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Chapter 8: Q94P (page 210)

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

Expert verified

The forward force exerted on the ship at speed.

Step by step solution

01

Given data:

Maximum power of the diesel power plant,P=92MW=92×106W

The velocity of the ship, v = 32.5 knot

02

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

03

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

The velocity of the ship is,

v=32.5×1.852×518=16.71m/s

The maximum power of the ship is defined as below.

P=FvF=Pv

Substitute known values in the above equation.

F=92.106W16.71m/s=5.5×106N

Hence, the forward force exerted on the ship at speed 32.5 knot is 5.5×106N.

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