Orifice Diameter for a Target Flow
The diameter that passes a target flow under a steady head.
Example
You enter
- Target discharge (cfs) 1.5
- Head to orifice center (ft) 4
- Discharge coefficient Cd 0.6
You get
- Orifice area 0.156 ft^2
- Required orifice diameter 5.34 in
Details, formula, and sources
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.
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.
The orifice equation is a public closed-form hydraulics result; the discharge coefficients are in the standard references.
Estimate. AHJ and licensed professional govern.
Field names used by the API: q_cfs, h_ft, cd, a_ft2, d_in
- Inverse sizing A = Q / (Cd sqrt(2 g h)); d = sqrt(4 A / pi), head to the orifice centroidorifice hydraulics
- Coefficient Cd about 0.6 sharp-edged (0.8 short tube, 0.98 rounded), enteredstandard hydraulics tables
- Steady head small orifice under a steady head; the falling-head drain time is separatescope of this tile