Explanation Draw the T − s diagram for regenerative gas turbine cycle as shown in Figure (1). Write the temperature and pressure relation for the isentropic process 1-2s. T 2 s = T 4 s = T 1 r p ( k − 1 ) / k (I) Here, the pressure ratio is r p , the specific heat ratio is k , temperature at state 1 is T 1 , temperature at state 2 s for isentropic process is T 2 s and temperature at state 4 s for isentropic process is T 4 s . Write the expression for the efficiency of the compressor ( η C ) . η C = c p ( T 2 s − T 1 ) c p ( T 2 − T 1 ) T 2 = T 1 + T 2 s − T 1 η C (II) Here, the specific heat at constant pressure is c p . Write the temperature and pressure relation ratio for the expansion process 6-7 s . T 7 s = T 6 ( 1 r p ) ( k − 1 ) / k (III) Here, temperature at state 7 s for isentropic process is T 7 s and temperature at state 6 for is T 6 . Write the expression for the efficiency of the turbine ( η T ) . η T = c p ( T 6 − T 7 ) c p ( T 6 − T 7 s ) T = T 6 + η T ( T 6 − T 7 s ) (IV) Here, temperature at state 7 is T 7 , Write the expression for the work input per unit mass flow rate to the compressor ( w C,in ) . w C,in = 2 c p ( T 2 − T 1 ) (V) Write the expression for the work done per unit mass flow rate by the turbine ( w T,out ) . w T,out = 2 c p ( T 6 − T 7 ) (VI) Write the expression for to calculate the net work output per unit mass flow rate by the regenerative gas-turbine cycle ( w net ) . w net = w T,out − w C,in (VII) Write the expression to calculate the mass flow rate ( m ˙ ) . m ˙ = W ˙ net w net (VIII) Here, the power produced by the engine is W ˙ net . Write the expression to calculate the rate of heat addition to the cycle ( Q ˙ i n ) . Q ˙ i n = m ˙ c p ( T 6 − T 5 ) + m ˙ c p ( T 8 − T 7 ) = m ˙ c p [ ( T 6 − T 5 ) + ( T 8 − T 7 ) ] (IX) Here, the specific heat at constant pressure is c p , temperature at state 8 for is T 8 . Conclusion. From Table A-2Ea, “Ideal-gas specific heats of various common gases”, obtain the following values for air at room temperature. k = 1.4 c p = 0.24 Btu / lbm ⋅ R Substitute 60 ° F for T 1 , 3 for r p , and 1.4 for k in Equation (I). T 2 s = 60 ° F ( 3 ) 1.4 − 1 / 1.4 = ( 60 + 460 ) R ( 3 ) 0.4 / 1.4 = 711.7 R Substitute 60 ° F for T 1 , 711.7 R for T 2 s , and 88 % for η C in Equation (II). T 2 = 60 ° F + 711.7 R − 60 ° F 88 % = ( 60 + 460 ) R + 711.7 R − ( 60 + 460 ) R 88 ( 1 / 100 ) = 737.8 R Substitute 940 ° F for T 6 , 3 for r p and 1.4 for k in Equation (III). T 7 s = 940 ° F ( 1 3 ) 1.4 − 1 / 1.4 = ( 940 + 460 ) R ( 1 3 ) 0

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Author: CENGEL

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Chapter 9.12, Problem 128P

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The change in the rate of heat addition to the cycle.

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Chapter 9.12, Problem 127P

Chapter 9.12, Problem 129P

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The change in the rate of heat addition to the cycle.

Solution Summary: The author explains the change in the rate of heat addition to the cycle. The temperature and pressure relation for the isentropic process is shown in Figure (1).

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The change in the rate of heat addition to the cycle.

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