A force of 9 N is required to tow a 150 ship model at 48 kmh

A force of 9 N is required to tow a 1:50 ship model at 4.8 km/h. Assuming the same water in towing basin and sea, calculate the corresponding speed and force in the prototype if the flow is dominated by: (a) density and gravity (equal Froude numbers), (b) density and surface tension (equal Weber numbers), and (c) density and viscosity (equal Reynolds number).

Solution

a) Fr = V / sqrt (gL)

V1 / sqrt (g*L1) = V2 / sqrt (g*L2)

4.8 / sqrt (g*1) = V2 / sqrt (g*50)

V2 = 33.94 km/h

Force F = K*A*V^2

F1 / F2 = A1*V1^2 / (A2*V2^2)

9 / F2 = (1/50)^2 * (4.8/33.94)^2

F2 = 1125000 N

b)

We = rho*V^2 / sigma

For same We, rho and sigma we get V1^2 = V2^2

or V1 = V2 = 4.8 km/h

Force F = K*A*V^2

F1 / F2 = A1*V1^2 / (A2*V2^2)

9 / F2 = (1/50)^2 * (4.8/4.8)^2

F2 = 22500 N

c)

Re = V*L/neu

For same Re and neu we get

V1 * L1 = V2 *L2

V2 = 4.8*1 / 50

V2 = 0.096 km/h

Force F = K*A*V^2

F1 / F2 = A1*V1^2 / (A2*V2^2)

9 / F2 = (1/50)^2 * (4.8/0.096)^2

F2 = 9 N