Question

The efficiency of Brayton cycle corresponding to maximum net work obtained for \( T_{\text{max}} = 900 \, \text{K} \) and \( T_{\text{min}} = 400 \, \text{K} \) is given by

• A. 3(3)3%
• B. 4(4)4%
• C. 55.5%
• D. 66.6%

Hint:

The ideal Brayton cycle efficiency is \( \eta = 1 - \frac{T_{\text{min}}}{T_{\text{max}}} \), but verify the problem’s interpretation if the answer differs.

Answer 1

The Correct Option is A

Step 1: Condition for Maximum Net Work
For a Brayton cycle, the maximum net work output occurs when the pressure ratio is optimized such that the compressor exit temperature \( T_2 \) and the turbine exit temperature \( T_4 \) satisfy: \[ T_2 = T_4 = \sqrt{T_{\text{max}} \cdot T_{\text{min}}} \] Step 2: Calculate Intermediate Temperature
Given: \[ T_{\text{max}} = 900 \, \text{K}, \quad T_{\text{min}} = 400 \, \text{K} \] Substitute these values: \[ T_2 = T_4 = \sqrt{900 \times 400} = \sqrt{360000} = 600 \, \text{K} \] Step 3: Determine the Efficiency
The thermal efficiency of the Brayton cycle under maximum net work conditions is: \[ \eta = 1 - \frac{T_{\text{min}}}{T_2} = 1 - \frac{400}{600} = 1 - \frac{2}{3} = \frac{1}{3} \approx 33.3% \] Alternatively: \[ \eta = 1 - \sqrt{\frac{T_{\text{min}}}{T_{\text{max}}}} = 1 - \sqrt{\frac{400}{900}} = 1 - \frac{2}{3} = 33.3% \] Step 4: Select the Correct Option
The efficiency is 33.3%, which corresponds to option {1}.