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Orifice Flow: Why a Hole Passes Only 61% of What It Should

Work out the flow through an orifice from the head above it, with the discharge coefficient, the drain time and the square-root law behind both.

ft

Vertical distance from the free surface down to the middle of the hole. Only the vertical drop counts — the width of the tank changes nothing.

in

Flow goes with the square of this, so a hole twice as wide passes four times as much.

The discharge coefficient, and the single biggest lever on the answer. It is about the edge, not the size.

ft²

For the drain time. A straight-sided tank only; a cone or a sphere empties on a different curve.

Specific gravity. It does not change the velocity at all — a light liquid and a heavy one leave the same hole at the same speed — but it changes the mass going through.

h

For the total volume figures, if the head were somehow held constant.

Flow through the opening

117.37gal/min

Cd × area × √(2gh), converted to US gallons a minute. This is the number a pump, a drain or a spillway has to match.

Area of the opening
0.021817ft²

πr² with the radius in feet. A 2 in hole is a small fraction of a square foot, which is why the flow numbers below are smaller than intuition expects.

Ideal exit velocity
19.65ft/s

√(2gh) — Torricelli. It is exactly the speed a stone reaches falling that far, because it is the same energy trade, and it does not depend on the liquid.

Actual exit velocity
19.26ft/s

The velocity coefficient is close to one for any ordinary liquid — almost nothing is lost to friction at the hole. The missing flow is lost to area, not to speed.

What the arithmetic alone would predict
192.4gal/min

Area times velocity, with no coefficient. Every hole in the real world passes less than this, and a sharp-edged one passes barely three fifths of it.

Share of the ideal flow that never arrives
39.0%

The vena contracta, almost entirely. The jet necks down to about 62% of the hole area a fraction of a diameter downstream, and that waist is the real opening.

Flow if the entry were rounded instead
188.56gal/min

A bell-mouthed entry runs at 0.98 because the water never has to turn a corner. Rounding an edge is cheaper than enlarging a hole and, from a sharp edge, buys more.

The same flow in litres a minute
444.3L/min
And in cubic metres an hour
26.657m³/h
Mass leaving each minute
978.8lb/min

Volume is the same for any thin liquid at this head; only the weight changes. Diesel leaves a hole exactly as fast as water and weighs 13% less doing it.

Volume over the run, at constant head
7,042gal

Only true if something keeps the level up. On a closed tank the head falls as it empties and the real figure is lower.

Flow at half the head
82.99gal/min

Not half the flow — about 71% of it, because the head enters as a square root. This is the whole reason a tank drains fast at first and crawls at the end.

Head needed to double the flow
24ft

Four times the depth for twice the flow. Widening the hole by 41% does the same thing and is usually far easier than raising a tank.

Diameter that would double the flow instead
2.83in

Area goes with the square of the diameter, so √2 across is twice the flow. Against four times the head, this is nearly always the cheaper move.

Time to empty the vessel
38 min

Twice as long as running at the starting flow the whole way, because the head is falling throughout. The second half of the depth takes far longer than the first.

Time to get down to half depth
11 min

Under a third of the total. Emptying the top half of a tank is quick; it is the shallow end that takes the afternoon.

Volume held above the opening
2,244gal

How to use this calculator

  1. Enter the vertical distance from the free surface down to the middle of the hole into the Head above the opening field.
  2. Type the width of the hole into the Diameter of the opening field.
  3. Select the physical geometry of the entry edge from the Shape of the edge dropdown menu.
  4. Input the horizontal surface area of the container into the Surface area of the vessel field to compute drain times.
  5. Choose the working liquid from the Liquid selection to adjust mass calculations correctly.
  6. Optionally enter a duration in the Run this many hours field to calculate total volume transferred.

The Physics Behind Torricelli's Law

When liquid escapes through an opening at the base of a container, its speed is governed by Torricelli's law. Discovered by Evangelista Torricelli in the seventeenth century, this principle states that the ideal exit velocity of a fluid under the influence of gravity matches the speed of an object dropped from the same vertical height. The formula v = √(2gh) uses a gravitational constant of 32.174 ft/s² and the vertical head above the opening. Only the vertical drop matters; a wide reservoir and a narrow standpipe of identical depth produce the exact same initial discharge speed.

However, real fluids never achieve that ideal theoretical velocity. As liquid converges toward a hole, the streamlines bend and contract past the opening, creating a pinched zone known as the vena contracta. This contraction reduces both the effective cross-sectional area and the actual flow rate. To account for these hydraulic losses, the discharge coefficient is introduced as a multiplier that scales down the ideal figures to match reality.

Understanding the Discharge Coefficient

The discharge coefficient, designated as Cd, is the single most critical lever in any orifice flow calculator. It bundles together frictional resistance and contraction effects. A standard sharp-edged hole punched in a flat plate has a Cd of approximately 0.61, meaning nearly forty percent of the theoretical energy is lost to turbulence and flow contraction.

By contrast, an inward-projecting pipe stub known as a Borda mouthpiece restricts flow further, dropping the coefficient to 0.50. If the entry is carefully rounded and bell-mouthed, fluid streamlines glide smoothly into the opening without separation, pushing the coefficient up to 0.98. This single adjustment demonstrates why two holes of identical diameter can pass radically different volumes based entirely on the geometry of their edges.

Edge GeometryTypical CdFlow EfficiencyCommon Application
Sharp-edged hole in a plate0.6161%Standard tank drain, sluice gates
Pipe stub projecting inward (Borda)0.5050%Penetration sleeves, stub connections
Short pipe running full0.8282%Outlet nozzles, short drain pipes
Rounded, bell-mouthed entry0.9898%Hydraulic test orifices, high-efficiency intakes

Calculating Flow Rates and Tank Drain Times

To determine the actual volumetric flow rate through a hole, the calculation multiplies the discharge coefficient by the cross-sectional area of the opening and the actual exit velocity. The resulting gpm through an orifice scales non-linearly with head pressure. Because flow relies on the square root of the head, quadrupling the liquid height only doubles the discharge rate. If an operator needs to double the flow by altering the hardware instead, they must either quadruple the head pressure or increase the diameter by a factor of the square root of two.

When evaluating a tank drain time calculator, the geometry of the vessel dictates the mathematical integration required. For a straight-sided vertical tank, the head drops continuously as the vessel empties, slowing the discharge rate progressively. The drain time formula accounts for this decaying head by integrating the instantaneous flow over the changing volume from the starting height down to zero. Conical, spherical, or horizontal cylindrical tanks empty on entirely different curves because their cross-sectional area changes at every vertical increment.

Limitations and When to Seek Professional Engineering

The arithmetic powering this tool assumes steady, incompressible fluid flow under standard gravity. It breaks down when applied to highly viscous fluids like heavy crude oil or thick molasses, where internal friction dominates over inertial forces. Similarly, if the downstream side of the orifice is submerged or subject to backpressure, the effective head drops, rendering free-discharge equations inaccurate.

For critical municipal water supplies, high-pressure industrial hydraulics, or safety-relief sizing on pressurized vessels, theoretical estimations are insufficient. In those regulated scenarios, consult a licensed mechanical or hydraulic engineer who can perform physical testing or utilize advanced computational fluid dynamics to verify safety margins.

The formula

ideal velocity v = √(2gh), with g = 32.174 ft/s² — the same speed as a free fall of hflow Q = Cd × A × v, where Cd is 0.61 for a sharp edge and 0.98 for a rounded oneflow follows √h, so four times the head is twice the flowdrain time for a straight-sided tank = 2 A √h ÷ (Cd a √(2g))

Frequently asked questions

Why does doubling the head pressure not double the flow rate?

Fluid velocity through an opening is proportional to the square root of the head, as demonstrated by Torricelli's law. Because of this square-root relationship, you must quadruple the vertical height of the liquid column to achieve double the volumetric discharge rate.

Does the width or shape of the tank affect how fast water leaves the hole?

No, the horizontal dimensions of a straight-sided vessel have zero impact on the exit velocity of the fluid stream. Only the vertical distance from the free surface down to the center of the orifice determines the pressure driving the liquid through.

What is the difference between a sharp-edged hole and a rounded entry?

A sharp-edged hole causes fluid streamlines to contract severely inside the opening, resulting in a discharge coefficient around 0.61. A smooth, bell-mouthed entry eliminates this turbulence and contraction, allowing nearly frictionless flow with a coefficient of 0.98.

How does specific gravity change the calculations?

Specific gravity does not affect fluid velocity or volumetric output in gallons per minute, as light and heavy liquids fall at the same rate under gravity. However, it directly scales the mass flow rate, meaning denser fluids like seawater yield a heavier mass output per minute than ethanol.

Can I use these formulas for tanks that are not straight-sided?

No, the standard drain time equations on this page apply exclusively to vertical vessels with constant horizontal cross-sectional areas. Containers like spheres, cones, or horizontal cylinders change their cross-section as they empty, requiring specialized calculus to determine drain duration.

Sources

Last reviewed . Results are for general guidance and are not professional advice.