---
title: "Open cycle gas turbine on heavy oil"
subject: "Power Plant Equipment"
chapter: 6
key: "semester_7/power_plant_equipment:gt-heavy-oil-capacity"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-heavy-oil-capacity
site: BCT Engineering
---

# Open cycle gas turbine on heavy oil

Power Plant Equipment · Chapter 6: Gas Turbine Power Plant

- Asked in: 2073 Shrawan, 10 marks
- Video: [Example 9.1: Calculating the power output from a gas turbine unit working on Open Brayton's cycle](https://www.youtube.com/watch?v=w_smCBy-iYc) — Dr. Arshad Ali Khan Official
- 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-heavy-oil-capacity
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-6

## Question

An open cycle gas turbine plant uses heavy oil as fuel. The maximum pressure and temperature in the cycle are 500kPa and 650°C. The pressure and temperature of air entering into the compressor are 10⁵Pa and 27°C. The exit pressure of the turbine is also 10⁵Pa. Assuming isentropic efficiencies of compressor and turbine to be 80% and 85% respectively, find the thermal efficiency of the cycle. Take Cp (for air and gas) = 1kJ/kg°C and γ (for air and gases) = 1.4. If the plant consumes 5 kg of fuel per sec, find the power generating capacity of the plant.

## Short answer

Thermal efficiency = 17.39%. Assuming fuel CV 42 MJ/kg, with 5 kg/s fuel consumption, plant power capacity = 0.1739 × 5 × 42000 = 36.5 MW.

## Answer

**Assumptions:** calorific value of the heavy oil is not given; a typical value **CV = 42 000 kJ/kg** is assumed. Heat released by the fuel equals the heat received by the gases in the combustion chamber.

**Data:** $p_2 = p_3 = 500$ kPa, $T_3 = 650 + 273 = 923$ K, $p_1 = p_4 = 100$ kPa, $T_1 = 27 + 273 = 300$ K, $\eta_c = 0.80$, $\eta_t = 0.85$, $c_p = 1$ kJ/kg K, $\gamma = 1.4$; $r_p = 5$, $5^{0.2857} = 1.5838$.

#### Compressor
$$
\begin{aligned}
T_{2s} &= 300 \times 1.5838 = 475.15\ \text{K} \\
T_2 &= 300 + \frac{475.15 - 300}{0.80} = 518.93\ \text{K} \\
W_c &= 1 \times (518.93 - 300) = 218.93\ \text{kJ/kg}
\end{aligned}
$$

#### Turbine
$$
\begin{aligned}
T_{4s} &= \frac{923}{1.5838} = 582.77\ \text{K} \\
T_4 &= 923 - 0.85(923 - 582.77) = 633.80\ \text{K} \\
W_t &= 1 \times (923 - 633.80) = 289.20\ \text{kJ/kg}
\end{aligned}
$$

#### Thermal efficiency
$$
\begin{aligned}
W_{net} &= 289.20 - 218.93 = 70.26\ \text{kJ/kg} \\
Q_s &= c_p (T_3 - T_2) = 923 - 518.93 = 404.07\ \text{kJ/kg} \\
\eta_{th} &= \frac{70.26}{404.07} = 0.1739 = 17.39\%
\end{aligned}
$$

#### Power generating capacity
Heat supplied by fuel:
$$
\dot Q = \dot m_f \times CV = 5 \times 42\,000 = 210\,000\ \text{kW}
$$
$$
P = \eta_{th} \times \dot Q = 0.1739 \times 210\,000 = 36\,518\ \text{kW} \approx 36.5\ \text{MW}
$$
Corresponding air–fuel ratio (from $\dot m_f CV = (\dot m_a + \dot m_f)c_p(T_3 - T_2)$): $\dfrac{\dot m_a}{\dot m_f} = \dfrac{42\,000 - 404.07}{404.07} = 102.9$, i.e. about 515 kg/s of air.

**Answer:** Thermal efficiency = **17.39 %**; power capacity ≈ **36.5 MW** (for CV = 42 MJ/kg; in general $P = 0.1739 \times 5 \times CV$).

## Check yourself

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

1. How is the pressure ratio found for this open-cycle plant?
   A. 500/100 = 5, since the turbine exhausts at 100 kPa
   B. 650/27
   C. 500 x 100
   D. 500 - 100
   Answer: A. Compressor inlet and turbine exit are both 100 kPa and maximum pressure is 500 kPa, so rp = 5. (Key point: Pressure ratio)

2. Heat supplied in the combustion chamber per kg is:
   A. cp(T3 - T1)
   B. cp(T3 - T2) with T2 the actual compressor exit
   C. cp(T4 - T1)
   D. cp(T3 - T4)
   Answer: B. Heat is added between the actual compressor exit (518.93 K) and T3 = 923 K, giving 404.07 kJ/kg. (Key point: Heat supplied per kg)

3. What is the thermal efficiency from Wt = 289.20 and Wc = 218.93 kJ/kg?
   A. 70.26 / 289.20 = 24.3 %
   B. 70.26 / 404.07 = 17.39 %
   C. 218.93 / 404.07 = 54 %
   D. 289.20 / 404.07 = 71.6 %
   Answer: B. Efficiency = Wnet/Qs = 70.26/404.07 = 17.39 %. (Key point: Thermal efficiency)

4. How is the plant power capacity calculated?
   A. Efficiency x fuel rate x calorific value
   B. Fuel rate x Wnet
   C. Fuel rate / calorific value
   D. Efficiency / (fuel rate x CV)
   Answer: A. Heat input = 5 kg/s x CV (42 000 kJ/kg assumed) = 210 MW, times 0.1739 gives about 36.5 MW. (Key point: Plant power from fuel)

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