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"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": {}
}
]
}
|
