span class="nf">A</span><span class="p">(</span><span class="n">x</span><span class="p">):</span> + <span class="k">if</span><span class="p">(</span><span class="n">x</span><span class="o">==</span><span class="s">"CS1"</span><span class="p">):</span> <span class="c">#Since cs1 and Math1 are the two computers under attack</span> + <span class="k">print</span> <span class="s">"A("</span><span class="p">,</span><span class="n">x</span><span class="p">,</span><span class="s">") is true."</span> + <span class="k">else</span><span class="p">:</span> + <span class="k">if</span><span class="p">(</span><span class="n">x</span><span class="o">==</span><span class="s">"MATH1"</span><span class="p">):</span> <span class="c">#Since CS1 and MATH1 are the two computers under attack</span> + <span class="k">print</span> <span class="s">"A("</span><span class="p">,</span><span class="n">x</span><span class="p">,</span><span class="s">") is true."</span> + <span class="k">else</span><span class="p">:</span> + <span class="k">print</span><span class="s">"A("</span><span class="p">,</span><span class="n">x</span><span class="p">,</span><span class="s">") is false."</span> +<span class="k">print</span> <span class="s">"Systems under attack are CS1 and MATH1. The truth values for the same are calculated using functions."</span> +<span class="n">A</span><span class="p">(</span><span class="n">x1</span><span class="p">)</span><span class="c">#Function call </span> +<span class="n">A</span><span class="p">(</span><span class="n">x2</span><span class="p">)</span> +<span class="n">A</span><span class="p">(</span><span class="n">x3</span><span class="p">)</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Systems under attack are CS1 and MATH1. The truth values for the same are calculated using functions. +A( CS1 ) is true. +A( CS2 ) is false. +A( MATH1 ) is true. +</pre> +</div> +</div> + +</div> +</div> + +</div> + </div> + </div> +</body> +</html> diff --git a/sample_notebooks/pratikgandhi/Chapter1.ipynb b/sample_notebooks/pratikgandhi/Chapter1.ipynb new file mode 100644 index 00000000..8d47558e --- /dev/null +++ b/sample_notebooks/pratikgandhi/Chapter1.ipynb @@ -0,0 +1,2189 @@ +<!DOCTYPE html> +<html> +<head> + +<meta charset="utf-8" /> +<title>Chapter1_1</title> + +<script src="https://cdnjs.cloudflare.com/ajax/libs/require.js/2.1.10/require.min.js"></script> +<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script> +<script src="https://rawgit.com/fossee/custom/static/custom.js"></script> + +<style type="text/css"> + /*! +* +* Twitter Bootstrap +* +*//*! normalize.css v3.0.2 | MIT License | git.io/normalize */html{font-family:sans-serif;-ms-text-size-adjust:100%;-webkit-text-size-adjust:100%}body{margin:0}article,aside,details,figcaption,figure,footer,header,hgroup,main,menu,nav,section,summary{display:block}audio,canvas,progress,video{display:inline-block;vertical-align:baseline}audio:not([controls]){display:none;height:0}[hidden],template{display:none}a{background-color:transparent}a:active,a:hover{outline:0}abbr[title]{border-bottom:1px dotted}b,strong{font-weight:bold}dfn{font-style:italic}h1{font-size:2em;margin:.67em 0}mark{background:#ff0;color:#000}small{font-size:80%}sub,sup{font-size:75%;line-height:0;position:relative;vertical-align:baseline}sup{top:-0.5em}sub{bottom:-0.25em}img{border:0}svg:not(:root){overflow:hidden}figure{margin:1em 40px}hr{-moz-box-sizing:content-box;box-sizing:content-box;height:0}pre{overflow:auto}code,kbd,pre,samp{font-family:monospace,monospace;font-size:1em}button,input,optgroup,select,textarea{color:inherit;font:inherit;margin:0}button{overflow:visible}button,select{text-transform:none}button,html input[type="button"],input[type="reset"],input[type="submit"]{-webkit-appearance:button;cursor:pointer}button[disabled],html input[disabled]{cursor:default}button::-moz-focus-inner,input::-moz-focus-inner{border:0;padding:0}input{line-height:normal}input[type="checkbox"],input[type="radio"]{box-sizing:border-box;padding:0}input[type="number"]::-webkit-inner-spin-button,input[type="number"]::-webkit-outer-spin-button{height:auto}input[type="search"]{-webkit-appearance:textfield;-moz-box-sizing:content-box;-webkit-box-sizing:content-box;box-sizing:content-box}input[type="search"]::-webkit-search-cancel-button,input[type="search"]::-webkit-search-decoration{-webkit-appearance:none}fieldset{border:1px solid #c0c0c0;margin:0 2px;padding:.35em .625em .75em}legend{border:0;padding:0}textarea{overflow:auto}optgroup{font-weight:bold}table{border-collapse:collapse;border-spacing:0}td,th{padding:0}/*! 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directory as the html file --> +<!--link rel="stylesheet" href="custom.css"--> +<link rel="stylesheet" href="https://rawgit.com/fossee/custom/static/custom.css"> + +<!-- Loading mathjax macro --> +<!-- Load mathjax --> + <script src="https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS_HTML"></script> + <!-- MathJax configuration --> + <script type="text/x-mathjax-config"> + MathJax.Hub.Config({ + tex2jax: { + inlineMath: [ ['$','$'], ["\\(","\\)"] ], + displayMath: [ ['$$','$$'], ["\\[","\\]"] ], + processEscapes: true, + processEnvironments: true + }, + // Center justify equations in code and markdown cells. Elsewhere + // we use CSS to left justify single line equations in code cells. + displayAlign: 'center', + "HTML-CSS": { + styles: {'.MathJax_Display': {"margin": 0}}, + linebreaks: { automatic: true } + } + }); + </script> + <!-- End of mathjax configuration --> + +</head> +<body> + <div tabindex="-1" id="notebook" class="border-box-sizing"> + <div class="container" id="notebook-container"> + +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h1 id="Ch-1-:-Introduction">Ch-1 : Introduction<a class="anchor-link" href="#Ch-1-:-Introduction">¶</a></h1> +</div> +</div> +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.1-page-1">exa 1.1 page 1<a class="anchor-link" href="#exa-1.1-page-1">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [25]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> +<span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span> +<span class="n">B</span><span class="o">=</span><span class="mi">100</span> <span class="c">#W(8Bulb)</span> +<span class="n">F</span><span class="o">=</span><span class="mi">60</span> <span class="c">#W(2Fan)</span> +<span class="n">L</span><span class="o">=</span><span class="mi">100</span> <span class="c">#W(2Light)</span> +<span class="n">LoadConnected</span><span class="o">=</span><span class="mi">8</span><span class="o">*</span><span class="n">B</span><span class="o">+</span><span class="mi">2</span><span class="o">*</span><span class="n">F</span><span class="o">+</span><span class="mi">2</span><span class="o">*</span><span class="n">L</span> <span class="c">#W</span> +<span class="k">print</span> <span class="s">"(a) Connected Load = </span><span class="si">%0.2f</span><span class="s"> W "</span><span class="o">%</span><span class="k">LoadConnected</span> +<span class="c">#12 midnight to 5am</span> +<span class="n">demand1</span><span class="o">=</span><span class="mi">1</span><span class="o">*</span><span class="n">F</span> <span class="c">#W</span> +<span class="c">#5am to 7am</span> +<span class="n">demand2</span><span class="o">=</span><span class="mi">2</span><span class="o">*</span><span class="n">F</span><span class="o">+</span><span class="mi">1</span><span class="o">*</span><span class="n">L</span> <span class="c">#W</span> +<span class="c">#7am to 9am</span> +<span class="n">demand3</span><span class="o">=</span><span class="mi">0</span> <span class="c">#W</span> +<span class="c">#9am to 6pm</span> +<span class="n">demand4</span><span class="o">=</span><span class="mi">2</span><span class="o">*</span><span class="n">F</span> <span class="c">#W</span> +<span class="c">#6pm to midnight</span> +<span class="n">demand5</span><span class="o">=</span><span class="mi">2</span><span class="o">*</span><span class="n">F</span><span class="o">+</span><span class="mi">4</span><span class="o">*</span><span class="n">B</span> <span class="c">#W</span> +<span class="n">DEMAND</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="n">demand1</span><span class="p">,</span><span class="n">demand2</span><span class="p">,</span> <span class="n">demand3</span><span class="p">,</span> <span class="n">demand4</span><span class="p">,</span> <span class="n">demand5</span><span class="p">])</span> +<span class="n">max_demand</span><span class="o">=</span><span class="nb">max</span><span class="p">(</span><span class="n">DEMAND</span><span class="p">)</span> +<span class="k">print</span> <span class="s">"(b) Maximum demand = </span><span class="si">%0.2f</span><span class="s"> W "</span><span class="o">%</span><span class="k">max_demand</span> +<span class="n">df</span><span class="o">=</span><span class="n">max_demand</span><span class="o">/</span><span class="n">LoadConnected</span> <span class="c">#demand factor</span> +<span class="k">print</span> <span class="s">"(c) Demand factor = </span><span class="si">%0.3f</span><span class="s">"</span><span class="o">%</span><span class="k">df</span> +<span class="n">E</span><span class="o">=</span><span class="n">demand1</span><span class="o">*</span><span class="mi">5</span><span class="o">+</span><span class="n">demand2</span><span class="o">*</span><span class="mi">2</span><span class="o">+</span><span class="n">demand3</span><span class="o">*</span><span class="mi">2</span><span class="o">+</span><span class="n">demand4</span><span class="o">*</span><span class="mi">9</span><span class="o">+</span><span class="n">demand5</span><span class="o">*</span><span class="mi">6</span> <span class="c">#Wh</span> +<span class="n">E</span><span class="o">=</span><span class="n">E</span><span class="o">/</span><span class="mi">1000</span> <span class="c">#kWh</span> +<span class="k">print</span> <span class="s">"(d) Energy consumed during 24 hours = </span><span class="si">%0.2f</span><span class="s"> kWh "</span><span class="o">%</span><span class="k">E</span> +<span class="n">Edash</span><span class="o">=</span><span class="n">LoadConnected</span><span class="o">*</span><span class="mi">24</span><span class="o">/</span><span class="mi">1000</span> <span class="c">#kWh</span> +<span class="k">print</span> <span class="s">"(e) Energy consumed during 24 hours if all devices are used = </span><span class="si">%0.2f</span><span class="s"> kWh"</span><span class="o">%</span><span class="k">Edash</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>(a) Connected Load = 1120.00 W +(b) Maximum demand = 520.00 W +(c) Demand factor = 0.464 +(d) Energy consumed during 24 hours = 4.94 kWh +(e) Energy consumed during 24 hours if all devices are used = 26.88 kWh +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.2-page-3">exa 1.2 page 3<a class="anchor-link" href="#exa-1.2-page-3">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [26]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span> +<span class="n">LoadA</span><span class="o">=</span><span class="mf">2.5</span><span class="o">*</span><span class="mi">1000</span> <span class="c">#W</span> +<span class="c">#12 midnight to 5am</span> +<span class="n">d1A</span><span class="o">=</span><span class="mi">100</span> <span class="c">#W</span> +<span class="c">#5am to 6am</span> +<span class="n">d2A</span><span class="o">=</span><span class="mf">1.1</span><span class="o">*</span><span class="mi">1000</span> <span class="c">#W</span> +<span class="c">#6am to 8am</span> +<span class="n">d3A</span><span class="o">=</span><span class="mi">200</span> <span class="c">#W</span> +<span class="c">#8am to 5pm</span> +<span class="n">d4A</span><span class="o">=</span><span class="mi">0</span> <span class="c">#W</span> +<span class="c">#5pm to 12 midnight</span> +<span class="n">d5A</span><span class="o">=</span><span class="mi">500</span> <span class="c">#W</span> +<span class="n">LoadB</span><span class="o">=</span><span class="mi">3</span><span class="o">*</span><span class="mi">1000</span> <span class="c">#W</span> +<span class="c">#11 pm to 7am</span> +<span class="n">d1B</span><span class="o">=</span><span class="mi">0</span> <span class="c">#W</span> +<span class="c">#7 am to 8 am</span> +<span class="n">d2B</span><span class="o">=</span><span class="mi">300</span> <span class="c">#W</span> +<span class="c">#8 am to 10 am</span> +<span class="n">d3B</span><span class="o">=</span><span class="mi">1</span><span class="o">*</span><span class="mi">1000</span> <span class="c">#W</span> +<span class="c">#10 am to 6 pm</span> +<span class="n">d4B</span><span class="o">=</span><span class="mi">200</span> <span class="c">#W</span> +<span class="c">#6 pm to 11 pm</span> +<span class="n">d5B</span><span class="o">=</span><span class="mi">600</span> <span class="c">#W</span> +<span class="n">DEMAND_A</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="n">d1A</span><span class="p">,</span> <span class="n">d2A</span><span class="p">,</span> <span class="n">d3A</span><span class="p">,</span> <span class="n">d4A</span><span class="p">,</span> <span class="n">d5A</span><span class="p">])</span> <span class="c">#W</span> +<span class="n">DEMAND_B</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="n">d1B</span><span class="p">,</span> <span class="n">d2B</span><span class="p">,</span> <span class="n">d3B</span><span class="p">,</span> <span class="n">d4B</span><span class="p">,</span> <span class="n">d5B</span><span class="p">])</span> <span class="c">#W</span> +<span class="n">max_demand_A</span><span class="o">=</span><span class="nb">max</span><span class="p">(</span><span class="n">DEMAND_A</span><span class="p">)</span> <span class="c">#W</span> +<span class="n">max_demand_B</span><span class="o">=</span><span class="nb">max</span><span class="p">(</span><span class="n">DEMAND_B</span><span class="p">)</span> <span class="c">#W</span> +<span class="n">df_A</span><span class="o">=</span><span class="n">max_demand_A</span><span class="o">/</span><span class="n">LoadA</span> <span class="c">#demand factor</span> +<span class="n">df_B</span><span class="o">=</span><span class="n">max_demand_B</span><span class="o">/</span><span class="n">LoadB</span> <span class="c">#demand factor</span> +<span class="k">print</span> <span class="s">"Demand factor of consumer A & B are : </span><span class="si">%0.2f</span><span class="s"> & </span><span class="si">%0.2f</span><span class="s">"</span><span class="o">%</span><span class="p">(</span><span class="n">df_A</span><span class="p">,</span><span class="n">df_B</span><span class="p">)</span> +<span class="n">gd_factor</span><span class="o">=</span><span class="p">(</span><span class="n">max_demand_A</span><span class="o">+</span><span class="n">max_demand_B</span><span class="p">)</span><span class="o">/</span><span class="n">max_demand_A</span> +<span class="k">print</span> <span class="s">"Group diversity factor = </span><span class="si">%0.3f</span><span class="s">"</span><span class="o">%</span><span class="k">gd_factor</span> +<span class="n">E_A</span><span class="o">=</span><span class="n">d1A</span><span class="o">*</span><span class="mi">5</span><span class="o">+</span><span class="n">d2A</span><span class="o">*</span><span class="mi">1</span><span class="o">+</span><span class="n">d3A</span><span class="o">*</span><span class="mi">2</span><span class="o">+</span><span class="n">d4A</span><span class="o">*</span><span class="mi">9</span><span class="o">+</span><span class="n">d5A</span><span class="o">*</span><span class="mi">7</span> <span class="c">#Wh</span> +<span class="n">E_B</span><span class="o">=</span><span class="n">d1B</span><span class="o">*</span><span class="mi">8</span><span class="o">+</span><span class="n">d2B</span><span class="o">*</span><span class="mi">1</span><span class="o">+</span><span class="n">d3B</span><span class="o">*</span><span class="mi">2</span><span class="o">+</span><span class="n">d4B</span><span class="o">*</span><span class="mi">8</span><span class="o">+</span><span class="n">d5B</span><span class="o">*</span><span class="mi">5</span> <span class="c">#Wh</span> +<span class="n">E_A</span><span class="o">=</span><span class="n">E_A</span><span class="o">/</span><span class="mi">1000</span> <span class="c">#kWh</span> +<span class="n">E_B</span><span class="o">=</span><span class="n">E_B</span><span class="o">/</span><span class="mi">1000</span> <span class="c">#kWh</span> +<span class="k">print</span> <span class="s">"Energy consumed by A & B during 24 hours = </span><span class="si">%0.2f</span><span class="s"> & </span><span class="si">%0.2f</span><span class="s"> kWh "</span><span class="o">%</span><span class="p">(</span><span class="n">E_A</span><span class="p">,</span><span class="n">E_B</span><span class="p">)</span> +<span class="n">Emax_A</span><span class="o">=</span><span class="n">max_demand_A</span><span class="o">*</span><span class="mi">24</span><span class="o">/</span><span class="mi">1000</span> <span class="c">#kWh</span> +<span class="n">Emax_B</span><span class="o">=</span><span class="n">max_demand_B</span><span class="o">*</span><span class="mi">24</span><span class="o">/</span><span class="mi">1000</span> <span class="c">#kWh</span> +<span class="k">print</span> <span class="s">"Maximum energy consumer A & B can consume during 24 hours = </span><span class="si">%0.2f</span><span class="s"> & </span><span class="si">%0.2f</span><span class="s"> kWh "</span><span class="o">%</span><span class="p">(</span><span class="n">Emax_A</span><span class="p">,</span><span class="n">Emax_B</span><span class="p">)</span> +<span class="n">ratio_A</span><span class="o">=</span><span class="n">E_A</span><span class="o">/</span><span class="n">Emax_A</span> +<span class="n">ratio_B</span><span class="o">=</span><span class="n">E_B</span><span class="o">/</span><span class="n">Emax_B</span> +<span class="k">print</span> <span class="s">"Ratio of actual energy to maximum energy of consumer A & B : </span><span class="si">%0.4f</span><span class="s"> & </span><span class="si">%0.4f</span><span class="s">"</span><span class="o">%</span><span class="p">(</span><span class="n">ratio_A</span><span class="p">,</span><span class="n">ratio_B</span><span class="p">)</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Demand factor of consumer A & B are : 0.44 & 0.33 +Group diversity factor = 1.909 +Energy consumed by A & B during 24 hours = 5.50 & 6.90 kWh +Maximum energy consumer A & B can consume during 24 hours = 26.40 & 24.00 kWh +Ratio of actual energy to maximum energy of consumer A & B : 0.2083 & 0.2875 +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.3-page-6">exa 1.3 page 6<a class="anchor-link" href="#exa-1.3-page-6">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [27]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="n">n1</span><span class="o">=</span><span class="mi">600</span> <span class="c">#No. of apartments</span> +<span class="n">L1</span><span class="o">=</span><span class="mi">5</span> <span class="c">#kW#Each Apartment Load</span> +<span class="n">n2</span><span class="o">=</span><span class="mi">20</span> <span class="c">#No. of general purpose shops</span> +<span class="n">L2</span><span class="o">=</span><span class="mi">2</span> <span class="c">#kW#Each Shop Load</span> +<span class="n">df</span><span class="o">=</span><span class="mf">0.8</span> <span class="c">#demand factor</span> +<span class="c">#1 Floor mill</span> +<span class="n">L3</span><span class="o">=</span><span class="mi">10</span> <span class="c">#kW#Load</span> +<span class="n">df3</span><span class="o">=</span><span class="mf">0.7</span> <span class="c">#demand factor</span> +<span class="c">#1 Saw mill</span> +<span class="n">L4</span><span class="o">=</span><span class="mi">5</span> <span class="c">#kW#Load</span> +<span class="n">df4</span><span class="o">=</span><span class="mf">0.8</span> <span class="c">#demand factor</span> +<span class="c">#1 Laundry</span> +<span class="n">L5</span><span class="o">=</span><span class="mi">20</span> <span class="c">#kW#Load</span> +<span class="n">df5</span><span class="o">=</span><span class="mf">0.65</span> <span class="c">#demand factor</span> +<span class="c">#1 Cinema</span> +<span class="n">L6</span><span class="o">=</span><span class="mi">80</span> <span class="c">#kW#Load</span> +<span class="n">df6</span><span class="o">=</span><span class="mf">0.5</span> <span class="c">#demand factor</span> +<span class="c">#Street lights</span> +<span class="n">n7</span><span class="o">=</span><span class="mi">200</span> <span class="c">#no. of tube lights</span> +<span class="n">L7</span><span class="o">=</span><span class="mi">40</span> <span class="c">#W#Load of each light</span> +<span class="c">#Residential Load</span> +<span class="n">df8</span><span class="o">=</span><span class="mf">0.5</span> <span class="c">#demand factor</span> +<span class="n">gdf_r</span><span class="o">=</span><span class="mi">3</span> <span class="c">#group diversity factor</span> +<span class="n">pdf_r</span><span class="o">=</span><span class="mf">1.25</span> <span class="c">#peak diversity factor</span> +<span class="c">#Commertial Load</span> +<span class="n">gdf_c</span><span class="o">=</span><span class="mi">2</span> <span class="c">#group diversity factor</span> +<span class="n">pdf_c</span><span class="o">=</span><span class="mf">1.6</span> <span class="c">#peak diversity factor</span> +<span class="c">#Solution :</span> +<span class="c">#Maximum demand of each apartment</span> +<span class="n">dmax_1a</span><span class="o">=</span><span class="n">L1</span><span class="o">*</span><span class="n">df8</span> <span class="c">#kW</span> +<span class="c">#Maximum demand of 600 apartment</span> +<span class="n">dmax_a</span><span class="o">=</span><span class="n">n1</span><span class="o">*</span><span class="n">dmax_1a</span><span class="o">/</span><span class="n">gdf_r</span> <span class="c">#kW</span> +<span class="c">#demand of apartments at system peak time</span> +<span class="n">d_a_sp</span><span class="o">=</span><span class="n">dmax_a</span><span class="o">/</span><span class="n">pdf_r</span> <span class="c">#kW</span> +<span class="c">#Maximum Commercial demand</span> +<span class="n">dmax_c</span><span class="o">=</span><span class="p">(</span><span class="n">n2</span><span class="o">*</span><span class="n">L2</span><span class="o">*</span><span class="n">df</span><span class="o">+</span><span class="n">L3</span><span class="o">*</span><span class="n">df3</span><span class="o">+</span><span class="n">L4</span><span class="o">*</span><span class="n">df4</span><span class="o">+</span><span class="n">L5</span><span class="o">*</span><span class="n">df5</span><span class="o">+</span><span class="n">L6</span><span class="o">*</span><span class="n">df6</span><span class="p">)</span><span class="o">/</span><span class="n">gdf_c</span> <span class="c">#kW</span> +<span class="c">#Commercial demand at system peak time</span> +<span class="n">d_c_sp</span><span class="o">=</span><span class="n">dmax_c</span><span class="o">/</span><span class="n">pdf_c</span> <span class="c">#kW</span> +<span class="c">#demand of street light at system peak time</span> +<span class="n">d_sl_sp</span><span class="o">=</span><span class="n">n7</span><span class="o">*</span><span class="n">L7</span><span class="o">/</span><span class="mi">1000</span> <span class="c">#kW</span> +<span class="c">#Increase in system peak demand</span> +<span class="n">DI</span><span class="o">=</span><span class="n">d_a_sp</span><span class="o">+</span><span class="n">d_c_sp</span><span class="o">+</span><span class="n">d_sl_sp</span> <span class="c">#kW</span> +<span class="k">print</span> <span class="s">"Increase in system peak demand </span><span class="si">%0.2f</span><span class="s"> kW"</span> <span class="o">%</span><span class="k">DI</span>, +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Increase in system peak demand 438.00 kW +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.4-page-12">exa 1.4 page 12<a class="anchor-link" href="#exa-1.4-page-12">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [28]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#12 to 5 am</span> +<span class="n">L1</span><span class="o">=</span><span class="mi">20</span> <span class="c">#MW</span> +<span class="n">t1</span><span class="o">=</span><span class="mi">5</span> <span class="c">#hours</span> +<span class="c">#5 to 9 am</span> +<span class="n">L2</span><span class="o">=</span><span class="mi">40</span> <span class="c">#MW</span> +<span class="n">t2</span><span class="o">=</span><span class="mi">4</span> <span class="c">#hours</span> +<span class="c">#9 to 6 pm</span> +<span class="n">L3</span><span class="o">=</span><span class="mi">80</span> <span class="c">#MW</span> +<span class="n">t3</span><span class="o">=</span><span class="mi">9</span> <span class="c">#hours</span> +<span class="c">#6 to 10 pm</span> +<span class="n">L4</span><span class="o">=</span><span class="mi">100</span> <span class="c">#MW</span> +<span class="n">t4</span><span class="o">=</span><span class="mi">4</span> <span class="c">#hours</span> +<span class="c">#10 to 12 am</span> +<span class="n">L5</span><span class="o">=</span><span class="mi">20</span> <span class="c">#MW</span> +<span class="n">t5</span><span class="o">=</span><span class="mi">2</span> <span class="c">#hours</span> +<span class="c">#Energy Poduced in 24 hours</span> +<span class="n">E</span><span class="o">=</span><span class="n">L1</span><span class="o">*</span><span class="n">t1</span><span class="o">+</span><span class="n">L2</span><span class="o">*</span><span class="n">t2</span><span class="o">+</span><span class="n">L3</span><span class="o">*</span><span class="n">t3</span><span class="o">+</span><span class="n">L4</span><span class="o">*</span><span class="n">t4</span><span class="o">+</span><span class="n">L5</span><span class="o">*</span><span class="n">t5</span> <span class="c">#MWh</span> +<span class="k">print</span> <span class="s">"Energy Supplied by the plant in 24 hours = </span><span class="si">%0.2f</span><span class="s"> MWh"</span> <span class="o">%</span><span class="k">E</span> +<span class="n">LF</span><span class="o">=</span><span class="n">E</span><span class="o">/</span><span class="mi">24</span> <span class="c">#%#Load Factor</span> +<span class="k">print</span> <span class="s">"Load Factor = </span><span class="si">%0.2f</span><span class="s"> </span><span class="si">%%</span><span class="s"> "</span><span class="o">%</span><span class="k">LF</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Energy Supplied by the plant in 24 hours = 1420.00 MWh +Load Factor = 59.17 % +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.5-page-13">exa 1.5 page 13<a class="anchor-link" href="#exa-1.5-page-13">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [29]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> +<span class="n">C</span><span class="o">=</span><span class="mi">125</span> <span class="c">#MW#Installed Capacity</span> +<span class="c">#12 to 5 am</span> +<span class="n">L1</span><span class="o">=</span><span class="mi">20</span> <span class="c">#MW</span> +<span class="n">t1</span><span class="o">=</span><span class="mi">5</span> <span class="c">#hours</span> +<span class="c">#5 to 9 am</span> +<span class="n">L2</span><span class="o">=</span><span class="mi">40</span> <span class="c">#MW</span> +<span class="n">t2</span><span class="o">=</span><span class="mi">4</span> <span class="c">#hours</span> +<span class="c">#9 to 6 pm</span> +<span class="n">L3</span><span class="o">=</span><span class="mi">80</span> <span class="c">#MW</span> +<span class="n">t3</span><span class="o">=</span><span class="mi">9</span> <span class="c">#hours</span> +<span class="c">#6 to 10 pm</span> +<span class="n">L4</span><span class="o">=</span><span class="mi">100</span> <span class="c">#MW</span> +<span class="n">t4</span><span class="o">=</span><span class="mi">4</span> <span class="c">#hours</span> +<span class="c">#10 to 12 am</span> +<span class="n">L5</span><span class="o">=</span><span class="mi">20</span> <span class="c">#MW</span> +<span class="n">t5</span><span class="o">=</span><span class="mi">2</span> <span class="c">#hours</span> +<span class="c">#Energy Poduced in 24 hours</span> +<span class="n">E</span><span class="o">=</span><span class="n">L1</span><span class="o">*</span><span class="n">t1</span><span class="o">+</span><span class="n">L2</span><span class="o">*</span><span class="n">t2</span><span class="o">+</span><span class="n">L3</span><span class="o">*</span><span class="n">t3</span><span class="o">+</span><span class="n">L4</span><span class="o">*</span><span class="n">t4</span><span class="o">+</span><span class="n">L5</span><span class="o">*</span><span class="n">t5</span> <span class="c">#MWh</span> +<span class="n">LF</span><span class="o">=</span><span class="n">E</span><span class="o">/</span><span class="mi">24</span> <span class="c">#%#Load Factor</span> +<span class="n">CF</span><span class="o">=</span><span class="n">LF</span><span class="o">/</span><span class="n">C</span> <span class="c">#%#Capacity Factor</span> +<span class="k">print</span> <span class="s">"Capacity Factor = </span><span class="si">%0.3f</span><span class="s">"</span> <span class="o">%</span><span class="k">CF</span> +<span class="n">UF</span><span class="o">=</span><span class="mi">100</span><span class="o">/</span><span class="n">C</span> <span class="c">#%#Utilisation Factor</span> +<span class="k">print</span> <span class="s">"Utilisation Factor = </span><span class="si">%.1f</span><span class="s">"</span> <span class="o">%</span><span class="k">UF</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Capacity Factor = 0.473 +Utilisation Factor = 0.8 +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.6-page-13">exa 1.6 page 13<a class="anchor-link" href="#exa-1.6-page-13">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [30]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="o">%</span><span class="k">matplotlib</span> inline +<span class="kn">from</span> <span class="nn">matplotlib.pyplot</span> <span class="kn">import</span> <span class="n">plot</span><span class="p">,</span> <span class="n">show</span><span class="p">,</span> <span class="n">title</span><span class="p">,</span> <span class="n">xlabel</span><span class="p">,</span> <span class="n">ylabel</span><span class="p">,</span> <span class="n">subplot</span> +<span class="kn">from</span> <span class="nn">numpy</span> <span class="kn">import</span> <span class="n">array</span> +<span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span> +<span class="c">#12 to 5 am & 10 to 12 am</span> +<span class="n">L1</span><span class="o">=</span><span class="mi">20</span> <span class="c">#MW</span> +<span class="n">E1</span><span class="o">=</span><span class="n">L1</span><span class="o">*</span><span class="mi">24</span> <span class="c">#MWh</span> +<span class="c">#5 to 9 am</span> +<span class="n">L2</span><span class="o">=</span><span class="mi">40</span> <span class="c">#MW</span> +<span class="n">E2</span><span class="o">=</span><span class="n">E1</span><span class="o">+</span><span class="p">(</span><span class="n">L2</span><span class="o">-</span><span class="n">L1</span><span class="p">)</span><span class="o">*</span><span class="mi">17</span> <span class="c">#MWh</span> +<span class="c">#9 to 6 pm</span> +<span class="n">L3</span><span class="o">=</span><span class="mi">80</span> <span class="c">#MW</span> +<span class="n">E3</span><span class="o">=</span><span class="n">E2</span><span class="o">+</span><span class="p">(</span><span class="n">L3</span><span class="o">-</span><span class="n">L2</span><span class="p">)</span><span class="o">*</span><span class="mi">13</span> <span class="c">#MWh</span> +<span class="c">#6 to 10 pm</span> +<span class="n">L4</span><span class="o">=</span><span class="mi">100</span> <span class="c">#MW</span> +<span class="n">E4</span><span class="o">=</span><span class="n">E3</span><span class="o">+</span><span class="p">(</span><span class="n">L4</span><span class="o">-</span><span class="n">L3</span><span class="p">)</span><span class="o">*</span><span class="mi">4</span> <span class="c">#MWh</span> +<span class="c">#Plotting Energy load curve</span> +<span class="n">L</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="n">L1</span><span class="p">,</span><span class="n">L2</span><span class="p">,</span><span class="n">L3</span><span class="p">,</span><span class="n">L4</span><span class="p">])</span> <span class="c">#MW</span> +<span class="n">E</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="n">E1</span><span class="p">,</span><span class="n">E2</span><span class="p">,</span><span class="n">E3</span><span class="p">,</span><span class="n">E4</span><span class="p">])</span> <span class="c">#Mwh</span> +<span class="n">subplot</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">)</span> +<span class="n">plot</span><span class="p">(</span><span class="n">E</span><span class="p">,</span><span class="n">L</span><span class="p">)</span> +<span class="n">xlabel</span><span class="p">(</span><span class="s">'Energy(MWh)'</span><span class="p">)</span> +<span class="n">ylabel</span><span class="p">(</span><span class="s">'Load(MW)'</span><span class="p">)</span> +<span class="n">title</span><span class="p">(</span><span class="s">'Energy Load Curve'</span><span class="p">)</span> +<span class="c">#Energy Supplied</span> +<span class="c">#Upto 5am</span> +<span class="n">t1</span><span class="o">=</span><span class="mi">5</span> <span class="c">#hours</span> +<span class="n">E1</span><span class="o">=</span><span class="n">L1</span><span class="o">*</span><span class="n">t1</span> <span class="c">#MWh</span> +<span class="c">#Upto 9am</span> +<span class="n">t2</span><span class="o">=</span><span class="mi">4</span> <span class="c">#hours</span> +<span class="n">E2</span><span class="o">=</span><span class="n">E1</span><span class="o">+</span><span class="n">L2</span><span class="o">*</span><span class="n">t2</span> <span class="c">#MWh</span> +<span class="c">#Upto 6pm</span> +<span class="n">t3</span><span class="o">=</span><span class="mi">9</span> <span class="c">#hours</span> +<span class="n">E3</span><span class="o">=</span><span class="n">E2</span><span class="o">+</span><span class="n">L3</span><span class="o">*</span><span class="n">t3</span> <span class="c">#MWh</span> +<span class="c">#Upto 10pm</span> +<span class="n">t4</span><span class="o">=</span><span class="mi">4</span> <span class="c">#hours</span> +<span class="n">E4</span><span class="o">=</span><span class="n">E3</span><span class="o">+</span><span class="n">L4</span><span class="o">*</span><span class="n">t4</span> <span class="c">#MWh</span> +<span class="c">#Upto 12pm</span> +<span class="n">t4</span><span class="o">=</span><span class="mi">2</span> <span class="c">#hours</span> +<span class="n">E4</span><span class="o">=</span><span class="n">E3</span><span class="o">+</span><span class="n">L4</span><span class="o">*</span><span class="n">t4</span> <span class="c">#MWh</span> +<span class="c">#Plotting Mass curve</span> +<span class="n">T</span><span class="o">=</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">4</span><span class="p">]</span> <span class="c">#MW</span> +<span class="n">E</span><span class="o">=</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="n">E1</span><span class="p">,</span><span class="n">E2</span><span class="p">,</span><span class="n">E3</span><span class="p">,</span><span class="n">E4</span><span class="p">]</span> <span class="c">#Mwh</span> +<span class="n">subplot</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">)</span> +<span class="n">plot</span><span class="p">(</span><span class="n">T</span><span class="p">,</span><span class="n">E</span><span class="p">)</span> +<span class="n">ylabel</span><span class="p">(</span><span class="s">'Ener2y(MWh)'</span><span class="p">)</span> +<span class="n">xlabel</span><span class="p">(</span><span class="s">'0-1: 12-5am 1-2: 5-9am 2-3: 9-6pm 3-4: 6-10pm above4: 10-12pm'</span><span class="p">)</span> +<span class="n">title</span><span class="p">(</span><span class="s">'Mass Curve'</span><span class="p">)</span> +<span class="n">show</span><span class="p">()</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> + + +<div class="output_png output_subarea "> +<img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAZMAAAEZCAYAAABSN8jfAAAABHNCSVQICAgIfAhkiAAAAAlwSFlz +AAALEgAACxIB0t1+/AAAIABJREFUeJztnXmYVNW1t9+fCCgikwOgoOAURVERVDQOxCGaGDXROMUB +HHKTGDWjcfoSyb03iUOiiTF6TaIBVBSNQzRxQiKJM4KAKBBBBUEZxAFRVIZe3x97F3266Oqu7qrq +c6p7vc9TT5864+/sqq511tp7ryUzw3Ecx3FKYYO0BTiO4zjVjxsTx3Ecp2TcmDiO4zgl48bEcRzH +KRk3Jo7jOE7JuDFxHMdxSsaNieNkHEmjJP1P2jocpyHcmDiZQNI8SSslrUi8rktbV2NIqpG0XYUv 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class="n">dmax</span><span class="o">=</span><span class="mi">40</span> <span class="c">#MW#Maximum demand</span> +<span class="n">CF</span><span class="o">=</span><span class="mf">0.5</span> <span class="c">#Capacity Factor</span> +<span class="n">UF</span><span class="o">=</span><span class="mf">0.8</span> <span class="c">#Utilisation Factor</span> +<span class="n">LF</span><span class="o">=</span><span class="n">CF</span><span class="o">/</span><span class="n">UF</span> <span class="c">#/Load Factor</span> +<span class="k">print</span> <span class="s">"(a) Load Factor : </span><span class="si">%0.3f</span><span class="s">"</span><span class="o">%</span><span class="k">LF</span> +<span class="n">C</span><span class="o">=</span><span class="n">dmax</span><span class="o">/</span><span class="n">UF</span> <span class="c">#MW#Plant Capacity</span> +<span class="k">print</span> <span class="s">"(b) Plant Capacity = </span><span class="si">%0.2f</span><span class="s"> MW "</span><span class="o">%</span><span class="k">C</span> +<span class="n">RC</span><span class="o">=</span><span class="n">C</span><span class="o">-</span><span class="n">dmax</span> <span class="c">#MW#Reserve Capacity</span> +<span class="k">print</span> <span class="s">"(c) Reserve Capacity = </span><span class="si">%0.02f</span><span class="s"> MW "</span><span class="o">%</span><span class="k">RC</span> +<span class="n">p</span><span class="o">=</span><span class="n">dmax</span><span class="o">*</span><span class="n">LF</span><span class="o">*</span><span class="mi">24</span><span class="o">*</span><span class="mi">365</span> <span class="c">#MWh#Annual Energy Production</span> +<span class="k">print</span> <span class="s">"(d) Annual Energy Production = </span><span class="si">%0.2f</span><span class="s"> MWh "</span><span class="o">%</span><span class="k">p</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>(a) Load Factor : 0.625 +(b) Plant Capacity = 50.00 MW +(c) Reserve Capacity = 10.00 MW +(d) Annual Energy Production = 219000.00 MWh +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.8-page-14">exa 1.8 page 14<a class="anchor-link" href="#exa-1.8-page-14">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [32]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> +<span class="o">%</span><span class="k">matplotlib</span> inline +<span class="kn">from</span> <span class="nn">matplotlib.pyplot</span> <span class="kn">import</span> <span class="n">plot</span><span class="p">,</span> <span class="n">show</span><span class="p">,</span> <span class="n">title</span><span class="p">,</span> <span class="n">xlabel</span><span class="p">,</span> <span class="n">ylabel</span><span class="p">,</span> <span class="n">subplot</span> +<span class="c">#from numpy import array</span> +<span class="n">L1</span><span class="o">=</span><span class="mi">50</span> <span class="c">#MW#Initial</span> +<span class="n">t1</span><span class="o">=</span><span class="mi">5</span> <span class="c">#hours</span> +<span class="n">L2</span><span class="o">=</span><span class="mi">50</span> <span class="c">#MW#5am</span> +<span class="n">t2</span><span class="o">=</span><span class="mi">4</span> <span class="c">#hours</span> +<span class="n">L3</span><span class="o">=</span><span class="mi">100</span> <span class="c">#MW#9am</span> +<span class="n">t3</span><span class="o">=</span><span class="mi">9</span> <span class="c">#hours</span> +<span class="n">L4</span><span class="o">=</span><span class="mi">100</span> <span class="c">#MW#6pm</span> +<span class="n">t4</span><span class="o">=</span><span class="mi">2</span> <span class="c">#hours</span> +<span class="n">L5</span><span class="o">=</span><span class="mi">150</span> <span class="c">#MW#8pm</span> +<span class="n">t5</span><span class="o">=</span><span class="mi">2</span> <span class="c">#hours</span> +<span class="n">L6</span><span class="o">=</span><span class="mi">80</span> <span class="c">#MW#10pm</span> +<span class="n">t6</span><span class="o">=</span><span class="mi">2</span> <span class="c">#hours</span> +<span class="n">L7</span><span class="o">=</span><span class="mi">50</span> <span class="c">#MW</span> +<span class="c">#Energy Required in 24 hours</span> +<span class="n">E</span><span class="o">=</span><span class="n">L1</span><span class="o">*</span><span class="n">t1</span><span class="o">+</span><span class="p">(</span><span class="n">L2</span><span class="o">+</span><span class="n">L3</span><span class="p">)</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">t2</span><span class="o">+</span><span class="p">(</span><span class="n">L3</span><span class="o">+</span><span class="n">L4</span><span class="p">)</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">t3</span><span class="o">+</span><span class="p">(</span><span class="n">L4</span><span class="o">+</span><span class="n">L5</span><span class="p">)</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">t4</span><span class="o">+</span><span class="p">(</span><span class="n">L5</span><span class="o">+</span><span class="n">L6</span><span class="p">)</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">t5</span><span class="o">+</span><span class="p">(</span><span class="n">L6</span><span class="o">+</span><span class="n">L1</span><span class="p">)</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">t6</span> <span class="c">#MWh</span> +<span class="k">print</span> <span class="s">"Energy required in one day = </span><span class="si">%0.2f</span><span class="s"> MWh "</span><span class="o">%</span><span class="k">E</span> +<span class="n">DLF</span><span class="o">=</span><span class="n">E</span><span class="o">/</span><span class="n">L5</span><span class="o">/</span><span class="mi">24</span><span class="o">*</span><span class="mi">100</span> <span class="c">#%#Daily Load Factor</span> +<span class="k">print</span> <span class="s">"Daily Load Factor = </span><span class="si">%0.2f</span><span class="s"> </span><span class="si">%%</span><span class="s">"</span> <span class="o">%</span><span class="k">DLF</span> +<span class="c">#Plotting load curve</span> +<span class="n">T</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">4</span><span class="p">,</span><span class="mi">5</span><span class="p">,</span><span class="mi">6</span><span class="p">])</span> <span class="c">#Slots</span> +<span class="n">L</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="n">L1</span><span class="p">,</span><span class="n">L2</span><span class="p">,</span><span class="n">L3</span><span class="p">,</span><span class="n">L4</span><span class="p">,</span><span class="n">L5</span><span class="p">,</span><span class="n">L6</span><span class="p">,</span><span class="n">L7</span><span class="p">])</span> <span class="c">#MW</span> +<span class="n">plot</span><span class="p">(</span><span class="n">T</span><span class="p">,</span><span class="n">L</span><span class="p">)</span> +<span class="n">ylabel</span><span class="p">(</span><span class="s">'Load(MW)'</span><span class="p">)</span> +<span class="n">xlabel</span><span class="p">(</span><span class="s">'0-1: 12-5am 1-2: 5-9am 2-3: 9-6pm 3-4: 6-8pm 4-5:8-10pm 5-6 :10-12pm'</span><span class="p">)</span> +<span class="n">title</span><span class="p">(</span><span class="s">'Chronological Load Curve'</span><span class="p">)</span> +<span class="n">show</span><span class="p">()</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Energy required in one day = 2060.00 MWh +Daily Load Factor = 57.22 % +</pre> +</div> +</div> + +<div class="output_area"><div class="prompt"></div> + + +<div class="output_png output_subarea "> +<img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAZEAAAEZCAYAAABWwhjiAAAABHNCSVQICAgIfAhkiAAAAAlwSFlz +AAALEgAACxIB0t1+/AAAIABJREFUeJzt3XucHFWZ//HPlxDUgBIQ5RokiihRUGBB0EVHFBVcEsD1 +gooGVBTWoAg/CYoSXUREvBEFVlgi4hJABAQJCiKjsFwFQrhKQC4JSLjI1bgQyPP745xJOp3unp6e +7q6+fN+v17zSXV1d/VTPpJ465zl1ShGBmZlZI1YpOgAzM+teTiJmZtYwJxEzM2uYk4iZmTXMScTM +zBrmJGJmZg1zErGGSZoh6bSi4wCQtFTSq5uwnTmS9h7lNqZKuny0sYwyhkFJnyoyBusPTiJWk6SP +SvqzpKclPZgPsm/LL/fcRUYRsWtEtCwxStokJ7xW/98Lavx+JG0m6ZeSHpH0hKSbJB3Uhrisx/gP +xqqS9CXgB8CRwCuBCcBPgN2GVhnBtlZteoDWEEmvAa4B7gPeGBHjgQ8C2wAvbWB7Y5oboXUTJxGr +SNKawDeAAyLivIj4Z0S8EBEXRsT0vFoAq0k6VdJTkm6RtE3JNu6V9GVJ84CnJY2RNFnSrZIel3SZ +pNeXrX9wPit+QtIZkl5U8vpnJM2X9JikX0tav1rskn4u6eG8za9KUn5tFUnfy2fgf5X0+dKWQXk3 +UP7M2/L+3Sppq7x8uqS7Spbv3oTvfANJ5+f9my/p0yWvbSfpqvy9PShppqSxJa/vLOmO/L3NJCX4 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input_prompt">In [33]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="o">%</span><span class="k">matplotlib</span> inline +<span class="kn">from</span> <span class="nn">matplotlib.pyplot</span> <span class="kn">import</span> <span class="n">plot</span><span class="p">,</span> <span class="n">show</span><span class="p">,</span> <span class="n">title</span><span class="p">,</span> <span class="n">xlabel</span><span class="p">,</span> <span class="n">ylabel</span><span class="p">,</span> <span class="n">subplot</span> +<span class="k">print</span> <span class="s">"load duration curve in fig1"</span> +<span class="k">print</span> <span class="s">"the energy consumed upto different times is as "</span> +<span class="n">a</span><span class="o">=</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">5</span> <span class="p">,</span><span class="mi">9</span> <span class="p">,</span><span class="mi">18</span><span class="p">,</span> <span class="mi">20</span><span class="p">,</span> <span class="mi">22</span><span class="p">,</span> <span class="mi">24</span><span class="p">]</span> <span class="c">#time in matrix format</span> +<span class="n">b</span><span class="o">=</span><span class="p">[</span><span class="mi">50</span><span class="p">,</span> <span class="mi">50</span> <span class="p">,</span><span class="mi">100</span> <span class="p">,</span><span class="mi">100</span> <span class="p">,</span><span class="mi">150</span> <span class="p">,</span><span class="mi">80</span> <span class="p">,</span><span class="mi">50</span><span class="p">]</span> <span class="c">#load in matrix format</span> + +<span class="n">z</span> <span class="o">=</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span> +<span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="nb">range</span> <span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">6</span><span class="p">):</span> + <span class="n">z</span><span class="p">[</span><span class="n">x</span><span class="p">]</span><span class="o">=</span><span class="p">((</span><span class="n">b</span><span class="p">[</span><span class="n">x</span><span class="p">]</span><span class="o">+</span><span class="n">b</span><span class="p">[</span><span class="n">x</span><span class="o">+</span><span class="mi">1</span><span class="p">])</span><span class="o">/</span><span class="mi">2</span><span class="p">)</span><span class="o">*</span><span class="p">(</span><span class="n">a</span><span class="p">[</span><span class="n">x</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span><span class="o">-</span><span class="n">a</span><span class="p">[</span><span class="n">x</span><span class="p">])</span> + +<span class="n">et</span><span class="o">=</span><span class="mi">0</span> +<span class="n">q</span> <span class="o">=</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">7</span><span class="p">)</span> +<span class="n">m</span> <span class="o">=</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">7</span><span class="p">)</span> +<span class="n">ett</span> <span class="o">=</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span> +<span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">6</span><span class="p">):</span> + <span class="n">et</span><span class="o">=</span><span class="n">et</span><span class="o">+</span><span class="n">z</span><span class="p">[</span><span class="n">x</span><span class="p">]</span> + <span class="n">A</span><span class="o">=</span><span class="n">a</span><span class="p">[(</span><span class="n">x</span><span class="p">)]</span> + <span class="n">ett</span><span class="p">[</span><span class="n">x</span><span class="p">]</span><span class="o">=</span><span class="n">et</span> + <span class="n">q</span><span class="p">[</span><span class="n">x</span><span class="p">]</span><span class="o">=</span><span class="n">a</span><span class="p">[</span><span class="n">x</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span> + <span class="k">print</span> <span class="s">"</span><span class="se">\n</span><span class="s">from mid night upto </span><span class="si">%d</span><span class="s">,energy=</span><span class="si">%d</span><span class="s">MWh"</span><span class="o">%</span><span class="p">(</span><span class="n">A</span><span class="p">,</span><span class="n">et</span><span class="p">)</span> +<span class="n">n</span> <span class="o">=</span> <span class="nb">sorted</span><span class="p">(</span><span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">b</span><span class="p">)),</span> <span class="n">key</span><span class="o">=</span><span class="k">lambda</span> <span class="n">k</span><span class="p">:</span> <span class="n">b</span><span class="p">[</span><span class="n">k</span><span class="p">],</span> <span class="n">reverse</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span> +<span class="n">m</span> <span class="o">=</span> <span class="nb">sorted</span><span class="p">(</span><span class="n">b</span><span class="p">,</span> <span class="n">reverse</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span> +<span class="k">print</span> <span class="s">"energy curve in fig 2"</span> +<span class="n">t</span><span class="o">=</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mf">3.88</span><span class="p">,</span> <span class="mf">15.88</span> <span class="p">,</span><span class="mf">19.88</span><span class="p">,</span> <span class="mi">23</span><span class="p">]</span> +<span class="n">k</span> <span class="o">=</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span> +<span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">6</span><span class="p">):</span> + <span class="n">k</span><span class="p">[</span><span class="n">j</span><span class="p">]</span><span class="o">=</span><span class="n">a</span><span class="p">[(</span><span class="n">j</span><span class="o">+</span><span class="mi">1</span><span class="p">)]</span> +<span class="n">M</span> <span class="o">=</span><span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">5</span><span class="p">)</span> +<span class="c">#rearranging for mass curve</span> +<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">5</span><span class="p">):</span> + <span class="n">M</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">=</span> <span class="n">m</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> +<span class="n">Q</span> <span class="o">=</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span> +<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">6</span><span class="p">):</span> + <span class="n">Q</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">=</span> <span class="n">q</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> + + +<span class="n">subplot</span><span class="p">(</span><span class="mi">121</span><span class="p">)</span> +<span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">M</span><span class="p">)</span> +<span class="n">title</span><span class="p">(</span><span class="s">"load duration"</span><span class="p">)</span> +<span class="n">xlabel</span><span class="p">(</span><span class="s">"hours"</span><span class="p">)</span> +<span class="n">ylabel</span><span class="p">(</span><span class="s">"MW"</span><span class="p">)</span> +<span class="n">subplot</span><span class="p">(</span><span class="mi">122</span><span class="p">)</span> +<span class="n">plot</span><span class="p">(</span><span class="n">Q</span><span class="p">,</span><span class="n">ett</span><span class="p">)</span> +<span class="n">title</span><span class="p">(</span><span class="s">"energy curve"</span><span class="p">)</span> +<span class="n">xlabel</span><span class="p">(</span><span class="s">"time"</span><span class="p">)</span> +<span class="n">ylabel</span><span class="p">(</span><span class="s">"MWh"</span><span class="p">)</span> +<span class="n">show</span><span class="p">()</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>load duration curve in fig1 +the energy consumed upto different times is as + +from mid night upto 0,energy=250MWh + +from mid night upto 5,energy=550MWh + +from mid night upto 9,energy=1450MWh + +from mid night upto 18,energy=1700MWh + +from mid night upto 20,energy=1930MWh + +from mid night upto 22,energy=2060MWh +energy curve in fig 2 +</pre> +</div> +</div> + +<div class="output_area"><div class="prompt"></div> + + +<div class="output_png output_subarea "> +<img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAYcAAAEZCAYAAAB8culNAAAABHNCSVQICAgIfAhkiAAAAAlwSFlz +AAALEgAACxIB0t1+/AAAIABJREFUeJzt3XvcVXP6//HXuxJSVEJK5BDJoZooCd3jNA3jOEhICDOM +4zjV8NDtO8PQMGp+DjOGklNEDlEoxq0ocigiRESlgxRFDh2u3x+fdde2730f23uvfbiej8f9aO21 +9lrrWt1r39den6PMDOeccy5RvbgDcM45l3s8OTjnnKvAk4NzzrkKPDk455yrwJODc865Cjw5OOec +q8CTQxpImiPpkAwct1TS/bV4/1pJO6U7jirON05S32ydzzmXPQ3iDqBAWPSTiePmBEmlwM5mti4Z +mNkR8UXknMskf3JwSPIvCS5WkurHHUOyYv9ceHJIM0kbSxoiaX70c6ukhtG2ppKekbRY0lJJT0tq +nbDvjpJelrRc0nigRTXnukLSl5LmSToraVuZpP4Jr8+QNCnh9VpJ50v6GPgoWjdU0heSvpX0pqQD 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+</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.10-page-15">exa 1.10 page 15<a class="anchor-link" href="#exa-1.10-page-15">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [34]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> +<span class="n">E</span><span class="o">=</span><span class="mi">438</span><span class="o">*</span><span class="mi">10</span><span class="o">**</span><span class="mi">4</span> <span class="c">#kWh</span> +<span class="n">LF</span><span class="o">=</span><span class="mi">20</span> <span class="c">#% annual</span> +<span class="n">CF</span><span class="o">=</span><span class="mi">15</span> <span class="c">#%#Capacity Factor</span> +<span class="n">Lmax</span><span class="o">=</span><span class="n">E</span><span class="o">/</span><span class="p">(</span><span class="n">LF</span><span class="o">/</span><span class="mi">100</span><span class="p">)</span><span class="o">/</span><span class="mi">24</span><span class="o">/</span><span class="mi">365</span> <span class="c">#kW</span> +<span class="n">Lmax</span><span class="o">=</span><span class="n">Lmax</span><span class="o">/</span><span class="mi">1000</span> <span class="c">#MW</span> +<span class="n">C</span><span class="o">=</span><span class="n">Lmax</span><span class="o">/</span><span class="n">CF</span><span class="o">*</span><span class="n">LF</span> <span class="c">#MW#Plant Capacity</span> +<span class="k">print</span> <span class="s">"Plant Capacity = </span><span class="si">%0.2f</span><span class="s"> MW "</span><span class="o">%</span><span class="k">C</span> +<span class="n">RC</span><span class="o">=</span><span class="n">C</span><span class="o">-</span><span class="n">Lmax</span> <span class="c">#MW#Reserve Capacity</span> +<span class="k">print</span> <span class="s">"Reserve Capacity = </span><span class="si">%0.2f</span><span class="s"> MW"</span><span class="o">%</span><span class="k">RC</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Plant Capacity = 3.33 MW +Reserve Capacity = 0.83 MW +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.11-page-16">exa 1.11 page 16<a class="anchor-link" href="#exa-1.11-page-16">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [35]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="n">L1</span><span class="o">=</span><span class="mi">10000</span> <span class="c">#kW</span> +<span class="n">L2</span><span class="o">=</span><span class="mi">6000</span> <span class="c">#kW</span> +<span class="n">L3</span><span class="o">=</span><span class="mi">8000</span> <span class="c">#kW</span> +<span class="n">L4</span><span class="o">=</span><span class="mi">7000</span> <span class="c">#kW</span> +<span class="n">df</span><span class="o">=</span><span class="mf">1.5</span> <span class="c">#diversity factor</span> +<span class="n">LF</span><span class="o">=</span><span class="mi">65</span> <span class="c">#%#Load Factor</span> +<span class="n">Dinc</span><span class="o">=</span><span class="mi">60</span> <span class="c">#%#Increase in maximum demand</span> +<span class="n">L</span><span class="o">=</span><span class="n">L1</span><span class="o">+</span><span class="n">L2</span><span class="o">+</span><span class="n">L3</span><span class="o">+</span><span class="n">L4</span> <span class="c">#kW#Sum </span> +<span class="n">L</span><span class="o">=</span><span class="n">L</span><span class="o">/</span><span class="mi">1000</span> <span class="c">#MW</span> +<span class="n">Dmax</span><span class="o">=</span><span class="n">L</span><span class="o">/</span><span class="n">df</span> <span class="c">#MW</span> +<span class="k">print</span> <span class="s">"Maximum demand on station = </span><span class="si">%0.3f</span><span class="s"> MWh "</span> <span class="o">%</span><span class="k">Dmax</span> +<span class="n">E</span><span class="o">=</span><span class="n">Dmax</span><span class="o">*</span><span class="mi">365</span><span class="o">*</span><span class="mi">24</span><span class="o">*</span><span class="n">LF</span><span class="o">/</span><span class="mi">100</span> <span class="c">#MWh#Annual Energy</span> +<span class="k">print</span> <span class="s">"Annual Energy Supplied = </span><span class="si">%0.0f</span><span class="s"> MWh "</span><span class="o">%</span><span class="k">E</span> +<span class="n">Din_max</span><span class="o">=</span><span class="n">Dinc</span><span class="o">/</span><span class="mi">100</span><span class="o">*</span><span class="n">Dmax</span> <span class="c">#MW</span> +<span class="n">C</span><span class="o">=</span><span class="n">Dmax</span><span class="o">+</span><span class="n">Din_max</span> <span class="c">#MW</span> +<span class="k">print</span> <span class="s">"Installed Capacity = </span><span class="si">%0.3f</span><span class="s"> MW"</span><span class="o">%</span><span class="k">C</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Maximum demand on station = 20.667 MWh +Annual Energy Supplied = 117676 MWh +Installed Capacity = 33.067 MW +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.12-page-16">exa 1.12 page 16<a class="anchor-link" href="#exa-1.12-page-16">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [36]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="o">%</span><span class="k">matplotlib</span> inline +<span class="kn">from</span> <span class="nn">matplotlib.pyplot</span> <span class="kn">import</span> <span class="n">plot</span><span class="p">,</span> <span class="n">title</span><span class="p">,</span> <span class="n">xlabel</span><span class="p">,</span> <span class="n">ylabel</span><span class="p">,</span> <span class="n">show</span> +<span class="c">#Arranging data for Load Duration Curve</span> +<span class="c">#week days 5-9pm load</span> +<span class="n">L1</span><span class="o">=</span><span class="mi">350</span> <span class="c">#MW</span> +<span class="n">t1</span><span class="o">=</span><span class="mi">4</span><span class="o">*</span><span class="mi">5</span> <span class="c">#hours</span> +<span class="c">#week days 8-12am & 1-5pm load</span> +<span class="n">L2</span><span class="o">=</span><span class="mi">250</span> <span class="c">#MW</span> +<span class="n">t2</span><span class="o">=</span><span class="n">t1</span><span class="o">+</span><span class="mi">8</span><span class="o">*</span><span class="mi">5</span> <span class="c">#hours</span> +<span class="c">#saturday & sunday 5-9pm load</span> +<span class="n">L3</span><span class="o">=</span><span class="mi">200</span> <span class="c">#MW</span> +<span class="n">t3</span><span class="o">=</span><span class="n">t2</span><span class="o">+</span><span class="mi">4</span><span class="o">*</span><span class="mi">2</span> <span class="c">#hours</span> +<span class="c">#All days 150MW load</span> +<span class="n">L4</span><span class="o">=</span><span class="mi">150</span> <span class="c">#MW</span> +<span class="n">t4</span><span class="o">=</span><span class="n">t3</span><span class="o">+</span><span class="mi">6</span><span class="o">*</span><span class="mi">5</span><span class="o">+</span><span class="mi">15</span><span class="o">*</span><span class="mi">2</span> <span class="c">#hours</span> +<span class="c">#All days 100MW load</span> +<span class="n">L5</span><span class="o">=</span><span class="mi">100</span> <span class="c">#MW</span> +<span class="n">t5</span><span class="o">=</span><span class="n">t4</span><span class="o">+</span><span class="mi">6</span><span class="o">*</span><span class="mi">5</span><span class="o">+</span><span class="mi">5</span><span class="o">*</span><span class="mi">2</span> <span class="c">#hours</span> +<span class="n">A</span><span class="o">=</span><span class="mi">31600</span> <span class="c">#Total Load Curve Area</span> +<span class="n">LF</span><span class="o">=</span><span class="n">A</span><span class="o">/</span><span class="n">L1</span><span class="o">/</span><span class="mi">24</span><span class="o">/</span><span class="mi">7</span><span class="o">*</span><span class="mi">100</span> <span class="c">#%#Weekly load factor</span> +<span class="k">print</span> <span class="s">"Weekly Load factor = </span><span class="si">%0.2f</span><span class="s"> </span><span class="si">%%</span><span class="s">"</span><span class="o">%</span><span class="k">LF</span> +<span class="k">print</span> <span class="s">"Load Duration Curve is shown in figure."</span> +<span class="c">#Load Duration Curve</span> +<span class="n">L</span><span class="o">=</span><span class="p">[</span><span class="n">L1</span> <span class="p">,</span><span class="n">L2</span><span class="p">,</span> <span class="n">L3</span><span class="p">,</span> <span class="n">L4</span><span class="p">,</span> <span class="n">L5</span><span class="p">]</span> <span class="c">#MW</span> +<span class="n">T</span><span class="o">=</span><span class="p">[</span><span class="n">t1</span> <span class="p">,</span><span class="n">t2</span> <span class="p">,</span><span class="n">t3</span> <span class="p">,</span><span class="n">t4</span> <span class="p">,</span><span class="n">t5</span><span class="p">]</span> <span class="c">#hours</span> +<span class="n">plot</span><span class="p">(</span><span class="n">T</span><span class="p">,</span><span class="n">L</span><span class="p">)</span> +<span class="n">title</span><span class="p">(</span><span class="s">'Load Duration Curve'</span><span class="p">)</span> +<span class="n">xlabel</span><span class="p">(</span><span class="s">'Time(Hours)'</span><span class="p">)</span> +<span class="n">ylabel</span><span class="p">(</span><span class="s">'Load(MW)'</span><span class="p">)</span> +<span class="n">show</span><span class="p">()</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Weekly Load factor = 53.74 % +Load Duration Curve is shown in figure. +</pre> +</div> +</div> + +<div class="output_area"><div class="prompt"></div> + + +<div class="output_png output_subarea "> +<img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAYoAAAEZCAYAAACJjGL9AAAABHNCSVQICAgIfAhkiAAAAAlwSFlz +AAALEgAACxIB0t1+/AAAIABJREFUeJzt3XmYVOWZ9/HvDxHUoBI1QVBQEiVKjFFHGbOorTMuyYwa 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+<span class="n">ratio1</span><span class="o">=</span><span class="mf">0.87</span> <span class="c">#daily peak load to monthly peak load</span> +<span class="n">ratio2</span><span class="o">=</span><span class="mf">0.78</span> <span class="c">#monthly peak load to annually peak load</span> +<span class="n">LF_annual</span><span class="o">=</span><span class="n">LF</span><span class="o">*</span><span class="n">ratio1</span><span class="o">*</span><span class="n">ratio2</span> <span class="c">#Annual Load Factor</span> +<span class="k">print</span> <span class="s">"Annual Load Factor : </span><span class="si">%0.4f</span><span class="s">"</span> <span class="o">%</span><span class="k">LF_annual</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Annual Load Factor : 0.5598 +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.14-page-19">exa 1.14 page 19<a class="anchor-link" href="#exa-1.14-page-19">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [38]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="c">#Transformer1</span> +<span class="n">Lm</span><span class="o">=</span><span class="mi">300</span> <span class="c">#kW</span> +<span class="n">df_m</span><span class="o">=</span><span class="mf">0.6</span> <span class="c">#demand factor</span> +<span class="n">Lc</span><span class="o">=</span><span class="mi">100</span> <span class="c">#kW#Commercial Load</span> +<span class="n">df_c</span><span class="o">=</span><span class="mf">0.5</span> <span class="c">#demand factor</span> +<span class="c">#Transformer2</span> +<span class="n">Lr2</span><span class="o">=</span><span class="mi">500</span> <span class="c">#kW#Residential Load</span> +<span class="n">df_Lr2</span><span class="o">=</span><span class="mf">0.4</span> <span class="c">#demand factor</span> +<span class="c">#Transformer3</span> +<span class="n">Lr3</span><span class="o">=</span><span class="mi">400</span> <span class="c">#kW</span> +<span class="n">df_Lr3</span><span class="o">=</span><span class="mf">0.5</span> <span class="c">#demand factor</span> +<span class="c">#Diversity factors</span> +<span class="n">df1</span><span class="o">=</span><span class="mf">2.3</span> +<span class="n">df2</span><span class="o">=</span><span class="mf">2.5</span> +<span class="n">df3</span><span class="o">=</span><span class="mi">2</span> +<span class="n">DF</span><span class="o">=</span><span class="mf">1.4</span> <span class="c">#Diversity factor between transformers</span> +<span class="c">#Solution :</span> +<span class="k">print</span> <span class="s">"Part(a)"</span> +<span class="n">Lp1</span><span class="o">=</span><span class="p">(</span><span class="n">Lm</span><span class="o">*</span><span class="n">df_m</span><span class="o">+</span><span class="n">Lc</span><span class="o">*</span><span class="n">df_c</span><span class="p">)</span><span class="o">/</span><span class="n">df1</span> <span class="c">#kW#Peak load on Transformer1</span> +<span class="k">print</span> <span class="s">"Peak load on Transformer1 = </span><span class="si">%0.2f</span><span class="s"> kW"</span><span class="o">%</span><span class="k">Lp1</span> +<span class="n">Lp2</span><span class="o">=</span><span class="n">Lr2</span><span class="o">*</span><span class="n">df_Lr2</span><span class="o">/</span><span class="n">df2</span> <span class="c">#kW#Peak load on Transformer2</span> +<span class="k">print</span> <span class="s">"Peak load on Transformer2 = </span><span class="si">%0.2f</span><span class="s"> kW"</span><span class="o">%</span><span class="k">Lp2</span> +<span class="n">Lp3</span><span class="o">=</span><span class="n">Lr3</span><span class="o">*</span><span class="n">df_Lr3</span><span class="o">/</span><span class="n">df3</span> <span class="c">#kW#Peak load on Transformer3</span> +<span class="k">print</span> <span class="s">"Peak load on Transformer3 = </span><span class="si">%0.2f</span><span class="s"> kW "</span><span class="o">%</span><span class="k">Lp3</span> +<span class="k">print</span> <span class="s">"Part(b)"</span> +<span class="n">LpF</span><span class="o">=</span><span class="p">(</span><span class="n">Lp1</span><span class="o">+</span><span class="n">Lp2</span><span class="o">+</span><span class="n">Lp3</span><span class="p">)</span><span class="o">/</span><span class="n">DF</span> <span class="c">#Peak load on feeder</span> +<span class="k">print</span> <span class="s">"Peak load on feeder = </span><span class="si">%0.2f</span><span class="s"> kW "</span><span class="o">%</span><span class="k">LpF</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Part(a) +Peak load on Transformer1 = 100.00 kW +Peak load on Transformer2 = 80.00 kW +Peak load on Transformer3 = 100.00 kW +Part(b) +Peak load on feeder = 200.00 kW +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.16-page-23">exa 1.16 page 23<a class="anchor-link" href="#exa-1.16-page-23">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [39]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="n">L</span><span class="o">=</span><span class="p">[</span><span class="mi">20</span><span class="p">,</span> <span class="mi">25</span><span class="p">,</span> <span class="mi">30</span> <span class="p">,</span><span class="mi">25</span> <span class="p">,</span><span class="mi">35</span> <span class="p">,</span><span class="mi">20</span><span class="p">]</span> <span class="c">#MW</span> +<span class="n">T</span><span class="o">=</span><span class="p">[</span><span class="mi">6</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">2</span> <span class="p">,</span><span class="mi">4</span> <span class="p">,</span><span class="mi">4</span> <span class="p">,</span><span class="mi">4</span><span class="p">]</span> <span class="c">#Hours</span> +<span class="n">Lmax</span><span class="o">=</span><span class="nb">max</span><span class="p">(</span><span class="n">L</span><span class="p">)</span> <span class="c">#MW</span> +<span class="k">print</span> <span class="s">"(a) Maximum demand = </span><span class="si">%0.2f</span><span class="s"> MW "</span><span class="o">%</span><span class="k">Lmax</span> +<span class="n">E</span><span class="o">=</span><span class="n">L</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">*</span><span class="nb">sum</span><span class="p">(</span><span class="n">T</span><span class="p">)</span><span class="o">+</span><span class="p">(</span><span class="n">L</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">-</span><span class="n">L</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span><span class="o">*</span><span class="n">T</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">+</span><span class="p">(</span><span class="n">L</span><span class="p">[</span><span class="mi">2</span><span class="p">]</span><span class="o">-</span><span class="n">L</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span><span class="o">*</span><span class="n">T</span><span class="p">[</span><span class="mi">2</span><span class="p">]</span><span class="o">+</span><span class="p">(</span><span class="n">L</span><span class="p">[</span><span class="mi">3</span><span class="p">]</span><span class="o">-</span><span class="n">L</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span><span class="o">*</span><span class="n">T</span><span class="p">[</span><span class="mi">3</span><span class="p">]</span><span class="o">+</span><span class="p">(</span><span class="n">L</span><span class="p">[</span><span class="mi">4</span><span class="p">]</span><span class="o">-</span><span class="n">L</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span><span class="o">*</span><span class="n">T</span><span class="p">[</span><span class="mi">4</span><span class="p">]</span><span class="o">+</span><span class="p">(</span><span class="n">L</span><span class="p">[</span><span class="mi">5</span><span class="p">]</span><span class="o">-</span><span class="n">L</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span><span class="o">*</span><span class="n">T</span><span class="p">[</span><span class="mi">5</span><span class="p">]</span> <span class="c">#MWh</span> +<span class="n">E</span><span class="o">=</span><span class="n">E</span><span class="o">*</span><span class="mi">1000</span> <span class="c">#kWh</span> +<span class="k">print</span> <span class="s">"(b) Units generated per day = %0.e kWh "</span><span class="o">%</span><span class="k">E</span> +<span class="n">Lavg</span><span class="o">=</span><span class="n">E</span><span class="o">/</span><span class="nb">sum</span><span class="p">(</span><span class="n">T</span><span class="p">)</span> <span class="c">#/kWh</span> +<span class="n">Lavg</span><span class="o">=</span><span class="n">Lavg</span><span class="o">/</span><span class="mi">1000</span> <span class="c">#/MW</span> +<span class="k">print</span> <span class="s">"(c) Average Load = </span><span class="si">%0.2f</span><span class="s"> MW "</span><span class="o">%</span><span class="k">Lavg</span> +<span class="n">LF</span><span class="o">=</span><span class="n">Lavg</span><span class="o">/</span><span class="n">Lmax</span><span class="o">*</span><span class="mi">100</span> <span class="c">#%</span> +<span class="k">print</span> <span class="s">"(d) Load Factor = </span><span class="si">%0.2f</span><span class="s"> </span><span class="si">%%</span><span class="s">"</span> <span class="o">%</span><span class="k">LF</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>(a) Maximum demand = 35.00 MW +(b) Units generated per day = 6e+05 kWh +(c) Average Load = 25.00 MW +(d) Load Factor = 71.43 % +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.17-page-24">exa 1.17 page 24<a class="anchor-link" href="#exa-1.17-page-24">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [40]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="kn">from</span> <span class="nn">math</span> <span class="kn">import</span> <span class="n">acos</span><span class="p">,</span> <span class="n">sin</span> +<span class="n">pf</span><span class="o">=</span><span class="mf">0.8</span> <span class="c">#power factor</span> +<span class="n">delf</span><span class="o">=</span><span class="mi">1</span> <span class="c">#%#drop in frequency(delf/f)</span> +<span class="c">#delP=-2*(sind(theta))**2*delf</span> +<span class="n">theta</span><span class="o">=</span><span class="n">acos</span><span class="p">(</span><span class="n">pf</span><span class="p">)</span> <span class="c">#degree</span> +<span class="n">delP_BY_delf</span><span class="o">=-</span><span class="mi">2</span><span class="o">*</span><span class="n">sin</span><span class="p">(</span><span class="n">theta</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span> <span class="c">#increase in load wrt frequency</span> +<span class="k">print</span> <span class="s">"1</span><span class="si">%%</span><span class="s"> drop in frequency, Increased in Load = </span><span class="si">%0.2f</span><span class="s"> </span><span class="si">%%</span><span class="s">"</span><span class="o">%-</span><span class="n">delP_BY_delf</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>1% drop in frequency, Increased in Load = 0.72 % +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.18-page-24">exa 1.18 page 24<a class="anchor-link" href="#exa-1.18-page-24">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [41]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="n">Lmax</span><span class="o">=</span><span class="mi">100</span> <span class="c">#MW</span> +<span class="n">LF</span><span class="o">=</span><span class="mi">40</span> <span class="c">#%#Load Factor</span> +<span class="n">Lavg</span><span class="o">=</span><span class="n">Lmax</span><span class="o">*</span><span class="n">LF</span><span class="o">/</span><span class="mi">100</span> <span class="c">#MW</span> +<span class="n">E</span><span class="o">=</span><span class="n">Lavg</span><span class="o">*</span><span class="mi">24</span><span class="o">*</span><span class="mi">365</span> <span class="c">#MWh</span> +<span class="k">print</span> <span class="s">"Energy generated in a year = %0.f MWh "</span><span class="o">%</span><span class="k">E</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>Energy generated in a year = 350400 MWh +</pre> +</div> +</div> + +</div> +</div> + +</div> +<div class="cell border-box-sizing text_cell rendered"> +<div class="prompt input_prompt"> +</div> +<div class="inner_cell"> +<div class="text_cell_render border-box-sizing rendered_html"> +<h2 id="exa-1.19-page-25">exa 1.19 page 25<a class="anchor-link" href="#exa-1.19-page-25">¶</a></h2> +</div> +</div> +</div> +<div class="cell border-box-sizing code_cell rendered"> +<div class="input"> +<div class="prompt input_prompt">In [42]:</div> +<div class="inner_cell"> + <div class="input_area"> +<div class=" highlight hl-ipython2"><pre><span class="kn">from</span> <span class="nn">math</span> <span class="kn">import</span> <span class="n">sqrt</span> +<span class="n">V</span><span class="o">=</span><span class="mi">400</span> <span class="c">#V</span> +<span class="n">s1</span><span class="o">=</span><span class="mf">0.03</span> <span class="c">#initial slip</span> +<span class="n">delV</span><span class="o">=</span><span class="mi">1</span> <span class="c">#%#/Voltage Drop</span> +<span class="n">R1</span><span class="o">=</span><span class="mf">0.290</span> <span class="c">#ohm/phase</span> +<span class="n">R2</span><span class="o">=</span><span class="mf">0.15</span> <span class="c">#ohm/phase</span> +<span class="n">X</span><span class="o">=</span><span class="mf">0.7</span> <span class="c">#ohm/phase(X1+X2)</span> +<span class="c">#V1**2*s1=V2**2*s2 for speed independent torque</span> +<span class="c">#taking for calculating s2</span> +<span class="n">V1</span><span class="o">=</span><span class="mi">1</span> <span class="c">#V </span> +<span class="n">V2</span><span class="o">=</span><span class="n">V1</span><span class="o">-</span><span class="n">V1</span><span class="o">*</span><span class="n">delV</span><span class="o">/</span><span class="mi">100</span> <span class="c">#V</span> +<span class="n">s2</span><span class="o">=</span><span class="n">V1</span><span class="o">**</span><span class="mi">2</span><span class="o">/</span><span class="n">V2</span><span class="o">**</span><span class="mi">2</span><span class="o">*</span><span class="n">s1</span> <span class="c">#slip</span> +<span class="n">I2ByI1</span><span class="o">=</span><span class="n">sqrt</span><span class="p">((</span><span class="n">R1</span><span class="o">+</span><span class="n">R2</span><span class="o">/</span><span class="n">s1</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span><span class="o">+</span><span class="n">X</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span><span class="o">/</span><span class="n">sqrt</span><span class="p">((</span><span class="n">R1</span><span class="o">+</span><span class="n">R2</span><span class="o">/</span><span class="n">s2</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span><span class="o">+</span><span class="n">X</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span><span class="o">*</span><span class="p">(</span><span class="n">V2</span><span class="o">/</span><span class="n">V1</span><span class="p">)</span> +<span class="n">delI</span><span class="o">=</span><span class="p">(</span><span class="n">I2ByI1</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span><span class="o">*</span><span class="mi">100</span> <span class="c">#%#Current Increase</span> +<span class="k">print</span> <span class="s">"1</span><span class="si">%%</span><span class="s"> drop in Voltage increases current by = </span><span class="si">%0.2f</span><span class="s"> </span><span class="si">%%</span><span class="s">"</span><span class="o">%</span><span class="k">delI</span> +<span class="c">#P=(R1+R2/s)*I**2</span> +<span class="n">P2ByP1</span><span class="o">=</span><span class="p">(</span><span class="n">R1</span><span class="o">+</span><span class="n">R2</span><span class="o">/</span><span class="n">s2</span><span class="p">)</span><span class="o">/</span><span class="p">(</span><span class="n">R1</span><span class="o">+</span><span class="n">R2</span><span class="o">/</span><span class="n">s1</span><span class="p">)</span><span class="o">*</span><span class="n">I2ByI1</span><span class="o">**</span><span class="mi">2</span> <span class="c">#ratio</span> +<span class="n">delP</span><span class="o">=</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">P2ByP1</span><span class="p">)</span><span class="o">*</span><span class="mi">100</span> <span class="c">#%#Power Decrease</span> +<span class="k">print</span> <span class="s">"1</span><span class="si">%%</span><span class="s"> drop in Voltage decreases power input by = </span><span class="si">%0.1f</span><span class="s"> </span><span class="si">%%</span><span class="s">"</span><span class="o">%</span><span class="k">delP</span> +<span class="c">#Answer in the textbook is not accurate.</span> +</pre></div> + +</div> +</div> +</div> + +<div class="output_wrapper"> +<div class="output"> + + +<div class="output_area"><div class="prompt"></div> +<div class="output_subarea output_stream output_stdout output_text"> +<pre>1% drop in Voltage increases current by = 0.86 % +1% drop in Voltage decreases power input by = 0.2 % +</pre> +</div> +</div> + +</div> +</div> + +</div> + </div> + </div> +</body> +</html> |
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