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diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch4.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch4.ipynb new file mode 100755 index 00000000..0e734b53 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch4.ipynb @@ -0,0 +1,415 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:cba11d6ad27a555b5f3aecc639d6f99831950fba74d42eb631521eded4620d06" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Second Law of Thermodynamics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1 Page No : 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "Q1 = 250.0;#Heat absorbed in Kcal\n", + "T1 = (260+273.0);#Temperature at which engine absorbs heat\n", + "T0 = (40+273.0);#Temperature at which engine discards heat\n", + "#To Calculate work output, heat rejected, entropy change of system,surronding & total change in entropy and the efficiency of the heat engine\n", + "\n", + "#(i)Calculation of work output\n", + "W = (Q1*((T1-T0)/T1));#Work done umath.sing equations 4.7 & 4.9 given on page no 98\n", + "print \"i)The work output of the heat engine is %f Kcal\"%(W);\n", + "\n", + "#(ii)Calculation of heat rejected\n", + "Q2 = (Q1*T0)/T1;\n", + "print \" ii)The heat rejected is %f Kcal\"%(Q2);\n", + "\n", + "#(iii)Calculation of entropy\n", + "del_S1 = -(Q1/T1);#Change in the entropy of source in Kcal/Kg K\n", + "del_S2 = Q2/T0;#Change in the entropy of math.sink in Kcal/Kg K\n", + "del_St = del_S1+del_S2;#Total change in entropy in Kcal/Kg K\n", + "print \" iii)Total change in entropy is %d confirming that the process is reversible\"%(del_St);\n", + "\n", + "#(iv)Calculation of efficiency\n", + "n = (W/Q1)*100;\n", + "print \" iv)The efficiency of the heat engine is %f percent\"%(n);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The work output of the heat engine is 103.189493 Kcal\n", + " ii)The heat rejected is 146.810507 Kcal\n", + " iii)Total change in entropy is 0 confirming that the process is reversible\n", + " iv)The efficiency of the heat engine is 41.275797 percent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 Page No : 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "T1 = 373.0;#Temperature of the saturated steam in K\n", + "T2 = 298.0;#Temperature of the saturated water in K\n", + "#To calculate the total change in entropy and hence determine the reversibility of the process\n", + "\n", + "#del_H = del_Q+(V*del_P)\n", + "#del_H =del_Q; math.since it is a consmath.tant pressure process\n", + "\n", + "#From steam table,\n", + "#enthalpy of saturated steam at 373K is\n", + "H1 = 6348.5;# in Kcal/Kg\n", + "#enthalpy of saturated liquid water at 373K is\n", + "H2 = 99.15;#in Kcal/Kg\n", + "Q = H2-H1;#heat rejected in Kcal/Kg\n", + "del_S1 = Q/T1;#change in entropy of the system in Kcal/Kg K\n", + "del_S2 = Q/T2;#change in entropy of the surronding in Kcal/Kg K\n", + "del_St = del_S1+del_S2;#total change in the entropy in Kcal/Kg K\n", + "if(del_St == 0):\n", + " print \"Process is reversible\";\n", + "else:\n", + " print \"Process is irreversible\";\n", + "#end\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Process is irreversible\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3 Page No : 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "Cp = 0.09;#specific heat of metal block in Kcal/Kg K\n", + "m = 10.0;#mass of metal block in Kg\n", + "T1 = 323.0;#initial temperature of the block in K\n", + "T2 = 298.0;#final temperature of the block in K\n", + "#consmath.tant pressure process\n", + "#To find out entropy change of block,air and total entropy change\n", + "\n", + "#(i)To calculate the entropy change of block\n", + "del_S1 = m*Cp*math.log(T2/T1);\n", + "print \"i)Entropy change of block is %f Kcal/Kg K\"%(del_S1);\n", + "\n", + "#(ii)To calculate the entropy change of air\n", + "Q = m*Cp*(T1-T2);#heat absorbed by air = heat rejected by block in Kcal\n", + "del_S2 = (Q/T2);\n", + "print \" ii)Entropy change of air is %f Kcal/Kg K\"%(del_S2);\n", + "\n", + "#(iii)To calculate the total entropy change\n", + "del_St = del_S1+del_S2;\n", + "print \" iii)Total entropy change is %f Kcal/Kg K\"%(del_St);\n", + "if(del_St == 0):\n", + " print \" Process is reversible\";\n", + "else:\n", + " print \" Process is irreversible\";\n", + "#end\n", + "#end \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)Entropy change of block is -0.072503 Kcal/Kg K\n", + " ii)Entropy change of air is 0.075503 Kcal/Kg K\n", + " iii)Total entropy change is 0.003000 Kcal/Kg K\n", + " Process is irreversible\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4 Page No : 94" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "m1 = 10 #mass of metal block in Kg\n", + "m2 = 50 #mass of water in Kg\n", + "Cp1 = 0.09 #Specific heat of metal block in Kcal/Kg K\n", + "Cp2 = 1 #Specific heat of water in Kcal/Kg K\n", + "T1 = 50 #Initial temperature of block in deg celsius\n", + "T2 = 25 #Final temperature of block in deg celsius\n", + "\n", + "#To calculate the total change in entropy\n", + "#Heat lost by block = Heat gained by water\n", + "Tf = ((m1*Cp1*T1)+(m2*Cp2*T2))/((m1*Cp1)+(m2*Cp2)) #final temperature of water in deg celsius\n", + "Tf1 = Tf+273.16 #final temperature in K\n", + "del_S1 = m1*Cp1*math.log(Tf1/(T1+273)) #change in entropy of the block in Kcal/K\n", + "del_S2 = m2*Cp2*math.log(Tf1/(T2+273)) #change in entropy of the block in Kcal/K\n", + "del_St = del_S1+del_S2\n", + "print \"The total change entropy is \",\n", + "print \"%.6f\" %del_St,\n", + "print \"Kcal/K\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total change entropy is 0.030226 Kcal/K\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5 Page No : 96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "#Air at 20 deg celsius\n", + "#P1 = 250;initial pressure in atm\n", + "#P2 = 10;final pressure after throttling in atm\n", + "\n", + "#To calculate the entropy change\n", + "#According to the given conditions from figure4.5(page no 103)\n", + "S1 = -0.38;#initial entropy in Kcal/Kg K\n", + "S2 = -0.15;#final entroy in Kcal/Kg K\n", + "del_S = S2-S1;\n", + "print \"Change in entropy for the throttling process is %f Kcal/Kg K\"%(del_S);\n", + "#From figure 4.6(page no 104), the final temperature is -10 deg celsius\n", + "#end \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in entropy for the throttling process is 0.230000 Kcal/Kg K\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7 Page No : 101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "#Basis: 1 hour\n", + "m = 10.0;#mass of air in Kg\n", + "T = 293.0;#Consmath.tant temperature throughout the process in K\n", + "#P1 = 1;#Initial pressure in atm\n", + "#P2 = 30;#Final pressure in atm\n", + "#According to the given data and umath.sing the graph or figure A.2.7 given in page no 105\n", + "S1 = 0.02;#Initial entropy in Kcal/Kg\n", + "S2 = -0.23;#Final entropy in Kcal/Kg\n", + "H1 = 5.0;#Initial enthalpy in Kcal/Kg\n", + "H2 = 3.0;#Final enthalpy in Kcal/Kg\n", + "\n", + "W = -((H2-H1)+T*(S2-S1))*m*(427/(3600*75.0));\n", + "print \"The horse power of the compressor is %f hp\"%(W);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The horse power of the compressor is 1.190065 hp\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8 Page No : 104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "#Basis: 1 Kg of steam\n", + "#P1 = 30;Intial pressure in Kgf/cm**2\n", + "#P2 = 3;Final pressure in Kgf/cm**2\n", + "#T = 300;#Operating temperature\n", + "#From figure A.2.8, \n", + "H1 = 715.0;#Initial enthalpy of steam in Kcal/Kg\n", + "H2 = 625.0;#Final enthalpy of steam in Kcal/Kg\n", + "S1 = 1.56;#Initial entropy of steam in Kcal/Kg K\n", + "S2 = 1.61;#Final entropy of steam in Kcal/Kg K\n", + "Q = -1.0;#heat loss in Kcal/Kg\n", + "To = 298;#The lowest surronding temperature in K\n", + "\n", + "#To calculate the effectiveness of the process\n", + "W = (-(H2-H1)+Q);#Actual work output by the turbine in Kcal\n", + "#The maximum or available work can be calculated from equation 4.14\n", + "del_B = -((H2-H1)-(To*(S2-S1)));# Maximum work that can be obtained in Kcal\n", + "E = (W/del_B)*100.0;\n", + "print \"The effectiveness of the process is %f percent\"%(E);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The effectiveness of the process is 84.842707 percent\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9 Page No : 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "m = 1.0;#mass of liquid water in Kg\n", + "T1 = 1350.0;#initial temperature in deg celsius\n", + "T2 = 400.0;#final temperature in deg celsius\n", + "Cp = 1.0;#Specific heat of water in Kcal/Kg K\n", + "Cpg = 0.2;#Specific heat of combustion gases in Kcal/Kg K\n", + "Hv = 468.35;#Heat of vapourisation at 14 Kgf/cm**2 and 194.16 deg celsius in Kcak/Kg\n", + "To = 298.0;#Surronding temperature\n", + "Tb = 194.16+273;#Boiling point of liquid water\n", + "\n", + "#To Calculate the maximum work obtained and the entropy change\n", + "#(i)Calculation of maximum work\n", + "#Q = del_H = m*Cp*(T2-T1); gas can be assumed to cool at consmath.tant pressure\n", + "#From equation 4.14 (page no 110)\n", + "del_B = -((m*Cpg*(T2-T1))-(To*m*Cp*math.log((T2+273)/(T1+273))));\n", + "print \"i)The maximum work that can be obtained is %f Kcal/Kg of gas\"%(del_B);\n", + "\n", + "#(ii)To Calculate the change in entropy\n", + "del_S =(m*Cp*math.log(Tb/To))+((m*Hv)/Tb);\n", + "print \"ii)The entropy change per Kg of water is %f\"%(del_S);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The maximum work that can be obtained is -72.325299 Kcal/Kg of gas\n", + "ii)The entropy change per Kg of water is 1.452126\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
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