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diff --git a/Fundamentals_Of_Thermodynamics/Chapter14.ipynb b/Fundamentals_Of_Thermodynamics/Chapter14.ipynb new file mode 100755 index 00000000..6654362c --- /dev/null +++ b/Fundamentals_Of_Thermodynamics/Chapter14.ipynb @@ -0,0 +1,295 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b6f84d510db3595ba71244d9d44f7e9987d44c62ba0d0520a72da2384551019e"
+ },
+ "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": [
+ "Ex14.4: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
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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