How this instrument works
The flow rate through a submerged opening is built from two ideas stacked on top of each other. First, Torricelli's theorem: a particle of fluid at depth h below the free surface has, by energy conservation, the same speed it would reach falling freely through that height, v = √(2gh). Second, multiply that speed by the opening's cross-sectional area to get a volume per second — the same logic used for flow through a pipe. Chain the two together and the theoretical discharge is A√(2gh); every extra unit of head adds speed to the jet, but only as its square root, because the potential energy driving the jet grows linearly while kinetic energy grows with the square of speed.
Real fluid falls short of that ideal number, which is why the discharge coefficient Cd sits in front of the formula. As fluid approaches a sharp-edged hole from every direction, it cannot turn the corner instantly, so the jet keeps contracting for a short distance past the opening — the vena contracta — before it reaches its narrowest, fastest cross-section. That narrowing alone accounts for most of the shortfall; wall friction and turbulence trim a little more. For a plain sharp-edged circular orifice, Cd typically lands between 0.60 and 0.62, a range measured and re-measured since the nineteenth century rather than assumed.
The formula assumes the opening is small next to the head above it, so every point on the orifice sits at essentially the same depth, and that the reservoir empties slowly enough for the flow to be treated as steady rather than a fast transient. It also assumes free discharge to open air. A submerged outlet, where fluid exits into more fluid rather than atmosphere, needs the difference in surface elevations on the two sides in place of a single depth — the same shape of equation, but h becomes Δh.
- Set Discharge coefficient Cd — 0.61 is the standard starting point for a plain sharp-edged circular hole; raise it toward 0.95–0.98 for a smoothly rounded, bell-mouthed entrance.
- Enter Orifice area, the actual open cross-section fluid passes through, not the pipe or tank cross-section it is cut into.
- Enter Head, the vertical depth of the free fluid surface above the centre of the orifice, in metres or centimetres.
- Read Flow rate through the orifice — switch its unit between litres per second and US gallons per minute as needed.
Worked example — a 5 cm² hole under 2 metres of head
A tank wall has a sharp-edged circular hole of 5 cm², which is 0.0005 m², sitting 2 metres below the water surface, and the discharge coefficient for that kind of edge is 0.61. The ideal Torricelli velocity is √(2 × 9.80665 × 2) = 6.2631 m/s, so the idealized discharge would be A times that speed, 0.0005 × 6.2631 = 0.0031316 m³/s. Multiplying by Cd brings it down to the real figure: Q = 0.61 × 0.0005 × 6.2631 = 0.00191024984361 m³/s.
Converted to the readout units, that is 1.9102 litres per second — roughly enough to fill a ten-litre bucket in five seconds. Notice the real flow is only 61% of the idealized number, exactly Cd, because Cd is defined as the ratio of actual to ideal discharge. A stormwater engineer sizing the orifice in a detention-pond outlet structure runs this same arithmetic in reverse: pick a target release rate, then solve for the opening that produces it at the pond's design head.
Questions
Why is the actual flow always less than A√(2gh) predicts?
Because that expression is the idealized, frictionless Torricelli velocity multiplied by the full geometric area, and real fluid cannot achieve either assumption. The jet contracts past a sharp edge (the vena contracta) to a smaller effective area, and viscous friction trims the speed a little further. The discharge coefficient Cd folds both losses into one measured multiplier, typically 0.60–0.62 for a sharp-edged hole.
What discharge coefficient should I use if I don't know it?
Use 0.61 as a default for a plain sharp-edged circular orifice — it is the most commonly published figure and close to the calculator's own default. A rounded, bell-mouthed entrance behaves much closer to the ideal and can reach 0.95–0.98, while a re-entrant tube projecting into the tank drops Cd toward 0.5. When precision matters, the value should come from a calibration test on the actual opening.
Why does doubling the head not double the flow rate?
Because head enters the formula under a square root. Quadrupling h only doubles √h, and so only doubles Q; the earlier vector in this calculator's own test set confirms it exactly — 2 m of head gives 1.910 L/s while 8 m gives 3.820 L/s, four times the head for twice the flow. It is the same square-root relationship that governs how fast a falling object's speed builds with distance.
Does this formula work for a submerged orifice, not just free discharge?
Not directly as written. Free discharge to open air uses the depth below the surface as h. A submerged orifice, where fluid exits into another body of fluid rather than air, uses the difference in surface elevation between the two sides, Δh, in place of h — the same square-root shape, driven by the pressure difference rather than a single depth.
Is head measured to the top, bottom, or centre of the orifice?
The centre. The formula assumes the orifice is small enough relative to the head that pressure is effectively uniform across the opening, so the centreline depth stands in for the whole area. For a large opening relative to its depth, that assumption weakens and a more careful integration across the opening's height gives a more accurate figure.
How is this different from an orifice plate flow meter in a pipeline?
A pipeline orifice meter, standardized under ISO 5167, infers flow from the pressure drop across a plate inside a full pipe, using a discharge coefficient tied to the pipe-to-hole diameter ratio and the Reynolds number. This calculator instead handles free or submerged discharge driven by a static head of fluid above the opening — same underlying physics, a different geometry and a different empirical coefficient.