Resistance + Propulsion: Scaling laws.

Resistance + Propulsion: Scaling laws

A 6 m model of a 180 m long ship is towed in a model basin at a speed of 1.61 m/s. The towing pull is 20 N. The wetted surface of the model is 4 m2. Estimate the corresponding speed for the ship in knots and the effective power PE, assuming resistance coefficients to be independent of scale for simplicity.

Solution

The scale is:        

[pic 1]

The wetted surface at full scale is:        

[pic 2] m2

The Froude number is:        

[pic 3]

The Froude number gives the full-scale speed:        

[pic 4] m/s

1 kn = 0.5144 m/s. Thus the speed of the full-scale ship is [pic 5]kn.

A simple scaling law assumes that the resistance coefficients remain constant. (More accurate prediction methods have a slight speed dependence of the frictional resistance coefficient and introduce a correlation coefficient.) Then:

[pic 6] kN

The effective power is:

[pic 7] kW

Resistance + Propulsion: Simple scaling laws

1. A ship of 15000 t displacement has speed 20 kn. Determine the corresponding speed for a similar ship of displacement 14000 t using geometric and Froude similarity.
2. A ship of 140 m length has speed 15 kn. Determine the corresponding speed for a 7 m model.
3. A ship of 5000 t displacement has speed 18 kn, length 120 m. The towing tank model has 6 m length. At which speed should the model be tested? What is the ratio of the ship-to-model PE at this speed?

Solution

1. Let's denote the ship with 15000 t displacement with index 1, the other with index 2. The lengths L of the ships correlate to the displacements Δ following geometric similarity:

[pic 8]

The speeds follow from Froude similarity:

[pic 9] kn

1. We keep Froude similarity:

[pic 10]kn = 1.73 m/s

1. We keep Froude similarity:

[pic 11]kn = 2.07 m/s

The effective power follows from PE = R⋅V. The resistance is expressed as [pic 12]; thus it scales with λ3, as the area scales with λ2 and the speed in Froude similarity scales with λ0.5. Thus the power scales with λ3.5:

                [pic 13]

Resistance + Propulsion: Power estimate following simple scaling laws

A ship (L = 122.00 m, B = 19.80 m, T = 7.33 m, Δ = 8700 t) has the following power requirements:

Estimate the power requirements for a similar ship with 16250 t displacement at speed 19.5 knots.

Solution

We convert the speed of the new ship of 19.5 kn to the corresponding speed for the original ship, following Froude and geometric similarity:

[pic 14]kn

For this speed, the power for the original ship is linearly interpolated to:

[pic 15] kW

This is scaled to the new ship:

[pic 16] kW

Explanation: Power scales with speed V3 and area A, which gives for the speed with Froude similarity a scale with length scale λ1.5 and for the area λ2, thus together λ3.5. If we use the displacement instead of the length, we divide by λ3. Hence the exponent of 3.5/3.

Resistance + Propulsion: Ship vs Car - Scaling and resistance prediction

A car has a wind resistance value of CW = 0.3. The car has a frontal area of 2 m2. The air density is 1.23 kg/m3 and the car weighs 1200 kg.

Now compare the efficiency of a ship to that of the car. A ferry has a resistance of 2000 kN at a speed of 25 kn. The ferry has a mass of 16500 t, length Lpp = 190 m, and a wetted surface of 4400 m2.

1. Compute the necessary power for car speeds of 10, 20, 30, 40, 100 and 200 km/h.
2. What scale has to be taken to get a ship model of same mass as the car?
3. What is the model speed if Froude similarity is kept?
4. What is the model resistance if the correlation allowance is cA = 0?
5. How fast could the car go with the same power the ship model needs in towing at design Froude number?

νsea = 1.16⋅10-6 m2/s, νfresh = 1.14⋅10-6 m2/s, ρsea = 1026 kg/m3

Solution

1. The force follows from:

[pic 17]

The reference area is here the frontal area. Thus:

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