Question

The drag on a submarine moving below the free surface is to be determined by a test on a 1/16 scale model in a water tunnel. The velocity of the prototype in sea water (rho = 1015 kg/m3, ν = 1.4 × 10-6 m2/s ) is 2 m/s. The test is done in pure water at 20°C. Determine (a) the speed of the water in the water tunnel for dynamic similitude and (b) the ratio of the drag force on the model to the drag force on the prototype.

Answer 1

To solve the problem it is necessary to take into account the concepts related to the Reynolds Number and the Force of drag on the bodies subjected to a Fluid.

The Reynolds number for the Prototype and the Model must therefore be preserved,

`Re_p = Re_m`

`(V_mL_m)/(\upsilon_m) = (V_pL_p)/(\upsilon_p)`

Re-arrange for the speed of the model we have,

`V_m = (L_p)/(L_m)(\upsilon_m)/(\upsilon_p)V_p`

Our values at 20°C would be given of the table of Physical Properties of water where

`\upsilon_m=1*10^(-6)m^2/s`

`\rho_m = 998kg/m^3`

While for the values previous given we have

`V_p = 2m/s`

`\upsilon_m=1*10^(-6)m^2/s`

`\upsilon_p=1.4*10^(-6)`

And we have a Ratio between the prototype and the model of 16:1, then

`V_m = (L_p)/(L_m)(\upsilon_m)/(\upsilon_p)V_p`

`V_m = (16)/(1)(1*10^(-6))/(1.4*10^(-6))*2`

`V_m = 22.857m/s`

PART B) To calculate the ratio of the drag force now we have to,

`(F_(DM))/(F_(DP)) = (L_m)/(L_p)^2(V_m)/(V_p)^2(\rho_m)/(\rho_p)`

Replacing with our values we have,

`(F_(DM))/(F_(DP)) = (1)/(16)^2(22.857)/(2)^2(998)/(1015)`

`(F_(DM))/(F_(DP)) = 0.5016`

Therefore the ratio of drag force for prototype and model is 0.5016