---
title: "Rocket engine thrust and specific impulse calculation"
subject: "Turbo Machines"
chapter: 5
key: "semester_7/turbo_machines:rocket-thrust-specific-impulse-numerical"
source: https://www.bctengineer.com/q/semester_7/turbo_machines/rocket-thrust-specific-impulse-numerical
site: BCT Engineering
---

# Rocket engine thrust and specific impulse calculation

Turbo Machines · Chapter 5: Theoretical Jet Engine

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/turbo_machines/rocket-thrust-specific-impulse-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-5

## Question

A rocket engine consumes 120 kg/s of propellant and the exhaust leaves the nozzle at 2800 m/s. The nozzle exit area is 0.5 m² and the exit pressure is 80 kPa. Find (a) the thrust and effective exhaust velocity at sea level (ambient 101.3 kPa), (b) the specific impulse at sea level and in vacuum, and (c) the propulsive efficiency at sea level when the rocket moves at 1500 m/s and the corresponding thrust power. Take $g_0 = 9.81$ m/s².

## Short answer

Sea-level thrust is 325.4 kN and effective exhaust velocity is 2711.2 m/s. Specific impulse is 276.4 s at sea level and 319.4 s in vacuum. Propulsive efficiency is 0.847 and thrust power is 488025 kW.

## Answer

**Data:** $\dot m_p = 120$ kg/s, $C_e = 2800$ m/s, $A_e = 0.5$ m², $p_e = 80$ kPa.

#### (a) Sea-level thrust
$$
F = \dot m_p C_e + (p_e - p_a)A_e = 120 \times 2800 + (80000 - 101300) \times 0.5
$$
$$
F = 336000 - 10650 = 325350\ \text{N} = 325.4\ \text{kN}
$$
The nozzle is over-expanded ($p_e < p_a$), so the pressure term is negative.
$$
C_{eff} = \frac{F}{\dot m_p} = \frac{325350}{120} = 2711.2\ \text{m/s}
$$

#### (b) Specific impulse
$$
I_{sp,SL} = \frac{C_{eff}}{g_0} = \frac{2711.2}{9.81} = 276.4\ \text{s}
$$
In vacuum $p_a = 0$:
$$
F_{vac} = 336000 + 80000 \times 0.5 = 376000\ \text{N}, \qquad C_{eff,vac} = 3133.3\ \text{m/s}, \qquad I_{sp,vac} = \frac{3133.3}{9.81} = 319.4\ \text{s}
$$

#### (c) Propulsive efficiency and thrust power
$$
\frac{v}{C_{eff}} = \frac{1500}{2711.2} = 0.5533
$$
$$
\eta_p = \frac{2(v/C_{eff})}{1 + (v/C_{eff})^2} = \frac{2 \times 0.5533}{1 + 0.3061} = 0.847
$$
$$
P_T = F v = 325350 \times 1500 = 488025\ \text{kW}
$$

**Answer:** $F = 325.4$ kN, $C_{eff} = 2711.2$ m/s; $I_{sp} = 276.4$ s (sea level), 319.4 s (vacuum); $\eta_p = 0.847$; $P_T = 488025$ kW.

## Check yourself

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

1. Why is the pressure term negative at sea level here?
   A. Exit pressure 80 kPa is below ambient 101.3 kPa (over-expanded nozzle)
   B. Exit pressure is above ambient
   C. The nozzle area is negative
   D. Ambient is zero
   Answer: A. (p_e - p_a)A_e = (80000 - 101300)(0.5) = -10650 N, so F = 336000 - 10650 = 325.35 kN. (Key point: Sea-level thrust)

2. The effective exhaust velocity at sea level is
   A. 2800 m/s
   B. 3133 m/s
   C. 2711 m/s
   D. 325 m/s
   Answer: C. C_eff = F/m_p = 325350/120 = 2711.2 m/s, lower than C_e because of the negative pressure thrust. (Key point: Effective exhaust velocity)

3. Specific impulse in vacuum is higher than at sea level because
   A. mass flow is larger
   B. exhaust velocity C_e increases
   C. g0 changes
   D. ambient pressure is zero so the pressure thrust is positive: 376 kN, Isp = 319.4 s
   Answer: D. In vacuum F = 336000 + 80000 x 0.5 = 376000 N, giving C_eff = 3133 m/s and Isp = 319.4 s versus 276.4 s at sea level. (Key point: Vacuum specific impulse)

4. Propulsive efficiency at v = 1500 m/s uses
   A. eta_p = v/C_e = 0.536
   B. eta_p = 2(v/C_eff)/(1 + (v/C_eff)^2) = 0.847 with C_eff = 2711 m/s
   C. eta_p = F/m_p
   D. eta_p = 1 - v/C_eff
   Answer: B. With v/C_eff = 0.5533, eta_p = 1.1066/1.3061 = 0.847. (Key point: Propulsive efficiency)

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