From c7fe425ef3c5e8804f2f5de3d8fffedf5e2f1131 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 15:58:05 +0530 Subject: added books --- Fundamentals_Of_Thermodynamics/Chapter2_6.ipynb | 379 ++++++++++++++++++++++++ 1 file changed, 379 insertions(+) create mode 100755 Fundamentals_Of_Thermodynamics/Chapter2_6.ipynb (limited to 'Fundamentals_Of_Thermodynamics/Chapter2_6.ipynb') diff --git a/Fundamentals_Of_Thermodynamics/Chapter2_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter2_6.ipynb new file mode 100755 index 00000000..08e192fe --- /dev/null +++ b/Fundamentals_Of_Thermodynamics/Chapter2_6.ipynb @@ -0,0 +1,379 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:96f5bd0fa542764d6a8f75d5a1a42e4b79791697476f4029d0fa0fea1de65bed" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter2:SOME CONCEPTS AND DEFINITIONS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2.1:pg-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 1\n", + "#weight of a person\n", + "\n", + "\n", + "m=1 #kg\n", + "g=9.75 #acc.due to gravity in m/s^2\n", + "F=m*g #weight of 1 kg mass in N\n", + "print\"\\n hence,weight of person is\",round(F,2),\" N\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " hence,weight of person is 9.75 N\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2.1E:pg-20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 2\n", + "#weight of a person\n", + "\n", + "\n", + "m=1 #lbm\n", + "g=32 #acc.due to gravity in ft/s^2\n", + "F=m*g/32.174 #weight of 1 lbm mass in lbf\n", + "print\"\\n hence,weight of person is\",round(F,4),\" lbf\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " hence,weight of person is 0.9946 lbf\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2.2:pg-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 3\n", + "#average volume and density\n", + "\n", + "Vliq=0.2 #volume of liquid in m^3\n", + "dliq=997 #density of liquid in kg/m^3\n", + "Vstone=0.12 #volume of stone in m^3\n", + "Vsand=0.15 #volume of sand in m^3\n", + "Vair=0.53 #vo;ume of air in m^3\n", + "mliq=Vliq*dliq #mass of liquid in kg\n", + "dstone=2750 #density of stone in kg/m^3\n", + "dsand=1500 #density of sand in kg/m^3\n", + "mstone=Vstone*dstone #volume of stone in m^3\n", + "msand=Vsand*dsand #volume of sand in m^3\n", + "Vtot=1 #total volume in m^3\n", + "dair=1.1 #density of air in kg/m^3\n", + "mair=Vair*dair #mass of air\n", + "mtot=mair+msand+mliq+mstone #total mass in kg\n", + "v=Vtot/mtot #specific volume in m^3/kg\n", + "d=1/v #overall density in kg/m^3\n", + "print\"\\n hence,average specific volume is\",round(v,6),\"m^3/kg\" \n", + "print\"\\n and overall density is\", round(d),\"kg/m^3\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " hence,average specific volume is 0.001325 m^3/kg\n", + "\n", + " and overall density is 755.0 kg/m^3\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2.3:pg-26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 4\n", + "#calculating the required force\n", + "import math\n", + "\n", + "Dcyl=0.1 #cylinder diameter in m\n", + "Drod=0.01 #rod diameter in m\n", + "Acyl=math.pi*Dcyl**2/4 #cross sectional area of cylinder in m^2\n", + "Arod=math.pi*Drod**2/4 #cross sectional area of rod in m^2\n", + "Pcyl=250000 #inside hydaulic pressure in Pa\n", + "Po=101000 #outside atmospheric pressure in kPa\n", + "g=9.81 #acc. due to gravity in m/s^2\n", + "mp=25 #mass of (rod+piston) in kg\n", + "F=Pcyl*Acyl-Po*(Acyl-Arod)-mp*g #the force that rod can push within the upward direction in N\n", + "print\"\\n hence,the force that rod can push within the upward direction is\",round(F,1),\" N\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " hence,the force that rod can push within the upward direction is 932.9 N\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2.4:pg-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 5\n", + "#Calculating atmospheric pressure \n", + "\n", + "dm=13534 #density of mercury in kg/m^3\n", + "H=0.750 #height difference between two columns in metres\n", + "g=9.80665 #acc. due to gravity in m/s^2\n", + "Patm=dm*H*g/1000 #atmospheric pressure in kPa\n", + "print\"\\n hence, atmospheric pressure is\",round(Patm,2),\"kPa\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " hence, atmospheric pressure is 99.54 kPa\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2.5:pg-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 6\n", + "#pressure inside vessel\n", + "\n", + "dm=13590 #density of mercury in kg/m^3\n", + "H=0.24 #height difference between two columns in metres\n", + "g=9.80665 #acc. due to gravity in m/s^2\n", + "dP=dm*H*g #pressure difference in Pa\n", + "Patm=13590*0.750*9.80665 #Atmospheric Pressure in Pa\n", + "Pvessel=dP+Patm #Absolute Pressure inside vessel in Pa\n", + "Pvessel=Pvessel/101325 #Absolute Pressure inside vessel in atm\n", + "print\"\\n hence, the absolute pressure inside vessel is\",round(Pvessel,3),\"atm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " hence, the absolute pressure inside vessel is 1.302 atm\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2.5E:pg-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 7\n", + "#pressure inside vessel\n", + "\n", + "dm=848 #density of mercury in lbm/ft^3\n", + "H=9.5 #height difference between two columns in inches\n", + "g=32.174 #acc. due to gravity in ft/s^2\n", + "dP=dm*H*g #pressure difference in lbf/in^2\n", + "Patm=848*9.5*32.174/(1728*32.174) #Atmospheric Pressure in lbf/in^2\n", + "Pvessel=dP+Patm #Absolute Pressure inside vessel in lbf/in^2\n", + "Pvessel=848*29.5*32.174/(1728*32.174)+Patm #Absolute Pressure inside vessel in lbf/in^2\n", + "print\"\\n hence, the absolute pressure inside vessel is\",round(Pvessel,2),\"lbf/in^2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " hence, the absolute pressure inside vessel is 19.14 lbf/in^2\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2.6:pg-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 8\n", + "#calculating pressure\n", + "\n", + "dg=750 #density of gaasoline in kg/m^3\n", + "dR=1206 #density of R-134a in kg/m^3\n", + "H=7.5 #height of storage tank in metres\n", + "g=9.807 #acc. due to gravity in m/s^2\n", + "dP1=dg*g*H/1000 #in kPa\n", + "Ptop1=101 #atmospheric pressure in kPa\n", + "P1=dP1+Ptop1\n", + "print\"hence,pressure at the bottom of storage tank if fluid is gasoline is\",round(P1,1),\"kPa\" \n", + "dP2=dR*g*H/1000 #in kPa\n", + "Ptop2=1000 #top surface pressure in kPa\n", + "P2=dP2+Ptop2\n", + "print\"\\n hence, pressure at the bottom of storage tank if liquid is R-134a is\",round(P2),\"kPa\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "hence,pressure at the bottom of storage tank if fluid is gasoline is 156.2 kPa\n", + "\n", + " hence, pressure at the bottom of storage tank if liquid is R-134a is 1089.0 kPa\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2.7:pg-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 9\n", + "#calculating balancing force\n", + "\n", + "Po=100#Outside atmospheric pressure in kPa\n", + "F1=25 #net force on the smallest piston in kN\n", + "A1=0.01 #cross sectional area of lower piston in m^2\n", + "P1=Po+F1/A1 #fluid pressure in kPa\n", + "d=900 #density of fluid in kg/m^3\n", + "g=9.81 #acc. due to gravity in m/s^2\n", + "H=6 #height of second piston in comparison to first one in m\n", + "P2=P1-d*g*H/1000 #pressure at higher elevation on piston 2 in kPa\n", + "A2=0.05 #cross sectional area of higher piston in m^3\n", + "F2=(P2-Po)*A2 #balancing force on second piston in kN\n", + "print\"\\n hence, balancing force on second larger piston is\",round(F2,1),\"N\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " hence, balancing force on second larger piston is 122.4 N\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit