Explanation The p - v diagram of the Diesel cycle is shown below. Figure-(1) The T - s diagram of the Diesel cycle is shown below. Figure-(2) Write the expression between, temperature and compression ratio for process 1-2. T 2 T 1 = ( r ) k − 1 (I) Here, the adiabatic index is k , the temperature at compressor outlet is T 2 , the compressor inlet is T 1 and the compression ratio is r . Write the expression for the cutoff ratio. r c = T 3 T 2 (II) Here, the temperature at point 3 is T 3 and at point 2 is T 2 , the volume of cylinder before expansion is V 3 and after compression is V 2 . Write the expression between, temperature and cutoff ratio for process 4-1. T 4 T 1 = ( r c ) k (III) Here, the temperature at point 4 is T 4 and at point 1 is T 1 and the cutoff ratio is r c . Write the expression the net developed. w n e t = m [ c p ( T 3 − T 2 ) − c v ( T 4 − T 1 ) ] (IV) Here, the mass of the air is m , Write the expression for the thermal efficiency of diesel cycle. η t h = w n e t m c p ( T 3 − T 2 ) (V) Write the expression for the inlet volume at point 1. v 1 = m R T 1 p 1 (VI) Here, the gas constant is R and mass of the air is m . Write the expression for compression ratio. r = v 1 v 2 (VII) Here, the volume at point 2 is v 2 . Write the expression for m e p . m e p = w n e t v 1 − v 2 (VIII) Here, the net work is w n e t . Conclusion: For r = 15 : Substitute 15 for r , 1.4 for k and 300 K for T 1 in Equation (I). T 2 300 K = ( 15 ) 1.4 − 1 T 2 = 300 K × ( 15 ) 0.4 T 2 = 886.25 K Substitute 886.25 K for T 2 and 1800 K for T 3 in Equation (II). r c = 1800 K 886.25 K = 2.03 Substitute 2.03 for r c , 1.4 for k and 300 K for T 1 in Equation (III). T 4 300 K = ( 2.03 ) 1.4 T 4 = 300 K × ( 2.03 ) 1.4 T 4 = 808.4 K Substitute 300 K for T 1 , 95 kPa for p 1 , 0.012 kg for m and 0.287 kJ / kg ⋅ K in Equation (VI). v 1 = 0.012 kg × 0.287 kJ / kg ⋅ K × 300 K 95 kPa = 1.0032 95 m 3 / kg = 0.0108 m 3 / kg Substitute 0.0108 m 3 / kg for v 1 and 15 for r in Equation (VII). 15 = 0.0108 m 3 / kg v 2 v 2 = 0.0108 m 3 / kg 15 v 2 = .00072 m 3 / kg Substitute 0.012 kg for m , 1.005 kJ / kg ⋅ K for c p , 0.718 kJ / kg ⋅ K for c v , 300 K for T 1 , 886.25 K for T 2 , 1800 K for T 3 and 808.4 K for T 4 in Equation (IV). W n e t = 0.012 kg × [ 1.005 kJ / kg ⋅ K ( 1800 K − 886.25 K ) − 0.718 kJ / kg ⋅ K × ( 808.4 K − 300 K ) ] = 0.012 kg × [ 918.32 kJ / kg − 365 kJ / kg ] = 6.63 kJ Thus, the net work developed is 6.63 kJ . Substitute 6.63 kJ for W n e t , 0.012 kg for m , 1.005 kJ / kg ⋅ K for c p , 886.25 K for T 2 , 1800 K for T 3 in Equation (V). η t h = 6.63 kJ 0.012 kg × 1.005 kJ / kg ⋅ K × ( 1800 K − 886.25 K ) = 0.6016 = 60.16 % Thus, the thermal efficiency is 60.16 % . Substitute 6.63 kJ for W n e t , 0.0108 m 3 / kg for v 1 and 0.00072 m 3 / kg for v 2 in Equation (VIII). m e p = 6.63 kJ 0.0108 m 3 / kg − 0.00072 m 3 / kg = 6.63 kJ 0.01008 m 3 / kg = 657.73 kPa Thus, the mean effective pressure is 657.73 kPa . For r = 18 : Substitute 18 for r , 1.4 for k and 300 K for T 1 in Equation (I). T 2 300 K = ( 18 ) 1.4 − 1 T 2 = 300 K × ( 18 ) 0.4 T 2 = 953 K Substitute 953 K for T 2 and 1800 K for T 3 in Equation (II). r c = 1800 K 953 K = 1.88 Substitute 1.88 for r c , 1.4 for k and 300 K for T 1 in Equation (III). T 4 300 K = ( 1.88 ) 1.4 T 4 = 300 K × ( 1.88 ) 1.4 T 4 = 726 K Substitute 300 K for T 1 , 95 kPa for p 1 , 0

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ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 9.14, Problem 29P

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The net work developed.

The thermal efficiency.

The mean effective pressure.

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Problem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and (y2), respectively. Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s]. Givens: y1 = 4.112 m y2 = 0.387 m b = 0.942 m Answers: ( 1 ) 1880.186 lit/s ( 2 ) 4042.945 lit/s ( 3 ) 2553.11 lit/s ( 4 ) 3130.448 lit/s

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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.