Explanation Draw the T − s diagram for ideal Stirling engine cycle as shown in Figure (1). Write the expression to calculate the mass of the air ( m ) at state 1. m = P 1 ν 1 R T 1 (I) Here, the pressure at state 1 is P 1 , mass of air is m , the gas constant is R , the temperature at state 1 is T 1 , volume at state 1 is ν 1 . Write the expression to calculate the pressure at state 2 ( P 2 ) at the end of the compression. P 2 = P 1 ( ν 1 ν 2 ) (II) Here, volume at state 2 is ν 2 and pressure at state 2 is P 2 . Write the expression for the change in entropy for the process 1-2. ( s 2 − s 1 ) = ( s 3 − s 4 ) = c p ln T 2 T 1 − R ln P 2 P 1 For process 1-2, T 1 and T 2 are equal. ( s 2 − s 1 ) = ( s 3 − s 4 ) = c p ln T 2 T 2 − R ln P 2 P 1 ( s 2 − s 1 ) = ( s 3 − s 4 ) = 0 − R ln P 2 P 1 ( s 2 − s 1 ) = ( s 3 − s 4 ) = − R ln P 2 P 1 (III) Here, entropy at state 1 is s 1 , entropy at state 2 is s 2 , entropy at state 3 is s 3 and entropy at state 4 is s 4 . The processes are reversible. Write the expression to calculate the external heat input to the cycle ( Q in ) . Q in = m T 1 ( s 2 − s 1 ) (IV) Write the expression to calculate the external heat rejection ( Q out ) . Q out = m T 3 ( s 3 − s 4 ) (V) Write the expression to calculate heat transfer between the working fluid and regenerator for each complete cycle ( Q regenerator ) . Q regenerator = m c v ( T 1 − T 4 ) (VI) Here, specific heat of air at constant volume is c v and temperature at state 4 is T 4 . Conclusion: From Table A-1E, “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant ( R ) for hydrogen as 5.3224 psia ⋅ ft 3 / lbm ⋅ R . From Table A-2Ea, “Ideal-gas specific heats of various common gases”, obtain the Following properties for hydrogen. k = 1.404 c v = 2.44 Btu / lbm ⋅ R c p = 3.43 Btu / lbm ⋅ R Substitute 400 psia for P 1 , 0.1 ft 3 for ν 1 , 5.3224 psia ⋅ ft 3 / lbm ⋅ R for R , and 640 ° F for T 3 in Equation (I). m = ( 400 psia ) ( 0.1 ft 3 ) ( 5.3224 psia ⋅ ft 3 / lbm ⋅ R ) ( 640 ° F ) = ( 400 psia ) ( 0.1 ft 3 ) ( 5

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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ISBN: 9781260048636

Author: CENGEL

Publisher: MCG

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Chapter 9.12, Problem 78P

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The amount of external heat addition, external heat rejection, and heat transfer between the working fluid and regenerator for each complete cycle.

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Chapter 9.12, Problem 77P

Chapter 9.12, Problem 79P

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CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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The amount of external heat addition, external heat rejection, and heat transfer between the working fluid and regenerator for each complete cycle.

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The amount of external heat addition, external heat rejection, and heat transfer between the working fluid and regenerator for each complete cycle.

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