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

$Re_p = Re_m$

$\frac{\rho_pV_pD_p}{\mu_p}=\frac{\rho_mV_mD_m}{\mu_m}$

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,

$\frac{\rho_pV_pD_p}{\mu_p}=\frac{\rho_mV_mD_m}{\mu_m}$

$\frac{(1.99)V_pD_p}{(2.51*10^{-5})}=\frac{(2.38*10^{-3})(183)(\frac{1}{15}D_p}{3.74*10^{-7}}$

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

$\frac{(1.99)V_p}{(2.51*10^{-5})}=\frac{(2.38*10^{-3})(183)(\frac{1}{15}}{3.74*10^{-7}}$

$V_p = 0.979233 ft/s$

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