Audited 31 Jul 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Orifice Flow Calculator

Estimate liquid flow through an orifice from Q = C_d·A·√(2gΔh): enter the discharge coefficient, opening area and head of liquid driving the flow.

Orifice Flow Calculator

m²
m/s²
m
Volumetric flow rate
0.0434
Volume of liquid passing through the orifice each second, in cubic meters per second; multiply by 1,000 for liters per second.
Model scope
Steady incompressible head-discharge model; the entered coefficient must match the geometry and Reynolds regime, and cavitation, submergence, viscosity, and compressibility are not solved.

Background.

Punch a hole in the side of a water tank and the jet obeys a rule Evangelista Torricelli worked out in the 1640s: the water leaves at the speed it would reach falling freely from the surface down to the hole, √(2gh). Two and a half meters of water above an opening drives a jet at about 7 m/s. Multiply that speed by the area of the hole and you have the flow an ideal, frictionless liquid would deliver.

Real orifices deliver less, and the discharge coefficient C_d accounts for the shortfall. Water converging on a sharp-edged hole cannot turn the corner instantly, so the jet keeps narrowing for a short distance beyond the opening, to a waist called the vena contracta, and a little speed is lost to friction at the edge. For a sharp-edged orifice the combined effect cuts the flow to roughly 62% of the ideal, the default here. A smoothly rounded, bell-mouthed entrance lets the stream follow the wall and brings the coefficient close to 1. Choosing it well matters more than anything else on the page, because the flow is directly proportional to it.

The head difference is the driving height: from the free water surface to the center of the orifice when the jet discharges into air, or between the two water surfaces when the orifice is submerged. Flow grows only with its square root, so quadrupling the head merely doubles the discharge.

This is the steady-flow model. It assumes the head holds constant, which a draining tank does not, and that the fluid is an incompressible liquid with negligible viscous effects. Gases at large pressure ratios, cavitating jets and very small or slow orifices fall outside it. For an orifice plate inside a pipe, ISO 5167-2 adds an approach-velocity factor that this simple equation leaves out, so the coefficient you enter must already allow for it.

What is orifice flow calculator?

Orifice flow is the volume of liquid passing each second through an opening in a wall, plate or tank under a steady head, given here in cubic meters per second. The theoretical figure is the opening's area times Torricelli's velocity √(2gΔh). The actual flow is that figure scaled by a discharge coefficient, found by experiment for each orifice shape, which captures how much the jet contracts and slows as it leaves.

How to use this calculator.

  1. Enter the discharge coefficient for your orifice. 0.62 suits a sharp-edged circular hole in a thin plate; for anything else take the value from the manufacturer, a calibration or a hydraulics handbook.
  2. Enter the orifice area in square meters, not the diameter. For a circular hole A = πd²/4, so a 5 cm hole is 0.0019635 m².
  3. Leave gravitational acceleration at standard gravity, 9.80665 m/s², unless you are working to a site value; the variation across Earth's surface changes the flow by less than 0.3%.
  4. Enter the head difference in meters: water surface to orifice center for free discharge, or upstream minus downstream level for a submerged orifice.
  5. Read the flow in m³/s. Multiply by 1,000 for liters per second or by 3,600 for cubic meters per hour.

The formula.

Q = C_d A sqrt(2gΔh)

Q = C_d × A × √(2gΔh). Q is the volumetric flow rate in m³/s, C_d the dimensionless discharge coefficient, A the orifice area in m², g the gravitational acceleration in m/s² and Δh the head difference in meters. The square-root term is Torricelli's ideal jet velocity, which follows from Bernoulli's equation when the potential energy of a fall through Δh turns entirely into kinetic energy. Multiplying by A gives the ideal flow, and C_d corrects it for the contraction of the jet and the friction at the edge. Flow is directly proportional to C_d and to A, so a 10% error in either is a 10% error in Q. It grows only with the square root of g and Δh, so doubling the head adds about 41%. Every input must be positive, and the result describes one instant at a constant head: a falling level, gas compressibility, cavitation and viscous effects at small openings are not modeled.

A worked example.

Example

A tank drains through a sharp-edged opening of 0.01 m², a circular hole about 11.3 cm across, with its water surface 2.5 m above the hole's center; C_d is 0.62 and g is 9.80665 m/s². First, 2 × 9.80665 × 2.5 = 49.03325, and its square root, 7.0024 m/s, is the ideal jet speed. Times the area, 7.0024 × 0.01 = 0.070024 m³/s would pass if the jet filled the whole opening without loss. Applying the coefficient, 0.62 × 0.070024 = 0.04341 m³/s, about 43.4 liters per second. Raising the water to 10 m, four times the head, only doubles the flow, to 0.0868 m³/s. Keeping 2.5 m but giving the hole a smoothly rounded entrance with a coefficient of 0.97 would pass 0.0679 m³/s.

discharge Coefficient0.62
head Difference M2.5
orifice Area M20.01
gravity Mps29.807

Frequently asked questions.

What discharge coefficient should I use?
It depends mostly on the shape of the edge. A sharp-edged hole in a thin plate sits close to the 0.62 default, while a smoothly rounded, bell-mouthed entrance approaches 1. For orifice plates in pipes, ISO 5167-2 gives an equation for the coefficient based on the diameter ratio and Reynolds number; for valves and nozzles, use the manufacturer's figure.
How long will my tank take to drain?
Longer than the starting flow suggests. As the level falls the head drops and the flow drops with its square root. For a tank with straight vertical sides, draining completely through a bottom orifice takes, under this model, twice as long as it would if the initial flow rate held all the way down.
My gauge reads pressure, not height. What do I enter?
Convert the pressure to head with Δh = Δp ÷ (ρg), using the density of the liquid that is flowing. For water at about 1,000 kg/m³, a 50 kPa difference is 5.10 m of head and 1 bar is about 10.2 m. A denser liquid gives less head for the same pressure.
Is this the same equation orifice-plate flow meters use?
The core is the same, but ISO 5167-2 also multiplies by 1 ÷ √(1 − β⁴), where β is the ratio of orifice diameter to pipe diameter, to account for the liquid already moving in the pipe. At β = 0.5 that factor is about 1.033, so leaving it out understates the flow by roughly 3% unless the coefficient you enter already includes it.
Can it be used for air or other gases?
Only roughly, and only when the pressure drop is a small fraction of the absolute pressure, so the gas density barely changes through the opening. The head must then be expressed as a column of the gas itself, Δp ÷ (ρg) with the gas density, not as water. Larger pressure ratios need an expansibility correction that this page does not apply.

How this page was produced

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Quanta Calculator
Primary sources
3 cited below
Method
Q = C_d A sqrt(2gΔh)
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