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diff --git a/Thermodynamics_An_Engineering_Approach/Chapter13.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter13.ipynb deleted file mode 100755 index 936d90ae..00000000 --- a/Thermodynamics_An_Engineering_Approach/Chapter13.ipynb +++ /dev/null @@ -1,453 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13: Gas Mixtures"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13-1 ,Page No.683"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given data\n",
- "mO2=3.0;#moles of oxygen\n",
- "mN2=5.0;#moles of nitrogen\n",
- "mCH4=12.0;#moles of methane\n",
- "\n",
- "#molecular masses in kg\n",
- "MO2=32.0;\n",
- "MN2=28.0;\n",
- "MCH4=16.0;\n",
- "\n",
- "#constants used\n",
- "Ru=8.314;#in kJ/kg - K\n",
- "\n",
- "#calculations\n",
- "\n",
- "#part - a\n",
- "mm=mO2+mN2+mCH4;\n",
- "mfO2=mO2/mm;\n",
- "mfN2=mN2/mm;\n",
- "mfCH4=mCH4/mm;\n",
- "print'mass fraction of oxygen is %f'%round(mfO2,2);\n",
- "print'mass fraction of nitrogen is %f'%round(mfN2,2);\n",
- "print'mass fraction of methane is %f'%round(mfCH4,2);\n",
- "\n",
- "#part - b\n",
- "NO2=mO2/MO2;\n",
- "NN2=mN2/MN2;\n",
- "NCH4=mCH4/MCH4;\n",
- "Nm=NO2+NN2+NCH4;\n",
- "yO2=NO2/Nm;\n",
- "yN2=NN2/Nm;\n",
- "yCH4=NCH4/Nm;\n",
- "print'mole fraction of oxygen is %f'%round(yO2,3);\n",
- "print'mole fraction of nitrogen is %f'%round(yN2,3);\n",
- "print'mole fraction of methane is %f'%round(yCH4,3);\n",
- "\n",
- "#part - c\n",
- "Mm=mm/Nm;\n",
- "print'average molecular mass %f kg/kmol'%round(Mm,1);\n",
- "Rm=Ru/Mm;\n",
- "print'gas constant of mixture %f kJ/kg - K'%round(Rm,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "mass fraction of oxygen is 0.150000\n",
- "mass fraction of nitrogen is 0.250000\n",
- "mass fraction of methane is 0.600000\n",
- "mole fraction of oxygen is 0.092000\n",
- "mole fraction of nitrogen is 0.175000\n",
- "mole fraction of methane is 0.734000\n",
- "average molecular mass 19.600000 kg/kmol\n",
- "gas constant of mixture 0.425000 kJ/kg - K\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13-2 ,Page No.687"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given data\n",
- "NN2=2.0;#moles of nitrogen\n",
- "NCO2=6.0;#moles of carbon dioxide\n",
- "Tm=300.0;#temperature of gases in K\n",
- "Pm=15000.0;#pressure of gases in kPa\n",
- "\n",
- "#constants used\n",
- "Ru=8.314;#in kJ/kmol - K\n",
- "\n",
- "#calculations\n",
- "\n",
- "#part - a\n",
- "Nm=NN2+NCO2;\n",
- "Vm=Nm*Ru*Tm/Pm;\n",
- "print'the volume of the tank on the basis of the ideal-gas equation of state %f m^3'%round(Vm,3);\n",
- "\n",
- "#part - b\n",
- "#from Table A-1\n",
- "#for nitrogen\n",
- "TcrN=126.2;\n",
- "PcrN=3390;\n",
- "#for Carbondioxide\n",
- "TcrC=304.2;\n",
- "PcrC=7390;\n",
- "yN2=NN2/Nm;\n",
- "yCO2=NCO2/Nm;\n",
- "Tcr=yN2*TcrN+yCO2*TcrC;\n",
- "Pcr=yN2*PcrN+yCO2*PcrC;\n",
- "Tr=Tm/Tcr;\n",
- "Pr=Pm/Pcr;\n",
- "#from Fig A-15b\n",
- "Zm=0.49;\n",
- "Vm=Zm*Nm*Ru*Tm/Pm;\n",
- "print'the volume of the tank on the basis Kay\u2019s rule %f m^3'%round(Vm,3);\n",
- "\n",
- "#part - c\n",
- "#for nitrogen\n",
- "TrN=Tm/TcrN;\n",
- "PrN=Pm/PcrN;\n",
- "#from Fig A-15b\n",
- "Zn=1.02;\n",
- "#for Carbondioxide\n",
- "TrC=Tm/TcrC;\n",
- "PcrC=Pm/PcrC;\n",
- "#from Fig A-15b\n",
- "Zc=0.3;\n",
- "Zm=yN2*Zn+yCO2*Zc;\n",
- "Vm=Zm*Nm*Ru*Tm/Pm;\n",
- "print'the volume of the tank on the basis compressibility factors and Amagat\u2019s law %f m^3'%round(Vm,3);\n",
- "\n",
- "#part - d\n",
- "VRN=(Vm/NN2)/(Ru*TcrN/PcrN);\n",
- "VRC=(Vm/NCO2)/(Ru*TcrC/PcrC);\n",
- "#from Fig A-15b\n",
- "Zn=0.99;\n",
- "Zc=0.56;\n",
- "Zm=yN2*Zn+yCO2*Zc;\n",
- "Vm=Zm*Nm*Ru*Tm/Pm;\n",
- "#When the calculations are repeated we obtain 0.738 m3 after the second iteration, 0.678 m3 after the third iteration, and 0.648 m3 after the fourth iteration.\n",
- "Vm=0.648;\n",
- "print'compressibility factors and Dalton\u2019s law the volume of the tank on the basis %f m^3'%round(Vm,3);\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the volume of the tank on the basis of the ideal-gas equation of state 1.330000 m^3\n",
- "the volume of the tank on the basis Kay\u2019s rule 0.652000 m^3\n",
- "the volume of the tank on the basis compressibility factors and Amagat\u2019s law 0.639000 m^3\n",
- "compressibility factors and Dalton\u2019s law the volume of the tank on the basis 0.648000 m^3\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13-3 ,Page No.691"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given data\n",
- "mN=4.0;#mass of nitrogen in kg\n",
- "T1N=20.0;#temperature of nitrogen in K\n",
- "P1N=150.0;#pressure of nitrogen in kPa\n",
- "mO=7.0;#mass of oxygen in kg\n",
- "T1O=40.0;#temperature of oxygen in K\n",
- "P1O=100.0;#pressure of oxygen in kPa\n",
- "\n",
- "#molecular masses in kg\n",
- "MO=32.0;\n",
- "MN=28.0;\n",
- "\n",
- "#constants used\n",
- "Ru=8.314;#in kJ/kg - K\n",
- "\n",
- "#from Table A-2a\n",
- "CvN=0.743;\n",
- "CvO=0.658;\n",
- "\n",
- "#calculations\n",
- "\n",
- "#part - a\n",
- "#Ein - Eout = dEsystem\n",
- "# (m*cv*dT)N2 + (m*cv*dT)= 0;\n",
- "Tm= (mN*CvN*T1N+ mO*CvO*T1O)/(mN*CvN+mO*CvO);\n",
- "print'the mixture temperature %f C'%round(Tm,1);\n",
- "\n",
- "#part - b\n",
- "NO=mO/MO;\n",
- "NN=mN/MN;\n",
- "Nm=NO+NN;\n",
- "VO=NO*Ru*(T1O+273)/P1O;\n",
- "VN=NN*Ru*(T1N+273)/P1N;#Exergy Destruction during Mixing of Ideal Gases\n",
- "Vm=VO+VN;\n",
- "Pm=Nm*Ru*(Tm+273)/Vm; \n",
- "print'the mixture pressure after equilibrium has been established %f kPa'%round(Pm,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the mixture temperature 32.200000 C\n",
- "the mixture pressure after equilibrium has been established 114.500000 kPa\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13-4 ,Page No.692"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#given data\n",
- "NO=3.0;#moles of oxygen \n",
- "NC=5.0;#moles of carbondioxide\n",
- "T0=25+273.0;#temperature of gasses in K\n",
- "\n",
- "#constants used\n",
- "Ru=8.314;#in kJ/kg - K\n",
- "\n",
- "#calculations\n",
- "Nm=NO+NC;\n",
- "yO=NO/Nm;\n",
- "yC=NC/Nm;\n",
- "#dSm= -Ru*(NO*log(yO)+NC*log(yC))\n",
- "Sm=-Ru*(NO*log(yO)+NC*log(yC));\n",
- "print'the entropy change %f kJ/K'%round(Sm);\n",
- "Xdestroyed=T0*Sm/1000;\n",
- "print'exergy destruction associated %f MJ'%round(Xdestroyed,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the entropy change 44.000000 kJ/K\n",
- "exergy destruction associated 13.100000 MJ\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13-5 ,Page No.694"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given data\n",
- "T1=220;#intial temperature in K\n",
- "T2=160;#final temperature in K\n",
- "Pm=10;#pressure of air in MPa\n",
- "yN=0.79;#mole fraction of nitrogen\n",
- "yO=0.21;#mole fractions of oxygen\n",
- "\n",
- "\n",
- "#critical properties\n",
- "#for Nitrogen\n",
- "TcrN=126.2;\n",
- "PcrN=3.39;\n",
- "#for Oxygen\n",
- "TcrO=154.8;\n",
- "PcrO=5.08;\n",
- "\n",
- "#constants used\n",
- "Ru=8.314;#in kJ/kg - K\n",
- "\n",
- "#from Tables A-18 & 19\n",
- "#at T1\n",
- "h1N=6391;\n",
- "h1O=6404;\n",
- "#for T2\n",
- "h2N=4648;\n",
- "h2O=4657;\n",
- "\n",
- "#calculations\n",
- "#part - a\n",
- "qouti=yN*(h1N-h2N)+yO*(h1O-h2O);\n",
- "print'the heat transfer during this process using the ideal-gas approximation %i kJ/kmol'%round(qouti);\n",
- "\n",
- "#part - b\n",
- "Tcrm=yN*TcrN+yO*TcrO;\n",
- "Pcrm=yN*PcrN+yO*PcrO;\n",
- "Tr1=T1/Tcrm;\n",
- "Tr2=T2/Tcrm;\n",
- "Pr=Pm/Pcrm;\n",
- "#at these values we get\n",
- "Zh1=1;\n",
- "Zh2=2.6\n",
- "qout=qouti-Ru*Tcrm*(Zh1-Zh2);\n",
- "print'the heat transfer during this process using Kay\u2019s rule %i kJ/kmol'%round(qout);\n",
- "\n",
- "#part - c\n",
- "#for nitrogen\n",
- "TrN1=T1/TcrN;\n",
- "TrN2=T2/TcrN;\n",
- "PrN=Pm/PcrN;\n",
- "#from Fig A-15b\n",
- "Zh1n=0.9;\n",
- "Zh2n=2.4;\n",
- "#for Oxygen\n",
- "TrO1=T1/TcrO;\n",
- "TrO2=T2/TcrO;\n",
- "PcrO=Pm/PcrO;\n",
- "#from Fig A-15b\n",
- "Zh1O=1.3;\n",
- "Zh2O=4.0;\n",
- "#from Eq 12-58\n",
- "h12N=h1N-h2N-Ru*TcrN*(Zh1n-Zh2n);# h1 - h2 for nitrogen\n",
- "h12O=h1O-h2O-Ru*TcrO*(Zh1O-Zh2O);# h1 - h2 for oxygen\n",
- "qout=yN*h12N+yO*h12O;\n",
- "print'the heat transfer during this process using Amagat\u2019s law %i kJ/kmol'%round(qout);\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the heat transfer during this process using the ideal-gas approximation 1744 kJ/kmol\n",
- "the heat transfer during this process using Kay\u2019s rule 3502 kJ/kmol\n",
- "the heat transfer during this process using Amagat\u2019s law 3717 kJ/kmol\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13-6 ,Page No.705"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#13.6 (d) answer not matching as float datatype is giving more accurate answer in comparison to textbook that has given approximate due to rounding off to two decimal places\n",
- "\n",
- "#given data\n",
- "mfs=0.0348;#salinity mass fraction\n",
- "mfw=1-mfs;\n",
- "T0=288.15;#temperature of sea water in K\n",
- "\n",
- "#constants used\n",
- "Mw=18;\n",
- "Ms=58.44;\n",
- "Rw=0.4615;\n",
- "pm=1028;\n",
- "Ru=8.314;\n",
- "\n",
- "#calculations\n",
- "#part - a\n",
- "Mm=1/((mfs/Ms)+(mfw/Mw));\n",
- "yw=mfw*Mm/Mw;\n",
- "ys=1-yw;\n",
- "print'the mole fraction of the water is %f'%round(yw,4);\n",
- "print'the mole fraction of the saltwater is %f'%round(ys,5);\n",
- "\n",
- "#part - b\n",
- "wmin=-Ru*T0*(ys*log(ys)+yw*log(yw));\n",
- "wm=wmin/Mm;\n",
- "print'the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts %f kJ'%round(wm,2);\n",
- "\n",
- "#part - c\n",
- "wmin=Rw*T0*log(1/yw);\n",
- "print'the minimum work input required to obtain 1 kg of fresh water from the sea %f kJ'%round(wmin,2);\n",
- "\n",
- "#part - d\n",
- "Pmin=pm*Rw*T0*log(1/yw);\n",
- "print'the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes %i kPa'%round(Pmin)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the mole fraction of the water is 0.989000\n",
- "the mole fraction of the saltwater is 0.010980\n",
- "the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts 7.850000 kJ\n",
- "the minimum work input required to obtain 1 kg of fresh water from the sea 1.470000 kJ\n",
- "the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes 1510 kPa\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |