Question

In an ideal Brayton power cycle operated for the gas turbine plant, if \(T_1\) is the minimum temperature, \(T_3\) is the maximum temperature and pressure ratio is \(r_p\), then the net work ratio is given as

• A. \(1 - \left(\frac{T_3}{T_1}\right)(r_p)^{(\gamma-1)/\gamma} \)
• B. \(1 - \left(\frac{T_1}{T_3}\right)(r_p)^{(\gamma-1)/\gamma} \)
• C. \(1 - \left(\frac{T_3}{T_1}\right)(r_p)^{\gamma/(\gamma-1)} \)
• D. \(1 - \left(\frac{T_1}{T_3}\right)(r_p)^{\gamma/(\gamma-1)} \)

Hint:

To verify your result quickly, think about extreme values: if the pressure ratio $r_p = 1$, no compression occurs, so compressor work is zero, meaning the work ratio must equal 1. Plugging $r_p = 1$ into option (B) gives $1 - \frac{T_1}{T_3}(1) = 1 - \frac{T_1}{T_3}$, which correctly shows the work balance at that limit. Also, remember that the isentropic exponent for pressure ratio is always $\frac{\gamma-1}{\gamma}$, which immediately eliminates options (C) and (D).

Answer 1

The Correct Option is B

Step 1: Non-dimensionalise the cycle using two ratios.
Let \(x = (r_p)^{(\gamma-1)/\gamma}\) be the pressure-ratio function, and let \(\tau = T_3/T_1\) be the overall temperature ratio of the cycle (maximum to minimum temperature). For the isentropic compression \(1 \to 2\) and isentropic expansion \(3 \to 4\):\[ T_2 = T_1 x, \qquad T_4 = \frac{T_3}{x} = \frac{T_1 \tau}{x} \]

Step 2: Write the compressor and turbine work in terms of \(x\) and \(\tau\).
\[ W_c = C_p(T_2 - T_1) = C_p T_1 (x-1) \]\[ W_t = C_p(T_3 - T_4) = C_p T_1 \tau \left(1 - \frac{1}{x}\right) = C_p T_1 \tau \left(\frac{x-1}{x}\right) \]

Step 3: Form the work ratio and cancel the common factor.
\[ r_w = 1 - \frac{W_c}{W_t} = 1 - \frac{T_1(x-1)}{T_1 \tau \left(\dfrac{x-1}{x}\right)} = 1 - \frac{x}{\tau} \]Putting \(\tau = T_3/T_1\) back in:\[ r_w = 1 - \frac{x \, T_1}{T_3} = 1 - \left(\frac{T_1}{T_3}\right)(r_p)^{(\gamma-1)/\gamma} \]\[ \boxed{r_w = 1 - \left(\frac{T_1}{T_3}\right)(r_p)^{(\gamma-1)/\gamma}} \]
This matches option (B).