(a) To determine The thermal efficiency of the cycle. Explanation Write the expression for the saturated temperature at state 2. T 2 s = T 1 ( p 2 p 1 ) k − 1 / k (I) Here, the specific heat ratio is k , the temperature at state 1 is T 1 , the saturated temperature at state 2 is T 2 s , the pressure at inlet of turbine is p 1 and the pressure at exit is p 2 . Write the expression for the temperature at state 2. T 2 = T 1 + ( T 2 s − T 1 η c ) (II) Here, the temperature at state 2 is T 2 and the compressor efficiency is η c . Write the expression for the saturated temperature at state 7. T 7 s = T 6 ( p 7 p 6 ) k − 1 / k (III) Here, the saturated temperature at state 7 is T 7 s , the pressure at state 6 is p 6 and the pressure at state 7 is p 7 . Write the expression for the temperature using first stage turbine efficiency. T 7 = T 6 − η t 1 ( T 6 − T 7 s ) (IV) Here, the thermal efficiency of the turbine at first stage is η t 1 , the temperature at state 6 is T 6 , the temperature at state 7 is T 7 . Write the expression for the temperature at first stage of regenerator. T 5 = T 4 − η r e g ( T 9 − T 4 ) (V) Here, the temperature at state 9 is T 9 , the thermal efficiency of the regenerator is η r e g and the temperature at first stage of regenerator is T 5 . Write the expression for the temperature using second stage regenerator. T 10 = T 9 − ( T 5 − T 4 ) (VI) Here, the temperature at second stage of regenerator is T 10 . Write the expression rate of total heat addition. Q ˙ i n = m ˙ c p [ ( T 6 − T 5 ) + ( T 8 − T 7 ) ] (VII) Here, the rate of heat transfer addition is Q ˙ i n , the specific heat constant at constant pressure is c p and the mass flow rate is m ˙ . Write the expression rate of heat rejection. Q ˙ o u t = m ˙ c p [ ( T 10 − T 1 ) + ( T 2 − T 3 ) ] (VIII) Here, the rate of heat rejection is Q ˙ o u t , the specific heat constant at constant pressure is c p and the mass flow rate is m ˙ . Write the expression for the overall thermal efficiency of the plant. η = 1 − Q ˙ o u t Q ˙ i n (IX) Here, the thermal efficiency of the plant is η . Conclusion: Substitute 300 K for T 1 , 1.4 for k , 316.23 kPa for p 2 and 100 kPa for p 1 in Equation (I). T 2 s = ( 300 K ) ( 316.23 kPa 100 kPa ) 1.4 − 1 1.4 = ( 300 K ) ( 3.1623 ) 0.2857 ≈ 416.85 K Substitute 300 K for T 1 , 0.8 for η c 1 and 416.85 K for T 2 s in Equation (II). T 2 = 300 K + ( 416.85 K − 300 K 0.8 ) = 300 K − 146.06 K = 446 .06 K Since temperature at state 1 and state 3 is same and, also the pressure ratio at state 4 and 3 is also same as the pressure ratio at state 2 and 1. So, the temperature at state 4 is also same as state 2. That is, T 4 = 446.06 K . Substitute 1400 K for T 6 , 1.4 for k , 100 kPa for p 7 and 316.23 kPa for p 6 in Equation (III). T 7 s = ( 1400 K ) ( 100 kPa 316.23 kPa ) 1 (b) To determine The back work ratio for the cycle. (c) To determine The net power developed in the cycle. (d) To determine The rate of the exergy destruction for the compressor. The rate of the exergy destruction for the turbine. The rate of the exergy destruction for the regenerator.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 9.14, Problem 81P

(a)

To determine

The thermal efficiency of the cycle.

(b)

To determine

The back work ratio for the cycle.

(c)

To determine

The net power developed in the cycle.

(d)

To determine

The rate of the exergy destruction for the compressor.

The rate of the exergy destruction for the turbine.

The rate of the exergy destruction for the regenerator.

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