Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Question
Chapter 13.3, Problem 97RP
To determine

The second law efficiency and the exergy destruction during the expansion process.

Expert Solution & Answer
Check Mark

Answer to Problem 97RP

The second law efficiency is 89.4%.

The exergy destruction during the expansion process is 79kJ/kg.

Explanation of Solution

Write the expression to obtain the mole number of O2 (NO2).

NO2=mO2MO2 (I)

Here, molar mass of O2 is MO2 and mass of O2 is mO2.

Write the expression to obtain the mole number of CO2 (NCO2).

NCO2=mCO2MCO2 (II)

Here, molar mass of CO2 is MCO2 and mass of CO2 is mCO2.

Write the expression to obtain the mole number of He (NHe).

NHe=mHeMHe (III)

Here, molar mass of He is MHe and mass of He is mHe.

Write the expression to obtain the mass fraction of O2 (mfO2).

mfO2=mO2mm (IV)

Write the expression to obtain the mass fraction of CO2 (mfCO2).

mfCO2=mCO2mm (V)

Write the expression to obtain the mass fraction of He (mfHe).

mfHe=mHemm (VI)

Write the expression to obtain the equation to calculate the mole number of the mixture (Nm).

Nm=NN2+NHe+NCH4+NC2H6 (VII)

Write the expression to obtain the molar mass of the gas mixture (Mm).

Mm=mmNm (VIII)

Write the expression to obtain the equation to calculate the constant-pressure specific heat of the mixture (cp).

cp=mfO2cp,O2+mfCO2cp,CO2+mfHecp,He (IX)

Here, constant pressure specific heat of O2, He,  and CO2 are cp,O2, cp,He, and cp,CO2 respectively.

Write the expression to obtain the gas constant of the mixture (R).

R=RuMm (X)

Here, the universal gas constant is Ru.

Write the expression to obtain the constant volume specific heat (cv).

cv=cpR (XI)

Write the expression to obtain the specific heat ratio (k).

k=cpcv (XII)

Write the expression to obtain the temperature at the end of the expansion for the isentropic process (T2s).

T2s=T1(P2P1)(k1)/k (XIII)

Write the expression to obtain the actual outlet temperature (T2).

T2=T1ηturb(T1T2s) (XIV)

Here, efficiency of the turbine is ηturb.

Write the expression to obtain the entropy change of the gas mixture (s2s1).

s2s1=cplnT2T1RlnP2P1 (XV)

Write the expression to obtain the actual work output (wout).

wout=cp(T1T2) (XVI)

Write the expression to obtain the reversible work output (wrev,out).

wrev,out=woutT0(s1s2) (XVII)

Write the expression to obtain the second law efficiency (ηII).

ηII=woutwrev,out (XVIII)

Write the expression to obtain the exergy destruction (xdest).

xdest=wrev,outwout (XIX)

Conclusion:

Refer Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the molar masses of O2, CO2 and He as 32kg/kmol, 44kg/kmol and 4kg/kmol respectively.

Substitute 0.1kg for mN2 and 32kg/kmol for MO2 in Equation (I).

NO2=0.1kg32kg/kmol=0.003125kmol

Substitute 1kg for mCO2 and 44kg/kmol for MCO2 in Equation (II).

NCO2=1kg(44kg/kmol)=0.02273kmol

Substitute 0.5kg for mHe and 4kg/kmol for MHe in Equation (III).

NHe=0.5lbm(4kg/kmol)=0.125kmol

Substitute 0.003125kmol for NO2, 0.02273kmol for NCO2 and 0.125kmol for NHe in Equation (VII).

Nm=0.003125kmol+0.02273kmol+0.125kmol=0.15086kmol

Substitute 1.6kg for mm and 0.15086kmol for Nm in Equation (X).

Mm=1.6kg0.15086kmol=10.61kg/kmol

Substitute 8.314kJ/KmolK for Ru and 10.61kg/kmol for Mm in Equation (IX).

R=8.314kJ/KmolK10.61kg/kmol=0.7836kJ/kgK

Substitute 0.1kg for mO2 and 1.6kg for mm in Equation (IV).

mfO2=0.1kg1.6kg=0.0625

Substitute 1kg for mCO2 and 1.6kg for mm in Equation (V).

mfCO2=1kg1.6kg=0.625

Substitute 0.5kg for mHe and 1.6kg for mm in Equation (VI).

mfHe=0.5kg1.6kg=0.3125

Refer Table A-2a, “Ideal gas specific heats of various common gases”, obtain the constant pressure specific heats of O2, He and CO2 as 0.918 kJ/kgK, 5.1926 kJ/kgK,  and 0.846 kJ/kgK.

Substitute 0.0625 for mfO2, 0.625 for mfCO2, 0.3125 for mfHe, 0.918 kJ/kgK for cp,O2, 5.1926 kJ/kgK for cp,He, and 0.846 kJ/kgK for cp,CO2 in Equation (IX).

cp=[(0.0625)(0.918 kJ/kgK)+(0.625)(0.846 kJ/kgK)++(0.3125)(5.1926 kJ/kgK)]=2.209 kJ/kgK

Substitute 0.7836 kJ/kgK for R and 2.209 kJ/kgK for cp in Equation (XI).

cv=2.209 kJ/kgK0.7836 kJ/kgK=1.425kJ/kgK

Substitute 2.209 kJ/kgK for cp and 1.425kJ/kgK for cv in Equation (XII).

k=2.209 kJ/kgK1.425kJ/kgK=1.550

Substitute 1.550 for k, 327°C for T1, 100kPa for P2, and 1000kPa for P1 in Equation (XIII).

T2s=(327°C)(100kPa1000kPa)(1.5501)/1.550=(327+273)K(100kPa1000kPa)(1.5501)/1.550=265K

Substitute 327°C for T1, 265K for T2s and 0.90 for ηturb in Equation (XIV).

T2=327°C0.90(327°C265K)=(327+273)K0.90((327+273)K265K)=299K

Substitute 2.209kJ/kgK for cp, 299K for T2, 327°C for T1, 0.7836kJ/kgK for R,  100kPa for P2, and 1000kPa for P1 in Equation (XV).

s2s1={(2.209kJ/kgK)ln(299K327°C)(0.7836kJ/kgK)ln(100kPa1000kPa)}={(2.209kJ/kgK)ln(299K(327+273)K)(0.7836kJ/kgK)ln(100kPa1000kPa)}=0.2658kJ/kgK

Substitute 2.209kJ/kgK for cp, 299K for T2 and 327°C for T1 in Equation (XVI).

wout=2.209kJ/kgK(327°C299K)=2.209kJ/kgK((327+273)K299K)=665kJ/kg

Substitute 665kJ/kg for wout, 25°C for T0 and (0.2658kJ/kgK) for (s1s2) in Equation (XVII).

wrev,out=665kJ/kg25°C(0.2658kJ/kgK)=665kJ/kg(25+273)K×(0.2658kJ/kgK)=744kJ/kg

Substitute 665kJ/kg for wout and 744kJ/kg for wrev,out in Equation (XVIII).

ηII=665kJ/kg744kJ/kg=0.894×100%=89.4%

Thus, the second law efficiency is 89.4%.

Substitute 665kJ/kg for wout and 744kJ/kg for wrev,out in Equation (XIX).

xdest=744kJ/kg665kJ/kg=79kJ/kg

Thus, the exergy destruction during the expansion process is 79kJ/kg.

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Chapter 13 Solutions

Thermodynamics: An Engineering Approach

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