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Unformatted text preview: he thermal efficiency is also given by TL 300 K ηth = 1 − = 1− TH 800 K = 0.625
The net work done by the cycle is Wnet , cycle = η th QH = 0.625(252.2 kJ ) = 157.6 kJ
b. Apply the first law, closed system, to the isentropic expansion process, 23. But the isentropic process is adiabatic, reversible; so, Q23 = 0. Chapter 7  50 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 751 Ein − E out = ∆E −W = ∆U W23 = − ∆U 23
Using the ideal gas relations, the work per unit mass is W23 = − mCv (T3 − T2 ) kJ )(300 − 800) K = − (1kg )(0.718 kg ⋅ K = 359.0 kJ
This is the work leaving the cycle in process 23. c. Using equation (634) ∆s34 = C p ln FG T IJ − R lnFG P IJ HT K H P K
4 4 3 3 But T4 = T3 = TL = 300 K, and we need to find P4 and P3. Consider process 12 where T1 = T2 = TH = 800 K, and, for ideal gases Chapter 7  51 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 752 m1 = m2 PV1 PV2 1 = 2 T1 T2 P2 = P 1 V1 3V1 1 3 = (800 kPa ) = 266.7 kPa
Consider process 23 where s3 = s2. T3 P3 = T2 P2
3 2 FG IJ H K FG T IJ P =P HT K
3 2 ( k −1)/ k k /( k −1) F 300K IJ = 266.7 kPa G H 800k K 1.4 /(1.4 −1) = 8.613 kPa Chapter 7  52 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 753 Now, consider process 41 where s4 = s1. T P4 = P 4 1 T1 k /( k −1) 300 K = 8000 kPa 800k = 25.834 kPa
Now, 1.4 /(1.4 −1) LMC lnF T I − R lnF P I OP ∆S = m GH T JK GH P JK Q N FG P IJ = − mR ln HPK kJ . F 25834kPa IJ = (1 kg )(0.287 ) lnG kg ⋅ K H 8.613kPa K
p 4 4 3 3 4 3 = −0.315
Extra Problem kJ K Use a second approach to find ∆S34 by noting that the temperature of process 34 is constant and applying the basic definition of entropy for an internally reversible process, dS = δQ/T. Chapter 7  53 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 754 Reversible SteadyFlow Work Isentropic, Steady Flow through Turbines, Pumps, and Compressors Consider a turbine, pump, compressor, or other steadyflow control volume, workproducing device. The general first law for the steadyflow control volume is Ein = Eout Vi 2 Ve2 Qnet + ∑ mi (hi + + gzi ) = Wnet + ∑ me (he + + gze ) 2 2 inlets exits
For a oneentrance, oneexit device undergoing an internally reversible process, this general equation of the conservation of energy reduces to, on a unit of mass basis δ wrev = δ qrev − dh − dke − dpe But δ qrev = T ds δ wrev = T ds − dh − dke − dp
Using the Gibb’s second equation, this becomes dh = T ds + v dP δ wrev = −v dP − dke − dpe
Integrating over the process, this becomes Chapter 7  54 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 755 wrev = − v dP − ∆ke − ∆pe
1 z 2 FG kJ IJ H kg K Neglecting changes in kinetic and potential energies, reversible work becomes wrev = − v dP
1 z 2 FG kJ IJ H kg K Based on the classical sign convention, this is the work done by the control volume. When work is done on the control volume such as compressors or pumps, the reversible work going into the control volume is wrev , in = v dP + ∆k...
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 Fall '08
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