(a) Explanation Write the expression for ideal gas constant of air. R = R ¯ M (I) Here, the universal gas constant is R ¯ and the molecular mass of the air is M . Write the expression for thermal efficiency of the different compression ratio. η r = 1 − 1 ( r ) k − 1 (II) Here, the compression ratio is r an he ratio of specific heat is k . Write the expression for temperature at point 2 for different compression ratio. ( T 2 ) r = ( r ) k − 1 × T 1 (III) Here, the temperature at state 1 is T 1 . Write the expression for work done for different compression ratio. ( W c y c l e m ) r = c v ( T 3 − ( T 2 ) i ) × η r (IV) Here, the specific heat at constant volume is c v and the temperature at state 3 is T 3 . Write the expression for specific volume at point 1. v 1 = R T 1 p 1 (V) Write the expression for mean effective pressure for different compression ratio. ( m e p ) r = W c y c l e / m v 1 ( 1 − 1 r ) (VI) Conclusion: Substitute 28.97 lb / mol for M and 1545 lbf ⋅ ft / lbmol ⋅ ° R for R ¯ in Equation (I). R = 1545 lbf ⋅ ft / lbmol ⋅ ° R 28.97 lb / mol = 53.343 lbf ⋅ ft / lb ° R Refer table A-20E “Ideal gas specific heat of some common gases” to obtain the specific heat ratio of air and specific heat at constant volume as k = 1.40 and c v = 0.171 Btu / lb ° R Substitute 6 for r and 1.4 for k in Equation (II). η 6 = 1 − 1 ( 6 ) 1.4 − 1 = 1 − 0.4883 = 0.5116 Thus, the thermal efficiency for compression ratio 6 is 0.5116 . Substitute 6 for r , 1.4 for k and 530 ° R for T 1 in Equation (III). ( T 2 ) 6 = ( 6 ) 1.4 − 1 ( 530 ° R ) = ( 6 ) 0.4 ( 530 ° R ) = 2.04767 ( 530 ° R ) = 1085.26 ° R Substitute 0.5116 for η 6 , 0.171 Btu / lb ° R for c v , 3000 ° R for T 3 and 1085.26 ° R for ( T 2 ) 6 in Equation (IV). ( W c y c l e m ) 6 = 0.171 Btu / lb ° R ( 3000 ° R − 1085.26 ° R ) × 0.5116 = 0.08748 Btu / lb ° R ( 1914.74 ° R ) = 167.5014 Btu / lb Thus, the net work for compression ratio 6 is 167.5014 Btu / lb . Substitute 53.343 ft ⋅ lbf / lb ⋅ ° R for R , 530 ° R for T 1 and 14.7 lbf / in 2 for p 1 in Equation (V). v 1 = ( 53.343 ft ⋅ lbf / lb ⋅ ° R ) ( 530 ° R ) 14.7 lbf / in 2 = 28271.79 ft ⋅ lbf / lb 14.7 lbf / in 2 ( 1 ft 2 144 in 2 ) = 13.35 ft 3 / lb Substitute 13.35 ft 3 / lb for v 1 , 6 for r and 167.5014 Btu / lb for W c y c l e / m in Equation (VI). ( m e p ) 6 = 167.5014 Btu / lb ( 13.35 ft 3 / lb ) ( 1 − 1 6 ) = 167.5014 Btu / lb ( 13.35 ft 3 / lb ) ( 5 6 ) ( 778 ft ⋅ lbf 1 Btu ) ( 1 ft 2 144 in 2 ) = 81.34 lbf / in 2 Thus, the mean effective pressure for compression ratio 6 is 81.34 lbf / in 2

8th Edition

ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 9.14, Problem 10P

(a)

To determine

The net work developed.

The thermal efficiency.

The mean effective pressure.

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The net work developed. The thermal efficiency. The mean effective pressure.

The net work developed. The thermal efficiency. The mean effective pressure.

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Chapter 9 Solutions

The net work developed.

The thermal efficiency.

The mean effective pressure.