{ "metadata": { "name": "", "signature": "sha256:85497d817768e6e1d81546cc4adcb855eed2a97589497e4f8a3f22d0bc4a6c3d" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "\n", "Chapter 23 : Calculation of Enthalpy Changes" ] }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 23.1 Page no. 686\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "%matplotlib inline\n", "from matplotlib.pyplot import *\n", "\n", "# Variables\n", "# Given\n", "x_Tl = [90,92,97,100] ;\t\t\t# Temperature of saturated liquid- [degree C]\n", "x_Tg = [100,102,107,110] ;\t\t\t# Temperature of saturated vapour- [degree C]\n", "y_Hl = [376.9,385.3,406.3,418.6] ;\t\t\t# Enthalpy change of saturated liquid -[kJ/kg]\n", "y_Hg = [2256.44,2251.2,2237.9,2229.86] ;\t\t\t# Enthalpy change of saturated vapour -[kJ/kg]\n", "\n", "# Results\n", "plot(x_Tl,y_Hl,x_Tg,y_Hg);\n", "show()\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Populating the interactive namespace from numpy and matplotlib\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ " Example 23.2 page no. 687\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math\n", "\n", "# Variables\n", "# Basis : 1 g mol\n", "R = 8.314 * 10**-3 ;\t\t\t# Ideal gas constant -[kJ/(g mol * K)]\n", "Hv = 30.20 ;\t\t\t# Experimental value of heat of vaporization of acetone -[kJ/g] \n", "\n", "# additional needed data for acetone from Appendix D\n", "T = 329.2 ;\t\t\t# Normal boiling point of acetone - [K]\n", "Tc = 508.0 ;\t\t\t# Critical temperature of acetone - [K]\n", "Pc = 47.0 ;\t\t\t# Critical presure of acetone -[atm]\n", "\n", "# Calculations and Results\n", "Tbc = T/Tc ;\t\t\t# variable required in etimation equations\n", "lnPc = math.log(Pc) ;\t\t\t# variable required in etimation equations\n", "\n", "B = 2940.46 ;\n", "C = -35.93 ;\n", "\n", "del_Hv1 = (R*B*T**2)/((C+T)**2) ;\t\t\t#Heat of vapourization -[kJ/g]\n", "d1 = (abs(Hv - del_Hv1)*100)/Hv ;\t\t\t# differece of experimental and calculated value -[%]\n", "print '(a) Heat of vapourization of acetone is %.2f kJ/g mol. And differece of experimental and calculated value is %.1f %% . '%(del_Hv1,d1);\n", "\n", "del_Hv2 = R*T*((3.978*Tbc - 3.938 +1.555*lnPc)/(1.07 - Tbc)) ;\t\t\t#Heat of vapourization -[kJ/g]\n", "d2 = (abs(Hv - del_Hv2)*100)/Hv ;\t\t\t# differece of experimental and calculated value -[%]\n", "print ' (b) Heat of vapourization of acetone is %.2f kJ/g mol. And differece of experimental and calculated value is %.1f %% . '%(del_Hv2,d2);\n", "\n", "\n", "del_Hv3 = 1.093*R*Tc*((Tbc*(lnPc-1))/(0.93-Tbc)) ;\t\t\t#Heat of vapourization -[kJ/g]\n", "d3 = (abs(Hv - del_Hv3)*100)/Hv ;\t\t\t# differece of experimental and calculated value -[%]\n", "print ' (c) Heat of vapourization of acetone is %.2f kJ/g mol. And differece of experimental and calculated value is %.1f %% . '%(del_Hv3,d3);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a) Heat of vapourization of acetone is 30.80 kJ/g mol. And differece of experimental and calculated value is 2.0 % . \n", " (b) Heat of vapourization of acetone is 30.01 kJ/g mol. And differece of experimental and calculated value is 0.6 % . \n", " (c) Heat of vapourization of acetone is 30.24 kJ/g mol. And differece of experimental and calculated value is 0.1 % . \n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "\n", "Example 23.3 Page no. 693\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Variables\n", "c = 2.675*10**4 #*.4536)/(1055*1.8) ;\n", "d = 42.27#*.4536)/(1055*1.8) ;\n", "e = 1.425*10**-2#*.4536)/(1055*1.8) ;\n", "# Calculations\n", "#Now convert Tk (Temperature in K) to TF (temperature in F) to get answer of form x + yT - zT**2,where\n", "x = c + d*460/1.8 - e*((460/1.8)**2) ;\n", "y = d/1.8;\n", "z = e/(1.8*1.8) ;\n", "\n", "# Results\n", "print 'The required answer is %.2e + (%.2e)T - (%.3e) T**2 Btu/(lb mol*F) , where T is in degree F . '%(x,y,z)\n", "\n", "print \"Note answer in textbook seems wrong by order of 10^-3\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The required answer is 3.66e+04 + (2.35e+01)T - (4.398e-03) T**2 Btu/(lb mol*F) , where T is in degree F . \n", "Note answer in textbook seems wrong by order of 10^-3\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ " Example 23.4 page no. 694\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Variables\n", "# Take all 18 experimenta data in an array Cp\n", "Cpi = [39.87,39.85,39.90,45.16,45.23,45.17,50.72,51.03,50.90,56.85,56.80,57.02,63.01,63.09,63.14,69.52,69.68,69.63] ;\t\t\t# Array of Cpi(Heat capacity) values\n", "# Take corresponding temperatures in array T\n", "Ti = [300,300,300,400,400,400,500,500,500,600,600,600,700,700,700,800,800,800] ;\t\t\t# array of Ti\n", "Ti_sqr = [300**2,300**2,300**2,400**2,400**2,400**2,500**2,500**2,500**2,600**2,600**2,600**2,700**2,700**2,700**2,800**2,800**2,800**2] ;\t\t\t# array of Ti**2\n", "Ti_cub = [300**3,300**3,300**3,400**3,400**3,400**3,500**3,500**3,500**3,600**3,600**3,600**3,700**3,700**3,700**3,800**3,800**3,800**3];\t\t\t# array of Ti**3\n", "Ti_qd = [300**4,300**4,300**4,400**4,400**4,400**4,500**4,500**4,500**4,600**4,600**4,600**4,700**4,700**4,700**4,800**4,800**4,800**4];\t\t\t# array of Ti**4\n", "Cpi_Ti = [39.87*300,39.85*300,39.90*300,45.16*400,45.23*400,45.17*400,50.72*500,51.03*500,50.90*500,56.85*600,56.80*600,57.02*600,63.01*700,63.09*700,63.14*700,69.52*800,69.68*800,69.63*800] ;\t\t\t# Array of Cpi(Heat capacity)*Ti values\n", "Cpi_Ti_sqr = [39.87*300**2,39.85*300**2,39.90*300**2,45.16*400**2,45.23*400**2,45.17*400**2,50.72*500**2,51.03*500**2,50.90*500**2,56.85*600**2,56.80*600**2,57.02*600**2,63.01*700**2,63.09*700**2,63.14*700**2,69.52*800**2,69.68*800**2,69.63*800**2] ;\t\t\t# Array of Cpi(Heat capacity)*Ti**2 values\n", "\n", "n = 18. ;\t\t\t# Number of data\n", "# Calculations\n", "\n", "from numpy import matrix\n", "# Solve equations (a),(b) & (c) simultaneously using matrix\n", "a = matrix([[n,sum(Ti),sum(Ti_sqr)],[sum(Ti),sum(Ti_sqr),sum(Ti_cub)],[sum(Ti_sqr),sum(Ti_cub),sum(Ti_qd)]]) ;\t\t\t# Matrix of coefficients of unknown\n", "b = matrix([[sum(Cpi)],[sum(Cpi_Ti)],[sum(Cpi_Ti_sqr)]]) ;\t\t\t# Matrix of constants\n", "x = (a)**-1 * b ;\t\t\t# Matrix of solutions a = x(1), b = x(2) , c = x(3) \n", "\n", "# Results\n", "print 'The solution is Cp = %.2f + %.3e T + %.2e T**2 .Therefore coefficients are as follows :'%(x[0],x[1],x[2])\n", "print ' a = %.2f. b = %.3e . c = %.2e .'%(x[0],x[1],x[2])\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The solution is Cp = 25.44 + 4.371e-02 T + 1.44e-05 T**2 .Therefore coefficients are as follows :\n", " a = 25.44. b = 4.371e-02 . c = 1.44e-05 .\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 23.5 page no : 695" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "from scipy.integrate import quad\n", "# Variables\n", "# Basis : 1 g mol of gas\n", "#Given\n", "T1 = 550. ;\t\t\t# Initial temperature - [degree F]\n", "T2 = 200. ;\t\t\t# Final temperature - [degree F]\n", "CO2 = 9.2/100 ;\t\t\t# Mole fraction \n", "CO = 1.5/100 ;\t\t\t# Mole fraction \n", "O2 = 7.3/100 ;\t\t\t# Mole fraction \n", "N2 = 82.0/100 ;\t\t\t#Mole fraction \n", "\n", "# Calculations\n", "# Additional data needed :\n", "a_N2 = 6.895;\t\t\t# constant\n", "b_N2 = 0.7624*10**-3;\t\t\t# coefficient of T\n", "c_N2 = -0.7009*10**-7;\t\t\t# coefficient of square T\n", "a_O2 = 7.104 ;\t\t\t# constant\n", "b_O2 = (0.7851*10**-3);\t\t\t# coefficient of T\n", "c_O2 = (-0.5528*10**-7); \t\t\t# coefficient of square T\n", "a_CO2 = 8.448;\t\t\t# constant\n", "b_CO2 = 5.757*10**-3;\t\t\t# coefficient of T\n", "c_CO2 = -21.59*10**-7;\t\t\t# coefficient of square T\n", "d_CO2 = 3.059*10**-10;\t\t\t# coefficient of cubic T\n", "a_CO = 6.865 ;\t\t\t# constant\n", "b_CO = 0.8024*10**-3;\t\t\t# coefficient of T\n", "c_CO = -0.7367*10**-7; \t\t\t# coefficient of square T\n", "\n", "# New coefficients after multiplying mole fraction of each component\n", "a1_N2 = 6.895*N2 ;\t\t\t# constant\n", "b1_N2 = N2*0.7624*10**-3; \t\t\t# coefficient of T\n", "c1_N2 = (-0.7009*10**-7)*N2; \t\t\t# coefficient of square T \n", "a1_O2 = 7.104*O2 ;\t\t\t# constant\n", "b1_O2 = (0.7851*10**-3)*O2;\t\t\t# coefficient of T\n", "c1_O2 = (-0.5528*10**-7)*O2; \t\t\t# coefficient of square T\n", "a1_CO2 = 8.448*CO2;\t\t\t# constant\n", "b1_CO2 = (5.757*10**-3)*CO2;\t\t\t# coefficient of T\n", "c1_CO2 = (-21.59*10**-7)*CO2; \t\t\t# coefficient of square T\n", "d1_CO2 = (3.059*10**-10)*CO2; \t\t\t# coefficient of cubic T\n", "a1_CO = 6.865*CO;\t\t\t# constant\n", "b1_CO = (0.8024*10**-3)*CO;\t\t\t# coefficient of T\n", "c1_CO = (-0.7367*10**-7)*CO; \t\t\t# coefficient of square T\n", "\n", "# Get net coefficients of T , square T and cubic T by adding them\n", "a_net = a1_N2+a1_CO2+a1_CO+a1_O2; \t\t\t#Net constant\n", "b_net = b1_N2+b1_CO2+b1_CO+b1_O2; \t\t\t#Net coefficient of T\n", "c_net = c1_N2+c1_CO2+c1_CO+c1_O2 ;\t\t\t#Net coefficient of square T\n", "d_net = d1_CO2;\t\t\t#Net coefficient of cubic T\n", "\n", "def f(T):\n", " return (a_net )+( b_net*T) + (c_net*(T**2)) + (d_net*(T**3))\n", " \n", "del_H = quad(f,T1,T2)[0] \t\t\t# Change in enthalpy of gas over given range-[Btu/lb mol gas]\n", "\n", "# Results\n", "print ' Change in enthalpy of gas over given range is %.0f Btu/lb mol gas . '%del_H\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Change in enthalpy of gas over given range is -2616 Btu/lb mol gas . \n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ " Example 23.6 page no. 700 \n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "\n", "# Solution \n", "#Given\n", "N2 = 1. ;\t\t\t# Moles of N2 - [kg mol]\n", "P = 100. ;\t\t\t# Pressure of gas - [kPa] \n", "T1 = 18. ;\t\t\t# Initial temperature - [degree C]\n", "T2 = 1100. ;\t\t\t# Final temperature - [degree C]\n", "\n", "# Calculations\n", "# In the book it is mentioned to use tables in Appendix D6 to calculate enthalpy change, we get \n", "H_T1 = 0.524;\t\t\t# Initial enthalpy -[kJ/kg mol]\n", "H_T2 = 34.715 ;\t\t\t# Final enthalpy - [kJ/kg mol]\n", "del_H = H_T2 - H_T1 ;\t\t\t# Change in enthalpy - [kJ/kg]\n", "\n", "# Results\n", "print ' Change in enthalpy of N2 over given range is %.3f kJ/kg mol N2 . '%del_H\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Change in enthalpy of N2 over given range is 34.191 kJ/kg mol N2 . \n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ " Example 23.7 page no. 701\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "#Given\n", "T1 = 640. ;\t\t\t# Initial temperature -[degree F]\n", "T2 = 480. ;\t\t\t# Final temperature -[degree F]\n", "P1 = 92. ;\t\t\t# Initial pressure -[psia]\n", "P2 = 52. ;\t\t\t# Final pressure - [psia]\n", "\n", "\n", "#From steam table\n", "#At 90 psia\n", "H1_600 = 1328.7 ;\t\t\t#H at 90 psia and 600 F-[Btu/lb]\n", "H1_700 = 1378.1 ;\t\t\t#H at 90 psia and 700 F-[Btu/lb]\n", "H2_600 = 1328.4 ;\t\t\t#H at 95 psia and 600 F-[Btu/lb]\n", "H2_700 = 1377.8 ;\t\t\t#H at 95 psia and 700 F-[Btu/lb]\n", "\n", "# Calculations\n", "H3_600 = H1_600+ ((H2_600-H1_600)/(95.-90))*(92-90);\t\t\t#H at 92 psia and 600 F-[Btu/lb]\n", "H3_700 = H1_700+ ((H2_700-H1_700)/(95.-90))*(92-90);\t\t\t#H at 92 psia and 700 F-[Btu/lb]\n", "H3_640 = H3_600+((H3_700-H3_600)/(700.-600))*(640-600);\t\t\t#H at 92 psia and 640 F-[Btu/lb]\n", "\n", "H1_450 = 1258.7 ;\t\t\t#H at 50 psia and 450 F-[Btu/lb]\n", "H1_500 = 1282.6 ;\t\t\t#H at 50 psia and 500 F-[Btu/lb]\n", "H2_450 = 1258.2 ;\t\t\t#H at 55 psia and 450 F-[Btu/lb]\n", "H2_500 = 1282.2 ;\t\t\t#H at 55 psia and 500 F-[Btu/lb]\n", "H3_450 = H1_450+ ((H2_450-H1_450)/(55.-50))*(52-50) ;\t\t\t#H at 52 psia and 450 F-[Btu/lb]\n", "H3_500 = H1_500+ ((H2_500-H1_500)/(55.-50))*(52-50);\t\t\t#H at 52 psia and 500 F-[Btu/lb]\n", "H3_480 = H3_450+((H3_500-H3_450)/(500.-450))*(480-450);\t\t\t# H at 52 psia and 480 F-[Btu/lb]\n", "del_H = H3_480 - H3_640;\t\t\t# Change in enthalpy - [Btu/lb]\n", "\n", "# Results\n", "print 'Change in enthalpy is %.1f Btu/lb .'%del_H\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Change in enthalpy is -75.5 Btu/lb .\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ " Example 23.8 page no. 702\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "\n", "# Solution \n", "\n", "# Variables\n", "W = 4. ;\t\t\t# Mass of water -[kg]\n", "Ti= 27.+273 ;\t\t\t# Initial temperature -[K]\n", "Pi = 200. ;\t\t\t# Initial pressure -[kPa]\n", "Pf = Pi ;\t\t\t# Final pressure -[kPa]\n", "V1 = 0.001004 ;\t\t\t# Specific volume at Ti -[cubic metre/kg]\n", "V2 = 1000. * V1 ;\t\t\t#Specific volume at final temperature(Tf) from given condition in problem - [cubic metre/kg]\n", "va = 0.9024 ;\t\t\t# Specific volume -[cubic metre/kg]\n", "Ta = 400. ;\t\t\t# [K]\n", "vb = 1.025 ;\t\t\t# Specific volume -[cubic metre/kg]\n", "Tb = 450. ;\t\t\t#[K]\n", "vf = V2 ;\t\t\t# Final specific volume -[cubic metre/kg]\n", " \n", "# Calculations\n", "m=(Tb - Ta)/(vb - va);\t\t\t# slope \n", "Tf=Ta + m*(vf - va) ;\t\t\t# Final temperature - [K]\n", "\n", "# Results\n", "print ' Final temperature is %.0f K.'%Tf\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Final temperature is 441 K.\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ " Example 23.9 page no. 704\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Solution \n", "\n", "# Variables\n", "mv = 1. ;\t\t\t# Mass of saturated vapour - [lb]\n", "P1 = 2. ;\t\t\t# Initial pressure -[atm]\n", "P2 = 20. ;\t\t\t# Final pressure -[atm]\n", "H_2 = 179. ;\t\t\t# Specific enthalpy at 2 atm - [Btu/lb]\n", "H_20 = 233. ;\t\t\t# Specific enthalpy at 20 atm - [Btu/lb]\n", "V_2 = 3.00 ;\t\t\t# Specific volume at 2 atm - [cubic feet/lb]\n", "V_20 = 0.30 ;\t\t\t# Specific volume at 20 atm - [cubic feet/lb]\n", "T_2 = 72. ;\t\t\t# Temperature at 2 atm -[degree F]\n", "T_20 = 239. ;\t\t\t# Temperature at 20 atm -[degree F]\n", "\n", "# Calculations\n", "del_H = H_20 - H_2 ;\t\t\t# Change in specific enthalpy -[Btu/lb] \n", "del_V = V_20 - V_2 ;\t\t\t# Change in specific volume -[cubic feet/lb] \n", "del_T = T_20 - T_2 ;\t\t\t# Change in temperature -[degree F]\n", "\n", "# Results\n", "print '(a) Change in specific enthalpy is %.0f Btu/lb.'%del_H\n", "print ' (b) Change in specific volume is %.2f cubic feet/lb.'%del_V\n", "print ' (c) Change in temperature is %.1f degree F.'%del_T\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a) Change in specific enthalpy is 54 Btu/lb.\n", " (b) Change in specific volume is -2.70 cubic feet/lb.\n", " (c) Change in temperature is 167.0 degree F.\n" ] } ], "prompt_number": 9 }, { "cell_type": "code", "collapsed": true, "input": [], "language": "python", "metadata": {}, "outputs": [], "prompt_number": 9 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }