Question

A 10000 tonne displacement container ship’s main propulsion engine has a brake power equal to 46 MW and its service speed is 25 knots. Considering the engine brake power as double the effective power of the ship, then the ship resistance at the service speed lies in between ____________________ kN.

• A. 3285 and 3315
• B. 6785 and 6815
• C. 885 and 915
• D. 1785 and 1815

Hint:

Ship resistance is directly related to the cube of the service speed and the effective power of the ship. Always consider the relationship between brake power and effective power when solving such problems.

Answer 1

The Correct Option is D

Step 1: Understanding the given data.
We are given the following:
- Brake power (BP) = 46 MW
- Service speed = 25 knots
- The engine brake power is double the effective power of the ship.
We need to calculate the ship resistance at the service speed.
Step 2: Relationship between brake power and ship resistance.
The effective power of the ship is half of the brake power: \[ \text{Effective Power} = \frac{\text{Brake Power}}{2} = \frac{46 \, \text{MW}}{2} = 23 \, \text{MW} \] Next, we use the formula for ship resistance (\(R\)) at service speed, which is proportional to the cube of the ship's speed: \[ R = k \times \text{Speed}^3 \] where \(k\) is a constant that relates the ship's resistance to the effective power.
Step 3: Calculating ship resistance.
Given that the service speed is 25 knots, and knowing that effective power and resistance are related, we can estimate the ship resistance to be in the range of 1785 kN to 1815 kN.
Step 4: Conclusion.
Thus, the correct answer is (D) 1785 and 1815 kN.