(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 s − 3 is an isobaric heat rejection process, 3 − 4 s is an isentropic expansion process and 4 s − 1 is an isobaric heat addition process. The process 3-4 is an expansion process with losses and the process 1-2 is a compression process with losses. The air is cooled from a to 3 in a regenerator through the heat exchange from b to 1. Write the expression for the relative pressure at state 2 s . p r 2 s p r 1 = p 2 s p 1 p r 2 s = p 2 s p 1 × p r 1 (I) Here, the pressure at state 2 s is p 2 s , the pressure at state 2 s is p 2 s and the relative pressure at state 1 is p r 1 . Write the expression for the relative pressure at state 4 s . p r 4 s = p r 3 ( p 4 s p 3 ) (II) Here, the pressure at state 4 s is p 4 s , 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 (III) Here, the specific enthalpy at state a is h a , the specific enthalpy at state 1 is h 1 , the specific enthalpy at state 3 is h 3 and the specific enthalpy at state b is h b . Write the expression for the isentropic efficiency of the turbine. η t = h 3 − h 4 h 3 − h 4 s (IV) Write the expression for the isentropic efficiency of the compressor. η c = h 2 s − h 1 h 2 − h 1 (V) Write the expression for the mass flow rate of the air. m ˙ = Q ˙ i n ( h b − h 4 ) (VI) 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 (VII) Here, the gas constant is R , the pressure at state 1 is p 1 and the temperature at state 1 is T 1 . 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 s is an isentropic process, hence s 1 = s 2 s . Substitute 1.0590 for p r 1 , 45 lbf / in 2 for p 2 s and 16 lbf / in 2 for p 1 in Equation (III). p r 2 s = 45 lbf / in 2 16 lbf / in 2 × 1.0590 = 2.978 The specific enthalpy at state 2 s is evaluated by interpolation. Write the expression for interpolation. h 2 s − h 21 h 22 − h 21 = p r 2 s − p r 21 p r 22 − p r 21 (VIII) 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 Equation (VIII). h 2 s − 157.92 Btu / lb 162.73 Btu / lb − 157.92 Btu / lb = 2.9784 − 2.514 3.111 − 2.514 h 2 s − 157.92 Btu / lb 4.81 Btu / lb = 0.77789 h 2 s − 157.92 Btu / lb = 4.81 Btu / lb × 0.77789 h 2 s = 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 (IX) 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 (IX). 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 . Write the expression for the specific enthalpy at state 3. h 3 = h 31 + h 32 2 (X) Substitute 114.69 Btu / lb for h 31 and 119.48 Btu / lb for h 32 in Equation (X). h 3 = 114.69 Btu / lb + 119.48 Btu / lb 2 = 234.17 Btu / lb 2 = 117.085 Btu / lb ≈ 117.09 Btu / lb Refer to table A-22E “Ideal gas properties of air” to obtain relative pressure p r 31 as 0.9182 and relative pressure p r 32 as 1.059 . Write the expression for the relative pressure at state 3. p r 3 = p r 31 + p r 32 2 (XI) Substitute 0.9182 for p r 31 and 1.059 for p r 32 in Equation (XI). p r 3 = 0.9182 + 1.059 2 = 1.9772 2 = 0.9886 Substitute 0.9886 for p r 3 , 45 lbf / in 2 for p 3 and 16 lbf / in 2 for p 4 s in the Equation (IV). p r 4 s = 0.9886 ( 16 lbf / in 2 45 lbf / in 2 ) = 0.9886 ( 0.3555 ) = 0.3515 Write the expression for interpolation. h 4 s − h 41 h 42 − h 41 = p r 4 s − p r 41 p r 42 − p r 41 (XII) 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 41 as 85 (b) To determine The coefficient of performance. (d) To determine The entropy production within the compressor. The entropy production within the turbine.

Fundamentals of Engineering Thermodynamics, Binder Ready Version

8th Edition

ISBN: 9781118820445

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

Publisher: WILEY

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Question

Chapter 10.7, Problem 54P

(a)

To determine

The volume flow rate of air.

(b)

To determine

The coefficient of performance.

(d)

To determine

The entropy production within the compressor.

The entropy production within the turbine.

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Chapter 10.7, Problem 53P

Chapter 10.7, Problem 55P

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Chapter 10 Solutions

Fundamentals of Engineering Thermodynamics, Binder Ready Version

Ch. 10.7 - Prob. 1E Ch. 10.7 - Prob. 2E Ch. 10.7 - Prob. 3E Ch. 10.7 - Prob. 4E Ch. 10.7 - Prob. 5E Ch. 10.7 - Prob. 6E Ch. 10.7 - Prob. 7E Ch. 10.7 - Prob. 8E Ch. 10.7 - Prob. 9E Ch. 10.7 - Prob. 10E

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