(a) Explanation The below figure shows the p − V diagram of the process. Figure-(1) Write the expression for work done in the process 1-2. W 1 − 2 = R T 1 ln ( p 1 p 2 ) (I) Here, the temperature at point 1 is T 1 , the pressure at point 1 is p 1 , the pressure at point 2 is p 2 . Write the expression for heat transfer in the process 1-2. q 1 − 2 = W 1 − 2 + Δ U (II) Here, the change in internal energy is Δ U . Write the expression for heat transfer for the process 2-3. q 2 − 3 = W 2 − 3 + u 3 − u 2 (III) Here, the work done in the process 2-3 is W 2 − 3 , the internal energy at point 3 is u 3 and the internal energy at point 2 is u 2 . Write the expression for work done in the process 3-4. W 3 − 4 = R T 3 ln ( p 2 p 1 ) (IV) Here, the temperature at point 3 is T 3 . Write the expression for heat transfer in the process 3-4. q 3 − 4 = W 3 − 4 + Δ U (V) Write the expression for heat transfer in the process 4-1. q 4 − 1 = W 4 − 1 + Δ U (VI) Conclusion: Substitute 385.96 ft ⋅ lbf / lb ° R for R , 560 ° R for T 1 , 15 bar for p 1 , 150 bar for p 2 in Equation (I). W 1 − 2 = ( 385.96 ft ⋅ lbf / lb ° R ) ( 560 ° R ) { ln ( 15 bar 150 bar ) } = ( 216137.66 ft ⋅ lbf / lb ) × ( − 2.3025 ) = ( − 497675.2158 ft ⋅ lbf / lb ) ( 1Btu 778 ft ⋅ lbf ) = − 639.6619 Btu / lb Thus, the work done in the process 1-2 is − 639.6619 Btu / lb . Since the process 1-2 isothermal process, so Δ U = 0 Btu / lb Substitute − 639.6619 Btu / lb for W 1 − 2 , 0 Btu / lb for Δ U in Equation (II). q 1 − 2 = − 639.6619 Btu / lb + 0 Btu / lb = − 639.6619 Btu / lb Thus, the heat transfer in the process 1-2 is − 639 (b) To determine The thermal efficiency of the cycle.

8th Edition

ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 9.14, Problem 103P

(a)

To determine

The work done in each process.

The heat transfer in each process.

(b)

To determine

The thermal efficiency of the cycle.

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Chapter 9.14, Problem 102P

Chapter 9.14, Problem 104P

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The work done in each process. The heat transfer in each process.