---
title: "Regenerative gas turbine: 1 to 6.9 bar"
subject: "Power Plant Equipment"
chapter: 6
key: "semester_7/power_plant_equipment:gt-regenerative-6-9-bar"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-regenerative-6-9-bar
site: BCT Engineering
---

# Regenerative gas turbine: 1 to 6.9 bar

Power Plant Equipment · Chapter 6: Gas Turbine Power Plant

- Asked in: 2072 Chaitra, 10 marks
- Video: [Numerical of Gas Turbine - Numerical 4](https://www.youtube.com/watch?v=Fs9Kr9ZCJr4) — BEST MECHANICAL ENGINEERING
- 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-regenerative-6-9-bar
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-6

## Question

In an open cycle regenerative gas turbine plant, the air enters the compressor at 1 bar and leaves at 6.9 bar. The temperature at the end of combustion chamber is 816°C. The isentropic efficiencies of compressor and turbine are 0.84 and 0.85 respectively. The regenerator effectiveness is 60%, determine (a) Thermal efficiency (b) Air rate (c) Work ratio [Take Cp = 1005 J/kg.K and γ = 1.4]

## Short answer

With inlet 305 K, use isentropic relations and efficiencies: compressor work 268.75 kJ/kg, turbine work 394.55 kJ/kg, net work 125.80 kJ/kg. Regenerator raises air to 646.81 K. Thermal efficiency is 28.31%, air rate 28.62 kg/kWh, work ratio 0.319.

## Answer

**Assumptions:** the inlet air temperature is not given; it is taken as **32 °C (305 K)**, the value used in the standard textbook version of this problem. Air-standard values $c_p = 1.005$ kJ/kg K and $\gamma = 1.4$ are used throughout; fuel mass neglected.

**Data:** $p_1 = 1$ bar, $p_2 = 6.9$ bar ($r_p = 6.9$), $T_3 = 816 + 273 = 1089$ K, $\eta_c = 0.84$, $\eta_t = 0.85$, $\varepsilon = 0.60$; $6.9^{0.2857} = 1.7365$.

```text
       +------ Regenerator ------+
       |  x                 y    |
 1     | 2                  4    |
 --> [C] --> [CC] --3--> [T] ----+
```

#### Compressor
$$
\begin{aligned}
T_{2s} &= 305 \times 1.7365 = 529.63\ \text{K} \\
T_2 &= 305 + \frac{529.63 - 305}{0.84} = 572.41\ \text{K} \\
W_c &= 1.005(572.41 - 305) = 268.75\ \text{kJ/kg}
\end{aligned}
$$

#### Turbine
$$
\begin{aligned}
T_{4s} &= \frac{1089}{1.7365} = 627.13\ \text{K} \\
T_4 &= 1089 - 0.85(1089 - 627.13) = 696.41\ \text{K} \\
W_t &= 1.005(1089 - 696.41) = 394.55\ \text{kJ/kg}
\end{aligned}
$$

#### Regenerator
$$
T_x = T_2 + \varepsilon(T_4 - T_2) = 572.41 + 0.6(696.41 - 572.41) = 646.81\ \text{K}
$$

#### (a) Thermal efficiency
$$
\begin{aligned}
W_{net} &= 394.55 - 268.75 = 125.80\ \text{kJ/kg} \\
Q_s &= 1.005(1089 - 646.81) = 444.40\ \text{kJ/kg} \\
\eta_{th} &= \frac{125.80}{444.40} = 0.2831 = 28.31\%
\end{aligned}
$$

#### (b) Air rate
Air needed per kWh of net output:
$$
\text{Air rate} = \frac{3600}{W_{net}} = \frac{3600}{125.80} = 28.62\ \text{kg/kWh}
$$

#### (c) Work ratio
$$
\text{Work ratio} = \frac{W_{net}}{W_t} = \frac{125.80}{394.55} = 0.319
$$

**Answer:** $\eta$ = **28.31 %**, air rate = **28.62 kg/kWh**, work ratio = **0.319**.

(If the inlet air is taken as 15 °C = 288 K instead, the same method gives $\eta$ = 30.79 %, air rate = 25.57 kg/kWh and work ratio = 0.357.)

## Check yourself

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

1. What was assumed since compressor inlet temperature was not given?
   A. T1 = 305 K (32 C), the standard textbook value
   B. T1 = 0 K
   C. T1 = 1089 K
   D. T1 equals the turbine exit
   Answer: A. The answer takes T1 = 305 K; a different value changes the results, e.g. 288 K gives 30.79 %. (Key point: Missing data assumption)

2. Effectiveness 0.6 means the air leaves the regenerator at:
   A. T4
   B. T2 + 0.6(T4 - T2) = 646.81 K
   C. 0.6 x T4
   D. T2 - 0.6(T4 - T2)
   Answer: B. Effectiveness = (Tx - T2)/(T4 - T2), so Tx = 572.41 + 0.6 x 124 = 646.81 K. (Key point: Regenerator exit temperature)

3. The heat supplied is based on:
   A. cp(T3 - T2)
   B. cp(T3 - Tx)
   C. cp(T4 - T1)
   D. cp(T3 - T4)
   Answer: B. Combustion only heats air from Tx to T3: 444.40 kJ/kg, giving 125.80/444.40 = 28.31 %. (Key point: Thermal efficiency with regeneration)

4. Air rate and work ratio are given by:
   A. Wnet/3600 and Wnet x Wt
   B. 3600/Wnet (kg/kWh) and Wnet/Wt
   C. 3600/Wt and Wc/Wt
   D. Wt/3600 and Wnet/Wc
   Answer: B. Air rate = 3600/125.80 = 28.62 kg/kWh; work ratio = 125.80/394.55 = 0.319. (Key point: Air rate and work ratio)

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