Discussion
An ideal Brayton cycle operates between maximum and minimum temperatures of \(T_{3}\) and \(T_{1}\), respectively. For constant values of \(T_{3}\) and \(T_{1}\), the pressure ratio (\(r_{p}\)) for maximum work output is
(\(\gamma\) is the specific heat ratio of air)
Answer: C
The optimum pressure ratio is \(\left(\frac{T_{3}}{T_{1}}\right)^{\frac{\gamma}{2(\gamma-1)}}\).
Write \(x = r_{p}^{(\gamma-1)/\gamma}\). Then \(T_{2} = T_{1}x\) and \(T_{4} = T_{3}/x\), and the net work per unit mass is
\[ \begin{aligned} w &= c_{p}\left(T_{3} - \frac{T_{3}}{x} - T_{1}x + T_{1}\right) \\ \frac{dw}{dx} &= c_{p}\left(\frac{T_{3}}{x^{2}} - T_{1}\right) = 0 \Rightarrow x = \sqrt{\frac{T_{3}}{T_{1}}} \end{aligned} \]
So \(r_{p} = (T_{3}/T_{1})^{\gamma/(2(\gamma-1))}\); at this ratio \(T_{2} = T_{4} = \sqrt{T_{1}T_{3}}\). The exponent \(\gamma/(\gamma-1)\) is the largest ratio the cycle can have, where the net work falls to zero, and the other two exponents are not stationary points.
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