Torricelli Tank Draining Time Calculator

How this Torricelli calculator works

This calculator estimates how long a liquid takes to drain from a tank through a small opening under gravity. It can also work in reverse. If you know any five of the six quantities, leave one field empty and the calculator rearranges Torricelli's law to solve for the missing value. That makes it useful for homework checks, quick design estimates, physics demonstrations, rain-barrel planning, and simple hydraulic sanity tests. The result is intentionally based on the classic ideal-flow model, so it is fast to use and easy to understand, but it should still be treated as an estimate rather than a final engineering certification.

The most common use is finding drain time. You enter the initial liquid height, the final liquid height, the tank cross-sectional area, the orifice area, and local gravity, then leave the time field blank. The same form can also solve for the initial height needed to achieve a certain drain time, the remaining height after a set interval, the effective tank area, the outlet area, or even the gravitational acceleration implied by a lab setup. Because the page solves algebraically, the same physical relationship sits underneath every mode of use.

The height measurements matter more than many people expect. Both $h_0$ and $h_1$ are measured vertically from the centerline of the outlet or orifice, not from the base of the container and not from the top rim. If you want the water level to fall all the way to the hole, set the final height to zero. If you are draining only partway, use the target liquid depth above the outlet. This detail is important because the available pressure head comes from the fluid column above the opening.

Use one consistent unit system all the way through. The labels here assume SI units: heights in meters, areas in square meters, gravity in meters per second squared, and time in seconds. You can work in feet or centimeters if you convert everything consistently before entering values. Mixing meters for one measurement with centimeters for another will quietly corrupt the answer. In practice, many surprising results come from unit mismatch rather than from the physics itself.

• Initial height h₀: liquid level above the outlet when the draining interval begins.
• Final height h₁: liquid level above the outlet when the interval ends.
• Tank area A_t: cross-sectional area of the tank, assumed constant as the level falls.
• Orifice area A_o: area of the drain opening that controls the outflow.
• Gravity g: local gravitational acceleration; on Earth, 9.81 m/s² is the usual value.
• Time t: the elapsed draining time between the two liquid heights.

Why Torricelli's Law Matters

Torricelli's seventeenth-century insight connects fluid height to efflux speed. It remains a staple of hydraulics because it converts a complicated draining process into a manageable relation grounded in energy conservation. Whether you are emptying a storage vessel, timing a classroom experiment, sizing a simple drain, or just building physical intuition, the law explains a familiar observation: tanks drain quickly at first and then slow down as the liquid level drops. That slowing is not random. It follows directly from the fact that outlet speed depends on the square root of the remaining head above the opening.

Deriving the draining formula

The instantaneous speed of fluid leaving a hole a distance $h$ below the surface is $v=\sqrt{2gh}$. For a small opening of area $A_o$ connected to a reservoir of cross-section $A_t$, the volumetric flow rate is $Q=A_{o}v$. Mass conservation links this outflow to the drop in fluid height:

Formula: A_t (d h) / (d t) = - Q

$$A_t\frac{dh}{dt}=-Q$$

Substituting $v$ and separating variables yields:

Formula: (d h) / sqrt(h) = - A_o / A_t sqrt(2 g) d t

$$\frac{dh}{\sqrt{h}}=-\frac{A_o}{A_t}\sqrt{2g}dt$$

Integrating from an initial height $h_0$ to a final height $h_1$ produces the general solution:

Formula: t = (2 A_t(sqrt(h_0) - sqrt(h_1))) / (A_o sqrt(2 g))

$$t=\frac{2A_t(\sqrt{h_0}-\sqrt{h_1})}{A_o\sqrt{2g}}$$

In plain language, that formula says drain time grows with tank area and shrinks with orifice area. Double the tank area and, all else equal, the drain time doubles because more liquid volume must be removed for the same drop in level. Double the outlet area and the time is cut roughly in half because fluid can leave faster. Gravity makes draining faster too, but only through a square-root effect, so changing $g$ has a milder influence than changing areas. The square-root terms in the heights are the reason the flow slows as the surface drops: losing the first half-meter of head matters more than losing the last few centimeters.

Step-by-step logic

The calculator follows the same practical sequence a student or engineer would use by hand, but it saves you from rearranging the algebra each time.

1. Measure or estimate the initial and final fluid heights relative to the orifice.
2. Enter the cross-sectional area of the tank and the opening.
3. Supply the local gravitational acceleration. On Earth, use 9.81 m/s² unless you have a different value.
4. Leave the quantity you want to solve for blank and press Compute.
5. The calculator rearranges Torricelli's formula algebraically to isolate the missing variable.

If the result looks unreasonable, check the basics first: make sure all areas and gravity are positive, make sure $h_0$ is at least as large as $h_1$, and make sure the units are consistent. Physically impossible combinations can produce non-useful outputs because the ideal equation assumes the situation itself is valid.

Worked example

Suppose a straight-sided tank has cross-sectional area $A_t=0.3$ m². It drains from an initial level $h_0=0.8$ m down to a final level $h_1=0.1$ m through an orifice of area $A_o=0.005$ m². Using $g=9.81$ m/s², the ideal Torricelli estimate gives a drain time of about 15.7 s. That number is shorter than many people expect because the outlet area is fairly large relative to the tank area. If you were modeling a real sharp-edged outlet, you would usually expect a longer actual time after accounting for discharge losses.

This is a good example of how to interpret the answer. The calculator is not saying the tank magically empties instantaneously; it is saying that, under the ideal assumptions of the formula, the available head and outlet size support a fast discharge. If your physical setup drains more slowly, the usual reasons are a smaller effective opening, a discharge coefficient below 1, viscosity, a partially obstructed outlet, or a tank shape whose area is not actually constant with height.

How to interpret the result

When the result is time, read it as the estimated duration for the liquid surface to fall from the starting level to the ending level under ideal gravity-driven flow. When the result is an area, read it as the area required for the model to match the other values you entered. That can be useful when back-solving an outlet size for a target drain time. When the result is an initial or final height, it tells you what fluid level is consistent with the rest of the scenario. Solving for gravity is less common in day-to-day design, but it can be helpful in laboratory demonstrations or planetary thought experiments.

One simple rule of thumb helps: because the formula contains square roots of height, tank drainage is not linear in time. A tank does not lose equal height in equal time intervals. Early in the process, with a taller liquid column above the outlet, the efflux speed is higher. Later, as the head decreases, the flow rate falls and the remaining liquid drains more slowly. So if the result seems to indicate that the last small amount takes a disproportionately long time, that is not a bug. It is part of the physics that Torricelli's law captures.

Comparison table

The table below shows how changing the outlet area changes the ideal drain time for a tank with $h_0=1$ m, $h_1=0$, $g=9.81$ m/s², and $A_t=0.5$ m². The pattern is the key takeaway: when you double the opening area, the ideal drain time is cut roughly in half.

Assumptions and limitations

This calculator assumes incompressible, non-viscous flow and a tank cross-section that stays constant while the fluid level falls. It also assumes the liquid surface area is much larger than the outlet area, so the velocity of the free surface itself can be neglected compared with the jet leaving the hole. Those assumptions are reasonable for many simple textbook and first-pass design problems, but real systems rarely match them perfectly. Viscosity, contraction at the orifice, turbulence, pipe attachments, and outlet geometry all reduce real discharge compared with the ideal prediction.

In practice, engineers often include a discharge coefficient C_d smaller than 1 to account for those losses. A common quick estimate for a sharp-edged hole is around 0.6, though the correct value depends on geometry and Reynolds number. Because this calculator does not include C_d directly, one rough workaround is to use an effective outlet area equal to C_d A_o. Doing that increases the predicted drain time and often brings the estimate closer to measured behavior. You should also be cautious with tapered tanks, submerged outlets, very viscous liquids, or situations where air entrainment or back pressure matters.

Another important limitation is geometry. If the tank narrows or widens noticeably with height, then $A_t$ is not constant and the simple integrated formula on this page is only an approximation. For those cases, a piecewise calculation or a more general differential-equation model is better. Even so, the ideal Torricelli relation remains a valuable baseline because it tells you what the system would do before real-world losses and shape effects are layered on top.

Practical applications

Even with its simplifications, Torricelli's law appears in many everyday and industrial contexts. It helps estimate how long a rain barrel or emergency reservoir takes to empty, how quickly a demonstration tank drains in a lab, how outlet size affects the pace of a process vessel, and why gravity-fed systems lose flow as the available head drops. Teachers like it because it links conservation of energy, fluid mechanics, and separable differential equations in one concrete example. Designers like it because it gives a quick answer before detailed computational or empirical work begins.

The law is also a good intuition builder. If you have ever noticed that the first part of draining feels fast and the last part feels slow, this formula explains why. The outflow does not depend on liquid depth in a linear way; it depends on the square root of depth. That means increasing the starting height helps, but with diminishing returns. Likewise, making the outlet larger is a very effective way to shorten the drain time, often more effective than making modest changes to the fluid head.

Related calculations and next steps

If this estimate is only your starting point, the next useful topics are discharge coefficients, Bernoulli-based pressure calculations, Reynolds number, hydrostatic pressure, and pipe-loss modeling. Those tools help when the outlet is attached to tubing, when the liquid is viscous, or when you need better agreement with measured data. A good workflow is to use this calculator first for a clean baseline, then refine the model only if the decision really depends on the extra accuracy.

Conclusion

Torricelli's formula is simple, but it captures the central idea behind gravity draining: more head means faster efflux, and the speed falls as the tank empties. This calculator turns that relationship into a quick practical tool. Enter five known values, leave one field blank, and use the result as an informed ideal estimate. If you later need a closer match to the real world, treat this answer as the baseline and then account for losses, coefficients, and non-ideal geometry. Used that way, the calculator is both a helpful teaching aid and a sensible first step in hydraulic planning.