Question:
In an ideal air-standard Brayton cycle, air enters the compressor at 100 kPa and 300 K. Thermal efficiency of the cycle is 50 %. The heat added to air is 1000 kJ/kg. Air has constant specific heat \( c_p = 1.0 \, \text{kJ/(kg-K)} \) and \( \gamma = 1.4 \). Air temperature, in K, at the turbine inlet is ..................
In an ideal air-standard Brayton cycle, air enters the compressor at 100 kPa and 300 K. Thermal efficiency of the cycle is 50 %. The heat added to air is 1000 kJ/kg. Air has constant specific heat \( c_p = 1.0 \, \text{kJ/(kg-K)} \) and \( \gamma = 1.4 \). Air temperature, in K, at the turbine inlet is ..................
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In an ideal Brayton cycle, the temperature at the turbine inlet can be found using the relationship between efficiency and temperatures at the compressor and turbine.
Updated On: Sep 4, 2025
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Solution and Explanation
- For an ideal Brayton cycle, the thermal efficiency \( \eta \) is given by:
\[
\eta = 1 - \left( \frac{T_1}{T_2} \right)^{\gamma - 1}
\]
where \( T_1 \) is the temperature at the compressor inlet, and \( T_2 \) is the temperature at the turbine inlet.
- Given that \( \eta = 50% = 0.5 \), \( T_1 = 300 \, \text{K} \), and \( \gamma = 1.4 \), we can solve for \( T_2 \): \[ 0.5 = 1 - \left( \frac{300}{T_2} \right)^{1.4 - 1} \] \[ 0.5 = 1 - \left( \frac{300}{T_2} \right)^{0.4} \] \[ \left( \frac{300}{T_2} \right)^{0.4} = 0.5 \] Taking the reciprocal and raising both sides to the power of \( 2.5 \), we find: \[ T_2 = 1595.00 \, \text{K} \text{ to } 1605.00 \, \text{K} \]
- Given that \( \eta = 50% = 0.5 \), \( T_1 = 300 \, \text{K} \), and \( \gamma = 1.4 \), we can solve for \( T_2 \): \[ 0.5 = 1 - \left( \frac{300}{T_2} \right)^{1.4 - 1} \] \[ 0.5 = 1 - \left( \frac{300}{T_2} \right)^{0.4} \] \[ \left( \frac{300}{T_2} \right)^{0.4} = 0.5 \] Taking the reciprocal and raising both sides to the power of \( 2.5 \), we find: \[ T_2 = 1595.00 \, \text{K} \text{ to } 1605.00 \, \text{K} \]
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