---
title: "Ideal turbojet at altitude: thrust and TSFC"
subject: "Turbo Machines"
chapter: 6
key: "semester_7/turbo_machines:ideal-turbojet-flight-numerical"
source: https://www.bctengineer.com/q/semester_7/turbo_machines/ideal-turbojet-flight-numerical
site: BCT Engineering
---

# Ideal turbojet at altitude: thrust and TSFC

Turbo Machines · Chapter 6: Gas Turbine Cycles of Aircraft Propulsion

- Asked in: year not given, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/turbo_machines/ideal-turbojet-flight-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-6

## Question

An ideal turbojet flies at Mach 0.8 at an altitude where the ambient temperature is 216.65 K and pressure is 22.63 kPa. The compressor pressure ratio is 12 and the maximum cycle temperature is 1400 K. Assume isentropic diffuser, compressor, turbine and nozzle, no pressure loss in the combustor, complete expansion in the nozzle, turbine work equal to compressor work, $c_p = 1.005$ kJ/kg K and $\gamma = 1.4$ throughout, and neglect the fuel mass in the thrust. The fuel has a lower calorific value of 43 000 kJ/kg. Find (a) the flight speed, (b) the stagnation temperature and pressure after the diffuser, (c) the temperature after the turbine, (d) the jet velocity, (e) the specific thrust, (f) the fuel-air ratio and TSFC, and (g) the propulsive, thermal and overall efficiencies. Also find the thrust for an air mass flow of 50 kg/s.

## Short answer

Flight speed is 236.0 m/s, jet velocity is 1039.0 m/s, and specific thrust is 803.0 N s/kg. Fuel-air ratio is 0.0211 with TSFC 94.6 kg/(kN h). Efficiencies are eta_p = 0.370, eta_th = 0.564, eta_o = 0.209, and thrust at 50 kg/s is 40.15 kN.

## Answer

**Data:** $Ma_0 = 0.8$, $T_0 = 216.65$ K, $p_0 = 22.63$ kPa, $r_c = 12$, $T_{04} = 1400$ K.

#### (a) Flight speed
$$
V = Ma_0\sqrt{\gamma R T_0} = 0.8\sqrt{1.4 \times 287 \times 216.65} = 236.0\ \text{m/s}
$$

#### (b) Diffuser exit (station 2)
$$
T_{02} = T_0\left(1 + 0.2 Ma_0^2\right) = 216.65 \times 1.128 = 244.4\ \text{K}, \qquad
p_{02} = p_0\left(\frac{T_{02}}{T_0}\right)^{3.5} = 34.50\ \text{kPa}
$$

#### (c) Compressor, turbine
$$
T_{03} = T_{02} r_c^{0.2857} = 244.4 \times 12^{0.2857} = 497.1\ \text{K}, \qquad p_{03} = 12 \times 34.50 = 413.9\ \text{kPa}
$$
Turbine work equals compressor work:
$$
T_{04} - T_{05} = T_{03} - T_{02} \;\Rightarrow\; T_{05} = 1400 - (497.1 - 244.4) = 1147.3\ \text{K}
$$
$$
p_{05} = p_{04}\left(\frac{T_{05}}{T_{04}}\right)^{3.5} = 413.9 \times \left(\frac{1147.3}{1400}\right)^{3.5} = 206.25\ \text{kPa}
$$

#### (d) Jet velocity
Complete expansion to $p_0$:
$$
T_6 = T_{05}\left(\frac{p_0}{p_{05}}\right)^{0.2857} = 610.2\ \text{K}
$$
$$
C_j = \sqrt{2c_p(T_{05} - T_6)} = \sqrt{2 \times 1005 \times 537.1} = 1039.0\ \text{m/s}
$$

#### (e) Specific thrust
$$
\frac{F}{\dot m_a} = C_j - V = 1039.0 - 236.0 = 803.0\ \text{N s/kg}
$$

#### (f) Fuel-air ratio and TSFC
$$
q_{in} = c_p (T_{04} - T_{03}) = 1.005 \times (1400 - 497.1) = 907.5\ \text{kJ/kg}, \qquad f = \frac{q_{in}}{LHV} = 0.0211
$$
$$
TSFC = \frac{f}{F/\dot m_a} \times 3600 \times 1000 = 94.6\ \text{kg/(kN h)}
$$

#### (g) Efficiencies
$$
\eta_p = \frac{2V}{C_j + V} = \frac{2 \times 236.0}{1039.0 + 236.0} = 0.370
$$
$$
\eta_{th} = \frac{(C_j^2 - V^2)/2}{q_{in}} = \frac{511.9}{907.5} = 0.564, \qquad \eta_o = \eta_p\eta_{th} = 0.209
$$

#### Thrust
$$
F = 50 \times 803.0 = 40.15\ \text{kN}
$$

**Answer:** $V = 236.0$ m/s; $C_j = 1039.0$ m/s; specific thrust = 803.0 N s/kg; $f = 0.0211$; TSFC = 94.6 kg/(kN h); $\eta_p = 0.370$, $\eta_{th} = 0.564$, $\eta_o = 0.209$; $F = 40.15$ kN.

## Check yourself

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

1. The stagnation temperature after the diffuser is found from
   A. T02 = T0 + V
   B. T02 = T0 r_c^0.2857
   C. T02 = T0 (1 + 0.2 Ma0^2) = 244.4 K
   D. T02 = T04
   Answer: C. Isentropic diffusion to rest gives T02/T0 = 1 + (gamma-1)/2 Ma^2 = 1.128. (Key point: Diffuser stagnation temperature)

2. How is T05 found?
   A. T05 = T04 / r_c
   B. T05 = T03
   C. T05 = T04 + (T03 - T02)
   D. T05 = T04 - (T03 - T02), because turbine work equals compressor work
   Answer: D. Equal work means equal temperature drop and rise: 1400 - (497.1 - 244.4) = 1147.3 K. (Key point: Turbine exit temperature)

3. The jet velocity of 1039 m/s is found from
   A. C_j = sqrt(2 cp (T05 - T6)) after complete expansion of the gas to ambient pressure
   B. C_j = sqrt(2 cp (T04 - T03))
   C. C_j = V + 803
   D. C_j = Ma0 a
   Answer: A. The nozzle expands isentropically from p05 to p0, giving T6 = 610.2 K and a drop of 537.1 K. (Key point: Jet velocity)

4. The fuel-air ratio is computed as
   A. f = cp (T05 - T02)/LHV
   B. f = cp (T04 - T03)/LHV = 907.5/43000 = 0.0211
   C. f = LHV / q_in
   D. f = C_j/V
   Answer: B. Heat added in the combustor per kg air divided by LHV. (Key point: Fuel-air ratio)

5. Why is propulsive efficiency (0.370) much lower than thermal efficiency (0.564)?
   A. Fuel mass is ignored
   B. The pressure ratio is too high
   C. The combustor has losses
   D. Jet velocity 1039 m/s is much larger than flight speed 236 m/s, so much jet kinetic energy is wasted
   Answer: D. eta_p = 2V/(Cj+V) = 0.370; the overall efficiency is then 0.370 x 0.564 = 0.209. (Key point: Efficiencies)

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