---
title: "Ideal gas turbine 100 kW: pressure ratio, work, efficiency"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 6
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:ideal-gas-turbine-100kw-pressure-ratio"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/ideal-gas-turbine-100kw-pressure-ratio
site: BCT Engineering
---

# Ideal gas turbine 100 kW: pressure ratio, work, efficiency

Fundamental of Thermodynamics & Heat Transfer · Chapter 6: Thermodynamic Cycles

- Asked in: 2071 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/ideal-gas-turbine-100kw-pressure-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6

## Question

An ideal gas turbine has a net power output 100 kW. The working medium is air. The minimum and maximum temperatures of the cycle are 30°C and 750°C respectively. If the turbine outlet temperature is 300°C, determine
a) Compressor pressure ratio.
b) Compressor work
c) Mass flow rate of air
d) Thermal efficiency
[Take Cp = 1005 J/kgK and γ = 1.4]

## Short answer

From T3/T4 = r_p^((γ-1)/γ), r_p = (1023.15/573.15)^3.5 = 7.60. Compressor work w_c = cp(T2 - T1) = 239.2 kJ/kg. Net work 213.0 kJ/kg gives mass flow rate ṁ = 100/213 = 0.469 kg/s. Efficiency η = 44.0%.

## Answer

Ideal Brayton cycle: 1-2 isentropic compression, 2-3 constant-pressure heat addition, 3-4 isentropic expansion, 4-1 constant-pressure rejection.

**Given:** W_net = 100 kW, T₁ = 30 + 273.15 = 303.15 K, T₃ = 750 + 273.15 = 1023.15 K, T₄ = 300 + 273.15 = 573.15 K, cp = 1.005 kJ/kg·K, γ = 1.4.

#### (a) Compressor pressure ratio
For the turbine: T₃/T₄ = r_p^((γ−1)/γ)

r_p^0.2857 = 1023.15/573.15 = 1.7851

r_p = 1.7851^3.5 = **7.60**

#### Compressor outlet temperature
T₂ = T₁ r_p^0.2857 = 303.15 × 1.7851 = 541.2 K

#### (b) Compressor work
w_c = cp (T₂ − T₁) = 1.005 × (541.2 − 303.15) = **239.2 kJ/kg**

#### (c) Mass flow rate
Turbine work: w_t = cp (T₃ − T₄) = 1.005 × (1023.15 − 573.15) = 452.2 kJ/kg
Net work: w_net = 452.2 − 239.2 = 213.0 kJ/kg

ṁ = Ẇ_net/w_net = 100/213.0 = **0.469 kg/s**

#### (d) Thermal efficiency
q_in = cp (T₃ − T₂) = 1.005 × (1023.15 − 541.2) = 484.4 kJ/kg

η = w_net/q_in = 213.0/484.4 = **44.0 %**

Check: η = 1 − 1/r_p^0.2857 = 1 − 1/1.7851 = 44.0 % ✓.

**Answer: r_p = 7.60; w_c = 239.2 kJ/kg; ṁ = 0.469 kg/s; η = 44.0 %.**

## Check yourself

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

1. Which relation gives r_p here?
   A. T4/T1 = r_p
   B. T2/T1 = T3/T1
   C. T3/T4 = r_p^0.2857
   D. T3/T1 = r_p
   Answer: C. Turbine isentropic: 1023.15/573.15 = 1.7851 = r_p^0.2857, so r_p = 7.60. (Key point: Pressure ratio relation)

2. Compressor work per kg is:
   A. cp(T3 - T2)
   B. cp(T4 - T1)
   C. cp(T3 - T4)
   D. cp(T2 - T1) = 239.2 kJ/kg
   Answer: D. T2 = 541.2 K. (Key point: Compressor work)

3. Mass flow is:
   A. Net power/w_net = 100/213.0 = 0.469 kg/s
   B. Power/q_in
   C. Power/w_comp
   D. Power x w_net
   Answer: A. w_net = 452.2 - 239.2 = 213.0 kJ/kg. (Key point: Mass flow)

4. Efficiency is:
   A. 60%
   B. 44.0%
   C. 77%
   D. 30%
   Answer: B. 1 - 1/1.7851 = 0.440. (Key point: Efficiency)

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