From 41f1f72e9502f5c3de6ca16b303803dfcf1df594 Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 4 Sep 2015 22:04:10 +0530 Subject: add/remove/update books --- .../Chapter15.ipynb | 686 --------------------- 1 file changed, 686 deletions(-) delete mode 100755 Thermodynamics_An_Engineering_Approach/Chapter15.ipynb (limited to 'Thermodynamics_An_Engineering_Approach/Chapter15.ipynb') diff --git a/Thermodynamics_An_Engineering_Approach/Chapter15.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter15.ipynb deleted file mode 100755 index c167fe73..00000000 --- a/Thermodynamics_An_Engineering_Approach/Chapter15.ipynb +++ /dev/null @@ -1,686 +0,0 @@ -{ - "metadata": { - "name": "" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 15: Chemical Reactions" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-1 ,Page No.755" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#given data\n", - "nO2i=20;#intial moles of air \n", - "nC8H18i=1;#intial moles octane\n", - "\n", - "#from Table A-1\n", - "Mair=29;\n", - "MC=12;\n", - "MH=2;\n", - "\n", - "#calculations\n", - "# Chemical Reaction\n", - "# C8H18 + 20(O2+3.76N2)= xCO2 + yH2O + zO2 + wN2\n", - "#by elemental balance of moles\n", - "x=8;\n", - "y=18/2;\n", - "z=20*2-2*x-y;\n", - "w=20*3.76;\n", - "print'kmoles of CO2 %i'%x;\n", - "print'kmoles of H2O %i'%y;\n", - "print'kmoles of O2 %f'%round(z,1);\n", - "print'kmoles of N2 %f'%round(w,1);\n", - "#thus equn becomes\n", - "# C8H18 + 20(O2+3.76N2)= 8CO2 + 9H2O + 7.5O2 +75.2N2\n", - "AF=nO2i*4.76*Mair/(x*MC + y*MH);\n", - "print'air-fuel ratio of combustion process %f kg air/kg fuel'%round(AF,1)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "kmoles of CO2 8\n", - "kmoles of H2O 9\n", - "kmoles of O2 15.000000\n", - "kmoles of N2 75.200000\n", - "air-fuel ratio of combustion process 24.200000 kg air/kg fuel\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-2 ,Page No.757" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#given data\n", - "P=100;#total pressure in kPa\n", - "\n", - "#from Table A-1\n", - "Mair=29.0;\n", - "MC=12.0;\n", - "MH=2.0;\n", - "\n", - "#calculations\n", - "#Chemical reaction\n", - "#C2H6 + 1.2at(1O2 + 3.76) =2CO2 + 3H2O + 0.2athO2 + (1.2*3.76)athN2\n", - "#ath is the stoichiometric coefficient for air\n", - "#Oxygen balance gives\n", - "# 1.2ath = 2 + 1.5 + 0.2ath\n", - "ath=(2+1.5)/(1.2-0.2);\n", - "AF=(1.2*ath)*4.76*Mair/(2*MC+3*MH);\n", - "print'air-fuel ratio of combustion process %f kg air/kg fuel'%round(AF,1);\n", - "#C2H6 + 4.2(O2 + 3.76N2) = 2CO2 + 3H2O + 0.7O2 + 15.79N2;\n", - "Nprod=2+3+0.7+15.79;\n", - "#for dew point water vapour condenses\n", - "Nv=3;\n", - "Pv=Nv/Nprod*P;\n", - "#at this Pv\n", - "Tdp=52.3;\n", - "print'the dew-point %f C'%Tdp\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "air-fuel ratio of combustion process 19.300000 kg air/kg fuel\n", - "the dew-point 52.300000 C\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-3 ,Page No.758" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#given data\n", - "P=101.325;#total pressure in kPa\n", - "RH=0.8;#realtive humidity\n", - "T1=20;#tempearture of air in C\n", - "\n", - "#from Table A-4\n", - "Psat=2.3392;\n", - "\n", - "#calculations\n", - "#consedering 1 kmol of fuel\n", - "# 0.72CH4 + 0.09H2 + 0.14N2 + 0.02O2 + 0.03CO2 + ath(O2 + 3.76N2) = xCO2 + yH2O + zN2\n", - "#element balance\n", - "x=0.72+0.03\n", - "y=(0.72*4+0.09*2)/2;\n", - "ath=x+y/2-0.02-0.03;\n", - "z=0.14+3.76*ath;\n", - "Pv=RH*Psat;\n", - "# Nv,air = Pv,air/Ptotal * Ntotal\n", - "Nvair=Pv/P*6.97/(1-(Pv/P));\n", - "#0.72CH4 + 0.09H2 + 0.14N2 + 0.02O2 + 0.03CO2 + 1.465(O2 + 3.76N2) + 0.131H20 = 0.75CO2 + 1.661H2O + 5.648N2\n", - "Pvprod=1.661/8.059*P;\n", - "#at this Pvprod\n", - "Tdp=60.9;\n", - "print'the dew-point %f C'%Tdp" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "the dew-point 60.900000 C\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-4 ,Page No.760" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#given data\n", - "Pprod=100;#total pressure in kPa\n", - "\n", - "#from Table A-1\n", - "Mair=29;\n", - "MC=12;\n", - "MH=2;\n", - "\n", - "#from Table A-4\n", - "Psat=3.1698;\n", - "\n", - "#calculations\n", - "#consedering 100 kmol of dry products\n", - "# xC8H18 + a (O2 + 3.76N2) = 10.02CO2 + 0.88C0 + 84.48N2 + bH20\n", - "#from mass balamces\n", - "a=83.48/3.76;\n", - "x=(0.88+10.02)/8;\n", - "b=18*x/2;\n", - "# 1.36C8H18 + 22.2 (O2 + 3.76N2) = 10.02CO2 + 0.88C0 + 84.48N2 + 12.24H20\n", - "# 1 mol conversion\n", - "# C8H18 + 16.32 (O2 + 3.76N2) = 7.37CO2 + 4.13C0 + 61.38N2 + 9H20\n", - "AF= 16.32*4.76*Mair/(8*MC+9*MH);\n", - "print'air-fuel ratio of combustion process %f kg air/kg fuel'%round(AF,2);\n", - "# C8H18 + ath (O2 + 3.76N2) = 8CO2 + 9H2O + 3.76athN2\n", - "ath=8+4.5;\n", - "Pth=16.32/ath*4.76/4.76*100;\n", - "print'percentage of theoretical air is %i'%round(Pth);\n", - "Nprod=7.37+0.65+4.13+61.98+9;\n", - "# Nv/Nprod = Pv/Pprod\n", - "Pv=Psat;\n", - "Nw= (Nprod*Pv-9*Pprod)/(Pv-Pprod);\n", - "print'the amount of H2O that condenses as the products %f kmol'%round(Nw,2)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "air-fuel ratio of combustion process 19.760000 kg air/kg fuel\n", - "percentage of theoretical air is 131\n", - "the amount of H2O that condenses as the products 6.570000 kmol\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-5 ,Page No.764" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#there is a difference in the answer due to approximation in the textbook\n", - "\n", - "#given data\n", - "T=25;#temperature of octane in C\n", - "\n", - "#from Table A-6\n", - "HCO2=-393520;\n", - "HH2O=-285830;\n", - "HC8H18=-249950;\n", - "\n", - "#calculations\n", - "# C8H18 + ath (O2 + 3.76N2) = 8CO2 + 9H2O + 3.76athN2\n", - "#N2 and O2 are stable elements, and thus their enthalpy of formation is zero\n", - "#hc = Hprod - Hreact\n", - "hc= 8*HCO2 + 9*HH2O - HC8H18;\n", - "print'the enthalpy of combustion of liquid octane %i kJ/kmol'%hc\n", - "print 'or %i kJ/kg C8H18'%round(hc/114,0)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "the enthalpy of combustion of liquid octane -5470680 kJ/kmol\n", - "or -47989 kJ/kg C8H18\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-6 ,Page No.767" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#given data\n", - "mfuel=0.05;#mass flow rate in kg/min\n", - "\n", - "#from Table A-1\n", - "Mair=29;\n", - "MC=12;\n", - "MH=2;\n", - "\n", - "#calculation\n", - "#stochiometric reaction\n", - "#C3H8 + ath(O2 + 3.76N2) = 3CO2 + 4H2O + 3.76athN2\n", - "#O2 balance\n", - "ath=3+5;\n", - "#50 percent excess air and some CO in the products\n", - "#C3H8 + 7.5(O2 + 3.76N2) = 2.7CO2 + 0.3CO + 4H2O + 2.65O2+ 28.2N2\n", - "AF=7.5*4.76*Mair/(3*MC+4*MH);\n", - "mair=AF*mfuel;\n", - "print'the mass flow rate of air %f kg air/min'%round(mair,2);\n", - "#from property tables\n", - "#C3H8 designated as p\n", - "hfp=-118910;\n", - "#oxygen as o\n", - "hfo=0;\n", - "ho280=8150;\n", - "ho298=8682;\n", - "ho1500=49292;\n", - "#nitrogen as n\n", - "hfn=0;\n", - "hn280=8141;\n", - "hn298=8669;\n", - "hn1500=47073;\n", - "#water as w\n", - "hfw=-241820;\n", - "hw298=9904;\n", - "hw1500=57999;\n", - "#carbondioxode as c\n", - "hfc=-393520;\n", - "hc298=9364;\n", - "hc1500=71078;\n", - "#carbon monoxide as co\n", - "hfco=-110530;\n", - "hco298=8669;\n", - "hco1500=47517;\n", - "qout=1*(hfp)+7.5*(hfo+ho280-ho298)+28.2*(hfn+hn280-hn298)-2.7*(hfc+hc1500-hc298)-0.3*(hfco+hco1500-hco298)-4*(hfw+hw1500-hw298)-2.65*(hfo+ho1500-ho298)-28.2*(hfn+hn1500-hn298);\n", - "#for kg of propane\n", - "qout=qout/44;\n", - "Qout=mfuel*qout/60;\n", - "print'the rate of heat transfer from the combustion chamber %f kW'%round(Qout,2)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "the mass flow rate of air 1.180000 kg air/min\n", - "the rate of heat transfer from the combustion chamber 6.890000 kW\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-7 ,Page No.769" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#the 0.175% error in last part is due to the approximation in the textbook\n", - "\n", - "#given data\n", - "Preact=1.0;#total pressure in kPa\n", - "Treact=77+460.0;#reaction temperature in R\n", - "Tprod=1800.0;#final temperature in R\n", - "\n", - "#constants used\n", - "Ru=1.986;\n", - "\n", - "#calculation\n", - "#CH4 + 3O2 = CO2 + 2H2O + O2\n", - "Nreact=4;\n", - "Nprod=4;\n", - "Pprod=Preact*Nprod/Nreact*Tprod/Treact;\n", - "print'the final pressure in the tank %f atm'%round(Pprod,2);\n", - "#from std. values of heat of formation and ideal gasses in Appendix\n", - "#CH4 as m\n", - "hfm=-32210.0;\n", - "#O2 as o\n", - "hfo=0;\n", - "h537o=3725.1;\n", - "h1800o=13485.8;\n", - "#water as w\n", - "hfw=-104040.0;\n", - "h537w=4528.0;\n", - "h1800w=15433.0\n", - "#carbondioxide as c\n", - "hfc=-169300.0;\n", - "h537c=4027.5;\n", - "h1800c=18391.5;\n", - "Qout=1*(hfm-Ru*Treact)+3*(hfo-Ru*Treact)-1*(hfc+h1800c-h537c-Ru*Tprod)-2*(hfw+h1800w-h537w-Ru*Tprod)-1*(hfo+h1800o-h537o-Ru*Tprod);\n", - "print'the heat transfer during this process %i Btu/lbmol'%round(Qout)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "the final pressure in the tank 3.350000 atm\n", - "the heat transfer during this process 309269 Btu/lbmol\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-8 ,Page No.771" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#this invovles EES hence the below code explains a approach with approximation\n", - "\n", - "#calculations\n", - "\n", - "#part - a\n", - "#C8H18 + 12.5 (O2 + 3.76N2) = 8CO+ 9H2O + 47N2\n", - "#from std. values of heat of formation and ideal gasses in Appendix\n", - "#octane as oc\n", - "hfoc=-249950.0;\n", - "#oxygen as o\n", - "hfo=0;\n", - "h298o=8682.0;\n", - "#nitrogen as n\n", - "hfn=0;\n", - "h298n=8669.0;\n", - "#water as w\n", - "hfw=-241820.0;\n", - "h298w=9904.0;\n", - "#carbondioxide as c\n", - "hfc=-393520.0;\n", - "h298c=9364.0;\n", - "#x refers to 8hCO2 + 9hH20 + 47hN2\n", - "xac=1*(hfoc)+8*(h298c-hfc)+9*(h298w-hfw)+47*(h298n-hfn);\n", - "#from EES the Tprod is determined by trial and error\n", - "#at 2400K\n", - "x2400=5660828.0;\n", - "#at 2350K\n", - "x2350=5526654.0;\n", - "#the actual value of x is xac and T can be determined by interpolation\n", - "Tprod=(xac-x2350)*(2400.0-2350.0)/(x2400-x2350)+2350.0;\n", - "print'adiabatic flame temperature for complete combustion with 100 percent theoretical air %i K'%round(Tprod);\n", - "\n", - "#part - b\n", - "#C8H18 + 50 (O2 + 3.76N2) = 8CO+ 9H2O + 37.5O2 + 188N2\n", - "#solved similarly using EES and approximation and interpolation\n", - "#similarly we can solve the part - c \n", - "#the above concept is applied\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "adiabatic flame temperature for complete combustion with 100 percent theoretical air 2395 K\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-9 ,Page No.776" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#from Table A-26E\n", - "#Gibbs function of formation at 77\u00b0F\n", - "gfc=0;#for carbon\n", - "gfo=0;#for oxygen\n", - "gfco=-169680;#for carbondioxide\n", - "\n", - "#calculations\n", - "# C + O2 = CO2\n", - "Wrev=1*gfc+1*gfo-1*gfco;\n", - "print'the reversible work for this process %i Btu'%round(Wrev) " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "the reversible work for this process 169680 Btu\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-10 ,Page No.777" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "from math import log\n", - "\n", - "#given values\n", - "T0=298;#combustion chamber temperature in K\n", - "\n", - "#contansts used \n", - "Ru=8.314;#in kJ/kmol K\n", - "\n", - "#calculations\n", - "# CH4 + 3(O2 + 3.76N2) = CO2 + 2H2O + O2 + 11.28N2\n", - "#from std. values of heat of formation and ideal gasses in Appendix\n", - "#methane as m\n", - "hfm=-74850;\n", - "#oxygen as o\n", - "hfo=0;\n", - "h298o=8682;\n", - "#nitrogen as n\n", - "hfn=0;\n", - "h298n=8669;\n", - "#water as w\n", - "hfw=-241820;\n", - "h298w=9904;\n", - "#carbondioxide as c\n", - "hfc=-393520;\n", - "h298c=9364;\n", - "#x refers to hCO2 + 2hH2O + 11.28hN2\n", - "xac=1*(hfm)+1*(h298c-hfc)+2*(h298w-hfw)+11.28*(h298n-hfn);\n", - "#from EES the Tprod is determined by trial and error\n", - "Tprod=1789;\n", - "print'the temperature of the products %i K'%round(Tprod);\n", - "#entropy calculations by using table A-26\n", - "#Si = Ni*(si - Ruln yiPm\n", - "#reactants\n", - "Sm=1*(186.16-Ru*log(1*1));\n", - "So=3*(205.04-Ru*log(0.21*1));\n", - "Sn=11.28*(191.61-Ru*log(.79*1));\n", - "Sreact=Sm+So+Sn;\n", - "#products\n", - "Nt=1+2+1+11.28;#total moles\n", - "yc=1/Nt;\n", - "yw=2/Nt;\n", - "yo=1/Nt;\n", - "yn=11.28/Nt;\n", - "Sc=1*(302.517-Ru*log(yc*1));\n", - "Sw=2*(258.957-Ru*log(yw*1));\n", - "So=1*(264.471-Ru*log(yo*1));\n", - "Sn=11.28*(247.977-Ru*log(yn*1));\n", - "Sprod=Sc+Sw+So+Sn;\n", - "Sgen=Sprod-Sreact;\n", - "print'exergy destruction %i kJ/kmol - K'%round(Sgen);\n", - "Xdestroyed=T0*Sgen/1000;#factor of 1000 for converting kJ to MJ\n", - "print'%i MJ/kmol'%round(Xdestroyed);\n", - "#This process involves no actual work. Therefore, the reversible work and energy destroyed are identical\n", - "Wrev=Xdestroyed;\n", - "print'the reversible work %i MJ/kmol'%round(Wrev)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "the temperature of the products 1789 K\n", - "exergy destruction 966 kJ/kmol - K\n", - "288 MJ/kmol\n", - "the reversible work 288 MJ/kmol\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15-11 ,Page No.778" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "from math import log\n", - "\n", - "#given values\n", - "Tsurr=298;#temperature of surroundings in K\n", - "\n", - "#contansts used \n", - "Ru=8.314;#in kJ/kmol K\n", - "\n", - "#calculations\n", - "\n", - "#part - a\n", - "# CH4 + 3(O2 + 3.76N2) = CO2 + 2H2O + O2 + 11.28N2\n", - "#The amount of water vapor that remains in the products is determined as in Example 15\u20133\n", - "Nv=0.43;#moles of water vapour\n", - "Nw=1.57;#moles of water in liquid\n", - "#hf values\n", - "#methane as m\n", - "hfm=-74850;\n", - "#carbondioxide as c\n", - "hfc=-393520;\n", - "#water vapour as v\n", - "hfv=-241820;\n", - "#water in liquid as w\n", - "hfw=-285830;\n", - "Qout=1*hfm-1*hfc-Nv*hfv-Nw*hfw;\n", - "print'Qout = %i kJ/kmol'%round(Qout)\n", - "\n", - "#part - b\n", - "#entropy calculations by using table A-26\n", - "#Si = Ni*(si - Ruln yiPm\n", - "#reactants\n", - "Sm=1*(186.16-Ru*log(1*1));\n", - "So=3*(205.04-Ru*log(0.21*1));\n", - "Sn=11.28*(191.61-Ru*log(.79*1));\n", - "Sreact=Sm+So+Sn;\n", - "#products\n", - "Nt=Nv+1+1+11.28;#total moles\n", - "yw=1;\n", - "yc=1/Nt;\n", - "yv=Nv/Nt;\n", - "yo=1/Nt;\n", - "yn=11.28/Nt;\n", - "Sw=Nw*(69.92-Ru*log(yw*1));\n", - "Sc=1*(213.80-Ru*log(yc*1));\n", - "Sv=Nv*(188.83-Ru*log(yv*1));\n", - "So=1*(205.04-Ru*log(yo*1));\n", - "Sn=11.28*(191.61-Ru*log(yn*1));\n", - "Sprod=Sc+Sw+So+Sn+Sv;\n", - "Sgen=Sprod-Sreact+Qout/Tsurr;\n", - "print'Sgen = %i kJ/kmol - K'%round(Sgen);\n", - "Xdestroyed=Tsurr*Sgen/1000;#factor of 1000 for converting kJ to MJ\n", - "print'exergy destruction %i MJ/kmol'%round(Xdestroyed);\n", - "#This process involves no actual work. Therefore, the reversible work and energy destroyed are identical\n", - "Wrev=Xdestroyed;\n", - "print'the reversible work %i MJ/kmol'%round(Wrev)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Qout = 871406 kJ/kmol\n", - "Sgen = 2746 kJ/kmol - K\n", - "exergy destruction 818 MJ/kmol\n", - "the reversible work 818 MJ/kmol\n" - ] - } - ], - "prompt_number": 4 - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit