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| author | hardythe1 | 2015-04-07 15:58:05 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-04-07 15:58:05 +0530 |
| commit | 92cca121f959c6616e3da431c1e2d23c4fa5e886 (patch) | |
| tree | 205e68d0ce598ac5caca7de839a2934d746cce86 /Fundamentals_Of_Thermodynamics/chapter10.ipynb | |
| parent | b14c13fcc6bb6d01c468805d612acb353ec168ac (diff) | |
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diff --git a/Fundamentals_Of_Thermodynamics/chapter10.ipynb b/Fundamentals_Of_Thermodynamics/chapter10.ipynb new file mode 100755 index 00000000..e985ae83 --- /dev/null +++ b/Fundamentals_Of_Thermodynamics/chapter10.ipynb @@ -0,0 +1,474 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:498eddefb3635fb861c09bdd97718d2b72fb04f4d3b083cedbf9f6de28a92b4e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10:IRREVERSIBILITY AND AVAILABILITY"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.1:pg-386"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1\n",
+ "#Calculating reversible work\n",
+ "#Form the Steam Tables,the inlet and the exit state properties are \n",
+ "\n",
+ "hi=171.95 #initial specific heat of enthalpy in kJ/kg\n",
+ "si=0.5705 #initial specific entropy in kJ/kg-K\n",
+ "se=2.1341 #final specific entropy in kJ/kg-K\n",
+ "he=765.34 #final specific heat of enthalpy in kJ/kg-K\n",
+ "m=5 #mass flow rate of feedwater in kg/s\n",
+ "q1=900/m #heat added by one of the sources in kJ/kg\n",
+ "q2=he-hi-q1 #second heat transfer in kJ/kg\n",
+ "To=25+273.3 #Temp. of the surroundings in K\n",
+ "T1=100+273.2 #temp. of reservoir of one of the source in K\n",
+ "T2=200+273.2 #temp. of reservoir of second source in K\n",
+ "wrev=To*(se-si)-(he-hi)+q1*(1-To/T1)+q2*(1-To/T2) #reversible work in kJ/kg\n",
+ "print\"\\n Hence, the irreversibility is\",round(wrev),\"kJ/kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Hence, the irreversibility is 62.0 kJ/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.2:pg-387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2\n",
+ "#Calculating reversible work\n",
+ "#Form the Steam Tables,the inlet and the exit state properties are\n",
+ "\n",
+ "hi=298.6 #initial specific heat of enthalpy in kJ/kg\n",
+ "si=6.8631 #initial specific entropy in kJ/kg-K\n",
+ "se=7.4664 #final specific entropy in kJ/kg-K\n",
+ "he=544.7 #final specific heat of enthalpy in kJ/kg-K\n",
+ "q=-50 #heat lost to surroundings in kJ/kg\n",
+ "w=hi-he+q #work in kJ/kg\n",
+ "To=25+273.2 #Temp. of the surroundings in K\n",
+ "P1=100 #Pressure of ambient air in kPa\n",
+ "P2=1000 #Final pressure of air after compression in kPa\n",
+ "R=0.287 #Universal gas constant in kJ/kg-K\n",
+ "wrev=To*(se-si-R*log(P2/P1))-(he-hi)+q*(1-To/To)#reversible work for the given change of state in kJ/kg\n",
+ "i=wrev-w #irreversibility in kJ/kg\n",
+ "print\"\\n Hence,the irreversibility is\",round(i,1),\"kJ/kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Hence,the irreversibility is 32.8 kJ/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.2E:pg-388"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2E\n",
+ "#Calculating reversible work\n",
+ "#Form the Steam Tables,the inlet and the exit state properties are\n",
+ "\n",
+ "hi=129.18 #initial specific heat of enthalpy in Btu/lbm\n",
+ "si=1.6405 #initial specific entropy in Btu/lbm R\n",
+ "se=1.7803 #final specific entropy in Btu/lbm R\n",
+ "he=231.20 #final specific heat of enthalpy in Btu/lbm\n",
+ "q=-22 #heat lost to surroundings in Btu/lbm\n",
+ "w=hi-he+q #work in Btu/lbm\n",
+ "To=539.7 #Temp. of the surroundings in R\n",
+ "P1=14.7 #Pressure of ambient air in lbf/in^2\n",
+ "P2=150 #Final pressure of air after compression in lbf/in^2\n",
+ "R=0.06855 #Universal gas constant in Btu/lbm R\n",
+ "wrev=To*(se-si-R*log(P2/P1))-(he-hi)+q*(1-To/To)#reversible work for the given change of state in Btu/lbm\n",
+ "i=wrev-w #irreversibility in Btu/lbm\n",
+ "print\"\\n Hence,the irreversibility is\",round(i,2),\"Btu/lbm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Hence,the irreversibility is 11.52 Btu/lbm\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.3:pg-390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3\n",
+ "#Calculating reversible work and irreversibility\n",
+ "#Form the Steam Tables at state 1\n",
+ "\n",
+ "u1=1243.5 #initial specific internal energy in kJ/kg\n",
+ "s1=4.4819 #initial specific entropy in kJ/kg-K\n",
+ "v1=28.895 #initial specific volume in m^3/kg\n",
+ "v2=2*v1 #final specific volume in kg/m^3\n",
+ "u2=u1 #initial specific internal energy in kJ/kg\n",
+ "#These two independent properties, v2 and u2 , fix state 2.The final temp. is calculated by interplotation using the data for T2=5C and v2,x=0/3928 and u=948.5 kJ/kg. For T2=10C and v2, x=0.5433 and u=1317 kJ/kg\n",
+ "T2=9.1+273.2 #final temp. in K\n",
+ "x2=0.513 #quality in final state\n",
+ "s2=4.644 #final specific entropy in kJ/kg\n",
+ "V1=1 #volume of part of A in m^3\n",
+ "m=V1/v1 #mass flow rate in kg/s\n",
+ "To=20+273.2 #Room temperature in K\n",
+ "Wrev=To*m*(s2-s1) #reversible work in kJ\n",
+ "I=Wrev #irreversibility of the process\n",
+ "print\"\\n The irreversibility is\",round(I,3),\"kJ\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The irreversibility is 1.645 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.4:pg-391"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4\n",
+ "#calculating the final mass in the tank and the irreversibility in the process\n",
+ "#From the ammonia tables, the initial and line state properties are\n",
+ "\n",
+ "V=1#volume of the rigid tank in m^3\n",
+ "v1=0.6995#initial specific volume in m^3/kg\n",
+ "u1=1369.5#initial specific internal energy in kJ/kg\n",
+ "s1=5.927#initial specific entropy in kJ/kg K\n",
+ "h1=134.41#initial specific heat of enthalpy in kJ/kg\n",
+ "si=0.5408#in kJ/kg K\n",
+ "m1=V/v1#initial mass in kg\n",
+ "x2=0.007182\n",
+ "v2=(0.001534 + x2 * 0.41684)#in m^3/kg\n",
+ "v2=round(v2,7)\n",
+ "s2=0.5762#final specific entropy in kJ/kg\n",
+ "m2=V/v2#the final mass in kg\n",
+ "mi=m2-m1#in kg\n",
+ "mi=round(mi,3)\n",
+ "T=293.15# in K\n",
+ "S2gen=(m2*s2-m1*s1-mi*si)#in kJ/kg\n",
+ "S2gen=round(S2gen,3)\n",
+ "Icv=T*S2gen#in kJ\n",
+ "print\"\\n the final mass is \",round(m2,2),\"kg\"\n",
+ "print\"\\n the irreversiblity is\",round(Icv,3),\"kJ\"\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " the final mass is 220.86 kg\n",
+ "\n",
+ " the irreversiblity is 34.885 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.5:pg-396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5\n",
+ "#calculating the availability per kilogram of the steam entering and at both points at which steam leaves the turbine,\n",
+ "#the isentropic efficiency,the second-law efficiency for this process\n",
+ "#For the ideal isentropic turbine,\n",
+ "\n",
+ "h0=104.9#in kJ/kg\n",
+ "s0=0.3674#entropy at state0 in kJ/kgK\n",
+ "m1=30#mass at state1 in kg\n",
+ "phi1=1109.6#in kJ/kg\n",
+ "h1=3115.3#enthalpy at state1 in kJ/kg\n",
+ "m2=5#mass at state2 in kg\n",
+ "phi2=755.3#in kJ/kg\n",
+ "h2=2855.4#enthalpy at state2 For the actual turbine in kJ/kg\n",
+ "m3=25#mass at state3 in kg\n",
+ "phi3=195.0#in kJ/kg\n",
+ "S2s=6.7428#entropy in kJ/kg K at state2\n",
+ "x2s=(6.7428-1.8606)/4.906# dryness factor at state2 \n",
+ "x2s=round(x2s,4)\n",
+ "h2s=640.2+(x2s*2108.5)#entropy at state2 for the ideal isentropic turbine\n",
+ "S3s=6.7428#entropy in kJ/kg K\n",
+ "x3s=(6.7428-0.7549)/7.2536# dryness factor at state3\n",
+ "x3s=round(x3s,4)\n",
+ "h3s=225.9+(x3s*2373.1)#enthalpy in kJ/kg\n",
+ "h3=2361.8#enthalpy at state3 For the actual turbine in kJ/kg\n",
+ "Ws=(m1*h1-m2*h2s-m3*h3s)#workdone for the ideal isentropic turbine in kW\n",
+ "W=(m1*h1-m2*h2-m3*h3)#workdone For the actual turbine in kW\n",
+ "W2=(m1*phi1-m2*phi2-m3*phi3)#workdone for the second ideal isentropic turbine in kW\n",
+ "eta1=W/Ws#The isentropic efficiency\n",
+ "eta2=W/W2#he second-law efficiency\n",
+ "print\"\\n the workdone for the ideal isentropic turbine is \",round(Ws),\"kW\"\n",
+ "print\"\\n the workdone for the actual turbine is \",round(W),\"kg\"\n",
+ "print\"\\n the workdone for the second ideal isentropic turbine is \",round(W2),\"kW\"\n",
+ "print\"\\n the isentropic efficiency is \",round(eta1,3)\n",
+ "print\"\\n the second-law efficiency is \",round(eta2,3)\n",
+ "#The answer of the x2s in the book is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " the workdone for the ideal isentropic turbine is 25145.0 kW\n",
+ "\n",
+ " the workdone for the actual turbine is 20137.0 kg\n",
+ "\n",
+ " the workdone for the second ideal isentropic turbine is 24637.0 kW\n",
+ "\n",
+ " the isentropic efficiency is 0.801\n",
+ "\n",
+ " the second-law efficiency is 0.817\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.6:pg-399"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6\n",
+ "#calculate the second-law efficiency for this process and the irreversibility per kilogram of water evaporated\n",
+ "\n",
+ "he1=2950#enthalpy in kJ/kg\n",
+ "hi1=632.2#enthalpy in kJ/kg\n",
+ "he2=599.5#enthalpy in kJ/kg\n",
+ "hi2=1199#enthalpy in kJ/kg\n",
+ "#r=mp/mh\n",
+ "r=(he1-hi1)/(hi2-he2)#the ratio of the mass flow of water\n",
+ "T0=298.15#in K\n",
+ "S1=1.8418#entropy at state1 in kJ/kgK\n",
+ "S2=7.0384#entropy at state2 in kJ/kgK\n",
+ "S3=7.8751#entropy at state3 in kJ/kgK\n",
+ "S4=7.3173#entropy at state4 in kJ/kgK\n",
+ "#phi=phi2-phi1\n",
+ "phi=(he1-hi1)-T0*(S2-S1)#The increase in availability of the water is, per kilogram of water in kJ/kg\n",
+ "#w=(mp2/mh2)*(phi3-phi4)\n",
+ "w=r*((hi2-he2)-T0*(S3-S4))#The decrease in availability of the products, per kilogram of water in kJ/kg\n",
+ "eta=phi/w#the second-law efficiency\n",
+ "#i=I/mh\n",
+ "i=phi+w#the process irreversibility per kilogram of water in kJ/kg\n",
+ "#e=I2/mh2\n",
+ "e=(T0*(S2-S1)+(T0*r*(S4-S3)))#The total irreversibility in kJ/kg\n",
+ "print\"\\n the second law efficiency is \",round(eta,3)\n",
+ "print\"\\n the total irreversibility is \",round(e,2),\"kJ/kg\"\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " the second law efficiency is 0.459\n",
+ "\n",
+ " the total irreversibility is 906.38 kJ/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.7:pg-403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7\n",
+ "#calculate the heater\u2019s second-law efficiency\n",
+ "\n",
+ "h0=104.87#enthalpy in kJ/kg\n",
+ "s0=0.3673#entropy in kJ/kg K\n",
+ "h1=171.97#in kJ/kg\n",
+ "he=765.25#in kJ/kg\n",
+ "s1=0.5705#in kJ/kg K\n",
+ "T0=298.2#in C\n",
+ "se=2.1341#in kJ/kg K\n",
+ "shi1=(h1-h0)-T0*(s1-s0)#in kJ/kg\n",
+ "shie=(he-h0)-T0*(se-s0)#in kJ/kg\n",
+ "T1=373.2#in C\n",
+ "T2=473.2#in C\n",
+ "q1=180#heat at state1 in kJ\n",
+ "q2=413.28#heat at state2 in kJ\n",
+ "phisource1=(1-T0/T1)*q1#in kJ/kg\n",
+ "phisource2=(1-T0/T2)*q2#in kJ/kg\n",
+ "phisource=phisource1+phisource2#in kJ/kg\n",
+ "Icv=phisource+shi1-shie#in kJ/kg\n",
+ "eta2law=(phisource-Icv)/phisource#the heater\u2019s second-law efficiency\n",
+ "print\"\\n the heater\u2019s second-law efficiency is\",round(eta2law,2)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " the heater\u2019s second-law efficiency is 0.67\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.8:pg-404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8\n",
+ "#calculate any heat transfer downward, and follow the flux of exergy, and find the exergy destruction in the process.\n",
+ "Qout=500.0#in W\n",
+ "Qin=500.0#in W\n",
+ "Tsurf=1000.0#in K\n",
+ "T0=298.15#in K\n",
+ "Sgen1=Qout/Tsurf#in W/K\n",
+ "phides=(T0*Sgen1)#in W\n",
+ "phitrans=((1-T0/Tsurf)*Qout)#in W\n",
+ "Ttop=500#in K\n",
+ "Sgen2=(Qout/Ttop)-(Qin/Tsurf)#in W/K\n",
+ "Te=phitrans-phides-((1-T0/Ttop)*Qout)#the exergy destruction in the process\n",
+ "print\"\\n the exergy destruction in the process is\",round(Te)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " the exergy destruction in the process is 0.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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