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UNIT-I1.Aheatenginereceivesheatattherateof1500kJ/minandgivesanoutputof8.2kW.Determinethethermalefficiencyandtherateofheatrejection.Given:Heatreceivedbyengine(HS)Workoutput(W)Required:thandHRSolution:=1500kJ/min=25kW=8.2kWth=Workoutput/HeatSupplied=8.2/25=0.328----AnsWorkdone=HS–HR8.2=25–HRHeatrejected(HR)=16.8kW2.Aheatenginereceivesheatattherateof1500kJ/minandgivesanoutputof10.2kW.Determinethethermalefficiencyandtherateofheatrejection.Given:Heatreceivedbyengine(HS)Workoutput(W)Required:thandHRSolution:=1500kJ/min=25kW=10.2kWth=Workoutput/HeatSupplied=8.2/25=0.328----AnsWorkdone=HS–HR8.2=25–HRHeatrejected(HR)=20.8kW----Ans3.0.04kgofCO(M=44)iscompressedfrom1bar,20oC,untilthepressureis9barandthevolumeisthen0.003m3.Calculatethechangeofentropy.TakeCpforCO2as0.88kJ/KandassumeCO2tobeaperfectgas.UNIT-I1.Aheatenginereceivesheatattherateof1500kJ/minandgivesanoutputof8.2kW.Determinethethermalefficiencyandtherateofheatrejection.Given:Heatreceivedbyengine(HS)Workoutput(W)Required:thandHRSolution:=1500kJ/min=25kW=8.2kWth=Workoutput/HeatSupplied=8.2/25=0.328----AnsWorkdone=HS–HR8.2=25–HRHeatrejected(HR)=16.8kW2.Aheatenginereceivesheatattherateof1500kJ/minandgivesanoutputof10.2kW.Determinethethermalefficiencyandtherateofheatrejection.Given:Heatreceivedbyengine(HS)Workoutput(W)Required:thandHRSolution:=1500kJ/min=25kW=10.2kWth=Workoutput/HeatSupplied=8.2/25=0.328----AnsWorkdone=HS–HR8.2=25–HRHeatrejected(HR)=20.8kW----Ans3.0.04kgofCO(M=44)iscompressedfrom1bar,20oC,untilthepressureis9barandthevolumeisthen0.003m3.Calculatethechangeofentropy.TakeCpforCO2as0.88kJ/KandassumeCO2tobeaperfectgas.UNIT-I1.Aheatenginereceivesheatattherateof1500kJ/minandgivesanoutputof8.2kW.Determinethethermalefficiencyandtherateofheatrejection.Given:Heatreceivedbyengine(HS)Workoutput(W)Required:thandHRSolution:=1500kJ/min=25kW=8.2kWth=Workoutput/HeatSupplied=8.2/25=0.328----AnsWorkdone=HS–HR8.2=25–HRHeatrejected(HR)=16.8kW2.Aheatenginereceivesheatattherateof1500kJ/minandgivesanoutputof10.2kW.Determinethethermalefficiencyandtherateofheatrejection.Given:Heatreceivedbyengine(HS)Workoutput(W)Required:thandHRSolution:=1500kJ/min=25kW=10.2kWth=Workoutput/HeatSupplied=8.2/25=0.328----AnsWorkdone=HS–HR8.2=25–HRHeatrejected(HR)=20.8kW----Ans3.0.04kgofCO(M=44)iscompressedfrom1bar,20oC,untilthepressureis9barandthevolumeisthen0.003m3.Calculatethechangeofentropy.TakeCpforCO2as0.88kJ/KandassumeCO2tobeaperfectgas.
Massofgas(m)Initialpressure(p1)=0.04kg=1.05barInitialtemperature(T1)=20oC=293KFinalpressure(p2)Finalvolume(V2)Required:SSolution:=9bar=0.003m3S2–S1=mRln[V2/V1]+mCvln[T2/T1]GeneralequationUniversalgasconstant(Ru)=8314J/kg-KCharacteristicgasconstant(R)=Ru/M=8314/44=188.95J/kg-KTofindmp1V1=mRT11x105xV1=0.04x188.95x293V1=0.02214494m3TofindT2p2V2=mRT29x105xV2=0.04x188.95xT2T2=357.24KTofindCvCp–Cv=R880–Cv=188.95Cv=691.05J/kg-KS2–S1=(0.04)(188.95)ln[0.003/0.02214494]+(0.04)(691.05)ln[357.24/293]=-9.62847J/kg-K---AnsUNIT-21.InaCarnotcycle,themaximumpressureandtemperaturearelimitedto18barand410oC.Theratioofisentropiccompressionis6andisothermalexpansionis1.5.Assumingthevolumeoftheairatthebeginningofisothermalexpansionas0.18m3,determine(i)thepressureandtemperatureatmainpoints(ii)changeinentropyduringisothermalexpansion(iii)meanthermalefficiencyofthecycle(iv)meaneffectivepressureofthecycleand(v)thetheoreticalpowerifthereare210workingcyclespermin.Given:Maximumpressure(p3)Maximumtemperature(T3)Ratioofisentropiccompression(V2/V3)=6=18bar=410oC=683K=T4Massofgas(m)Initialpressure(p1)=0.04kg=1.05barInitialtemperature(T1)=20oC=293KFinalpressure(p2)Finalvolume(V2)Required:SSolution:=9bar=0.003m3S2–S1=mRln[V2/V1]+mCvln[T2/T1]GeneralequationUniversalgasconstant(Ru)=8314J/kg-KCharacteristicgasconstant(R)=Ru/M=8314/44=188.95J/kg-KTofindmp1V1=mRT11x105xV1=0.04x188.95x293V1=0.02214494m3TofindT2p2V2=mRT29x105xV2=0.04x188.95xT2T2=357.24KTofindCvCp–Cv=R880–Cv=188.95Cv=691.05J/kg-KS2–S1=(0.04)(188.95)ln[0.003/0.02214494]+(0.04)(691.05)ln[357.24/293]=-9.62847J/kg-K---AnsUNIT-21.InaCarnotcycle,themaximumpressureandtemperaturearelimitedto18barand410oC.Theratioofisentropiccompressionis6andisothermalexpansionis1.5.Assumingthevolumeoftheairatthebeginningofisothermalexpansionas0.18m3,determine(i)thepressureandtemperatureatmainpoints(ii)changeinentropyduringisothermalexpansion(iii)meanthermalefficiencyofthecycle(iv)meaneffectivepressureofthecycleand(v)thetheoreticalpowerifthereare210workingcyclespermin.Given:Maximumpressure(p3)Maximumtemperature(T3)Ratioofisentropiccompression(V2/V3)=6=18bar=410oC=683K=T4Massofgas(m)Initialpressure(p1)=0.04kg=1.05barInitialtemperature(T1)=20oC=293KFinalpressure(p2)Finalvolume(V2)Required:SSolution:=9bar=0.003m3S2–S1=mRln[V2/V1]+mCvln[T2/T1]GeneralequationUniversalgasconstant(Ru)=8314J/kg-KCharacteristicgasconstant(R)=Ru/M=8314/44=188.95J/kg-KTofindmp1V1=mRT11x105xV1=0.04x188.95x293V1=0.02214494m3TofindT2p2V2=mRT29x105xV2=0.04x188.95xT2T2=357.24KTofindCvCp–Cv=R880–Cv=188.95Cv=691.05J/kg-KS2–S1=(0.04)(188.95)ln[0.003/0.02214494]+(0.04)(691.05)ln[357.24/293]=-9.62847J/kg-K---AnsUNIT-21.InaCarnotcycle,themaximumpressureandtemperaturearelimitedto18barand410oC.Theratioofisentropiccompressionis6andisothermalexpansionis1.5.Assumingthevolumeoftheairatthebeginningofisothermalexpansionas0.18m3,determine(i)thepressureandtemperatureatmainpoints(ii)changeinentropyduringisothermalexpansion(iii)meanthermalefficiencyofthecycle(iv)meaneffectivepressureofthecycleand(v)thetheoreticalpowerifthereare210workingcyclespermin.Given:Maximumpressure(p3)Maximumtemperature(T3)Ratioofisentropiccompression(V2/V3)=6=18bar=410oC=683K=T4
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