Rayleigh's method for orifice discharge

Relation

$Q$ depends on $d, H, \mu, \rho, g$. By Rayleigh's method:

$$Q = K\,d^{a}H^{b}\mu^{c}\rho^{e}g^{f}$$

Dimensions

$Q = L^3T^{-1}$, $d = H = L$, $\mu = ML^{-1}T^{-1}$, $\rho = ML^{-3}$, $g = LT^{-2}$.

There are 5 unknowns and 3 equations. Express in terms of $b$ and $c$:

$$e = -c,\qquad f = \frac{1-c}{2},\qquad a = \frac52 - \frac{3c}{2} - b$$

Substitute

$$Q = K\,d^{\,\frac52-\frac{3c}{2}-b}\,H^{b}\,\mu^{c}\,\rho^{-c}\,g^{\frac{1-c}{2}}$$

Collect terms with the same powers:

$$Q = K\,\sqrt g\,d^{5/2}\left(\frac Hd\right)^{b}\left(\frac{\mu}{\rho\,g^{1/2}d^{3/2}}\right)^{c}$$

Answer:

$$Q = \sqrt{g}\,d^{5/2}\ \phi\left[\frac Hd,\ \frac{\mu}{\rho\sqrt g\,d^{3/2}}\right]$$

Equivalently, $Q = d^2\sqrt{gH}\ \phi_1\!\left(H/d,\ Re\right)$. The dimensionless groups are the head ratio $H/d$ and a Reynolds-type number. For a given orifice, experiments give $Q = C_dA\sqrt{2gH}$.