Question

Consider the ideal regenerative Brayton cycle. Determine the pressure ratio that maximizes the thermal efficiency of the cycle and compare this value with the pressure ratio that maximizes the cycle net work. For the same maximumto- minimum temperature ratios, explain why the pressure ratio for maximum efficiency is less than the pressure ratio for maximum work.

Short answer

Answer: The pressure ratio for maximum efficiency is less than the pressure ratio for maximum work because increasing the pressure ratio reduces the back work ratio (ratio of compressor work to turbine work), which in turn optimizes the net work output. However, since both the heat input and net work increase with increasing pressure ratio, the increase in heat input is larger than the increase in net work, leading to a decrease in thermal efficiency.

Step-by-step solution

Define the Ideal Brayton Cycle

The ideal Brayton cycle is an air-standard cycle that consists of four processes: 1. Isentropic compression (Process 1-2) 2. Constant pressure heat addition (Process 2-3) 3. Isentropic expansion (Process 3-4) 4. Constant pressure heat rejection (Process 4-1)

Determine the pressure ratio for maximum thermal efficiency

Thermal efficiency (\(\eta_{th}\)) of a cycle is defined as the ratio of net work output to the heat input. For the Brayton cycle, it can be expressed as: \(\eta_{th} = 1 - \frac{T_1}{T_2} \cdot \frac{T_3 - T_4}{T_3 - T_2}\) The pressure ratio (\(r_p\)) is given as: \(r_p = \frac{P_2}{P_1} = \frac{P_3}{P_4}\) Using the ideal gas relations, we can express the temperature ratios in terms of the pressure ratio: \(\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(\gamma - 1)/\gamma} = r_p^{(\gamma - 1)/\gamma}\) \(\frac{T_4}{T_3} = \left(\frac{P_4}{P_3}\right)^{(\gamma - 1)/\gamma} = \frac{1}{r_p^{(\gamma - 1)/\gamma}}\) Now, substituting the expressions for the temperature ratios into the thermal efficiency formula, we get: \(\eta_{th} = 1 - \frac{1}{r_p^{(\gamma - 1)/\gamma}}\) To maximize \(\eta_{th}\), we need to find the value of \(r_p\) that maximizes this expression. Taking the derivative with respect to \(r_p\) and setting it to zero, we obtain the following result: \(r_p = \left( \frac{\gamma}{\gamma - 1} \right)^{\gamma/(\gamma - 1)}\)

Determine the pressure ratio for maximum net work

We can express the net work (\(W_{net}\)) for a Brayton cycle as the difference between the work done during expansion and the work required for compression: \(W_{net} = (T_3 - T_4) - (T_2 - T_1)\) Now, we can express the temperatures as before in terms of the pressure ratio: \(W_{net} = T_1 \left(\frac{T_3}{T_1} - r_p^{(\gamma - 1)/\gamma} \left(\frac{T_3}{T_1} - 1\right)\right)\) To maximize \(W_{net}\), we need to find the value of \(r_p\) that maximizes this expression. Taking the derivative with respect to \(r_p\) and setting it to zero, we obtain the following result: \(r_p = \left(\frac{2\gamma}{\gamma - 1}\right)^{\gamma/(\gamma - 1)}\)

Compare the pressure ratios and provide an explanation

Comparing the pressure ratios for maximum thermal efficiency and maximum net work, we find that: \(r_p(max \, \eta_{th}) = \left( \frac{\gamma}{\gamma - 1} \right)^{\gamma/(\gamma - 1)} < \left(\frac{2\gamma}{\gamma - 1}\right)^{\gamma/(\gamma - 1)} = r_p(max \, W_{net})\) The pressure ratio for maximum efficiency is less than the pressure ratio for maximum work. This is because increasing the pressure ratio reduces the back work ratio (ratio of compressor work to turbine work). When the back work ratio is minimized, the net work output is optimized. However, since both the heat input and net work increase with increasing pressure ratio, the increase in heat input is larger than the increase in net work, leading to a decrease in thermal efficiency. This is the reason why the pressure ratio for maximum efficiency is less than the pressure ratio for maximum work when the maximum-to-minimum temperature ratios are the same.