From 92cca121f959c6616e3da431c1e2d23c4fa5e886 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 15:58:05 +0530 Subject: added books --- Thermodynamics/Chapter6.ipynb | 1076 +++++++++++++++++++++++++++++++++++++++++ 1 file changed, 1076 insertions(+) create mode 100755 Thermodynamics/Chapter6.ipynb (limited to 'Thermodynamics/Chapter6.ipynb') diff --git a/Thermodynamics/Chapter6.ipynb b/Thermodynamics/Chapter6.ipynb new file mode 100755 index 00000000..2e760ba9 --- /dev/null +++ b/Thermodynamics/Chapter6.ipynb @@ -0,0 +1,1076 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:009f400f836702cde3e9c0f5d144589b0ec3eb912d883ee490ce810f95fed24c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 06:SECOND LAW OF THERMODYNAMICS AND ENTROPY" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1, Page No:259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "QH=500; # Heat supplied in kJ\n", + "QL=200; # Heat rejected in kJ\n", + "TH=720; # Resorvior Temperature in kelvin\n", + "TL=360; # Resorvior Temperature in kelvin\n", + "W=260; # Work developed in kJ\n", + "\n", + "#Calculation\n", + "e_max=1-TL/TH; # maximum efficiency\n", + "e_clamied=W/QH; # Efficiency clamied\n", + "\n", + "#Result\n", + "if e_clamiede_clamied:\n", + " print \"\\nThe machine is irreversible\"\n", + "else:\n", + " print \"\\nHere e_clamied > e_carnot so the cyclic machine is impossible.\"\n", + "\n", + "print \"It would be reversible if its thermal efficiency is equal to Carnot efficiency,\"\n", + "print \"and irreversible if it is less than Carnot efficiency.\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "e_carnot = 0.6\n", + "e_clamied= 0.615\n", + "\n", + "Here e_clamied > e_carnot so the cyclic machine is impossible.\n", + "It would be reversible if its thermal efficiency is equal to Carnot efficiency,\n", + "and irreversible if it is less than Carnot efficiency.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3, Page No:260" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# Air conditioning unit\n", + "TL=278; # Operating temperature in kelvin\n", + "TH=318; # Operating temperature in kelvin\n", + "\n", + "#Calculation\n", + "COP1=TL/(TH-TL); # COP of Air conditioning unit\n", + "QL=1; # For some calculation purpose\n", + "W1=QL/COP1; # Work input of Air conditioning unit\n", + "# Food refrigeration unit\n", + "TL=258; # Operating temperature in kelvin\n", + "TH=318; # Operating temperature in kelvin\n", + "COP2=TL/(TH-TL); # COP of Food refrigeration unit\n", + "W2=QL/COP2; # Work input of Food refrigeration unit\n", + "Wper=(W2-W1)/W1; # Increase in work input\n", + "\n", + "#Result\n", + "print \"Increase in work input = \",round(Wper*100,0),\"%\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Increase in work input = 62.0 %\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4, Page No:261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "#(a).Summer air conditioning (cooling)\n", + "TL=298; # Operating temperature in kelvin\n", + "TH=318; # Operating temperature in kelvin\n", + "q=0.75; # Heat Transfer from fabric of room per degree of temperature difference in kW\n", + "\n", + "#Calculation for (a)\n", + "QL=q*(TH-TL); # Heat Transfer from fabric of room\n", + "COPc=TL/(TH-TL); # COP of Air conditioning unit\n", + "W=QL/COPc; # Work input of Air conditioning unit\n", + "\n", + "#Result for (a)\n", + "print \"(a).Summer air conditioning (cooling)\",\"\\nWork input of Air conditioning unit = \",round(W,0),\"kW\"\n", + "\n", + "#Calculation for (b)\n", + "# (b).Winter air conditioning (recerse cycle heating)\n", + "TH=293; # Operating temperature in kelvin\n", + "TL=(-(-2*q*TH)-math.sqrt ((-2*q*TH)**2-(4*q*(q*TH**2-TH))))/(2*q);# Lowest outdoor Temperature by root\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).Winter air conditioning (recerse cycle heating)\",\"\\nLowest outdoor Temperature = \",round(TL,0),\"K\"\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).Summer air conditioning (cooling) \n", + "Work input of Air conditioning unit = 1.0 kW\n", + "\n", + "(b).Winter air conditioning (recerse cycle heating) \n", + "Lowest outdoor Temperature = 273.0 K\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5, Page No:263" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# (a).For the refrigerator \n", + "TL=258; # Operating temperature in kelvin\n", + "TH=313; # Operating temperature in kelvin\n", + "QL=3.5167; # Ton of refrigeration in kW\n", + "\n", + "#Calculation for(a)\n", + "COP=TL/(TH-TL); # COP of Refrigeration unit\n", + "W=QL/COP; # Power comsumption of refrigerator\n", + "\n", + "#Result for (a)\n", + "print \"(a).For the refrigerator\",\"\\nPower comsumption of refrigerator = \",round(W,2),\"kW\"\n", + "\n", + "#calculation for (b)\n", + "# (b). For the freezer\n", + "TL=248; # Operating temperature in kelvin\n", + "TH=313; # Operating temperature in kelvin\n", + "COP=TL/(TH-TL); # COP of Freezer unit\n", + "QL=W*COP; # Refrigeration produced\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b). For the freezer\",\"\\nRefrigeration produced = \",round(QL,3),\"kW\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).For the refrigerator \n", + "Power comsumption of refrigerator = 0.75 kW\n", + "\n", + "(b). For the freezer \n", + "Refrigeration produced = 2.86 kW\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6, Page No:277" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Psat=200;#Pressure of water in kPa\n", + "Tsat=393.38; # Saturation temperaure at Psat in kelvin\n", + "# (i).From the equation Tds=du+pdv \n", + "# Following are from steam table at Psat\n", + "ufg=2025; # specific internal energy of vapourization in kJ/kg\n", + "vg=0.8857; # specific volume in m^3/kg\n", + "vf=0.001061; # specific volume in m^3/kg\n", + "\n", + "#Calculation for (i)\n", + "sfg=(ufg/Tsat)+(Psat*(vg-vf)/Tsat); # specific entropy of vapourization\n", + "\n", + "#Result for (i)\n", + "print \"(i).From the equation Tds=du+pdv \",\"\\nspecific entropy of vapourization = \",round(sfg,4),\"kJ/kg K\"\n", + "\n", + "#Calculation for (ii)\n", + "# (ii).From the equation Tds=dh-vdp\n", + "hfg=2201.9; # Specific enthalpy of vapourization in kJ/kg\n", + "sfg=hfg/Tsat; # specific entropy of vapourization\n", + "\n", + "#Result for (ii)\n", + "print \"\\n(ii).From the equation Tds=dh-vdp \",\"\\nspecific entropy of vapourization = \",round(sfg,4),\"kJ/kg K\"\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i).From the equation Tds=du+pdv \n", + "specific entropy of vapourization = 5.5975 kJ/kg K\n", + "\n", + "(ii).From the equation Tds=dh-vdp \n", + "specific entropy of vapourization = 5.5974 kJ/kg K\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7, Page No:277" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "p1=1; # Pressure of steam at state 1 in bar\n", + "T=473; # Temperature of steam at state 1 in kelvin\n", + "\n", + "#Calculation for (i)\n", + "# (i).Pressure after compression\n", + "p2=1.5538; # Pressure after compression at (Psat)T from steam table in MPa\n", + "\n", + "#Result for (i)\n", + "print \"(i).Pressure after compression\",\"\\nPressure after compression = \",p2,\"MPa\"\n", + "\n", + "#Calcultion for (ii)\n", + "# (ii).Heat Transfer and work done during the process\n", + "# Following are from steam table \n", + "s2=6.4323; # specific entropy of steam at state 2 in kJ/kg K\n", + "s1=7.8343; # specific entropy of steam at state 1 in kJ/kg K\n", + "u2=2595.3; # specific internal energy of steam at state 2 in kJ/kg \n", + "u1=2658.1; # specific internal energy of steam at state 1 in kJ/kg \n", + "q=T*(s2-s1); # Heat transfer during the process\n", + "w=q-(u2-u1); # Work done during the process\n", + "\n", + "#Result for (ii)\n", + "print \"\\n(ii).Heat Transfer and work done during the process\",\"\\nHeat transfer during the process = \",round(q,0),\"kJ\"\n", + "print \"Work done during the process = \",round(w,1),\"kJ\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i).Pressure after compression \n", + "Pressure after compression = 1.5538 MPa\n", + "\n", + "(ii).Heat Transfer and work done during the process \n", + "Heat transfer during the process = -663.0 kJ\n", + "Work done during the process = -600.3 kJ\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8, Page No:278" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "p1=6; # Initial pressure of steam in MPa\n", + "T1=500; # Initial temperature of steam in degree celcius\n", + "p2=10; # Final pressure of steam in bar\n", + "# From steam tables\n", + "s1=6.8803; sf2=1.3026; sfg2=6.0568; # specific entropy in kJ/kg K\n", + "u1=3082.2; uf2=761.68; ufg2=1822; # specific internal energy in kJ/kg\n", + "v1=0.05665; vf2=0.001043; vg2=1.694; # specific volume in m^3/kg\n", + "\n", + "#Calculation\n", + "x2=(v1-vf2)/(vg2-vf2);# Quality of steam\n", + "u2=uf2+x2*ufg2; # specific internal energy in kJ/kg \n", + "s2=sf2+x2*sfg2; # specific entropy in kJ/kg K\n", + "s21=s2-s1; # Entropy change\n", + "q=u2-u1; # Heat transfer\n", + "\n", + "#Result\n", + "print \"Entropy change of the process = \",round(s21,3),\"kJ/kg\",\"\\nHeat transfer for the process =\",round(q,1),\"kJ\"\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Entropy change of the process = -5.379 kJ/kg \n", + "Heat transfer for the process = -2260.7 kJ\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10, Page No:280" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "p1=3; # initial pressure of air in bar\n", + "T1=200; # initial temperature of air in degree celcius\n", + "p2=1.5; # final pressure of air in bar\n", + "T2=105; # final temperature of air in degree celcius\n", + "Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K\n", + "R=0.287; # characteristic gas constant of air in kJ/kg K\n", + "\n", + "#Calculation\n", + "delta_s= Cpo*math.log ((T2+273)/(T1+273))- R*math.log (p2/p1) # change in entropy during irreversible process\n", + "#The value of p2 is taken as wrong in the textbook\n", + "\n", + "#Result\n", + "print \"change in entropy during irreversible process = \",round(delta_s,4),\"kJ/kg K (Answer in the textbook was wrong)\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "change in entropy during irreversible process = -0.0261 kJ/kg K (Answer in the textbook was wrong)\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11, Page No:281" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "p1=5; # Initial pressure of argon gas in bar\n", + "T1=30; # Initial temperature of argon gas in degree celcius\n", + "v1=1; # Initial volume of argon gas in m^3 by assumption\n", + "v2=2*v1; # Final volume of argon gas in m^3\n", + "R=8.3144/40; # Characteristic gas constant of argon gas in kJ/kg K\n", + "\n", + "#Calculation\n", + "p2=p1*(v1/v2); # Final pressure of argon gas\n", + "delta_s= R*math.log (v2/v1); # change in entropy (choosing the reversible isothermal path)\n", + "\n", + "#Result\n", + "print \"Final pressure of argon gas =\",p2,\"bar\"\n", + "print \"change in entropy (choosing the reversible isothermal path) = \",round(delta_s,4),\"kJ/kg K\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Final pressure of argon gas = 2.5 bar\n", + "change in entropy (choosing the reversible isothermal path) = 0.1441 kJ/kg K\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12, Page No:284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "p1=1; # Atmospheric pressure in bar\n", + "T1=348; # Atmospheric temperature in kelvin\n", + "V1=800; # Volume of air sucked into the cylinder in cm^3\n", + "p2=15; # pressure of air after compression in bar\n", + "V2=V1/8; # volume of air after compression in cm^3\n", + "p3=50; # pressure of air after heat addition in bar\n", + "Cvo=0.7165; # Specific heat at constant volme in kJ/kg K\n", + "R=0.287; # characteristic gas constant of air in kJ/kg K\n", + "\n", + "#Calculation for (a)\n", + "# (a).Index of compression process\n", + "n=math.log (p2/p1)/math.log (V1/V2); # Index of compression process\n", + "\n", + "#Result for (a)\n", + "print \"(a).Index of compression process\"\n", + "print \"Index of compression process = \",round(n,1),\" which is less than 1.4. The compression process is polytropic.\"\n", + "\n", + "#Calculation for (b)\n", + "# (b).Change in entropy of air during each process\n", + "m=(p1*10**2*V1*10**-6)/(R*T1); # Mass of air in cylinder\n", + "T2=T1*(p2/p1)*(V2/V1); # Temperature after compression\n", + "T3=T2*(p3/p2); # Temperature after heat addition\n", + "delta_s21=m*(Cvo*math.log (T2/T1)+R*math.log (V2/V1)); # change in entropy during compression\n", + "delta_s32=m*Cvo*math.log (T3/T2); #change in entropy during heat addition\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).Change in entropy of air during each process\"\n", + "print \"change in entropy during compression = (Error in textbook)\",round(delta_s21,6),\"kJ/K\"\n", + "print \"change in entropy during heat addition = (Error in textbook)\",round(delta_s32,6),\"kJ/K\"\n", + "\n", + "#Calculation for (c)\n", + "# (c).Heat transfer during polytropic compression process\n", + "k=1.4;# Index of isentropic preocess\n", + "Q=m*Cvo*((k-n)/(1-n))*(T2-T1); # Heat transfer during polytropic compression process\n", + "\n", + "#Result for (c)\n", + "print \"\\n(c).Heat transfer during polytropic compression process\"\n", + "print \"Heat transfer during polytropic compression process = (Error in textbook)\",round(Q,4),\"kJ\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).Index of compression process\n", + "Index of compression process = 1.3 which is less than 1.4. The compression process is polytropic.\n", + "\n", + "(b).Change in entropy of air during each process\n", + "change in entropy during compression = (Error in textbook) -0.000117 kJ/K\n", + "change in entropy during heat addition = (Error in textbook) 0.000691 kJ/K\n", + "\n", + "(c).Heat transfer during polytropic compression process\n", + "Heat transfer during polytropic compression process = (Error in textbook) -0.0565 kJ\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13, Page No:287" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "p1=0.3; # initial pressure of ateam in MPa\n", + "T1=350; # Initial temperature of steam in degree celcius\n", + "# following are the values taken from steam table for initial state \n", + "v1=0.9535; # specific volume in m^3/kg\n", + "u1=2886.2; # specific internal energy in kJ/kg\n", + "s1=7.868; # specific entropy in kJ/kg K\n", + "v2=2*v1; # final specific volume of steam\n", + "u2=u1;\n", + "# following are the values taken from steam table final state\n", + "T2=349; # Final temperature of steam in degree celcius\n", + "p2=0.167; # Final pressure of ateam in MPa\n", + "s2=8.164; # specific entropy in kJ/kg K\n", + "\n", + "#Calculation\n", + "delta_s=s2-s1; # Entropy generation\n", + "LW=(T1+T2)/2 * delta_s; # Lost work\n", + "\n", + "#Result\n", + "print \"Entropy Generation =\",round(delta_s,3),\"kJ/kg K\",\"\\nLost work = \",round(LW,1),\"kJ\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Entropy Generation = 0.296 kJ/kg K \n", + "Lost work = 103.5 kJ\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.15, Page No:291" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "m=1; # Mass of water in kg\n", + "T1=300; # Temperature of water in kelvin\n", + "C=4.1868; # Specific heat in kJ/kg K\n", + "# (a). Heat Transfer\n", + "T2=500; # Temperature of heat reservoir in kelvin\n", + "\n", + "#Calculation for (a)\n", + "Q=m*C*(T2-T1); # Heat transfer\n", + "del_Swater=m*C*math.log (T2/T1); # Entropy change of water\n", + "del_Sreservoir=-Q/T2; # Entropy change of reservoir\n", + "del_Suniverse=del_Swater+del_Sreservoir; # Entropy change of universe\n", + "\n", + "#Result for (a)\n", + "print \"(a).Heat Transfer\",\"\\nEntropy change of universe =\",round(del_Suniverse,4),\"kJ/K\"\n", + "\n", + "#Calculation for (b)\n", + "# (b).Heat Transfer in each reservoir\n", + "T2=400; # Temperature of intermediate reservoir in kelvin\n", + "T3=500; # Temperature of heat reservoir in kelvin\n", + "Q=m*C*(T3-T2); # Heat transfer\n", + "del_Swater=m*C*(math.log (T2/T1)+math.log (T3/T2)); # Entropy change of water\n", + "del_SreservoirI=-Q/T2; # Entropy change of reservoir I\n", + "del_SreservoirII=-Q/T3; # Entropy change of reservoir II\n", + "del_Suniverse=del_Swater+del_SreservoirI+del_SreservoirII; # Entropy change of universe\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).Heat Transfer in each reservoir\",\"\\nEntropy change of universe =\",round(del_Suniverse,5),\"kJ/K\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).Heat Transfer \n", + "Entropy change of universe = 0.464 kJ/K\n", + "\n", + "(b).Heat Transfer in each reservoir \n", + "Entropy change of universe = 0.25466 kJ/K\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.16, Page No:292" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "m=1; # Mass of saturated steam in kg\n", + "T=100; # Teamperature of steam in degree celcius\n", + "T0=303; # temperature of Surroundings in kelvin\n", + "hfg=2257; # Latent heat of evaporation in kJ/kg\n", + "sfg=6.048; # specific entropy in kJ/kg K\n", + "\n", + "#Calculation for (a)\n", + "# (a).Entropy change\n", + "Q=m*hfg; # Heat transfer\n", + "del_Ssystem=-m*sfg; # Change of entropy of system\n", + "del_Ssurr=Q/T0; # Change of entropy of surroundings\n", + "del_Suniverse=del_Ssystem+del_Ssurr; # Change of entropy of universe\n", + "\n", + "#Ressult for (a)\n", + "print \"(a).Entropy change\",\"\\nChange of entropy of system =\",del_Ssystem,\"kJ/K\"\n", + "print \"Change of entropy of surroundings =\",round(del_Ssurr,4),\"kJ/K\"\n", + "print \"Change of entropy of universe =\",round(del_Suniverse,4),\"kJ/K\"\n", + "\n", + "#Calculation for (b)\n", + "# (b).Effect of heat transfer\n", + "del_Suniverse=0; # process is reversible\n", + "del_Ssurr=del_Suniverse-del_Ssystem; #Change of entropy of surroundings\n", + "QH=hfg; # Heat transfer from the condensing steam to reversible heat engine\n", + "QL=T0*del_Ssurr; # Heat receiveded by the surroundins reversible heat engine\n", + "W=QH-QL; #work output of reversible heat engine\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).Effect of heat transfer\",\"\\nHeat transfer from the condensing steam to reversible heat engine =\",QH,\"kJ\"\n", + "print \"Heat receiveded by the surroundins reversible heat engine =\",round(QL,1),\"kJ\"\n", + "print \"work output of reversible heat engine =\",round(W,1),\"kJ\"\n", + "print \"Difference between QH & QL is converted into work output in a reversible cyclic process\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).Entropy change \n", + "Change of entropy of system = -6.048 kJ/K\n", + "Change of entropy of surroundings = 7.4488 kJ/K\n", + "Change of entropy of universe = 1.4008 kJ/K\n", + "\n", + "(b).Effect of heat transfer \n", + "Heat transfer from the condensing steam to reversible heat engine = 2257 kJ\n", + "Heat receiveded by the surroundins reversible heat engine = 1832.5 kJ\n", + "work output of reversible heat engine = 424.5 kJ\n", + "Difference between QH & QL is converted into work output in a reversible cyclic process\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.17, Page No:293" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "m=1; # Mass of ice in kg\n", + "T1=258;# Temperature of ice in kelvin\n", + "Tm=273; # Melting point of ice in kelvin\n", + "T2=303; # temperature of Surroundings in kelvin\n", + "Cpice=2.095; # Specific heat of ice in kJ/kg K\n", + "hsg=333.5; # Latent heat of fusion in kJ/kg\n", + "Cpw=4.1868; # Specific heat of water in kJ/kg K\n", + "\n", + "#Calculation for (a)\n", + "# (a).Change of entropy\n", + "Q=m*(Cpice*(Tm-T1)+hsg+Cpw*(T2-Tm));# Heat transfer\n", + "del_Ssystem=m*((Cpice*math.log (Tm/T1))+(hsg/Tm)+(Cpw*math.log (T2/Tm)));# Change of entropy of system\n", + "del_Ssurr=-Q/T2; # Change of entropy of surroundings\n", + "del_Suniverse=del_Ssystem+del_Ssurr; # Change of entropy of universe\n", + "\n", + "#Result for (a)\n", + "print \"(a).Entropy change\",\"\\nChange of entropy of system =\",round(del_Ssystem,4),\"kJ/K\"\n", + "print \"Change of entropy of surroundings =\",round(del_Ssurr,4),\"kJ/K\"\n", + "print \"Change of entropy of universe =\",round(del_Suniverse,4),\"kJ/K\"\n", + "\n", + "#Calculation for (b)\n", + "# (b).The minimum work of restoring water back to ice\n", + "QL=Q; # Refrigerating effect\n", + "W=T2*del_Ssystem-QL; # The minimum work of restoring water back to ice\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).The minimum work of restoring water back to ice = \",round(W,1),\"kJ\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).Entropy change \n", + "Change of entropy of system = 1.7765 kJ/K\n", + "Change of entropy of surroundings = -1.6189 kJ/K\n", + "Change of entropy of universe = 0.1576 kJ/K\n", + "\n", + "(b).The minimum work of restoring water back to ice = 47.8 kJ\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.18, Page No:297" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "TA=323;# Temperature at section A in kelvin\n", + "PA=125; # Pressure at section A in kPa\n", + "TB=287;# Temperature at section B in kelvin\n", + "PB=100; # Pressure at section B in kPa\n", + "Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K\n", + "R=0.287; # characteristic gas constant of air in kJ/kg K\n", + "\n", + "#Calculation\n", + "SBA=(Cpo*math.log (TB/TA))-(R*math.log (PB/PA)); # Change in entropy\n", + "\n", + "#Result\n", + "print \"Change in entropy from B to A =\",SBA,\"kJ/kg (Error in Textbook)\",\"\\nHence SA>SB. Therefore B to A\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in entropy from B to A = -0.054541503612 kJ/kg (Error in Textbook) \n", + "Hence SA>SB. Therefore B to A\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.19, Page No:298" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "p1=12.5; # Pressure of steam at inlet in MPa\n", + "T1=500; # Temperature of steam at inlet in degree celcius\n", + "V1=50; # Velocity of steam at inlet in m/s\n", + "p2=10; # Pressure of steam at outlet in kPa\n", + "V2=100; # Velocity of steam at outlet in m/s\n", + "# (a).Actual expansion\n", + "x2=0.85; # Quality of steam\n", + "# From steam table\n", + "h1=3341.8; hf2=191.83; hg2=2584.7; # specific enthalpy in kJ/kg \n", + "s1=6.4618; sf2=0.6493; sfg2=7.5009; # specific entropy in kJ/kg K\n", + "\n", + "#Calculation for (a)\n", + "h2a=(1-x2)*hf2+x2*hg2; # specific enthalpy in kJ/kg \n", + "wa=(h1-h2a)+((V1**2-V2**2)/2000); # Actual work output\n", + "\n", + "#Result for (a)\n", + "print \"(a).Actual work output of turbine = \",round(wa,2),\"kJ\"\n", + "\n", + "#Calculation for (b)\n", + "# (b).Reversible adiabatic expansion\n", + "x2s=(s1-sf2)/sfg2; # Quality of steam after reversible adiabatic expansion\n", + "h2s=(1-x2s)*hf2+x2s*hg2; # specific enthalpy in kJ/kg \n", + "ws=(h1-h2s)+((V1**2-V2**2)/2000); # Reversible adiabatic work output\n", + "L=ws-wa; # Lost of work\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).Reversible adiabatic expansion\",\"\\nReversible adiabatic work output = \",round(ws,1),\"kJ/kg\"\n", + "print \"Lost of work due to irreversibity of expansion process =\",round(L,1),\"kJ/kg\"\n", + "\n", + "#Calculation for (c)\n", + "# (c).Entropy Generation\n", + "s2a=sf2+x2*sfg2; # actual specific entropy in kJ/kg K\n", + "Sgen=s2a-s1; # Entropy generation\n", + "\n", + "#Reult for (c)\n", + "print \"\\n(c).Entropy Generation =\",round(Sgen,4),\"kJ/kg K\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).Actual work output of turbine = 1112.28 kJ\n", + "\n", + "(b).Reversible adiabatic expansion \n", + "Reversible adiabatic work output = 1292.0 kJ/kg\n", + "Lost of work due to irreversibity of expansion process = 179.7 kJ/kg\n", + "\n", + "(c).Entropy Generation = 0.5633 kJ/kg K\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.20, Page No:302" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "p1=0.1; # pressure at state 1 in MPa\n", + "p2=6; # Pressure at state 2 in MPa\n", + "# (a).Pump work for water\n", + "vf1=0.001043; # specific volume in m^3/kg\n", + "\n", + "#Calculation for (a)\n", + "wp=-vf1*(p2-p1)*10**3; # Pump work for water\n", + "\n", + "#Result for (a)\n", + "print \"(a).Pump work for water =\",round(wp,2),\"kJ\"\n", + "\n", + "\n", + "#Variable declaration for (b)\n", + "# (b).For steam\n", + "h1=2675.5;# specific enthalpy in kJ/kg \n", + "s1=7.3595;# specific entropy in kJ/kg K\n", + "# From superheated steam table\n", + "t2=675; # Temperature at state 2 in degree celcius\n", + "h2=3835.3;# specific enthalpy in kJ/kg \n", + "\n", + "#Calculation for (b)\n", + "wc=-(h2-h1); # Compressor work for steam\n", + "\n", + "#Result for (b)\n", + "print \"(b).Compressor work for steam =\",wc,\"kJ/kg\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).Pump work for water = -6.15 kJ\n", + "(b).Compressor work for steam = -1159.8 kJ/kg\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.21, Page No:303" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "# (a).Restoring to initial state by throttling process\n", + "T1=303; #Temperature of air at state 1 in kelvin\n", + "p1=1; #Pressure of air at state 1 in bar\n", + "p2=5; #Pressure of air at state 2 in bar\n", + "p3=1;#Pressure of air at state 3 in bar\n", + "T3=303; #Temperature of air at state 3 in kelvin\n", + "Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K\n", + "R=0.287; # characteristic gas constant of air in kJ/kg K\n", + "k=1.4; # Index of reversible adiabatic compression\n", + "\n", + "#Calculation for (a)\n", + "T2=T1*(p2/p1)**((k-1)/k); # Temperature after reversible adiabatic compression\n", + "w12=Cpo*(T2-T1); # Work of reversible adiabatic compression\n", + "s21=0; # Entropy change of air\n", + "s32=-R*math.log (p3/p2); # Entropy change \n", + "s31=s32; # Net entropy change of air\n", + "d_Ssurr=0; # Entropy change of surroundings because There is no heat transfer\n", + "d_Suniv=s31+d_Ssurr; # Net Entropy change of universe\n", + "\n", + "#Result for (a)\n", + "print \"(a).Restoring to initial state by throttling process\",\"\\nWork of reversible adiabatic compression = \",round(w12,1),\"kJ/kg\"\n", + "print \"Net Entropy change of universe = \",round(d_Suniv,4),\"kJ/kg K\"\n", + "\n", + "#Calculation for (b)\n", + "# (b).Restoring to initial state by by completing cycle\n", + "T0=298; # Temperature of surroundings in kelvin\n", + "d_Ssystem=0; # Entropy change of systrem is zero because it is cyclic process\n", + "q31=Cpo*(T2-T3); # Heat rejected to the surroundings\n", + "d_Ssurr=q31/T0; # Entropy change of surroundings\n", + "d_Suniv=d_Ssystem+d_Ssurr; # Increase in entropy of the universe\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).Restoring to initial state by by completing cycle\",\"\\nNet Entropy change of universe = \",round(d_Suniv,3),\"kJ/kg K\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).Restoring to initial state by throttling process \n", + "Work of reversible adiabatic compression = 177.5 kJ/kg\n", + "Net Entropy change of universe = 0.4619 kJ/kg K\n", + "\n", + "(b).Restoring to initial state by by completing cycle \n", + "Net Entropy change of universe = 0.596 kJ/kg K\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit