Thrust and efficiency of a water-jet ship

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