---
title: "Ideal gas turbine with regenerator at 1400 K"
subject: "Power Plant Equipment"
chapter: 6
key: "semester_7/power_plant_equipment:gt-ideal-regenerator-1400k"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-ideal-regenerator-1400k
site: BCT Engineering
---

# Ideal gas turbine with regenerator at 1400 K

Power Plant Equipment · Chapter 6: Gas Turbine Power Plant

- Asked in: 2076 Chaitra, 12 marks
- Video: [3112 Unit 3 Brayton with Regen, Reheat and Intercooling In Class Problem 9](https://www.youtube.com/watch?v=y2A4tLZutsE) — Regina Vrikkis
- 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-ideal-regenerator-1400k
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-6

## Question

Air enters the compressor of an ideal gas turbine power plant at 100kPa, 300K. The pressure ratio is 10. The turbine inlet temperature is 1400K. Determine the efficiency of the plant. Also determine the increment in efficiency if a regenerator with effectiveness 80% is used in the plant. [Take Cp=1.005 kJ/kgK and γ=1.4]

## Short answer

Ideal cycle efficiency = 48.21%. With 80% effective regenerator, air preheated to 695.94 K, heat input reduced, efficiency becomes 56.20%, an increase of 7.99 percentage points.

## Answer

**Data:** ideal (isentropic) compressor and turbine, $p_1 = 100$ kPa, $T_1 = 300$ K, $r_p = 10$, $T_3 = 1400$ K, $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$; $r_p^{0.2857} = 10^{0.2857} = 1.9307$.

#### Simple ideal Brayton cycle
$$
\begin{aligned}
T_2 &= T_1 r_p^{0.2857} = 300 \times 1.9307 = 579.21\ \text{K} \\
T_4 &= \frac{T_3}{r_p^{0.2857}} = \frac{1400}{1.9307} = 725.13\ \text{K} \\
W_c &= 1.005(579.21 - 300) = 280.61\ \text{kJ/kg} \\
W_t &= 1.005(1400 - 725.13) = 678.25\ \text{kJ/kg} \\
W_{net} &= 678.25 - 280.61 = 397.64\ \text{kJ/kg} \\
Q_s &= 1.005(1400 - 579.21) = 824.89\ \text{kJ/kg} \\
\eta &= \frac{397.64}{824.89} = 0.4821 = 48.21\%
\end{aligned}
$$
Check: $\eta = 1 - \dfrac{1}{r_p^{(\gamma-1)/\gamma}} = 1 - \dfrac{1}{1.9307} = 48.21\%$.

#### With regenerator ($\varepsilon = 0.8$)
$T_4 = 725.13$ K $> T_2 = 579.21$ K, so regeneration is useful.
$$
\begin{aligned}
T_x &= T_2 + \varepsilon(T_4 - T_2) = 579.21 + 0.8(145.92) = 695.94\ \text{K} \\
Q_s' &= 1.005(1400 - 695.94) = 707.58\ \text{kJ/kg} \\
\eta' &= \frac{W_{net}}{Q_s'} = \frac{397.64}{707.58} = 0.5620 = 56.20\%
\end{aligned}
$$
Net work is unchanged; only the fuel heat is reduced.

#### Increment in efficiency
$$
\Delta\eta = 56.20 - 48.21 = 7.99\ \text{percentage points}
$$
(a relative increase of $7.99/48.21 = 16.6\%$).

**Answer:** Efficiency without regenerator = **48.21 %**; with 80 % regenerator = **56.20 %**; increase ≈ **7.99 %** (points).

## Check yourself

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

1. For the ideal simple cycle with rp = 10 and gamma = 1.4, the efficiency depends on:
   A. Pressure ratio only, eta = 1 - 1/rp^0.2857
   B. Turbine inlet temperature only
   C. Both T3 and mass flow
   D. Compressor inlet pressure only
   Answer: A. For an ideal Brayton cycle eta = 1 - 1/rp^((gamma-1)/gamma) = 48.21 %, independent of T3. (Key point: Efficiency of ideal Brayton cycle)

2. Why is a regenerator useful in this cycle?
   A. Because T2 is greater than T3
   B. Because turbine exhaust T4 (725 K) is hotter than compressor exit T2 (579 K)
   C. Because it increases net work
   D. Because it reduces compressor work
   Answer: B. Heat can flow from exhaust to compressed air only if T4 > T2, which holds here. (Key point: Condition for regeneration)

3. What is the air temperature leaving the 80 % regenerator?
   A. T2 + 0.8(T4 - T2) = 695.94 K
   B. T4 - 0.8(T4 - T2) = 608.4 K
   C. 0.8 x T4 = 580 K
   D. T4 = 725.13 K
   Answer: A. Effectiveness = (Tx - T2)/(T4 - T2), so Tx = 579.21 + 0.8 x 145.92 = 695.94 K. (Key point: Regenerator effectiveness)

4. What changes when the regenerator is added?
   A. Net work rises and heat supplied is unchanged
   B. Both net work and heat supplied fall
   C. Net work is unchanged and heat supplied falls from 824.89 to 707.58 kJ/kg
   D. Net work falls and heat supplied is unchanged
   Answer: C. Compressor and turbine works are unchanged; only the fuel heat needed is cut, so efficiency rises from 48.21 % to 56.20 %, about 7.99 points. (Key point: Effect of regenerator on work and fuel)

## Related in Power Plant Equipment

- [Gas turbine with reheat and 90% regenerator](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-reheat-regenerator-1400k) — Power Plant Equipment, chapter 6
- [Gas turbine 800 kW with regenerator](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-800kw-regenerator) — Power Plant Equipment, chapter 6, asked 2×
- [Gas turbine with 78% effective regenerator](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-regenerator-78) — Power Plant Equipment, chapter 6
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- [Gas turbine 5:1: power, efficiency, work ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-5-1-work-ratio) — Power Plant Equipment, chapter 6
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