From 41f1f72e9502f5c3de6ca16b303803dfcf1df594 Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 4 Sep 2015 22:04:10 +0530 Subject: add/remove/update books --- Fundamentals_Of_Thermodynamics/Chapter2.ipynb | 314 -------------------------- 1 file changed, 314 deletions(-) delete mode 100755 Fundamentals_Of_Thermodynamics/Chapter2.ipynb (limited to 'Fundamentals_Of_Thermodynamics/Chapter2.ipynb') diff --git a/Fundamentals_Of_Thermodynamics/Chapter2.ipynb b/Fundamentals_Of_Thermodynamics/Chapter2.ipynb deleted file mode 100755 index 95850873..00000000 --- a/Fundamentals_Of_Thermodynamics/Chapter2.ipynb +++ /dev/null @@ -1,314 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:eb1b6239947f111d6ea8f1bcaafe0bb84b121b2561bbb9f5f8a8bdec77169fd2" - }, - "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": [ - "Ex2.2:pg-24" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 2\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 3\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 4\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 5\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": [ - "Ex2.6:pg-29" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 6\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 7\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 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit