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|
{
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"name": "",
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},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter11:Power and Refrigeration Systems\u2014With Phase Change"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.1:Pg-425"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Ques 1\n",
"#To determine the efficiency of Rankine cycle\n",
"\n",
"#1-Inlet state of pump\n",
"#2-Exit state of pump\n",
"P2=2000;#Exit pressure in kPa\n",
"P1=10;#Inlet pressure in kPa\n",
"v=0.00101;#specific weight of water in m^3/kg\n",
"wp=v*(P2-P1);#work done in pipe in kJ/kg\n",
"h1=191.8;#Enthalpy in kJ/kg from table\n",
"h2=h1+wp;#enthalpy in kJ/kg\n",
"#2-Inlet state for boiler\n",
"#3-Exit state for boiler\n",
"h3=2799.5;#Enthalpy in kJ/kg\n",
"#3-Inlet state for turbine\n",
"#4-Exit state for turbine\n",
"#s3=s4(Entropy remain same)\n",
"s4=6.3409;#kJ/kg\n",
"sf=0.6493;#Entropy at liquid state in kJ/kg\n",
"sfg=7.5009;#Entropy difference for vapor and liquid state in kJ/kg\n",
"x4=(s4-sf)/sfg;#x-factor\n",
"hfg=2392.8;#Enthalpy difference in kJ/kg for turbine\n",
"h4=h1+x4*hfg;#Enthalpy in kJ/kg\n",
"\n",
"nth=((h3-h2)-(h4-h1))/(h3-h2);\n",
"print\" Percentage efficiency =\",round(nth*100,1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Percentage efficiency = 30.3\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.2:Pg-429"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Ques 2\n",
"#To determine the efficiency of Rankine cycle\n",
"\n",
"#1-Inlet state of pump\n",
"#2-Exit state of pump\n",
"P2=4000;#Exit pressure in kPa\n",
"P1=10;#Inlet pressure in kPa\n",
"v=0.00101;#specific weight of water in m^3/kg\n",
"wp=v*(P2-P1);#work done in pipe in kJ/kg\n",
"h1=191.8;#Enthalpy in kJ/kg from table\n",
"h2=h1+wp;#enthalpy in kJ/kg\n",
"#2-Inlet state for boiler\n",
"#3-Exit state for boiler\n",
"h3=3213.6;#Enthalpy in kJ/kg from table\n",
"#3-Inlet state for turbine\n",
"#4-Exit state for turbine\n",
"#s3=s4(Entropy remain same)\n",
"s4=6.7690;#Entropy in kJ/kg from table\n",
"sf=0.6493;#Entropy at liquid state in kJ/kg from table\n",
"sfg=7.5009;#Entropy difference for vapor and liquid state in kJ/kg from table\n",
"x4=(s4-sf)/sfg;#x-factor\n",
"hfg=2392.8;#Enthalpy difference in kJ/kg for turbine\n",
"h4=h1+x4*hfg;#Enthalpy in kJ/kg\n",
"\n",
"nth=((h3-h2)-(h4-h1))/(h3-h2);\n",
"print\"Percentage efficiency =\",round(nth*100,1),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage efficiency = 35.3 %\n"
]
}
],
"prompt_number": 41
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.2E:Pg-431"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Ques 2\n",
"#To determine the efficiency of Rankine cycle\n",
"\n",
"#1-Inlet state of pump\n",
"#2-Exit state of pump\n",
"P2=600.0 ;#Exit pressure in lbf/in^2\n",
"P1=1.0;#Inlet pressure in lbf/in^2\n",
"v=0.01614;#specific weight of water in ft^3/lbm\n",
"wp=v*(P2-P1)*(144.0/778.0);#work done in pipe in Btu/lbm\n",
"h1=69.70;#Enthalpy in Btu/lbm from table\n",
"h2=h1+wp;#enthalpy in Btu/lbm\n",
"#2-Inlet state for boiler\n",
"#3-Exit state for boiler\n",
"h3=1407.6;#Enthalpy in Btu/lbm from table\n",
"#3-Inlet state for turbine\n",
"#4-Exit state for turbine\n",
"#s3=s4(Entropy remain same)\n",
"s4=1.6343;#Entropy in Btu/lbm from table\n",
"sf=1.9779;#Entropy at liquid state in Btu/lbm from table\n",
"sfg=1.8453;#Entropy difference for vapor and liquid state in Btu/lbm from table\n",
"x4=-(s4-sf)/sfg;#x-factor\n",
"hfg=1036.0;#Enthalpy difference in Btu/lbm for turbine\n",
"h4=1105.8-x4*hfg;#Enthalpy in Btu/lbm\n",
"wt=(h3-h4) #work done in turbine in Btu/lbm\n",
"\n",
"nth=((h3-h4)-wp)/(h3-h2);\n",
"print\"Percentage efficiency =\",round(nth*100,1),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage efficiency = 36.9 %\n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.3:Pg-433"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Ques 3\n",
"#To determine the efficiency of a cycle\n",
"\n",
"#1-Inlet state of pump\n",
"#2-Exit state of pump\n",
"P2=4000;#Exit pressure in kPa\n",
"P1=10;#Inlet pressure in kPa\n",
"v=0.00101;#specific weight of water in m^3/kg\n",
"wp=v*(P2-P1);#work done in pipe in kJ/kg\n",
"h1=191.8;#Enthalpy in kJ/kg from table\n",
"h2=h1+wp;#enthalpy in kJ/kg\n",
"#2-Inlet state for boiler\n",
"#3-Exit state for Boiler\n",
"h3=3213.6;#Enthalpy in kJ/kg from table\n",
"#3-Inlet state for high pressure turbine\n",
"#4-Exit state for high pressure turbine\n",
"#s3=s4(Entropy remain same)\n",
"s4=6.7690;#Entropy in kJ/kg from table\n",
"sf=1.7766;#Entropy at liquid state in kJ/kg from table\n",
"sfg=5.1193;#Entropy difference for vapor and liquid state in kJ/kg from table\n",
"x4=(s4-sf)/sfg;#x-factor\n",
"hf=604.7#Enthalpy of liquid state in kJ/kg\n",
"hfg=2133.8;#Enthalpy difference in kJ/kg for turbine\n",
"h4=hf+x4*hfg;#Enthalpy in kJ/kg\n",
"#5-Inlet state for low pressure turbine\n",
"#6-Exit pressure for low pressure turbine\n",
"sf=0.6493;#Entropy in liquid state in kJ/kg for turbine\n",
"h5=3273.4;#enthalpy in kJ/kg \n",
"s5=7.8985;#Entropy in kJ/kg\n",
"sfg=7.5009;#entropy diff in kJ/kg \n",
"x6=(s5-sf)/sfg;#x-factor\n",
"hfg=2392.8;#enthalpy difference for low pressure turbine in kj/kg\n",
"h6=h1+x6*hfg;#entropy in kg/kg\n",
"wt=(h3-h4)+(h5-h6);#work output in kJ/kg\n",
"qh=(h3-h2)+(h5-h4);\n",
"\n",
"nth=(wt-wp)/qh;\n",
"print\" Percentage efficiency =\",round(nth*100,1),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Percentage efficiency = 35.9 %\n"
]
}
],
"prompt_number": 43
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.4:Pg-438"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#ques4\n",
"#Efficiency of Refrigeration cycle\n",
"\n",
"#from previous examples\n",
"h1=191.8;#kJ/kg\n",
"h5=3213.6;#kg/kg\n",
"h6=2685.7;#kJ/kg\n",
"h7=2144.1;#kJ/kg\n",
"h3=604.7;#kJ/kg\n",
"#1-Inlet state of pump\n",
"#2-Exit state of pump\n",
"P2=400;#Exit pressure in kPa\n",
"P1=10; #Inlet pressure in kPa\n",
"v=0.00101;#specific weight of water in m^3/kg\n",
"wp1=v*(P2-P1);#work done for low pressure pump in kJ/kg\n",
"h1=191.8;#Enthalpy in kJ/kg from table\n",
"h2=h1+wp1;#enthalpy in kJ/kg\n",
"#5-Inlet state for turbine\n",
"#6,7-Exit state for turbine\n",
"y=(h3-h2)/(h6-h2);#extraction fraction\n",
"wt=(h5-h6)+(1-y)*(h6-h7);#turbine work in kJ/kg\n",
"#3-Inlet for high pressure pump\n",
"#4-Exit for high pressure pump\n",
"P3=400;#kPa\n",
"P4=4000;#kPa\n",
"v=0.001084;#specific heat for 3-4 process in m^3/kg\n",
"wp2=v*(P4-P3);#work done for high pressure pump\n",
"h4=h3+wp2;#Enthalpy in kJ/kg\n",
"wnet=wt-(1-y)*wp1-wp2;\n",
"qh=h5-h4;#Heat output in kJ/kg\n",
"nth=wnet/qh;\n",
"print\" Refrigerator Efficiency =\",round(nth*100,1),\"%\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Refrigerator Efficiency = 37.5 %\n"
]
}
],
"prompt_number": 44
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.5:Pg-443"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#ques5\n",
"#To determine thermal efficiency of cycle\n",
"\n",
"#5-Inlet state for turbine\n",
"#6-Exit state for turbine\n",
"#h-Enthalpy at a state \n",
"#s-Entropy at a state\n",
"#from steam table\n",
"h5=3169.1;#kJ/kg\n",
"s5=6.7235;#kJ/kg\n",
"s6s=s5;\n",
"sf=0.6493;#Entropy for liquid state in kJ/kg\n",
"sfg=7.5009;#Entropy difference in kJ/kg\n",
"hf=191.8;#kJ/kg\n",
"hfg=2392.8;#Enthalpy difference in kJ/kg\n",
"x6s=(s6s-sf)/sfg;#x-factor\n",
"h6s=hf+x6s*hfg;#kJ/Kg at state 6s\n",
"nt=0.86;#turbine efficiency given\n",
"wt=nt*(h5-h6s);\n",
"#1-Inlet state for pump\n",
"#2-Exit state for pump\n",
"np=0.80;#pump efficiency given\n",
"v=0.001009;#specific heat in m^3/kg\n",
"P2=5000;#kPa\n",
"P1=10;#kPa\n",
"wp=v*(P2-P1)/np;#Work done in pump in kJ/kg\n",
"wnet=wt-wp;#net work in kJ/kg\n",
"#3-Inlet state for boiler\n",
"#4-Exit state for boiler\n",
"h3=171.8;#in kJ/kg from table\n",
"h4=3213.6;#kJ/kg from table\n",
"qh=h4-h3;\n",
"nth=wnet/qh;\n",
"print \"Cycle Efficiency =\",round(nth*100,1),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Cycle Efficiency = 29.2 %\n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.5E:Pg-445"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#ques5\n",
"#To determine thermal efficiency of cycle\n",
"\n",
"#5-Inlet state for turbine\n",
"#6-Exit state for turbine\n",
"#h-Enthalpy at a state \n",
"#s-Entropy at a state\n",
"#from steam table\n",
"h5=1386.8;#Btu/lbm\n",
"s5=1.6248;#Btu/lbm\n",
"s6s=s5;\n",
"sf=1.9779;#Entropy for liquid state in Btu/lbm\n",
"sfg=1.8453;#Entropy difference in Btu/lbm\n",
"hf=1105.8;# Btu/lbm\n",
"hfg=1036.0;#Enthalpy difference in Btu/lbm\n",
"x6s=(s6s-sf)/sfg;#x-factor\n",
"h6s=hf+x6s*hfg;#Btu/lbm at state 6s\n",
"nt=0.86;#turbine efficiency given\n",
"wt=nt*(h5-h6s);\n",
"#1-Inlet state for pump\n",
"#2-Exit state for pump\n",
"np=0.80;#pump efficiency given\n",
"v=0.016;#specific heat in ft^3/lbm\n",
"P2=800.0;# lbf/in^2\n",
"P1=1.0;# lbf/in^2\n",
"wp=(v*(P2-P1)*144.0)/(np*778.0);#Work done in pump in Btu/lbm\n",
"wnet=wt-wp;#net work in Btu/lbm\n",
"#3-Inlet state for boiler\n",
"#4-Exit state for boiler\n",
"h3=65.1;#in Btu/lbm from table\n",
"h4=1407.6;# Btu/lbm from table\n",
"qh=h4-h3;\n",
"nth=wnet/qh;\n",
"print \"Cycle Efficiency =\",round(nth*100,2),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Cycle Efficiency = 30.48 %\n"
]
}
],
"prompt_number": 60
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.6:Pg-451"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#ques6\n",
"#to determine the rate of refrigeration\n",
"\n",
"# refer to fig 11.21 in book\n",
"mdot=0.03 # mass flow rate in Kg/s\n",
"T1=-20 # temperature in evaporator in celsius\n",
"T3=40 #temperature in evaporator in Celsius\n",
"P2=1017 # saturation pressure in KPa\n",
"\n",
"# from table of R-134a refrigerant\n",
"h1=386.1 # enthalpy at state 1 in kJ/kg,\n",
"S1=1.7395 # entropy at state 1 in kJ/kg.K\n",
"S2=S1 # isentropic process\n",
"T2=47.7# corresponding value to S2 in table of R-134a in degree celsius\n",
"h2=428.4 # corresponding value to S2 in table of R-134a in kJ/kg\n",
"wc=h2-h1 # work done in compressor in kJ/kg\n",
"h4=h3=256.5 #enthalpy at state 4 and 3 in kJ/kg\n",
"qL=h1-h4 #Heat rejected in kJ/kg\n",
"\n",
"B=qL/wc # COP\n",
"\n",
"print\" the COP of the plant is\",round(B,2)\n",
"print\" the refrigeration rate is\",round(mdot*qL,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" the COP of the plant is 3.06\n",
" the refrigeration rate is 3.89\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.7:Pg-454"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#ques7\n",
"#to determine the COP of cycle\n",
"\n",
"P1=125 # pressure at state 1 in kPa\n",
"P2=1.2 # pressure at state 2 in MPa\n",
"P3=1.19 # pressure at state 3 in MPa,\n",
"P4=1.16 # pressure at state 4 in MPa,\n",
"P5=1.15 # pressure at state 5 in MPa,\n",
"P6=P7=140 # pressure at state 6 and 7 in kPa,\n",
"P8=130 # pressure at state 8 in kPa,\n",
"T1=-10 #temperaure at state 1 in \u25e6C\n",
"T2=100 #temperaure at state 2 in \u25e6C\n",
"T3=80 #temperaure at state 3 in \u25e6C\n",
"T4=45 #temperaure at state 4 in \u25e6C\n",
"T5=40 #temperaure at state 5 in \u25e6C\n",
"T8=-20 #temperaure at state 8 in \u25e6C\n",
"q=-4 # heat transfer in kJ/Kg\n",
"\n",
"#x6=x7 quality condition given in question\n",
"\n",
"\n",
"# the following values are taken from table for refrigerant R-134a\n",
"h1=394.9 # enthalpy at state 1 in kJ/kg\n",
"h2=480.9 # enthalpy at state 2 in kJ/kg\n",
"h8=386.6 # enthalpy at state 8 in kJ/kg\n",
"wc=h2-h1-q # from first law\n",
"h5=h6=h7=256.4 # as x6=x7 and from table at state 5 in Kj/Kg\n",
"qL=h8-h7 # from first law \n",
"B=qL/wc # COP\n",
"print\" the COP of the plant is\",round(B,3)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" the COP of the plant is 1.447\n"
]
}
],
"prompt_number": 20
}
],
"metadata": {}
}
]
}
|