The second law efficiency and the exergy destruction during the expansion process.
Answer to Problem 97RP
The second law efficiency is 89.4%.
The exergy destruction during the expansion process is 79 kJ/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=cp−R (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)(k−1)/k (XIII)
Write the expression to obtain the actual outlet temperature (T2).
T2=T1−ηturb(T1−T2s) (XIV)
Here, efficiency of the turbine is ηturb.
Write the expression to obtain the entropy change of the gas mixture (s2−s1).
s2−s1=cplnT2T1−RlnP2P1 (XV)
Write the expression to obtain the actual work output (wout).
wout=cp(T1−T2) (XVI)
Write the expression to obtain the reversible work output (wrev,out).
wrev,out=wout−T0(s1−s2) (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,out−wout (XIX)
Conclusion:
Refer Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the molar masses of O2, CO2 and He as 32 kg/kmol, 44 kg/kmol and 4 kg/kmol respectively.
Substitute 0.1 kg for mN2 and 32 kg/kmol for MO2 in Equation (I).
NO2=0.1 kg32 kg/kmol=0.003125 kmol
Substitute 1 kg for mCO2 and 44 kg/kmol for MCO2 in Equation (II).
NCO2=1 kg(44 kg/kmol)=0.02273 kmol
Substitute 0.5 kg for mHe and 4 kg/kmol for MHe in Equation (III).
NHe=0.5 lbm(4 kg/kmol)=0.125 kmol
Substitute 0.003125 kmol for NO2, 0.02273 kmol for NCO2 and 0.125 kmol for NHe in Equation (VII).
Nm=0.003125 kmol+0.02273 kmol+0.125 kmol=0.15086 kmol
Substitute 1.6 kg for mm and 0.15086 kmol for Nm in Equation (X).
Mm=1.6 kg0.15086 kmol=10.61 kg/kmol
Substitute 8.314 kJ/Kmol⋅K for Ru and 10.61 kg/kmol for Mm in Equation (IX).
R=8.314 kJ/Kmol⋅K10.61 kg/kmol=0.7836 kJ/kg⋅K
Substitute 0.1 kg for mO2 and 1.6 kg for mm in Equation (IV).
mfO2=0.1 kg1.6 kg=0.0625
Substitute 1 kg for mCO2 and 1.6 kg for mm in Equation (V).
mfCO2=1 kg1.6 kg=0.625
Substitute 0.5 kg for mHe and 1.6 kg for mm in Equation (VI).
mfHe=0.5 kg1.6 kg=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/kg⋅K, 5.1926 kJ/kg⋅K, and 0.846 kJ/kg⋅K.
Substitute 0.0625 for mfO2, 0.625 for mfCO2, 0.3125 for mfHe, 0.918 kJ/kg⋅K for cp,O2, 5.1926 kJ/kg⋅K for cp,He, and 0.846 kJ/kg⋅K for cp,CO2 in Equation (IX).
cp=[(0.0625)(0.918 kJ/kg⋅K)+(0.625)(0.846 kJ/kg⋅K)++(0.3125)(5.1926 kJ/kg⋅K)]=2.209 kJ/kg⋅K
Substitute 0.7836 kJ/kg⋅K for R and 2.209 kJ/kg⋅K for cp in Equation (XI).
cv=2.209 kJ/kg⋅K−0.7836 kJ/kg⋅K=1.425 kJ/kg⋅K
Substitute 2.209 kJ/kg⋅K for cp and 1.425 kJ/kg⋅K for cv in Equation (XII).
k=2.209 kJ/kg⋅K1.425 kJ/kg⋅K=1.550
Substitute 1.550 for k, 327°C for T1, 100 kPa for P2, and 1000 kPa for P1 in Equation (XIII).
T2s=(327°C)(100 kPa1000 kPa)(1.550−1)/1.550=(327+273)K(100 kPa1000 kPa)(1.550−1)/1.550=265 K
Substitute 327°C for T1, 265 K for T2s and 0.90 for ηturb in Equation (XIV).
T2=327°C−0.90(327°C−265 K)=(327+273)K−0.90((327+273)K−265 K)=299 K
Substitute 2.209 kJ/kg⋅K for cp, 299 K for T2, 327°C for T1, 0.7836 kJ/kg⋅K for R, 100 kPa for P2, and 1000 kPa for P1 in Equation (XV).
s2−s1={(2.209 kJ/kg⋅K)ln(299 K327°C)−(0.7836 kJ/kg⋅K)ln(100 kPa1000 kPa)}={(2.209 kJ/kg⋅K)ln(299 K(327+273)K)−(0.7836 kJ/kg⋅K)ln(100 kPa1000 kPa)}=0.2658 kJ/kg⋅K
Substitute 2.209 kJ/kg⋅K for cp, 299 K for T2 and 327°C for T1 in Equation (XVI).
wout=2.209 kJ/kg⋅K(327°C−299 K)=2.209 kJ/kg⋅K((327+273)K−299 K)=665 kJ/kg
Substitute 665 kJ/kg for wout, 25°C for T0 and (−0.2658 kJ/kg⋅K) for (s1−s2) in Equation (XVII).
wrev,out=665 kJ/kg−25°C(0.2658 kJ/kg⋅K)=665 kJ/kg−(25+273)K×(−0.2658 kJ/kg⋅K)=744 kJ/kg
Substitute 665 kJ/kg for wout and 744 kJ/kg for wrev,out in Equation (XVIII).
ηII=665 kJ/kg744 kJ/kg=0.894×100%=89.4%
Thus, the second law efficiency is 89.4%.
Substitute 665 kJ/kg for wout and 744 kJ/kg for wrev,out in Equation (XIX).
xdest=744 kJ/kg−665 kJ/kg=79 kJ/kg
Thus, the exergy destruction during the expansion process is 79 kJ/kg.
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