Explanation For the given ideal Otto cycle with air as the working fluid, let T i , P i , v i , v r i , and u i be the temperature, pressure, specific volume, relative specific volume, and specific internal energy at the i t h state respectively. Write the expression of specific volume and relative specific volume relation for the isentropic compression process 1-2. v r 2 = v 2 v 1 v r 1 = 1 r v r 1 (I) Here, compression ratio is r . Write the pressure, temperature, and specific volume relation for isentropic compression process 1-2. P 2 v 2 T 2 = P 1 v 1 T 1 (II) Write the expression of heat addition for constant volume heat addition process 2-3 ( q 23 , in ) . q 23 , i n = u 3 − u 2 (III) Rearrange Equation (III), and substitute 750 kJ / kg for q 23 , in , 491.2 kJ / kg for u 2 . u 3 = u 2 + q 23 , in = ( 491.2 + 750 ) kJ / kg = 1 , 241.2 kJ / kg Write the expression of temperature, pressure, and specific volume relation for the constant volume heat addition process 2-3. P 3 v 3 T 3 = P 2 v 2 T 2 (IV) Write the expression of specific volume and relative specific volume relation for the isentropic expansion process 3-4. v r 4 = v 4 v 3 v r 3 v r 4 = v 1 v 2 v r 3 v r 4 = r v r 3 (V) Write the expression of heat rejected ( q out ) for constant volume heat rejection process 4-1. q out = u 4 − u 1 (VI) Write the ideal gas equation for air at state 1. v 1 = R T 1 P 1 (VII) Here, gas constant is R . Write the expression of total exergy destruction associated with the given Otto cycle ( x destroyed ) . x destroyed = T 0 ( q out T L − q in T H ) (VIII) Here, temperature of the surroundings is T 0 , the temperature of the heat source is T H , and the temperature of the sink is T L . Write the expression of exergy at the end of the power stroke ( ϕ 4 ) . ϕ 4 = ( u 4 − u 0 ) − T 0 ( s 4 − s 0 ) + P 0 ( v 4 − v 0 ) (IX) Here, specific entropy at state 4 is s 4 , and the specific entropy of the surroundings is s 0 , specific volume of the surroundings is v 0 , and specific internal energy of the surroundings is u 0 . Write the expression of change in specific entropy of air for the power stroke ( s 4 − s 0 ) . s 4 − s 0 = s 4 − s 1 s 4 − s 0 = s 4 ∘ − s 1 ∘ − R ln P 4 P 1 s 4 − s 0 = s 4 ∘ − s 1 ∘ − R ln T 4 v 1 T 1 v 4 s 4 − s 0 = s 4 ∘ − s 1 ∘ − R ln T 4 T 1 (X) Here, the entropy at state 4 as a function of temperature alone is s 4 ∘ , and the entropy at state 1 as a function of temperature alone is s 1 ∘ . Conclusion: Refer Table A-17,“Ideal gas properties of air”,obtain the properties of air at 300 K ( T 1 ) , which gives 214 .07 kJ / kg for u 1 , and 621.2 for v r 1 . Substitute 621.2 for v r 1 , and 8 for r in Equation (I). v r 2 = 1 8 ( 621.2 ) = 77.65 Refer Table A-17,“Ideal gas properties of air”,obtain the properties of air at 77.65 ( v r 2 ) , which gives 673.1 K for T 2 , and 491.2 kJ / kg for u 2 . Rearrange Equation (II), and substitute 8 for ( v 1 v 2 ) , 95 kPa for P 1 , 673.1 K for T 2 , and 300 K for T 1 . P 2 = v 1 v 2 T 2 T 1 P 1 = ( 8 ) ( 673

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Question

Chapter 9.12, Problem 138P

To determine

The total exergy destruction associated with the given Otto cycle and the exergy at the end of the power stroke.

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Chapter 9.12, Problem 137P

Chapter 9.12, Problem 139P

Chapter 9 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 9.12 - What are the air-standard assumptions?Ch. 9.12 - What is the difference between air-standard...Ch. 9.12 - How does the thermal efficiency of an ideal cycle,...Ch. 9.12 - What does the area enclosed by the cycle represent...Ch. 9.12 - Prob. 5P Ch. 9.12 - Prob. 6P Ch. 9.12 - Can the mean effective pressure of an automobile...Ch. 9.12 - Prob. 8P Ch. 9.12 - What is the difference between spark-ignition and...Ch. 9.12 - Prob. 10P

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