a) Explanation Draw the T − s diagram for the simple Brayton cycle as shown in Figure (1). Write the relation for the temperature and pressure relation for the isentropic process 1-2. T 2 = T 1 ( P 2 P 1 ) ( k − 1 ) / k = T 1 ( r p ) ( k − 1 ) / k (I) Here, relative pressure is r p , pressure at state 1 is P 1 , pressure at state 2 is P 2 , temperature at state 1 is T 1 , temperature at state 2 is T 2 and Specific heat ratio is k . Write the relation for the temperature and pressure relation for the isentropic process 3-4. T 4 = T 3 ( P 4 P 3 ) ( k − 1 ) / k T 4 = T 3 ( 1 r p ) ( k − 1 ) / k (II) Here, pressure at state 4 is P 4 , pressure at state 3 is P 3 , temperature at state 4 is T 4 and temperature at state 3 is T 3 . Write the expression for the heat added ( q i n ) in the combustion chamber. q i n = h 3 − h 2 = c p ( T 3 − T 2 ) (III) Here, specific enthalpy of air from the exit of compressor is h 2 , specific enthalpy of air from the exit of the combustion chamber is h 3 . Write the expression for the heat rejected ( q o u t ) to atmosphere. q o u t = h 4 − h 1 = c p ( T 4 − T 1 ) (IV) Here, specific enthalpy of air at the exit of the turbine is h 4 and specific enthalpy of air at the inlet to the compressor is h 1 . Write the expression for net work done ( w n e t ) per unit mass when the pressure ratio is 6. w n e t = q i n − q o u t (V) Write the expression for net work done ( w ' n e t ) per unit mass when the pressure ratio is 12. w ' n e t = q i n − q o u t (VI) Write the expression for the thermal efficiency ( η t h ) of the system when the pressure ratio is 6. η t h = w n e t q i n (VII) Write the expression for the thermal efficiency ( η ' t h ) of the system when the pressure ratio is 12. η ' t h = w ' n e t q i n (VIII) Write the expression for change in the net work ( Δ w n e t ) output per unit mass of the cycle. Δ w n e t = w ' n e t − w n e t (IX) Here, net work done for the initial cycle is w n e t and net work done for the newer cycle is w ' n e t . Conclusion: For relative pressure ( r p ) is 6. From the Table A-2, “Ideal-gas specific heats of various common gases” obtain the following properties for air. c p = 1.005 kJ / kg ⋅ K k = 1.4 Substitute 300 K for T 1 , 1.4 for k , and 6 for r p in Equation (I). T 2 = ( 300 K ) ( 6 ) 1.4 − 1 / 1.4 = ( 300 K ) ( 6 ) 0.4 / 1.4 = 500.6 K Substitute 1300 K for T 3 , 1.4 for k , and 6 for r p in Equation (II). T 4 = ( 1300 K ) ( 1 6 ) 1.4 − 1 / 1.4 = ( 1300 K ) ( 1 6 ) 0.4 / 1.4 = 779.1 K Substitute 1.005 kJ / kg ⋅ K for c p , 1300 K for T 3 and 500.6 K for T 2 in Equation (III). q i n = 1.005 kJ / kg ⋅ K ( 1300 K − 500.6 K ) = 803.4 kJ / kg Substitute 1.005 kJ / kg ⋅ K for c p , 779 b) To determine The change in thermal efficiency of the cycle.

9th Edition

ISBN: 9781260917055

Author: CENGEL

Publisher: MCG

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Chapter 9.12, Problem 171RP

To determine

The change in net work output per unit mass.

To determine

The change in thermal efficiency of the cycle.

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Chapter 9 Solutions

CENGEL'S 9TH EDITION OF THERMODYNAMICS:

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The change in net work output per unit mass.

Solution Summary: The author explains the change in net work output per unit mass and the temperature and pressure relation for the isentropic process 1-2.

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The change in net work output per unit mass.

The change in thermal efficiency of the cycle.

Chapter 9 Solutions