Explanation The below figure shows the p − v diagram of the process. Figure-(1) Here, the process 1-2 is isentropic compression, the process 2-3 is constant volume heat addition process , the process 3-4 isentropic expansion and the process 4-1 is constant volume heat rejection process. Write the expression for swept volume. V 1 − V 2 = π 4 × d 2 × l (I) Here, the bore diameter is d and the length of stroke is l . Write the expression for characteristics gas constant of air. R = R ¯ M (II) Here, the universal gas constant is R ¯ and the mass of the air is M . Write the expression for mass of air. m = p 1 V 1 R T 1 (III) Here, the temperature at state 1 is T 1 , the volume at state 1 is V 1 and the characteristics gas constant is R . Write the expression for relative volume at state 2. V r 2 = V 2 V 1 × V r 1 (IV) Write the expression for relative volume at point 4. V r 4 = V 4 V 3 × V r 3 (V) Here, the volume at state 4 is V 4 , and the volume at state 3 is V 3 . Write the expression for net work per cycle. W c y c l e = m [ ( u 3 − u 2 ) − ( u 4 − u 1 ) ] (VI) Here, the specific internal energy at state 1 is u 1 , the specific internal energy at state 2 is u 2 , the specific internal energy at state 3 is u 3 , and the specific internal energy at state 4 is u 4 . Write the expression for net power. W ˙ n e t = 4 N W c y c l e (VII) Here, the number of cycle per minute is N . Conclusion: Substitute 3.7 in for d and 3.4 in for l in Equation (I). V 1 − V 2 = π 4 × ( 3.7 in ) 2 × ( 3.4 in ) = ( 36.557 in 3 ) ( 1 ft 3 1728 in 3 ) = 0.02115 ft 3 (VII) Since the value of V 2 is 16% of the V 1 so V 2 = 0.16 V 1 . Substitute 0.16 V 1 for V 2 in Equation (VII). V 1 − 0.16 V 1 = 0.02115 ft 3 V 1 ( 1 − 0.16 ) = 0.02115 ft 3 0.84 V 1 = 0.02115 ft 3 V 1 = 0.02115 ft 3 0.84 V 1 = 0.0251785 ft 3 Substitute 1545 ft ⋅ lbf for R ¯ and 28.97 lb ⋅ ° R for M in Equation (II). R = 1545 ft ⋅ lbf 28.97 lb ⋅ ° R = 53.33 ft ⋅ lbf / lb ⋅ ° R Substitute 14.5 lbf / in 2 for p 1 , 0.0251785 ft 3 for V 1 , 53.33 ft ⋅ lbf / lb ⋅ ° R for R and 520 ° R for T 1 in Equation (III). m = ( 14.5 lbf / in 2 ) ( 0.0251785 ft 3 ) ( 53.33 ft ⋅ lbf / lb ⋅ ° R ) ( 520 ° R ) = ( 14

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832318

Author: MORAN

Publisher: WILEY

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Chapter 9.14, Problem 14P

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Chapter 9 Solutions

Fundamentals of Engineering Thermodynamics

Ch. 9.14 - Prob. 1E Ch. 9.14 - Prob. 2E Ch. 9.14 - Prob. 3E Ch. 9.14 - Prob. 4E Ch. 9.14 - Prob. 5E Ch. 9.14 - 6. What is the purpose of a rear diffuser on a...Ch. 9.14 - 7. What is the meaning of the octane rating that...Ch. 9.14 - Prob. 8E Ch. 9.14 - Prob. 9E Ch. 9.14 - 10. What is the purpose of the gas turbine–powered...

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The net work per cycle. The power developed by engine.

The net work per cycle. The power developed by engine.

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Chapter 9 Solutions

The net work per cycle.

The power developed by engine.