Chapter_7_2

625 the net work done by the cycle is wnet cycle th qh

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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 7-51 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 2-3. c. Using equation (6-34) ∆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 1-2 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 7-52 m1 = m2 PV1 PV2 1 = 2 T1 T2 P2 = P 1 V1 3V1 1 3 = (800 kPa ) = 266.7 kPa Consider process 2-3 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 7-53 Now, consider process 4-1 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 3-4 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 7-54 Reversible Steady-Flow Work Isentropic, Steady Flow through Turbines, Pumps, and Compressors Consider a turbine, pump, compressor, or other steady-flow control volume, work-producing device. The general first law for the steady-flow control volume is Ein = Eout Vi 2 Ve2 Qnet + ∑ mi (hi + + gzi ) = Wnet + ∑ me (he + + gze ) 2 2 inlets exits For a one-entrance, one-exit 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 7-55 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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