---
title: "Turbojet thrust, efficiencies and fuel consumption"
subject: "Turbo Machines"
chapter: 5
key: "semester_7/turbo_machines:turbojet-thrust-efficiencies-numerical"
source: https://www.bctengineer.com/q/semester_7/turbo_machines/turbojet-thrust-efficiencies-numerical
site: BCT Engineering
---

# Turbojet thrust, efficiencies and fuel consumption

Turbo Machines · Chapter 5: Theoretical Jet Engine

- Asked in: year not given, 8 marks
- Video: [Turbojets: Thermodynamics for Mechanical Engineers](https://www.youtube.com/watch?v=Ir4zlj45vRY) — Mike Schertzer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/turbo_machines/turbojet-thrust-efficiencies-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-5

## Question

A turbojet aircraft flies at 250 m/s. The engine takes in 40 kg/s of air and the fuel-air ratio is 0.015. The exhaust gases leave the propelling nozzle at 600 m/s relative to the engine, with the nozzle exit pressure equal to the ambient pressure. The lower calorific value of the fuel is 43 000 kJ/kg. Calculate (a) the net thrust, (b) the thrust power, (c) the propulsive efficiency, (d) the thermal efficiency and overall efficiency, and (e) the thrust specific fuel consumption.

## Short answer

Net thrust F = m_dot_j Cj - m_dot_a V = 14.36 kN. Thrust power P_T = F V = 3590 kW. Propulsive efficiency eta_p = P_T / (0.5 m_dot_j Cj^2 - 0.5 m_dot_a V^2) = 0.593, thermal efficiency eta_th = 0.235, overall eta_o = 0.139. TSFC = m_dot_f / F = 150.4 kg/(kN h).

## Answer

**Data:** $V = 250$ m/s, $\dot m_a = 40$ kg/s, $f = 0.015$, $C_j = 600$ m/s, $p_j = p_0$, $LHV = 43000$ kJ/kg.

Fuel flow $\dot m_f = f\dot m_a = 0.015 \times 40 = 0.60$ kg/s; jet mass flow $\dot m_j = 40 + 0.60 = 40.60$ kg/s.

#### (a) Net thrust
$$
F = \dot m_j C_j - \dot m_a V = 40.60 \times 600 - 40 \times 250 = 14360\ \text{N} = 14.36\ \text{kN}
$$
(Specific thrust $F/\dot m_a = 359.0$ N s/kg.)

#### (b) Thrust (propulsive) power
$$
P_T = F V = 14360 \times 250 = 3590\ \text{kW}
$$

#### (c) Propulsive efficiency
Rate of increase of kinetic energy of the working fluid:
$$
\Delta KE = \tfrac12 \dot m_j C_j^2 - \tfrac12 \dot m_a V^2 = \tfrac12 (40.60)(600^2) - \tfrac12 (40)(250^2) = 6058\ \text{kW}
$$
$$
\eta_p = \frac{P_T}{\Delta KE} = \frac{3590}{6058} = 0.593
$$
(The simple formula $2V/(C_j + V) = 0.588$ ignores the fuel mass; the value above is the exact one.)

#### (d) Thermal and overall efficiency
Energy input $= \dot m_f \times LHV = 0.60 \times 43000 = 25800$ kW.
$$
\eta_{th} = \frac{\Delta KE}{\dot m_f LHV} = \frac{6058}{25800} = 0.235, \qquad
\eta_o = \eta_p\eta_{th} = \frac{P_T}{\dot m_f LHV} = 0.139
$$

#### (e) Thrust specific fuel consumption
$$
TSFC = \frac{\dot m_f}{F} = \frac{0.60 \times 3600}{14.36} = 150.4\ \text{kg/(kN h)}
$$

**Answer:** $F = 14.36$ kN; $P_T = 3590$ kW; $\eta_p = 0.593$; $\eta_{th} = 0.235$; $\eta_o = 0.139$; TSFC = 150.4 kg/(kN h).

## Check yourself

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

1. Net thrust here is found by
   A. F = m_a (C_j - V) ignoring fuel, giving 14.0 kN
   B. F = m_j C_j - m_a V = 40.60 x 600 - 40 x 250 = 14.36 kN
   C. F = m_j C_j + m_a V
   D. F = m_f C_j
   Answer: B. The jet mass flow includes fuel (40.6 kg/s), and the nozzle is fully expanded so no pressure thrust. (Key point: Net thrust)

2. Thrust (propulsive) power is
   A. F x C_j = 8616 kW
   B. Delta KE of the stream = 6058 kW
   C. F x V = 14360 x 250 = 3590 kW
   D. m_f x LHV = 25800 kW
   Answer: C. Useful power is thrust times flight speed. (Key point: Thrust power)

3. Propulsive efficiency is
   A. thrust power divided by fuel energy
   B. kinetic energy rate divided by fuel energy
   C. jet velocity divided by flight velocity
   D. thrust power divided by the rate of increase of kinetic energy of the fluid, 0.593
   Answer: D. eta_p = 3590/6058 = 0.593; the simple 2V/(Cj+V) = 0.588 ignores fuel mass. (Key point: Propulsive efficiency definition)

4. How are thermal and overall efficiency related?
   A. eta_o = eta_p x eta_th = 0.593 x 0.235 = 0.139
   B. eta_o = eta_p + eta_th
   C. eta_o = eta_th / eta_p
   D. eta_o = eta_p only
   Answer: A. eta_th = Delta KE / (m_f LHV) = 0.235 and eta_o = thrust power / fuel energy = eta_p eta_th. (Key point: Thermal and overall efficiency)

5. TSFC is calculated as
   A. m_a / F
   B. m_f / F, here 0.60 x 3600 / 14.36 = 150.4 kg/(kN h)
   C. F / m_f
   D. m_f x LHV / F
   Answer: B. Thrust specific fuel consumption is fuel flow per unit thrust. (Key point: TSFC)

## Related in Turbo Machines

- [Ideal turbojet at altitude: thrust and TSFC](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aideal-turbojet-flight-numerical) — Turbo Machines, chapter 6
- [Rocket engine thrust and specific impulse calculation](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Arocket-thrust-specific-impulse-numerical) — Turbo Machines, chapter 5
- [Fuel-air ratio in a gas turbine combustor](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Acombustion-chamber-fuel-air-ratio-numerical) — Turbo Machines, chapter 3
- [Turbojet cycle, thrust and propulsive efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aturbojet-cycle-thrust-propulsive-efficiency) — Turbo Machines, chapter 6
- [Actual gas turbine cycle with component efficiencies](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aactual-brayton-cycle-numerical) — Turbo Machines, chapter 3
- [Convergent-divergent nozzle with efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Acd-nozzle-efficiency-exit-area-numerical) — Turbo Machines, chapter 4

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- [Efficiency ratio and specific fuel consumption](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Arelative-efficiency-sfc-numerical) — Internal Combustion Engines, chapter 8
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