Explanation Show the given diesel cycle on P − V diagram as in Figure (1). For the given air-standard diesel cycle, V i , T i , P i , h i , and u i are the volumes at i t h state, temperature at i t h state, pressure at i t h state, enthalpy at i t h state, and internal energy at i t h state respectively. Write the expression of temperature and volume relation for the isentropic compression process 1-2. T 2 = T 1 ( V 1 V 2 ) k − 1 (I) Here, specific heat ratio is k . Write the expression of temperature, pressure, and volume relation for the constant pressure heat addition process 2-3. P 3 V 3 T 3 = P 2 V 2 T 2 (II) Write the expression of temperature, and volume relation for the isentropic expansion process 3-4. T 4 = T 3 ( V 3 V 4 ) k − 1 (III) Write the expression of ideal gas equation for air at state 1 of the given diesel cycle. m = P 1 V 1 R T 1 (IV) Here, mass of air is m and gas constant is R . Write the expression of heat input to the cycle ( Q in ) . Q in = m ( h 3 − h 2 ) Q in = m c p ( T 3 − T 2 ) (V) Here, specific heat at constant pressure is c p . Write the expression of heat rejected from the cycle ( Q out ) . Q out = m ( u 4 − u 1 ) Q out = m c v ( T 4 − T 1 ) (VI) Here, specific heat at constant volume is c v . Write the expression of net work output per revolution by the cycle ( W net,out ) . W net,out = Q in − Q out (VII) Write the expression of power developed by the engine ( W ˙ net,out ) . W ˙ net,out = n ˙ W net,out (VIII) Here, speed of the engine is n ˙ . From the table of “Molar mass, gas constant and critical-point properties”, select the gas constant of air. R = 0 .287 kPa ⋅ m 3 / kg ⋅ K From the table of “Ideal-gas specific heats of various common gases”, select the specific heat of air. c p = 1.005 kJ / kg ⋅ K c v = 0.718 kJ / kg ⋅ K Conclusion: Substitute 343 K for T 1 , 22 for V 1 V 2 , and 1.4 for k in Equation (I). T 2 = ( 343 K ) ( 22 ) 1.4 − 1 = ( 343 K ) ( 22 ) 0.4 = 1181 K For process 2-3, the pressures P 2 and P 3 are equal, as the pressure remains constant. Substitute 1.8 for V 3 V 2 and 1181 K for T 2 in Equation (II). T 3 = ( 1.8 ) ( 1181 K ) = 2126 K Substitute 1

8th Edition

ISBN: 9780073398174

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 9.12, Problem 57P

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The amount of power the engine will deliver at 3500 rpm.

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Chapter 9 Solutions

Thermodynamics: An Engineering Approach

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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