Question 1.5: Prove that the drag force F on a partially submerged body is......
Prove that the drag force F on a partially submerged body is given by:
F= V^2l^2ρf\left(\frac{k}{l},\frac{lg}{V^2}\right)where V is the velocity of the body, l is the linear dimension, ρ, the fluid density, k is the rms height of surface roughness, and g is the gravitational acceleration.
Let the functional relation be:
F = f (V, l, k,ρ, g)
Or in the general form:
F = f (F, V, l, k, ρ, g) = 0
In the above equation, there are only two primary dimensions. Thus, m = 2. Taking V, l, and ρ as repeating variables, we get:
\prod{}_1 = (V)^a(l)^b(ρ)^c F
M^oL^oT^o = (LT^{-1})^a(L)^b(ML^{-3})^c(MLT^{-2})
Equating the powers of M, L, and T on both sides of the equation, for M, 0 = c + 1 or c = −1; for T, 0 = -a – 2 or a = -2; and for L, 0 = a + b- 3c + 1 or b = −2.
Therefore,
\prod{}_1 = (V)^{-2}(l)^{-2}(ρ)^{-1} F = \frac{F}{V^2l^2ρ}
Similarly,
\prod{}_1 = (V)^{d}(l)^{e}(ρ)^{f}(k)
Therefore,
M^0L^0T^0 = (LT^{-1})^d(L)^e(ML^{-3})^f(L)
for M, 0 = f or f = 0; for T, 0 = -d or d = 0; and for L, 0 = d + e – 3f + 1 or e = -1.
Thus,
\prod{}_2 =(V)^0(l)^{-1}(ρ)^0 k=\frac{k}{l}
and
\prod{}_3 =(V)^g(l)^{h}(ρ)^i (g)M^0L^0T^0 =(LT^{-1})^g(L)^h(ML^{-3})^i(LT^{-2})
Equating the exponents gives, for M, 0 = i or i = 0; for T, 0 = -g–2 or g = – 2; for L, 0 = g + h – 3i + 1 or h = 1.
Therefore; \prod{}_3 =(V)^{-2}(l)^{1}(ρ)^0 g=\frac{lg}{V^2}
Now the functional relationship may be written as:
f\left(\frac{F}{V^2l^2ρ},\frac{k}{l},\frac{lg}{V^2}\right) = 0
Therefore,
F = V^2 l^2 ρ f \left(\frac{k}{l},\frac{lg}{V^2}\right)