# Fluid Mechanics - similarity

1. May 1, 2009

### Pietair

1. The problem statement, all variables and given/known data
A body when tested in air of density 1.18 kg/m3 and dynamic viscosity 1.52x10^-5 Pas at a velocity of 35 m/s was found to produce a resistance of 250 N. A similarly shaped body 10 times longer than the original was tested in water of dynamic viscosity 1x10^-3 Pas. Determine the resistance for the larger body.

2. Relevant equations
In the previous question I have obtained via the Pi-theorem:
R / (rho x l^2 x v^2) = f ((rho x v x l)/μ)
which is correct.

3. The attempt at a solution
1:
rho air = 1.18 kg/m3
μair = 1.52x10^-5 Pas
v = 35 m/s
R = 250N
l

2:
rho water = 1000 kg/m3
μwater = 1x10^-3 Pas
v = ?
R = ?
10l

For dynamic similarity:
(rho1 x v1 x l1)/μ1 = (rho2 x v2 x l2)/μ2
(1.18 x 35 x l)/1.25x10^-5 = (1000 x v2 x 10l)/1x10^-3
(1.18 x 35)/1.25x10^-5 = (1000 x v2 x 10)/1x10^-3
I get: v2 = 0.2717 m/s

Complete similarity:
R1 / (rho1 x (l^2)1 x (v^2)1) = R2 / (rho2 x (l^2)2 x (v^2)2)
250 / (1.18 x l^2 x 35^2 = R2 / (1000 x 100l^2 x 0.2717^2)
250 / (1.18 x 35^2 = R2 / (1000 x 100 x 0.2717^2)
I get: R2 = 1276.8 N