Question

The luxury liner Queen Elizabeth 2 has a diesel-electric power plant with a maximum power of92 MWat a cruising speed of 32.5 knots. What forward force is exerted on the ship at this speed?(1 knot = 1.852 km / h).

Short answer

The forward force exerted on the ship at speed.

Step-by-step solution

Given data:

Maximum power of the diesel power plant,$P=92\mathrm{MW}=92\times {10}^6\mathrm{W}$

The velocity of the ship, v = 32.5 knot

To understand the concept:

Convert the velocity of the ship in meters per second using conversion factors. Then using the relation between power, force, and velocity, find the applied force on the ship.

Formula:

P = Fv

Calculate the forward force exerted on the ship at a speed of 32.5 knots :

The velocity of the ship is,

$$\begin{gathered} v=\left(32.5\times 1.852\times \frac{5}{18}\right) \\ =16.71\mathrm{m}/\mathrm{s} \end{gathered}$$

The maximum power of the ship is defined as below.

$$\begin{gathered} P=Fv \\ F=\frac{P}{v} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} F=\frac{92.{10}^6\mathrm{W}}{16.71\mathrm{m}/\mathrm{s}} \\ =5.5\times {10}^6\mathrm{N} \end{gathered}$$

Hence, the forward force exerted on the ship at speed 32.5 knot is $5.5\times {10}^6\mathrm{N}$.