Explanation The following figure shows the P - V diagram of the cycle. Figure-(1) In Figure-(1), the process 1 to 2 is isentropic compression, process 2 to 3 is constant volume heat addition, process 3 to 4 is isentropic expansion and process 4 to 1 is constant volume heat rejection. Write the expression for compression ratio. r = V 1 V 2 (I) Here, compression ratio is r , volume at state 1 is V 1 and volume at state 2 is V 2 . Write the expression for temperature at state 2. T 2 = T 1 ( V 1 V 2 ) k − 1 (II) Here, temperature at state 2 is T 2 , temperature at state 1 is T 1 and adiabatic index is k . Write the expression for pressure at state 2. P 2 = ( P 1 ) ( V 1 V 2 ) k (III) Here, pressure at state 2 is P 2 and pressure at state 1 is P 1 . Write the expression for temperature at state 4. T 4 = T 3 ( V 3 V 4 ) k − 1 (IV) Here, temperature at state 3 is T 3 , temperature at state 4 is T 4 . Write the expression for temperature at state 3. T 3 = ( P 3 P 2 ) ( T 2 ) (V) Here, pressure at state 3 is P 3 . Write the expression for mass of air. m = P 1 V 1 R T 1 (VI) Here, mass of the air is m and gas constant for air is T 1 . Write the expression for work done per cycle. W cycle = m c v ( ( T 1 − T 2 ) + ( T 3 − T 4 ) ) (VII) Here, work done per cycle is W cycle and specific heat at constant volume c v . Calculate the power developed by cycle. W ˙ = ( n ) ( N 2 ) ( W cycle ) (VIII) Here, power developed by cycle is W ˙ , number of cylinder is n and rotation in RPM is N . Calculate the mean effective pressure. P mean = W cycle V 1 ( 1 − V 2 V 1 ) (IX) Here, mean effective pressure is P mean . Conclusion: The adiabatic index for air 1.4 , specific heat at constant volume for air is 0.171 Btu / lb ⋅ ° R and gas constant for air is 53.33 ft ⋅ lbf / lb ⋅ ° R . Substitute 10 for V 1 V 2 in Equation (I). r = 10 Convert initial temperature T 1 at state 1 from Fahrenheit to Rankine. T 1 = 80 ° F = ( 80 ° F + 460 ) ° R = 540 ° R Substitute 540 ° R for T 1 , 10 for V 1 V 2 and 1.4 for k in Equation (II). T 2 = ( 540 ° R ) ( 10 ) 1.4 − 1 = ( 540 ° R ) ( 2.5118 ) = 1356.418 ° R Substitute 14.7 lbf / in 2 for P 1 , 10 for V 1 V 2 and 1.4 for k in Equation (III). P 2 = ( 14.7 lbf / in 2 ) ( 10 ) 1.4 = ( 14.7 lbf / in 2 ) ( 25.1188 ) = 369.247 lbf / in 2 Substitute 1080 lbf / in 2 for P 3 , 369.247 lbf / in 2 for P 2 and 1356.418 ° R for T 2 in Equation (V). T 3 = ( 1080 lbf / in 2 369.247 lbf / in 2 ) ( 1356.418 ° R ) = ( 2.92248 ) ( 1356.418 ° R ) ≈ 3967.348 ° R Substitute 3967.348 ° R for T 3 , 1 10 for V 3 V 4 and 1

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ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 9.14, Problem 6P

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The power developed by the engine.

The mean effective pressure of the cycle.

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The power developed by the engine. The mean effective pressure of the cycle.

The power developed by the engine. The mean effective pressure of the cycle.

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Chapter 9 Solutions

The power developed by the engine.

The mean effective pressure of the cycle.