(a) Explanation The figure below shows the T - s diagram of the given cycle. Figure-(1) The process 1-2 is an isentropic compression process, process 2-3 is an isobaric heat rejection process, 3-4 is an isentropic expansion process and 4-1 is an isobaric heat addition process. The air is cooled from a to 3 in a regenerator through the heat exchange from b to 1. Write the expression for the mass flow rate of the air. m ˙ = Q ˙ i n ( h b − h 4 ) (I) Here, the rate of heat input to the cycle is Q ˙ i n , the specific enthalpy at state 4 is h 4 and the specific enthalpy at state b is h b . Write the expression for the volumetric flow rate of the air. ( A V ) 1 = m ˙ R T 1 p 1 (II) Here, the gas constant is R , the pressure at state 1 is p 1 and the temperature at state 1 is T 1 . Write the expression for the relative pressure at state 2. p r 2 p r 1 = p 2 p 1 p r 2 = p 2 p 1 × p r 1 (III) Write the expression for the relative pressure at state 4. p r 4 = p r 3 ( p 4 p 3 ) (IV) Here, the pressure at state 4 is p 4 , the pressure at state 3 is p 3 and the relative pressure at state 3 is p r 3 . Write the expression for the energy balance equation in the regenerator. h a − h 3 = h 1 − h b h b = h 1 − h a + h 3 (V) Here, the specific enthalpy at state a is h a and the specific enthalpy at state b is h b . Conclusion: Refer to table A-22E “Ideal gas properties of air” to obtain the specific enthalpy as 119.48 Btu / lb , specific entropy as 0.58233 Btu / lb ⋅ ° R and relative pressure as 1.0590 with respect to 500 ° R . The process 1-2 is an isentropic process hence s 1 = s 2 . Substitute 1.0590 for p r 1 , 45 lbf / in 2 for p 2 and 16 lbf / in 2 for p 1 in Equation (III). p r 2 = 45 lbf / in 2 16 lbf / in 2 × 1.0590 = 2.978 The specific enthalpy at state 2 is evaluated by interpolation. Write the expression for interpolation. h 2 − h 21 h 22 − h 21 = p r 2 − p r 21 p r 22 − p r 21 (VI) Here, the specific enthalpy is h and the relative pressure is p r . Refer to table A-22E “Ideal gas properties of air” to obtain specific enthalpy h 21 as 157.92 Btu / lb with respect to relative pressure p r 21 as 2.514 and the specific enthalpy h 22 as 162.73 Btu / lb with respect to relative pressure p r 22 as 3.111 . Substitute 157.92 Btu / lb for h 21 , 2.514 for p r 21 , 162.73 Btu / lb for h 22 , 3.111 for p r 22 and 2.9784 for p r 2 in the Equation (VI). h 2 − 157.92 Btu / lb 162.73 Btu / lb − 157.92 Btu / lb = 2.9784 − 2.514 3.111 − 2.514 h 2 − 157.92 Btu / lb 4.81 Btu / lb = 0.77789 h 2 − 157.92 Btu / lb = 4.81 Btu / lb × 0.77789 h 2 = 160.67 Btu / lb The specific enthalpy at state a is determined by interpolation. Write the expression for interpolation. T a − T a 1 T a 2 − T a 1 = h a − h a 1 h a 2 − h a 1 (VII) Here, the temperature is T and the specific enthalpy is h . Refer to table A-22E “Ideal gas properties of air” to obtain specific enthalpy h a 1 as 129.06 Btu / lb with respect to temperature T a 1 as 540 ° R and specific enthalpy h a 2 as 133.86 Btu / lb with respect to temperature T a 2 as 560 ° R . Substitute 129.06 Btu / lb for h a 1 , 540 ° R for T a 1 , 133.86 Btu / lb for h a 2 , 560 ° R for T a 2 and 550 ° R for T a in Equation (VII). 550 ° R − 540 ° R 560 ° R − 540 ° R = h a − 129.06 Btu / lb 133.86 Btu / lb − 129.06 Btu / lb 10 20 = h a − 129.06 Btu / lb 4.8 Btu / lb h a − 129.06 Btu / lb = 0.5 × 4.8 Btu / lb h a = 131.46 Btu / lb Refer to table A-22E “Ideal gas properties of air” to obtain specific enthalpy h 31 as 114.69 Btu / lb and specific enthalpy h 32 as 119.48 Btu / lb (b) To determine The coefficient of performance.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 10.7, Problem 53P

(a)

To determine

The volumetric flow rate at the compressor inlet.

(b)

To determine

The coefficient of performance.

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Chapter 10.7, Problem 51P

Chapter 10.7, Problem 54P

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Fundamentals of Engineering Thermodynamics

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