A ship has a wetted surface area of 3200 ml . Calculate the power required to overcome frictional resistance at 17 knots if n = 1.825 and f = 0.424
The correct answer and explanation is:
To calculate the power required to overcome frictional resistance, we use the following formula: Power (P)=f⋅A⋅V3/n\text{Power (P)} = f \cdot A \cdot V^3 / n
Where:
- ff = coefficient of frictional resistance (given as 0.424)
- AA = wetted surface area (given as 3200 m² — assuming “ml” was a typo, as wetted surface area is measured in square meters)
- VV = ship speed in m/s (convert 17 knots to m/s)
- nn = propulsive efficiency or constant (given as 1.825)
Step 1: Convert knots to m/s
1 knot=0.51444 m/s1 \text{ knot} = 0.51444 \text{ m/s} 17 knots=17×0.51444=8.7455 m/s17 \text{ knots} = 17 \times 0.51444 = 8.7455 \text{ m/s}
Step 2: Plug values into the formula
P=0.424⋅3200⋅(8.7455)31.825P = \frac{0.424 \cdot 3200 \cdot (8.7455)^3}{1.825}
First, calculate (8.7455)3(8.7455)^3: 8.74553≈669.0598.7455^3 \approx 669.059
Now plug in: P=0.424⋅3200⋅669.0591.825P = \frac{0.424 \cdot 3200 \cdot 669.059}{1.825} P=0.424⋅3200⋅669.0591.825=0.424⋅2140988.81.825P = \frac{0.424 \cdot 3200 \cdot 669.059}{1.825} = \frac{0.424 \cdot 2140988.8}{1.825} P≈907857.251.825≈497478.49 wattsP \approx \frac{907857.25}{1.825} \approx 497478.49 \text{ watts}
✅ Final Answer:
497.48 kW\boxed{497.48 \text{ kW}}
🔍 Explanation (Approx. 300 words):
Frictional resistance is one of the major components of total resistance a ship must overcome to maintain motion through water. It is caused by the water’s viscosity acting against the wetted surface of the hull. The formula used here captures this by combining three key factors: the wetted surface area AA, the friction coefficient ff, and the cube of the velocity V3V^3. The reason velocity is cubed is because resistance increases significantly with speed — even small increases in speed greatly increase power requirements.
In this case, the ship’s wetted surface area is 3200 m², and it travels at a speed of 17 knots, which converts to about 8.7455 m/s. The frictional coefficient, which depends on hull surface smoothness and water properties, is given as 0.424. The constant n=1.825n = 1.825 may represent a form of efficiency or unit normalization.
By applying these values in the formula: P=f⋅A⋅V3nP = \frac{f \cdot A \cdot V^3}{n}
we find that the ship requires approximately 497.48 kilowatts of power to overcome just the frictional part of its total resistance. This doesn’t include wave-making resistance or air resistance, which would add to the total power demand.
This calculation is critical in naval architecture and marine engineering, as it helps in selecting appropriate engines, optimizing hull design, and improving fuel efficiency — all vital for operational cost savings and environmental compliance.