Question

The air standard efficiency of closed gas turbine cycle is given by: Where r = Pressure ratio for compression and turbine; and $\gamma$ = Isentropic index of air.

• A. $\eta = \{ 1 - \frac{1}{r^{(\frac{\gamma - 1}{\gamma})}} \}$
• B. $\eta = \{ 1 - \frac{1}{r^{(\gamma - 1)}} \}$
• C. $\eta = \{ 1 - r^{(\gamma - 1)} \}$
• D. $\eta = \{ r^{(\gamma - 1)} - 1 \}$

Hint:

Don't confuse the Brayton cycle (Gas Turbine) with the Otto cycle (Petrol Engine).

• Otto Cycle: Uses Volume Ratio ($r$) and the power is $(\gamma - 1)$.

• Brayton Cycle: Uses Pressure Ratio ($r$) and the power is $(\frac{\gamma - 1}{\gamma})$.
Just remember that pressure ratios always need that extra "$\gamma$" in the denominator of the exponent.

Answer 1

The Correct Option is A

1. Efficiency Derivation: The thermal efficiency ($\eta$) of a Brayton cycle is defined as: $$\eta = \frac{\text{Net Work}}{\text{Heat Added}} = 1 - \frac{\text{Heat Rejected}}{\text{Heat Added}}$$ For constant pressure heat addition ($Q_{in} = c_p(T_3 - T_2)$) and rejection ($Q_{out} = c_p(T_4 - T_1)$): $$\eta = 1 - \frac{T_4 - T_1}{T_3 - T_2}$$

2. Relation to Pressure Ratio ($r$): Using the isentropic relations for the compressor ($1 \to 2$) and turbine ($3 \to 4$): $$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} = r^{\frac{\gamma-1}{\gamma}} \quad \text{and} \quad \frac{T_3}{T_4} = \left(\frac{P_3}{P_4}\right)^{\frac{\gamma-1}{\gamma}} = r^{\frac{\gamma-1}{\gamma}}$$ Substituting these into the efficiency equation and simplifying leads to the final expression: $$\eta = 1 - \frac{1}{r^{\left(\frac{\gamma - 1}{\gamma}\right)}}$$

3. Significance: This formula shows that the efficiency of a gas turbine depends primarily on the

pressure ratio. Increasing the pressure ratio ($r$) or using a gas with a higher isentropic index ($\gamma$) will increase the theoretical efficiency of the cycle.