Explanation Draw P − v diagram for diesel cycle as shown in Figure (1). Write the relation between the specific volume, and relative specific volume for the isentropic compression process 1-2. v r 2 = v 2 v 1 v r 1 v r 2 = 1 r v r 1 (I) Here, the compression ratio is r , relative specific volume at state 1 is v r 1 , relative specific volume at state 2 is v r 2 , specific volume at state 1 is v 1 and specific volume at state 2 is v 2 . Write the ideal gas relation for the constant pressure heat addition process 2-3. P 3 v 3 T 3 = P 2 v 2 T 2 For the process 2-3, P 2 = P 3 v 3 T 3 = v 2 T 2 v 3 v 2 = T 3 T 2 (II) Here, pressure at state 3 is P 3 , pressure at state 2 is P 2 , specific volume at state 3 is v 3 , temperature at state 3 is T 3 and temperature at state 2 is T 2 . Write the expression of heat input to the cycle ( q i n ) . q i n = h 3 − h 2 (III) Here, enthalpy at state 3 is h 3 and enthalpy at state 2 is h 2 . Write the relation between the specific volume, and relative specific volume for the isentropic expansion process 3-4. v r 4 = v 4 v 3 v r 3 (IV) Here, relative specific volume at state 4 is v r 4 , relative specific volume at state 3 is v r 3 , specific volume at state 4 is v 4 and specific volume at state 3 is v 3 . Write the expression of heat rejected for constant volume heat rejection process 4-1 ( q out ) . q out = u 4 − u 1 (V) Here, internal energy at state 4 is u 4 and internal energy at state 1 is u 1 . Write the pressure and temperature relation for the process 4-1. P 4 P 1 = T 4 T 1 P 4 = P 1 T 4 T 1 (VI) Here, pressure at state 4 is P 4 , pressure at state 1 is P 1 and temperature at state 4 is T 4 . Write the expression of entropy change for the process 4-1 ( s 1 − s 4 ) . s 1 − s 4 = s 1 @ 580 R ∘ − s 4 @ 1531.8 R ∘ − R ln P 1 P 4 (VII) Write the expression of exergy destruction associated with the heat rejection process of the Diesel cycle ( x destroyed,41 ) . x destroyed,41 = T 0 ( s 1 − s 4 + q out T R ) (VIII) Here, the temperature of the surroundings is T 0 , and the temperature of the sink is T R . Write the expression of exergy at the end of the expansion stroke ( ϕ 4 ) . ϕ 4 = ( u 4 − u 0 ) − T 0 ( s 4 − s 0 ) + P 0 ( v 4 − v 0 ) (IX) Here, the specific entropy at state 4 is s 4 , the specific entropy of the surroundings is s 0 , pressure of the surroundings is P 0 , specific volume of the surroundings is v 0 , and specific internal energy of the surroundings is u 0 . Conclusion: Refer Table A-17E, “Ideal gas properties of air”, obtain the properties of air at 580 R ( T 1 ) , as 98.90 Btu / lbm for u 1 , and 120.70 for v r 1 . Substitute 18.2 for r and 120.70 for v r 1 in Equation (I). v r 2 = 1 18.2 120.70 = 6.632 Refer Table A-17E, “Ideal gas properties of air”, obtain the properties of air at 6.632 ( v r 2 ) as 1,725.8 R for T 2 , and 429.56 Btu / lbm for h 2 . Substitute 3,200 R for T 3 , and 1,725.8 R for T 2 in Equation (II). v 3 v 2 = 3 , 200 R 1 , 725.8 R = 1.854 Refer Table A-17E, “Ideal gas properties of air”, obtain the properties of air at 3200 R ( T 3 ) , as 849.48 Btu / lbm for h 3 , and 0

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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Author: CENGEL

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Chapter 9.12, Problem 144P

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The exergy destruction associated with the heat rejection process of the given Diesel cycle and the exergy at the end of the isentropic expansion process.

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Chapter 9.12, Problem 143P

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CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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The exergy destruction associated with the heat rejection process of the given Diesel cycle and the exergy at the end of the isentropic expansion process.

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The exergy destruction associated with the heat rejection process of the given Diesel cycle and the exergy at the end of the isentropic expansion process.

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