To solve this problem it is necessary to apply dynamic similarity relationships from the Reynolds model, for which you have the relationship
[tex]Re_p = Re_m[/tex]
[tex]\frac{\rho_pV_pD_p}{\mu_p}=\frac{\rho_mV_mD_m}{\mu_m}[/tex]
Where,
[tex]\rho[/tex]= 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:
[tex]\rho_m = 2.38*10^{-3}slugs/ft^3[/tex]
[tex]\mu_m = 3.74*10^{-7}lb\cdot s/ft^2[/tex]
From the properties of fluids (water) we have that the scale model properties would be
[tex]\rho_p = 1.99slugs/ft^3[/tex]
[tex]\mu_p = 2.51*10^{-5}lb\cdot s/ft^2[/tex]
Substituting the values given in the equation we have to,
[tex]\frac{\rho_pV_pD_p}{\mu_p}=\frac{\rho_mV_mD_m}{\mu_m}[/tex]
[tex]\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}}[/tex]
The term of the diameter is eliminated on both sides so the speed for the prototype would be:
[tex]\frac{(1.99)V_p}{(2.51*10^{-5})}=\frac{(2.38*10^{-3})(183)(\frac{1}{15}}{3.74*10^{-7}}[/tex]
[tex]V_p = 0.979233 ft/s[/tex]
Therefore the speed of the prototype to ensure Reynolds number similarity is 0.979233ft/s
