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The minimum work that must be supplied to the compressor and turbine due to irreversibilities.
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Answer to Problem 143P
The minimum work that must be supplied to the compressor due to irreversibilities is 40.1kJ/kg.
The minimum work that is developed by the turbine due to irreversibilities is 34.8kJ/kg.
It can be noted that the compressor is more sensitive to irreversibilities than the turbine.
Explanation of Solution
Show the simple Brayton cycle, with air as the working fluid on T−s diagram.
For the given Brayton cycle with air as the working fluid Ti, hi, and Pi are the temperature at ith state, specific enthalpy at ith state and pressure at ith state respectively.
Write the expression of temperature and pressure relation ratio for the compression process 1-2.
T2s=T1(P2P1)(k−1)/k (I)
Here, specific heat ratio is k.
Write the expression of efficiency of the compressor (ηC).
ηC=h2s−h1h2−h1
ηC=cp(T2s−T1)cp(T2−T1) (II)
Here, specific heat at constant pressure is cp.
Write the expression of temperature and pressure relation ratio for the expansion process 3-4.
T4s=T3(P4P3)(k−1)/k (III)
Write the expression of efficiency of the turbine (ηT).
ηT=h3−h4h3−h4s
ηT=cp(T3−T4)cp(T3−T4s) (IV)
For compression processes
Write the expression for the entropy change for the isentropic process 1-2s (s2s−s1).
s2s−s1=cplnT2sT1−RlnP2P1 (V)
Here, gas constant of air is R.
Write the expression for reversible work for the isentropic process 1-2s (wrev,1-2s).
wrev,1-2s=cp(T2s−T1)−T0(s2s−s1) (VI)
Here, temperature of the surroundings is T0.
Write the expression for the entropy change for the process 1-2 (s2−s1).
s2−s1=cplnT2T1−RlnP2P1 (VII)
Write the expression for the reversible work for the process 1-2 (wrev,1-2).
wrev,1-2=cp(T2−T1)−T0(s2−s1) (VIII)
Write the expression for the minimum work that must be supplied to the compressor due to irreversibilities (Δwrev,C).
Δwrev,C=wrev,1-2−wrev,1-2s (IX)
For expansion processes
Write the expression for the entropy change for the isentropic process 3-4s (s3−s4s).
s3−s4s=cplnT3T4s−RlnP3P4 (X)
Write the expression for the reversible work for the isentropic process 3-4s (wrev,3-4s).
wrev,3-4s=cp(T3−T4s)−T0(s3−s4s) (XI)
Write the expression for the entropy change for the process 3-4 (s3−s4).
s3−s4=cplnT3T4−RlnP3P4 (XII)
Write the expression for the reversible work for the process 3-4 (wrev,3-4).
wrev,3-4=cp(T3−T4)−T0(s3−s4) (XIII)
Write the expression for the minimum work that is developed by the turbine due to irreversibilities (Δwrev,T).
Δwrev,T=wrev,3-4s−wrev,3-4 (XIV)
Conclusion:
Substitute 288 K for T1, 12 for P2P1, and 1.4 for k in Equation (I).
T2s=(288 K)(12)1.4−1/1.4=585.8 K
Rearrange Equation (II), and solve for T2. Substitute 288 K for T1, 585.8 K for T2s, and 0.80 for ηC in Equation (II).
T2=T1+T2s−T1ηC=288 K+(585.8−288)K0.80=660.2 K
Substitute 873 K for T3, 112 for P4P3, and 1.4 for k in Equation (III).
T4s=(873 K)(112)1.4−1/1.4=429.2 K
Rearrange Equation (IV), and substitute 873 K for T3, 429.2 K for T4s, and 0.80 for ηT.
T4=T3−ηT(T3−T4s)=873 K−(0.80)(873 K−429.2 K)=518 K
Substitute 1.005 kJ/kg⋅K for cp, 585.8 K for T2s, 288 K for T1, 0.287 kJ/kg⋅K for R, and 12 for P2P1 in Equation (V).
s2s−s1=(1.005 kJ/kg⋅K)ln585.8 K288 K−(0.287 kJ/kg⋅K)ln(12)=0.0003998 kJ/kg⋅K
Substitute 1.005 kJ/kg⋅K for cp, 585.8 K for T2s, 288 K for T1, 288 K for T0, and 0.0003998 kJ/kg⋅K for (s2s−s1) in Equation (VI).
wrev,1-2s=(1.005 kJ/kg⋅K)(585.8−288) K−(288 K)(0.0003998 kJ/kg⋅K)=299.2 kJ/kg
Substitute 1.005 kJ/kg⋅K for cp, 660.2 K for T2, 288 K for T1, 0.287 kJ/kg⋅K for R, and 12 for P2P1 in Equation (VII).
s2−s1=(1.005 kJ/kg⋅K)ln660.2 K288 K−(0.287 kJ/kg⋅K)ln(12)=0.1206 kJ/kg⋅K
Substitute 1.005 kJ/kg⋅K for cp, 660.2 K for T2, 288 K for T1, 288 K for T0, and 0.1206 kJ/kg⋅K for (s2−s1) in Equation (VIII).
wrev,1-2=(1.005 kJ/kg⋅K)(660.2−288) K−(288 K)(0.1206 kJ/kg⋅K)=339.3 kJ/kg
Substitute 339.3 kJ/kg for wrev,1-2, and 299.2 kJ/kg for wrev,1-2s in Equation (IX).
Δwrev,C=(339.3−299.2) kJ/kg=40.1kJ/kg
Thus, the minimum work that must be supplied to the compressor due to irreversibilities is 40.1kJ/kg.
Substitute 1.005 kJ/kg⋅K for cp, 873 K for T3, 429.2 K for T4s, 0.287 kJ/kg⋅K for R, and 12 for P3P4 in Equation (X).
s3−s4s=(1.005 kJ/kg⋅K)ln873 K429.2 K−(0.287 kJ/kg⋅K)ln(12)=0.0003944 kJ/kg⋅K
Substitute 1.005 kJ/kg⋅K for cp, 873 K for T3, 429.2 K for T4s, 288 K for T0, and 0.0003944 kJ/kg⋅K for (s3−s4s) in Equation (XI).
wrev,3-4s=(1.005 kJ/kg⋅K)(873−429.2) K−(288 K)(0.0003944 kJ/kg⋅K)=445.9 kJ/kg
Substitute 1.005 kJ/kg⋅K for cp, 873 K for T3, 518.0 K for T4, 0.287 kJ/kg⋅K for R, and 12 for P3P4 in Equation (XII).
s3−s4=(1.005 kJ/kg⋅K)ln873 K518.0 K−(0.287 kJ/kg⋅K)ln(12)=−0.1885 kJ/kg⋅K
Substitute 1.005 kJ/kg⋅K for cp, 873 K for T3, 518.0 K for T4, 288 K for T0, and −0.1885 kJ/kg⋅K for (s3−s4) in Equation (XIII).
wrev,3-4=(1.005 kJ/kg⋅K)(873−518.0) K−(288 K)(−0.1885 kJ/kg⋅K)=411.1 kJ/kg
Substitute 445.9 kJ/kg for wrev,3-4s, and 411.1 kJ/kg for wrev,3-4 in Equation (XIV).
Δwrev,T=(445.9−411.1) kJ/kg=34.8kJ/kg
Thus, the minimum work that is developed by the turbine due to irreversibilities is 34.8kJ/kg.
It can be noted that the compressor is more sensitive to irreversibilities than the turbine.
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Thermodynamics: An Engineering Approach
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