Explanation Ideal Brayton cycle with regeneration. Draw T − s diagram for an ideal Brayton cycle with regeneration as shown in Figure (1). Consider, the pressure is P i , the specific volume is v i , the temperature is T i , the entropy is s i , the enthalpy is h i corresponding to i t h state. Write the expression of the temperature and pressure relation for the isentropic process 1-2. T 2 = T 1 r p ( k − 1 ) / k Here, pressure ratio is r p and specific heat ratio is k . Write the expression of temperature and pressure relation for the isentropic process 4-5. T 5 = T 4 ( 1 r p ) ( k − 1 ) / k Write the expression to calculate the thermal efficiency of ideal Brayton cycle with regeneration ( η th ) . η th = 1 − q out q in η th = 1 − c p ( T 6 − T 1 ) c p ( T 3 − T 5 ) η th = 1 − ( T 6 − T 1 ) ( T 3 − T 5 ) (I) Here, heat input to the cycle is q in , heat rejected by the cycle is q out and specific heat at constant pressure is c p . Conclusion: For an ideal regenerator of 100 % effectiveness, the temperature at state 4 and state 5 are equal ( T 4 = T 5 ) and temperature at state 6 and state 2 are equal ( T 6 = T 2 ) . Substitute T 2 for T 6 , and T 4 for T 5 in Equation (I)

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Chapter 9.12, Problem 104P

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An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator of

effectiveness 100%.

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Chapter 9.12, Problem 103P

Chapter 9.12, Problem 105P

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An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator of effectiveness 100% .

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An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator of effectiveness 100% .

Solution Summary: The author illustrates an expression for the thermal efficiency of an ideal Brayton cycle with a regenerator of effectiveness 100%.

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An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator of effectiveness 100%.

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Chapter 9 Solutions

An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator of

effectiveness 100%.