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Orifice Flow Calculator

Discharge through an orifice from upstream head and orifice area. The classical Torricelli formula plus a discharge coefficient for the contracted jet.

in
ft
ft²
cfs

Defaults: 6-inch round sharp-edged orifice, 4 ft of water head, Cd = 0.61.

$$ Q = C_d \, A \, \sqrt{2 g H} $$
Q discharge · Cd discharge coefficient · A orifice cross-section area · g gravitational acceleration · H head from water surface to centroid of orifice.

Discharge coefficient table

The discharge coefficient Cd = Cc · Cv combines the contraction coefficient (how much the jet shrinks below the orifice area — the vena contracta) and the velocity coefficient (velocity reduction due to friction).

Discharge coefficients Cd for common orifice and short-tube configurations
ConfigurationCdCcCv
Sharp-edged orifice, thin plate0.610.620.98
Slightly rounded entrance (r > 0.15 d)0.950.970.98
Well-rounded entrance (bellmouth)0.981.000.98
Borda mouthpiece (re-entrant tube)0.510.520.98
Standard short tube (length 2–3 d)0.821.000.82
Conical converging tube (13.5° total)0.940.990.95
Square-edged hole in thick wall (l/d ≈ 1)0.730.750.97
Vertical sluice gate (typical)0.55–0.65
Radial (Tainter) gate0.65–0.75
Drop inlet, square edge (FHWA HDS-5)0.60

Source: Brater, King, Lindell & Wei (1996), Handbook of Hydraulics, 7th ed., Table 4-1; USBR (1997) Water Measurement Manual; FHWA HDS-5.

Worked examples

Example 1 — Detention basin low-flow outlet (sharp-edged orifice)

Given: Round 4-inch sharp-edged orifice (Cd = 0.61) on a stormwater detention basin riser, 5.0 ft of head from WSE to orifice centroid.
Find: Discharge Q.
A = π·(4/12)²/4 = π·(0.333)²/4 = 0.0873 ft²
Q = 0.61 · 0.0873 · √(2 · 32.2 · 5.0) = 0.0532 · √322
Q = 0.0532 · 17.94
Q = 0.955 cfs ≈ 1.0 cfs

Example 2 — Tank drainage time

Given: 10-ft diameter cylindrical tank, initial depth 4.0 ft above orifice, drains through a 6-inch sharp-edged orifice in the bottom (Cd = 0.61). Drain to empty.
Find: Drain time.
Atank = π·(10)²/4 = 78.54 ft²; Aorifice = π·(0.5)²/4 = 0.1963 ft²
t = 2·Atank·(√H₁ − √H₂) / (Cd·Aorifice·√(2g))
t = 2·78.54·(√4.0 − 0) / (0.61·0.1963·√64.4)
t = 314.2 / (0.61·0.1963·8.025) = 314.2 / 0.961
t = 327 seconds ≈ 5.5 minutes

Free vs. submerged orifice

For a free orifice (jet discharges into atmosphere), H is measured from the water surface upstream to the orifice centroid. For a submerged orifice (downstream water level above the orifice), H is the difference in water surface elevations on either side of the orifice. The discharge equation is the same; the head definition changes.

Where this calculator applies

This is the standard equation for: gate openings on small dams, drop inlets and sluice gates, orifice plate flow meters in pipes, low-flow outlet works for stormwater detention basins, valve-gallery openings, and tank drainage time problems. For very large gates where velocity head upstream is non-negligible (drawdown over 10% of total head), use the more general energy equation rather than this simplified form.

Tank drainage time

If you're computing the time to drain a tank through an orifice, you can integrate dV/dt = −Q over the head range. For a constant-cross-section tank: t = 2Atank · (√H₁ − √H₂) / (Cd·A·√(2g)). That's a separate but related calculation.

Reference: Brater, E.F., King, H.W., Lindell, J.E., Wei, C.Y. (1996). Handbook of Hydraulics (7th ed.), Chapter 4. Original: Torricelli's law (1643).

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