---
title: "Thrust and efficiency of a water-jet ship"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_4/fluid_mechanics_mechanical:ship-water-jet-propulsion"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/ship-water-jet-propulsion
site: BCT Engineering
---

# Thrust and efficiency of a water-jet ship

Fluid Mechanics · Chapter 4: Basic Equations for Fluid Flow

- 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_4/fluid_mechanics_mechanical/ship-water-jet-propulsion
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-4

## Question

A ship is propelled by a water-jet system. A pump draws water from the sea through the bow, and discharges it astern at a velocity of 15 m/s relative to the ship at a rate of 2.5 m$^3$/s. The ship moves at 6 m/s. Find the thrust, the useful (propulsive) power, the power input to the jet, and the propulsive efficiency. Take the density of sea water as 1000 kg/m$^3$.

## Short answer

Thrust is 22.5 kN, propulsive power 135 kW, power input to the jet 236.25 kW, and propulsive efficiency 57.1%.

## Answer

#### Given data
$Q = 2.5\ \text{m}^3/\text{s}$, $V_j = 15\ \text{m/s}$ (relative to the ship, astern), $V_s = 6\ \text{m/s}$, $\rho = 1000\ \text{kg/m}^3$.

#### Principle
Relative to the ship, water enters the system with velocity $V_s$ (the ship's speed) and leaves with $V_j$. The thrust equals the rate of change of momentum:

$$
F = \rho Q (V_j - V_s)
$$

#### Thrust
$$
F = 1000 \times 2.5 \times (15 - 6) = 22\,500\ \text{N} = 22.5\ \text{kN}
$$

#### Useful power
$$
P_{out} = F V_s = 22\,500 \times 6 = 135\,000\ \text{W} = 135\ \text{kW}
$$

#### Power input to the jet (rate of kinetic energy gain of the water)
$$
\begin{aligned}
P_{in} &= \tfrac12 \rho Q (V_j^2 - V_s^2) \\
&= 0.5 \times 1000 \times 2.5 \times (225 - 36) \\
&= 236\,250\ \text{W} = 236.25\ \text{kW}
\end{aligned}
$$

#### Propulsive efficiency
$$
\eta_p = \frac{P_{out}}{P_{in}} = \frac{135}{236.25} = 0.571 = 57.1\%
$$

Check with the formula $\eta_p = \dfrac{2V_s}{V_j + V_s} = \dfrac{12}{21} = 0.571$.

The efficiency rises as the jet velocity approaches the ship speed, which means a larger mass flow at a smaller jet velocity is more efficient.

**Answer: $F = 22.5\ \text{kN}$, $P_{out} = 135\ \text{kW}$, $P_{in} = 236.25\ \text{kW}$, $\eta_p = 57.1\%$**

## Check yourself

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

1. Relative to the ship, water enters at the ship speed V_s and leaves at V_j. The thrust is:
   A. rho Q (V_j + V_s)
   B. rho Q (V_j - V_s) = 22.5 kN
   C. rho Q V_j = 37.5 kN
   D. rho Q V_s
   Answer: B. Thrust equals the rate of change of momentum: 1000 x 2.5 x (15 - 6) = 22,500 N. (Key point: Thrust formula)

2. The useful power and the input power to the jet are respectively:
   A. F V_s = 135 kW; 0.5 rho Q (V_j^2 - V_s^2) = 236.25 kW
   B. F V_j = 337.5 kW; 0.5 rho Q V_j^2 = 281 kW
   C. F V_s = 135 kW; 0.5 rho Q V_j^2 = 281 kW
   D. F V_j = 337.5 kW; 0.5 rho Q V_s^2 = 45 kW
   Answer: A. Useful power is thrust times ship speed; input is the rate of kinetic energy gain of the water. (Key point: Useful and input power)

3. The propulsive efficiency is 2 V_s/(V_j + V_s) = 57.1%. It increases when:
   A. V_j increases well above V_s
   B. V_j approaches V_s (larger flow, smaller jet speed)
   C. V_s is reduced to zero
   D. Q is reduced
   Answer: B. The efficiency tends to 1 as V_j approaches V_s, so a larger mass flow at lower jet velocity is more efficient. (Key point: Propulsive efficiency)

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