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diff --git a/Thermodynamics_for_Engineers/Chapter_14.ipynb b/Thermodynamics_for_Engineers/Chapter_14.ipynb deleted file mode 100755 index 2bdb47a9..00000000 --- a/Thermodynamics_for_Engineers/Chapter_14.ipynb +++ /dev/null @@ -1,394 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:57baf6d72b0852fbedefdd618b07a013f83f99bf6037a025a931e79491c28af0"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14 - Energies associated with chemical reactions"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 297"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the HHV and LHV at constant volume\n",
- "#Initalization of variables\n",
- "lhs=8.5 #moles of reactants\n",
- "rhs=6 #moles of CO2\n",
- "n=3. #moles of H2O\n",
- "R=1545. #Universal gas constant\n",
- "R2=18.016 #molar mass of water\n",
- "J=778. #Work conversion constant\n",
- "T=537. #R\n",
- "T2=1050.4 #R\n",
- "T3=991.3 #R\n",
- "Qhp=1417041. #Btu/mol\n",
- "#calculations\n",
- "Qhpv=(lhs-rhs)*R*T/J\n",
- "Qhv=Qhp-Qhpv\n",
- "hfg=(rhs-n)*R2*T2\n",
- "Qlp=Qhp-hfg\n",
- "Qlpv=(lhs-rhs-n)*R/J *T\n",
- "Qlv=Qlp-Qlpv\n",
- "Qhlv=(rhs-n)*R2*T3\n",
- "Qlv3=Qhv-Qhlv\n",
- "#results\n",
- "print '%s %d %s' %(\"Higher heating value at constant volume =\",Qhv,\"Btu/mol\")\n",
- "print '%s %d %s' %(\"\\n Lower heating value at constant pressure =\",Qlp,\"Btu/mol\")\n",
- "print '%s %d %s' %(\"\\n In case 1,Lower heating value at constant volume =\",Qlv,\" Btu/mol\")\n",
- "print '%s %d %s' %(\"\\n In case 2,Lower heating value at constant volume =\",Qlv3,\"Btu/mol\")\n",
- "print '%s' %(\"The answers might differ a bit from textbook due to rounding off error.\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Higher heating value at constant volume = 1414374 Btu/mol\n",
- "\n",
- " Lower heating value at constant pressure = 1360268 Btu/mol\n",
- "\n",
- " In case 1,Lower heating value at constant volume = 1360802 Btu/mol\n",
- "\n",
- " In case 2,Lower heating value at constant volume = 1360797 Btu/mol\n",
- "The answers might differ a bit from textbook due to rounding off error.\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 301"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in chemical energy during complete combustion and LHV at constant volume\n",
- "#Initalization of variables\n",
- "print '%s' %(\"From table 5-4,\")\n",
- "no=7.5\n",
- "n1=3.\n",
- "n2=6.\n",
- "Q=1360805. #Btu/mol\n",
- "#calculations\n",
- "Uo=337+no*85\n",
- "Uf=n1*104+n2*118\n",
- "delo= Q-(Uo-Uf)\n",
- "Uo2=1656+no*402\n",
- "Uf2=n1*490+n2*570\n",
- "Qv=Uo2-Uf2+delo\n",
- "#results\n",
- "print '%s %d %s' %(\"Change in chemical energy during complete combustion =\",delo,\"Btu/mol\")\n",
- "print '%s %d %s' %(\"\\n Lower heating value at constant volume =\",Qv,\"Btu/mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 5-4,\n",
- "Change in chemical energy during complete combustion = 1360850 Btu/mol\n",
- "\n",
- " Lower heating value at constant volume = 1360631 Btu/mol\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 302"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the heat removed in the process\n",
- "#Initalization of variables\n",
- "print '%s' %(\"From table 5-4,\")\n",
- "a=1 #moles of C6H6\n",
- "b=7.5 #moles of O2 in reactant\n",
- "c=1.875 #moles of excess O2\n",
- "d=35.27 #moles of N2\n",
- "e=3 #moles of H2O\n",
- "flow=40. #lb/min\n",
- "w=1360850. #Btu/mol\n",
- "#calculations\n",
- "U11=a*337\n",
- "U12=(b+c)*85\n",
- "U13=d*82\n",
- "U14=(a+b+c+d)*1066\n",
- "Ua1=U11+U12+U13+U14\n",
- "U21=c*2539\n",
- "U22=d*2416\n",
- "U23=e*3009\n",
- "U24=2*e*3852\n",
- "U25=(c+d+e+2*e)*1985\n",
- "Ua2=U21+U22+U23+U24+U25\n",
- "Q=Ua1+w-Ua2\n",
- "fuel=flow/(6*12+2.*e)\n",
- "Q2=Q*fuel\n",
- "#results\n",
- "print '%s %d %s' %(\"Heat removed =\",Q2,\"Btu/min\")\n",
- "print '%s' %(\"The answers might differ a bit from textbook due to rounding off error.\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 5-4,\n",
- "Heat removed = 615294 Btu/min\n",
- "The answers might differ a bit from textbook due to rounding off error.\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the furnace efficiency\n",
- "#Initalization of variables\n",
- "rate=10700. #lb/min\n",
- "t2=97.90 \n",
- "t1=33.05 \n",
- "r1=46. #lb/min\n",
- "#calculations\n",
- "print '%s' %(\"From steam tables,\")\n",
- "Hv=1417041.\n",
- "Qw=rate*(t2-t1)\n",
- "Q=r1/(12*6+6) *Hv\n",
- "eff=Qw/Q*100.\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Furnace efficiency =\",eff,\"percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From steam tables,\n",
- "Furnace efficiency = 83.0 percent\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 306"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the thermal efficiency\n",
- "#Initalization of variables\n",
- "rate=94. #lb/hr\n",
- "hp=197. #hp\n",
- "c=8.\n",
- "h=18.\n",
- "Lv=17730. #Btu/hr\n",
- "H=2368089. #Btu/hr\n",
- "#calculations\n",
- "amount=rate*c/12 +h\n",
- "amount=0.824\n",
- "Lvv=H-Lv\n",
- "eff=hp*2544/(amount*Lvv) *100\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Thermal efficiency =\",eff,\" percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal efficiency = 25.88 percent\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - Pg 306"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the thermal efficiency\n",
- "#Initalization of variables\n",
- "rate=94. #lb/hr\n",
- "hp=197. #hp\n",
- "c=8.\n",
- "h=18.\n",
- "mole=9.\n",
- "H=2350359. #Btu/hr\n",
- "#calculations\n",
- "amount=rate*c/12 +h\n",
- "amount=0.824\n",
- "Lvv=H-mole*18.016*1050.4\n",
- "eff=hp*2544/(amount*Lvv) *100\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Thermal efficiency =\",eff,\"percent\")\n",
- "print '%s' %(\"The answer in the textbook is a different due to rounding off error\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal efficiency = 27.90 percent\n",
- "The answer in the textbook is a different due to rounding off error\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - Pg 307"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the total available energy\n",
- "#Initalization of variables\n",
- "import math\n",
- "hv=14000. #Btu/lb\n",
- "ef=0.4\n",
- "tmin=80. #F\n",
- "tmid=300. #F\n",
- "m=13. #lb\n",
- "c=0.27\n",
- "tmean=2300. #F\n",
- "#calculations\n",
- "heat=ef*hv\n",
- "Qavail=heat*(tmean-tmin)/(tmean+460)\n",
- "Q=m*c*(tmean-tmid)\n",
- "Q2=Q- (tmin+460)*m*c*math.log((tmean+460)/(tmid+460))\n",
- "tot=Qavail+Q2\n",
- "#results\n",
- "print '%s %d %s' %(\"Total available energy =\",tot,\" Btu/lb of fuel\")\n",
- "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total available energy = 9079 Btu/lb of fuel\n",
- "The answer is a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - Pg 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the max amount of work available\n",
- "#Initalization of variables\n",
- "print '%s' %(\"From table 14-2,\")\n",
- "G1=55750. #Btu/mol\n",
- "co2=-169580. #Btu/mol\n",
- "h2o=-98290. #Btu/mol\n",
- "#calculations\n",
- "G2=6*co2+3*h2o\n",
- "avail=G1-G2\n",
- "#results\n",
- "print '%s %d %s' %(\"Max. amount of work =\",avail,\"Btu/mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 14-2,\n",
- "Max. amount of work = 1368100 Btu/mol\n"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
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