Explanation Draw the given ideal dual cycle on P − v diagram as in Figure (1) Write the ideal gas equation to calculate the mass of air. m = P 1 ν 1 R T 1 (I) Here, mass of air is m , , gas constant is R and volume at state 1 is v 1 . Write the temperature and volume relation for the isentropic compression process 1-2. T 2 = T 1 ( ν 1 ν 2 ) k − 1 = T 1 ( r ) k − 1 (II) Here, the compression ratio is r , the specific heat ratio is k , temperature at state 1 is T 1 , temperature at state 2 is T 2 , volume at state 1 is v 1 and volume at state 2 is v 2 . Write the expression for the heat addition to the process 2 − x ( Q x-2,in ) . Q 2-x,in = m ( u x − u 2 ) = m c v ( T x − T 2 ) (III) Here, the specific heat at constant volume is c v , internal energy at state 2 is u 2 , internal energy at state x is u x and temperature at state x is T x . Write the expression for the heat addition to the process x − 3 ( Q x − 3 ,in ) . Q x − 3 ,in = m c v ( h 3 − h x ) = m c v ( T 3 − T x ) (IV) Here, enthalpy at state 3 is h 3 , enthalpy at state x is h x and temperature at state 3 is T 3 . Write the ideal gas relation for the process x − 3 . P 3 v 3 T 3 = P x v x T x v 3 T 3 = v x T x v 3 v x = T 3 T x (V) Here, volume at state x is v x , volume at state 3 is v 3 pressure at state x is P x . P x and P 3 are equal for the process x − 3 . Write the expression for temperature and volume relation for isentropic expansion process3-4. T 4 = T 3 ( ν 3 ν 4 ) k − 1 (VI) Here, temperature at state 4 is T 4 and volume at state 4 is v 4 . Write the expression for the heat rejected from the cycle ( Q out ) . Q out = m ( u 4 − u 1 ) = m c v ( T 4 − T 1 ) (VII) Here, the specific heat at constant volume is c v . Write the expression for the thermal efficiency of the cycle ( η t h ) . η t h = 1 − Q o u t Q i n (VIII) Here, heat input is Q i n . Conclusion: From Table A-2E, “Ideal-gas specific heats of various common gases”, obtain the following properties for air at room temperature. c v = 0.171 Btu / lbm ⋅ R c p = 0.240 Btu / lbm ⋅ R k = 1.4 From Table A-1E, “Molar mass, gas constant, and critical-point properties”, obtain the Value of gas constant ( R ) for air at room temperature as 0.3704 psia ⋅ ft 3 / lbm ⋅ R . Substitute 14.7 psia for P 1 , 98 in 3 for ν 1 , 0.3704 psia ⋅ ft 3 / lbm ⋅ R for R , and 120 ° F for T 1 in Equation (I). m = ( 14.7 psia ) ( 98 in 3 ) ( 0.3704 psia ⋅ ft 3 / lbm ⋅ R ) ( 120 ° F ) = ( 14

Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684

9th Edition

ISBN: 9781260048667

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

Publisher: McGraw-Hill Education

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Chapter 9.12, Problem 166RP

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The thermal efficiency of the cycle.

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Chapter 9.12, Problem 165RP

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Chapter 9 Solutions

Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684

Ch. 9.12 - What are the air-standard assumptions?Ch. 9.12 - What is the difference between air-standard...Ch. 9.12 - Prob. 3P Ch. 9.12 - How does the thermal efficiency of an ideal cycle,...Ch. 9.12 - How are the combustion and exhaust processes...Ch. 9.12 - What does the area enclosed by the cycle represent...Ch. 9.12 - Prob. 7P Ch. 9.12 - Can the mean effective pressure of an automobile...Ch. 9.12 - What is the difference between spark-ignition and...Ch. 9.12 - Prob. 10P

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The thermal efficiency of the cycle.

Solution Summary: The author illustrates the thermal efficiency of the cycle by drawing the ideal gas equation to calculate the mass of air.

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