(a) Explanation The figure below shows the T − s diagram of an air-standard regenerative cycle. Figure-(1) The process 1 − 2 s is an isentropic expansion process, 2-3 is an isobaric heat addition process, 3 − 4 s is an isentropic compression process and 4-1 is an isobaric heat rejection process. The process 1-2 is a compression process with losses and the process 3-4 is an expansion process with losses. The air in process 2 s − x is preheated through the exhaust air from 4 s − y . Write the expression for the efficiency of the regenerator. η r e g = h x − h 2 h 4 − h 2 h x = h 2 + η r e g ( h 4 − h 2 ) (I) The heat lost by the exhaust gases is utilized by the regenerator. Write the expression for the steady flow energy equation for the regenerator. Q ˙ + m ˙ ( h 2 + h 4 ) = W ˙ + m ˙ ( h x + h y ) (II) Here, the specific enthalpy at state 2 is h 2 , the specific enthalpy at state 4 is h 4 , the specific enthalpy at state x is h x , the specific enthalpy at state y is h y , the rate of work transfer through the regenerator is W ˙ , the mass flow rate is m ˙ and the rate of heat transfer through the regenerator is Q ˙ . There is no heat transfer and work transfer through the regenerator hence, W ˙ and Q ˙ is zero. Substitute 0 for W ˙ and 0 for Q ˙ in the Equation (II). 0 + m ˙ ( h 2 + h 4 ) = 0 + m ˙ ( h x + h y ) h y = h 2 + h 4 − h x (III) Conclusion: Substitute 688.27 kJ / kg for h 2 , 0.7 for η r e g and 955.74 kJ / kg for h 4 in Equation (I). h x = 688.27 kJ / kg + 0.7 ( 955.74 kJ / kg − 688.27 kJ / kg ) = 688.27 kJ / kg + 0.7 ( 267.47 kJ / kg ) = 688.27 kJ / kg + 187.229 kJ / kg ≈ 875.5 kJ / kg Thus, the specific enthalpy at state x is 875.5 kJ / kg . Refer to table A-22 “Ideal Gas Properties of Air” and obtain the below properties to determine the temperature at state x using interpolation. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (IV) Here, the variables denote by x is specific enthalpy and the variable denoted by y is temperature. Show the specific enthalpy as in Table A-22. Specific enthalpy Temperature 866.08 kJ / kg 840 K 875.5 kJ / kg ? 888.27 kJ / kg 860 K Substitute 866.08 kJ / kg for x 1 , 875 (b) To determine The back work ratio of the cycle. (c) To determine The rate of heat transfer to the working fluid passing through the combustor. (d) To determine The thermal efficiency of the cycle.

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ISBN: 2818440116926

Author: MORAN

Publisher: WILEY CONS

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Chapter 9.14, Problem 61P

(a)

To determine

The specific enthalpy of each stream exiting from the regenerator.

The temperature of each stream exiting from the regenerator.

(b)

To determine

The back work ratio of the cycle.

(c)

To determine

The rate of heat transfer to the working fluid passing through the combustor.

(d)

To determine

The thermal efficiency of the cycle.

Expert Solution & Answer

Chapter 9 Solutions

FUNDAMENTALS OF ENGINEERING THERMODYNAM

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The specific enthalpy of each stream exiting from the regenerator. The temperature of each stream exiting from the regenerator.

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The specific enthalpy of each stream exiting from the regenerator. The temperature of each stream exiting from the regenerator.

The back work ratio of the cycle.

The rate of heat transfer to the working fluid passing through the combustor.

The thermal efficiency of the cycle.

Chapter 9 Solutions

The specific enthalpy of each stream exiting from the regenerator.

The temperature of each stream exiting from the regenerator.