Question

A model of a submarine, 1 : 15 scale, is to be tested at 183 ft/s in a wind tunnel with standard sea-level air, while the prototype will be operated in seawater. Determine the speed of the prototype to ensure Reynolds number similarity. Use for air and for seawater.

Answer 1

To solve this problem it is necessary to apply dynamic similarity relationships from the Reynolds model, for which you have the relationship

`Re_p = Re_m`

`(\rho_pV_pD_p)/(\mu_p)=(\rho_mV_mD_m)/(\mu_m)`

Where,

`\rho`= Density

V = Velocity

D = Diameter

\mu = Dynamic viscosity

Here the number p refers to the scale model, that is the prototype and m refers the real model.

For the conditions presented we have through the properties of gases (air for this case) in query tables that:

`\rho_m = 2.38*10^(-3)slugs/ft^3`

`\mu_m = 3.74*10^(-7)lb\cdot s/ft^2`

From the properties of fluids (water) we have that the scale model properties would be

`\rho_p = 1.99slugs/ft^3`

`\mu_p = 2.51*10^(-5)lb\cdot s/ft^2`

Substituting the values given in the equation we have to,

`(\rho_pV_pD_p)/(\mu_p)=(\rho_mV_mD_m)/(\mu_m)`

`((1.99)V_pD_p)/((2.51*10^(-5)))=((2.38*10^(-3))(183)((1)/(15)D_p)/(3.74*10^(-7))`

The term of the diameter is eliminated on both sides so the speed for the prototype would be:

`((1.99)V_p)/((2.51*10^(-5)))=((2.38*10^(-3))(183)((1)/(15))/(3.74*10^(-7))`

`V_p = 0.979233 ft/s`

Therefore the speed of the prototype to ensure Reynolds number similarity is 0.979233ft/s