{ "metadata": { "name": "", "signature": "sha256:4d90868c7316d94fc2ffdc05c5691234f77f3a832d71da7eb2108eaff0a215bc" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter14:THERMODYNAMIC RELATIONS" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex14.1:Pg-567" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#ques1\n", "#to determine the sublimation pressure of water\n", "import math\n", "#from table in appendix B.1.5\n", "T1=213.2;#K, Temperature at state 1\n", "P2=0.0129;#kPa, pressure at state 2\n", "T2=233.2;#K, Temperature at state 2\n", "hig=2838.9;#kJ/kg, enthalpy of sublimation \n", "R=.46152;#Gas constant \n", "#using relation log(P2/P1)=(hig/R)*(1/T1-1/T2) \n", "P1=P2*math.exp(-hig/R*(1/T1-1/T2));\n", "print\" Sublimation Pressure \",round(P1,5),\"kPa\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Sublimation Pressure 0.00109 kPa\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex14.1E:Pg-567\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#ques1\n", "#to determine the sublimation pressure of water\n", "import math\n", "#from table in appendix B.1.5\n", "T1=-70+460.7;# R, Temperature at state 1\n", "P2=0.0019 # lbf/in^2 pressure at state 2\n", "T2=-40+460.7;# R, Temperature at state 2\n", "hig=1218.7;#Btu/lbm, enthalpy of sublimation \n", "R=85.67;#Gas constant \n", "#using relation log(P2/P1)=(hig/R)*(1/T1-1/T2) \n", "P1=P2*math.exp(-hig*778/R*(1/T1-1/T2));\n", "print\" Sublimation Pressure \",round(P1,5),\"lbf/in^2\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Sublimation Pressure 0.00025 lbf/in^2\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex14.4:Pg-579" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#ques4\n", "#Volume expansivity, Isothermal and Adiabatic compressibility\n", "\n", "#known data\n", "ap=5*10**-5;#K^-1 Volume expansivity\n", "bt=8.6*10**-12;#m^2/N, Isothermal compressibility\n", "v=0.000114;#m^3/kg, specific volume\n", "P2=100*10**6;#pressure at state 2 in kPa\n", "P1=100;#pressure at state 1 in kPa\n", "w=-v*bt*(P2**2-P1**2)/2;#work done in J/kg\n", "#q=T*ds and ds=-v*ap*(P2-P1)\n", "#so q=-T*v*ap*(P2-P1)\n", "T=288.2;#Temperature in K\n", "q=-T*v*ap*(P2-P1);#heat in J/kg\n", "du=q-w;#change in internal energy in J/kg\n", "print\" Change in internal energy =\",round(du,3),\"J/kg\"\n", "\n", "#the answer is correct within given limts\n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Change in internal energy = -159.372 J/kg\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex14.5:Pg-586" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#ques5\n", "#adiabatic steady state processes\n", "\n", "#from table A.2\n", "P1=20;#pressure at state 1 in MPa\n", "P2=2;#pressure at state 2 in MPa\n", "T1=203.2;#Temperature at state 1 in K\n", "Pr1=P1/3.39;#Reduced pressure at state 1\n", "Pr2=P2/3.39;#Reduced pressure at state 2\n", "Tr1=T1/126.2;#Reduced temperature\n", "#from compressibility chart h1*-h1=2.1*R*Tc\n", "#from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)\n", "#h2*-h2=0.5*R*Tc\n", "#this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.5*R*Tc\n", "R=0.2968;#gas constant for given substance\n", "Tc=126.2;#K, Constant temperature\n", "Cp=1.0416;#heat enthalpy at constant pressure in kJ/kg\n", "T2=146;#temperature at state 2\n", "dh=-1.6*R*Tc+Cp*(T1-T2);#\n", "print\" Enthalpy change =\",round(dh,3),\"kJ/kg \\n\"\n", "print\" Since Enthalpy change is nearly \",-round(dh),\"kJ/kg so Temperature =\",round(T2,3),\"K\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Enthalpy change = -0.35 kJ/kg \n", "\n", " Since Enthalpy change is nearly 0.0 kJ/kg so Temperature = 146.0 K\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex14.6:Pg-589" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#ques6\n", "#isothermal steady state processes\n", "import math\n", "#from table A.2\n", "P1=8;#pressure at state 1 in MPa\n", "P2=0.5;#pressure at state 2 in MPa\n", "T1=150.0;#Temperature at state 1 in K\n", "Pr1=P1/3.39;#Reduced pressure at state 1\n", "Pr2=P2/3.39;#Reduced pressure at state 2\n", "Tr1=T1/126.2;#Reduced temperature\n", "T2=125.0;#temperature at state 2\n", "#from compressibility chart h1*-h1=2.1*R*Tc\n", "#from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)\n", "#h2*-h2=0.5*R*Tc\n", "#this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.15*R*Tc\n", "R=0.2968;#gas constant for given substance\n", "Tc=126.2;#K, Constant temperature\n", "Cp=1.0416;#heat enthalpy at constant pressure in kJ/kg\n", "dh=(2.35)*R*Tc+Cp*(T2-T1);#\n", "print\" Enthalpy change =\",round(dh),\"kJ/kg\"\n", "#change in entropy \n", "#ds= -(s2*-s2)+(s2*-s1*)+(s1*-s1)\n", "#s1*-s1=1.6*R\n", "#s2*-s2=0.1*R\n", "#s2*-s1*=Cp*log(T2/T1)-R*log(P2/P1)\n", "#so\n", "ds=1.6*R-0.1*R+Cp*math.log(T2/T1)-R*math.log(P2/P1);\n", "print\" Entropy Change =\",round(ds,3),\"kJ/kg.K \"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Enthalpy change = 62.0 kJ/kg\n", " Entropy Change = 1.078 kJ/kg.K \n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex14.7:Pg-596" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#ques7\n", "#percent deviation using specific volume calculated by kays rule and vander waals rule\n", "import math\n", "\n", "#a-denotes C02\n", "#b-denotes CH4\n", "T=310.94;#Temperature of mixture K\n", "P=86.19;#Pressure of mixture in MPa\n", "#Tc- critical Temperature\n", "#Pc-critical pressure\n", "Tca=304.1;#K\n", "Tcb=190.4;#K\n", "Pca=7.38;#MPa\n", "Pcb=4.60;#MPa\n", "Ra=0.1889;#gas constant for a in kJ/kg.K\n", "Rb=0.5183;#gas constant for b in kJ/kg.K\n", "xa=0.8;#fraction of CO2\n", "xb=0.2;#fraction of CH4\n", "Rm=xa*Ra+xb*Rb;#mean gas constant in kJ/kg.K\n", "Ma=44.01;#molecular mass of a\n", "Mb=16.043;#molecular mass of b\n", "#1.Kay's rule\n", "ya=xa/Ma/(xa/Ma+xb/Mb);#mole fraction of a\n", "yb=xb/Mb/(xa/Ma+xb/Mb);#mole fraction of b\n", "Tcm=ya*Tca+yb*Tcb;#mean critical temp in K\n", "Pcm=ya*Pca+yb*Tcb;#mean critical pressure n MPa\n", "#therefore pseudo reduced property of mixture\n", "Trm=T/Tcm;\n", "Prm=P/Pcm;\n", "Zm=0.7;#Compressiblity from generalised compressibility chart\n", "vc=Zm*Rm*T/P/1000;#specific volume calculated in m^3/kg\n", "ve=0.0006757;#experimental specific volume in m^3/kg\n", "pd1=(ve-vc)/ve*100;#percent deviation\n", "print\" Percentage deviation in specific volume using Kays rule =\",round(pd1,1),\"percent \\n\"\n", "\n", "#2. using vander waals equation\n", "#values of vander waals constant\n", "Aa=27*(Ra**2)*(Tca**2)/(64*Pca*1000);\n", "Ba=Ra*Tca/(8*Pca*1000);\n", "Ab=27*(Rb**2)*(Tcb**2)/(64*Pcb*1000);\n", "Bb=Rb*Tcb/(8*Pcb*1000);\n", "#mean vander waals constant\n", "Am=(xa*math.sqrt(Aa)+xb*math.sqrt(Ab))**2;\n", "Bm=(xa*Ba+xb*Bb);\n", "#using vander waals equation we get cubic equation \n", "#solving we get\n", "vc=0.0006326;#calculated specific volume in m^3/kg\n", "pd2=((ve-vc)/ve)*100;\n", "print\" Percentage deviation in specific volume using vander waals eqn =\",round(pd2,1),\"percent\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Percentage deviation in specific volume using Kays rule = 4.8 percent \n", "\n", " Percentage deviation in specific volume using vander waals eqn = 6.4 percent\n" ] } ], "prompt_number": 9 } ], "metadata": {} } ] }