---
title: "Gas turbine 5:1: power, efficiency, work ratio"
subject: "Power Plant Equipment"
chapter: 6
key: "semester_7/power_plant_equipment:gt-5-1-work-ratio"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-5-1-work-ratio
site: BCT Engineering
---

# Gas turbine 5:1: power, efficiency, work ratio

Power Plant Equipment · Chapter 6: Gas Turbine Power Plant

- Asked in: 2079 Baisakh, 8 marks
- Video: [Gas turbine power plant analysis](https://www.youtube.com/watch?v=v_vdcp9mv8k) — Prof Dr Mahesh Shelar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-5-1-work-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-6

## Question

The gas turbine has an overall pressure ratio of 5:1 and the maximum cycle temperature of 550°C. The turbine drives the compressor and an electric generator. The mechanical efficiency of the drive being 0.97. The ambient temperature is 20° C and the isentropic efficiencies of the compressor and turbine are 0.8 and 0.83 respectively. Calculate the power output in kW for an air flow of 15 kg/s. Calculate also thermal efficiency and the work ratio. Neglect changes in kinetic energy and the loss of pressure in combustion chamber.

## Short answer

Net work after mechanical loss = 38.16 kJ/kg; power = 0.97×15×38.16 = 555.3 kW. Thermal efficiency = 12.01%, work ratio = 0.151.

## Answer

**Assumptions:** no gas properties are given, so air-standard values $c_p = 1.005$ kJ/kg K and $\gamma = 1.4$ are used throughout ($\frac{\gamma-1}{\gamma} = 0.2857$, $5^{0.2857} = 1.5838$). Fuel mass is neglected. The mechanical efficiency 0.97 applies to the net shaft power delivered to the generator.

**Data:** $r_p = 5$, $T_3 = 550 + 273 = 823$ K, $T_1 = 20 + 273 = 293$ K, $\eta_c = 0.80$, $\eta_t = 0.83$, $\dot m = 15$ kg/s, $\eta_{mech} = 0.97$.

#### Compressor
$$
\begin{aligned}
T_{2s} &= 293 \times 1.5838 = 464.06\ \text{K} \\
T_2 &= 293 + \frac{464.06 - 293}{0.80} = 506.82\ \text{K} \\
W_c &= 1.005(506.82 - 293) = 214.89\ \text{kJ/kg}
\end{aligned}
$$

#### Turbine
$$
\begin{aligned}
T_{4s} &= \frac{823}{1.5838} = 519.63\ \text{K} \\
T_4 &= 823 - 0.83(823 - 519.63) = 571.20\ \text{K} \\
W_t &= 1.005(823 - 571.20) = 253.06\ \text{kJ/kg}
\end{aligned}
$$

#### Power output
$$
\begin{aligned}
W_{net} &= 253.06 - 214.89 = 38.16\ \text{kJ/kg} \\
P &= \eta_{mech}\,\dot m\,W_{net} = 0.97 \times 15 \times 38.16 = 555.3\ \text{kW}
\end{aligned}
$$

#### Thermal efficiency
$$
\begin{aligned}
Q_s &= c_p (T_3 - T_2) = 1.005(823 - 506.82) = 317.76\ \text{kJ/kg} \\
\eta_{th} &= \frac{W_{net}}{Q_s} = \frac{38.16}{317.76} = 0.1201 = 12.01\%
\end{aligned}
$$
(Overall, including the drive loss: $0.97 \times 12.01 = 11.65\%$.)

#### Work ratio
$$
\text{Work ratio} = \frac{W_{net}}{W_t} = \frac{38.16}{253.06} = 0.151
$$

**Answer:** Power output ≈ **555 kW**, thermal efficiency ≈ **12.0 %**, work ratio ≈ **0.151**.

(If gas properties $c_p = 1.15$ kJ/kg K, $\gamma = 1.333$ are used for the expansion, as in some textbooks, the results become about 657 kW, 12.4 % and 0.174.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the actual compressor exit temperature T2 found from the isentropic value T2s?
   A. T2 = T1 + (T2s - T1) x 0.80
   B. T2 = T1 + (T2s - T1) / 0.80
   C. T2 = T2s x 0.80
   D. T2 = T2s / 0.80
   Answer: B. Compressor efficiency = ideal rise / actual rise, so the actual rise is the isentropic rise divided by 0.80, giving T2 = 506.82 K. (Key point: Actual compressor exit temperature)

2. How is the actual turbine exit temperature T4 obtained?
   A. T4 = T3 - (T3 - T4s) / 0.83
   B. T4 = T4s x 0.83
   C. T4 = T3 - 0.83 (T3 - T4s)
   D. T4 = T3 / 0.83
   Answer: C. Turbine efficiency = actual drop / ideal drop, so the actual drop is 0.83 times the isentropic drop, giving T4 = 571.20 K. (Key point: Actual turbine exit temperature)

3. How is the power delivered to the generator calculated?
   A. P = m x (Wt - Wc) x 0.97
   B. P = m x Wt x 0.97 - Wc
   C. P = m x Wnet / 0.97
   D. P = m x Wt x Wc x 0.97
   Answer: A. Net specific work (Wt - Wc = 38.16 kJ/kg) is multiplied by mass flow and the mechanical efficiency to get about 555 kW. (Key point: Net output and mechanical efficiency)

4. With Wt = 253.06, Wc = 214.89 and heat supplied 317.76 kJ/kg, which pair is correct?
   A. Efficiency 12.0 %, work ratio 0.151
   B. Efficiency 15.1 %, work ratio 0.120
   C. Efficiency 12.0 %, work ratio 0.849
   D. Efficiency 84.9 %, work ratio 0.151
   Answer: A. Efficiency = Wnet/Qs = 38.16/317.76 = 12.0 %; work ratio = Wnet/Wt = 0.151. The work ratio is small because the compressor takes most of the turbine work. (Key point: Thermal efficiency and work ratio)

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