To determine Minimum temperature and the maximum thermal efficiency of this cycle Explanation The given cycle is the simple Rankine cycle and operated at the boiler and condenser pressures of 3 MPa and 30 kPa respectively. The simple Rankine cycle involves following processes and following conditions: Processes: 1. Process 1 to 2 – Isentropic compression in a Pump 2. Process 2 to 3 – Heat addition at constant pressure in a Boiler 3. Process 3 to 4 – Isentropic expansion in a Turbine 4. Process 4 to 1 – Heat rejection at constant pressure in a Condenser Conditions: The pressure is constant for the process 2 to 3 and the process 4 to 1. P 2 = P 3 and P 4 = P 1 The entropy is constant for the process 1 to 2 and the process 3 to 4. s 1 = s 2 and s 3 = s 4 Here, water is the working fluid. Write the formula for work done by the pump during process 1-2. w p,in = v 1 ( P 2 − P 1 ) (I) Here, the specific volume is v , the pressure is P . Write the formula for enthalpy ( h ) at state 2. h 2 = h 1 + w p,in (II) Write the formula for heat in ( q in ) and heat out ( q out ) of the cycle. q in = h 3 − h 2 (III) q out = h 4 − h 1 (IV) Write the formula for thermal efficiency of the cycle ( η th ) . η th = 1 − q out q in (V) It is given that the quality of water at the exit of the turbine (state 4) is 85 %. x 4 = 80 % × 1 100 = 0.80 The enthalpy and entropy at state 4 is expressed as follows. h 4 = h f , 4 + x 4 h f g , 4 (VI) s 4 = s f , 4 + x 4 s f g , 4 (VII) Here, the enthalpy is h , the entropy is s , the quality of the water is x , the suffix f indicates the fluid condition, the suffix f g indicates the change of vaporization phase; the suffix 4 indicates the state 4. At state 1: The water enters the condenser as a saturated liquid at the pressure of 5 psia . Hence, the enthalpy at state 1 is as follows. h 1 = h f @ 5 psia Refer Table A-5E, “Saturated water-Pressure table”. The enthalpy ( h 1 ) and specific volume ( v 1 ) at state 1 corresponding to the pressure of 5 psia is 130.18 Btu/lbm and 0.01641 ft 3 /lbm respectively. At state 4: Refer Table A-5E, “Saturated water-Pressure table”. Obtain the following properties corresponding to the pressure of 5 psia . h f , 4 = 130.18 Btu/lbm h f g , 4 = 1000.5 Btu/lbm s f , 4 = 0.23488 Btu/lbm ⋅ R s f g , 4 = 1.60894 Btu/lbm ⋅ R Conclusion: Substitute 0.01641 ft 3 /lbm for v 1 , 2500 psia for P 2 and 5 psia for P 1 in Equation (I). w p,in = ( 0.01641 ft 3 /lbm ) ( 2500 psia − 5 psia ) = ( 0.01641 ft 3 /lbm ) ( 2500 psia − 5 psia ) ( 1 Btu 5.404 psia ⋅ ft 3 ) = 7.58 Btu / lbm Substitute 130.18 Btu / lbm for h 1 and 7.58 Btu / lbm for w p,in in Equation (II). h 2 = 130.18 Btu / lbm + 7.58 Btu / lbm = 137.76 Btu / lbm Substitute 130.18 Btu / lbm for h f , 4 , 0.80 for x 4 and 1000.5 Btu / lbm for h f g , 4 in Equation (VI). h 4 = 130.18 Btu / lbm + ( 0.80 ) ( 1000

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Chapter 10.9, Problem 15P

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Minimum temperature and the maximum thermal efficiency of this cycle

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Ch. 10.9 - Why is the Carnot cycle not a realistic model for...Ch. 10.9 - Prob. 2P Ch. 10.9 - A steady-flow Carnot cycle uses water as the...Ch. 10.9 - A steady-flow Carnot cycle uses water as the...Ch. 10.9 - Consider a steady-flow Carnot cycle with water as...Ch. 10.9 - Consider a simple ideal Rankine cycle with fixed...Ch. 10.9 - Consider a simple ideal Rankine cycle with fixed...Ch. 10.9 - Consider a simple ideal Rankine cycle with fixed...Ch. 10.9 - How do actual vapor power cycles differ from...Ch. 10.9 - The entropy of steam increases in actual steam...

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