{ "metadata": { "name": "", "signature": "sha256:a88db765bb23172043955029f8f04edaf8e710f3cbd46a5474827f42fa8a65ec" }, "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": {} } ] }