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

Concept: The work ratio ($r_w$) in a gas turbine power plant cycle is defined as the ratio of the net useful work output to the total work produced by the turbine component: \[ r_w = \frac{W_{\text{net}}}{W_t} = \frac{W_t - |W_c|}{W_t} = 1 - \frac{|W_c|}{W_t} \] Let's analyze the components using standard Brayton cycle notations:

• $T_1$: Compressor inlet temperature (minimum cycle temperature)

• $T_2$: Compressor outlet temperature

• $T_3$: Turbine inlet temperature (maximum cycle temperature)

• $T_4$: Turbine exhaust temperature
Assuming constant specific heats, the compressor work input and turbine work output are given by: \[ |W_c| = C_p(T_2 - T_1), \quad W_t = C_p(T_3 - T_4) \]

Step 1: Set up the temperature equations using the pressure ratio \(r_p\).
For isentropic compression ($1 \rightarrow 2$) and expansion ($3 \rightarrow 4$) processes, we apply the ideal gas temperature-pressure relations: \[ \frac{T_2}{T_1} = (r_p)^{\frac{\gamma-1}{\gamma}} \quad \Rightarrow \quad T_2 = T_1(r_p)^{\frac{\gamma-1}{\gamma}} \] \[ \frac{T_3}{T_4} = (r_p)^{\frac{\gamma-1}{\gamma}} \quad \Rightarrow \quad T_4 = \frac{T_3}{(r_p)^{\frac{\gamma-1}{\gamma}}} \]

Step 2: Substitute these expressions into the work ratio equation.
\[ r_w = 1 - \frac{C_p(T_2 - T_1)}{C_p(T_3 - T_4)} = 1 - \frac{T_2 - T_1}{T_3 - T_4} \] Factor out $T_1$ from the numerator and $T_3$ from the denominator: \[ r_w = 1 - \frac{T_1 \left( \frac{T_2}{T_1} - 1 \right)}{T_3 \left( 1 - \frac{T_4}{T_3} \right)} \] Notice from the isentropic relations that: \[ \frac{T_2}{T_1} = (r_p)^{\frac{\gamma-1}{\gamma}} \quad \text{and} \quad \frac{T_4}{T_3} = \frac{1}{(r_p)^{\frac{\gamma-1}{\gamma}}} \] Let's define $x = (r_p)^{\frac{\gamma-1}{\gamma}}$ to make simplification easier: \[ r_w = 1 - \frac{T_1 (x - 1)}{T_3 \left(1 - \frac{1}{x}\right)} = 1 - \frac{T_1 (x - 1)}{T_3 \left(\frac{x - 1}{x}\right)} \] Canceling the common term $(x - 1)$ from both the numerator and denominator simplifies the equation to: \[ r_w = 1 - \frac{T_1 \cdot x}{T_3} \]

Step 3: Re-substitute the value of \(x\).
Replacing $x$ back with its full pressure-ratio form: \[ r_w = 1 - \left(\frac{T_1}{T_3}\right)(r_p)^{\frac{\gamma-1}{\gamma}} \] This matches Option (B).