Question
Download Solution PDFThe pressure ratio and maximum temperature of a Brayton cycle are 5:1 and 928 K respectively. Air enters the compressor at 1 bar and 300 K. What will be the network output of the cycle per kg of air flow?
[\(C_p = 1.0 \, \text{kJ/kg-K}, \gamma = 1.41, 5^{\frac{(0.41)}{(1.41 ) }} = 1.6\) ]
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The Brayton cycle is an ideal gas turbine cycle that consists of isentropic compression, constant-pressure heat addition, isentropic expansion, and constant-pressure heat rejection. The net work output is the difference between turbine and compressor work.
Given:
Pressure ratio, \( r_p = 5 \)
Inlet temperature, \( T_1 = 300~K \)
Maximum temperature, \( T_3 = 928~K \)
\( C_p = 1.0~\text{kJ/kg·K},~\gamma = 1.41 \)
Calculation:
\( T_2 = T_1 \times r_p^{\frac{\gamma - 1}{\gamma}} = 300 \times 5^{\frac{0.41}{1.41}} \approx 300 \times 1.6 = 480~K \)
\( T_4 = T_3 \times r_p^{-\frac{\gamma - 1}{\gamma}} = 928 \div 1.6 = 580~K \)
Net work output per kg of air:
\( W_{net} = C_p \left[ (T_3 - T_4) - (T_2 - T_1) \right] \)
\( W_{net} = 1.0 \times \left[ (928 - 580) - (480 - 300) \right] = 348 - 180 = 168~\text{kJ/kg} \)
Last updated on Jul 15, 2025
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