Orifice Diameter for a Target Flow
The inverse of the orifice-discharge tile: the diameter that passes a target flow under a steady head, from Q = Cd A sqrt(2 g h) solved for A = Q / (Cd sqrt(2 g h)) and d = sqrt(4 A / pi), Cd about 0.6 sharp-edged, head to the orifice center. To release 1.5 cfs under a 4 ft head takes a 5.34 in orifice (0.156 ft^2); because the flow scales with the square root of head, the required area scales as 1/sqrt(h), so a 4x head shrinks the diameter to 0.71x - the sizing side of a detention-outlet or restrictor plate. Free/submerged, steady head, small orifice; the falling-head time-to-drain is separate. A design aid; the engineer of record governs.
Formula and source
A = Q / (Cd sqrt(2 g h)) (g = 32.2 ft/s^2); d = 12 sqrt(4 A / pi) in.
The orifice discharge equation Q = Cd A sqrt(2 g h) solved for the orifice diameter, the inverse of the orifice-discharge tile, a standard hydraulics result, by name.
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