Explanation Draw the T − s diagram for the simple ideal Brayton cycle as shown in Figure (1). Write the expression to calculate the specific volume at state 1 v 1 . v 1 = R T 1 P 1 (I) Here, gas constant is R , temperature at state 1 is T 1 and pressure at state 1 is P 1 . Write the expression to calculate the mass flow rate. m ˙ = A 1 V 1 v 1 (II) Here, area at which argon enters the compressor is A 1 and initial velocity is V 1 . Write the temperature and pressure relation for the isentropic process 1-2. T 2 = T 1 ( P 2 P 1 ) ( k − 1 ) / k (III) Here, the specific heat ratio is k , temperature at state 1 is T 1 , temperature at state 2 is T 2 and pressure at state 2 is P 2 . Write the temperature and pressure relation for the isentropic process 3-4. T 4 = T 3 ( P 4 P 3 ) ( k − 1 ) / k (IV) Here, temperature at state 4 is T 4 , temperature at state 3 is T 3 , pressure at state 4 is P 4 . And pressure at state 4 is P 4 . Write the expression to calculate the net power produced by the cycle ( W ˙ n e t ) . W ˙ n e t = m ˙ c p [ T 3 − T 4 − T 2 − T 1 ] (V) Here, heat capacity of air is c p , and mass flow rate of air is m ˙ a i r . Write the expression to calculate the heat addition rate to the cycle ( Q ˙ i n ) . Q ˙ i n = m ˙ a i r c p ( T 3 − T 2 ) (VI) Write the expression to calculate the rate of heat rejection in the cycle ( Q ˙ o u t ) . Q ˙ o u t = Q ˙ i n − W ˙ net (VII) Write the expression to calculate the rate of entropy generated by the cycle ( S ˙ g e n ) S ˙ g e n = Q ˙ o u t T L − Q ˙ i n T H (VII) Here, the high temperature of the hot or higher state is T H , and the temperature of cold or lower state is T L . Conclusion: From Table A-1E, “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant ( R ) for argon as 0.2686 psia ⋅ ft 3 / lbm ⋅ R . From Table A-2Ea, “Ideal-gas specific heats of various common gases”, obtain the value of specific heat ratio ( k ) for argon as 1.667 . Substitute 0.2686 psia ⋅ ft 3 / lbm ⋅ R for R , 80 ° F for T 1 and 15 psia for P 1 in Equation (I). v 1 = 0.2686 psia ⋅ ft 3 / lbm ⋅ R ( 80 ° F ) 15 psia = 0.2686 psia ⋅ ft 3 / lbm ⋅ R ( 80 + 460 ) R 15 psia = 9.670 ft 3 / lbm Substitute 3 ft 3 for A 1 , 200 ft / s for V 1 and 9.670 ft 3 / lbm for v 1 in Equation (II). m ˙ = 3 ft 3 × 200 ft / s 9.670 ft 3 / lbm = 62.05 lbm / s Substitute 150 psia 6 for P 2 , , 15 psia for P 1 , 80 ° F for T 1 and 1.667 for k in Equation (III). T 2 = 80 ° F ( 150 psia 15 psia ) ( 1.667 − 1 ) / 1.667 = ( 80 + 460 ) R ( 150 psia 15 psia ) ( 0

9th Edition

ISBN: 9781259822674

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

Publisher: McGraw-Hill Education

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In physics, entropy means the amount of disorder or the degree of randomness associated with a thermodynamic system. The system is not self-organized when the entropy of the system is high. For instance, when a system is at an elevated temperature, its m…

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Chapter 9.12, Problem 146P

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The rate of entropy generated by the cycle.

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Chapter 9.12, Problem 145P

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