Tank Draining Simulator Back
Fluid Dynamics Simulator

Tank Draining Simulator — Torricelli's Law

Visualizes the falling water level of a tank draining through a bottom orifice using Torricelli's law. Vary the cross-section, orifice area, and discharge coefficient to learn the behavior of unsteady outflow.

Parameters
Tank cross-section A
Orifice area a
cm²
Initial level h₀
m
Discharge coefficient Cd
Elapsed time t
%

"Elapsed time t" is the percentage of the total drain time.

Results
Current level h
Outflow velocity v
Current flow rate Q
Drain time
Tank and Water Level

Blue = current water level / Yellow arrow at the bottom = outflow from the orifice

Water Level over Time h(t)

Horizontal axis = elapsed time t / Vertical axis = water level h (yellow dot = current time)

Theory & Key Formulas

Treats the unsteady outflow when a tank of cross-section A drains through an orifice of area a.

Outflow velocity and flow rate from Torricelli's law (Cd is the discharge coefficient):

$$v = C_d \sqrt{2gh}, \qquad Q = a\,v = C_d\,a\,\sqrt{2gh}$$

Governing equation for the water level (conservation of mass):

$$A\,\frac{dh}{dt} = -Q = -C_d\,a\,\sqrt{2gh}$$

Solving this, the level decreases as a quadratic function of time:

$$h(t) = \left(\sqrt{h_0} - \frac{C_d\,a}{A}\sqrt{\frac{g}{2}}\;t\right)^{2}$$

The drain time is found by setting h(t) = 0 in the equation above:

$$t_{empty} = \frac{A}{C_d\,a}\sqrt{\frac{2h_0}{g}}$$

What is the Tank Draining Simulator

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When you pull the plug in a bathtub, the flow is strong at first and then gradually slows down. Why is that?
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Good observation. Roughly speaking, it is because the speed of the outflow is determined by the "water level height." According to Torricelli's law, the outflow velocity is $v = \sqrt{2gh}$. The higher the level h, the faster; as it gets lower, the slower. Try raising the "Initial level h₀" in the simulator above and pressing "Animate drain." The initial momentum is completely different.
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It really is! But why does it become the same formula as the velocity of a falling object? $\sqrt{2gh}$ is the free-fall formula, right?
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Sharp. Actually, applying Bernoulli's equation between the "water surface" and the "orifice exit" shows that the potential energy $gh$ is converted directly into kinetic energy $v^2/2$. That is why it becomes the same formula as free fall. From the energy-conservation viewpoint, it equals the kinetic energy of water that has fallen by height h.
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I see! But when I raise the "discharge coefficient Cd" from 0.62 to 1.0, the drain time becomes much shorter. What is Cd?
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Cd is the ratio of the actual flow rate to the theoretical flow rate. With a sharp-edged orifice, the flow contracts sharply just after leaving the hole — the "vena contracta" — so the effective cross-section is reduced, and there is viscous friction too. So a realistic Cd is around 0.6. Conversely, with a smooth nozzle it rises above 0.95. When you design drain piping in practice, estimating this Cd correctly governs the accuracy of your drain-time prediction.

Frequently Asked Questions

The governing equation A dh/dt = -Cd·a·sqrt(2gh) is a separable differential equation. Dividing both sides by sqrt(h) and integrating makes sqrt(h) a linear function of time, so h decreases as a quadratic function of time. This is why the graph is a downward-convex parabola: the slope (flow rate) is maximum when full and becomes gentler as it approaches empty.
The drain time is inversely proportional to the orifice area a. Doubling the diameter quadruples the area, so the drain time becomes about one quarter. In this simulator, changing the orifice area from 10 cm² to 40 cm² shows that the drain time is reduced to about one quarter.
No. This simulator assumes a cylindrical (prismatic) tank whose cross-section A is constant regardless of the water level. For conical or spherical tanks, the cross-section becomes a function of the level A(h), so the governing equation A(h)·dh/dt = -Cd·a·sqrt(2gh) must be integrated anew, and the water level does not change as a quadratic function of time.
The h in Torricelli's law is the "height from the free surface to the center of the orifice." If the orifice is partway up the side, the outflow stops when the water level drops to the height of the orifice. This simulator treats the most basic case where the orifice is at the bottom of the tank, so h coincides with the water level measured from the tank bottom.

Real-World Applications

Water supply and drainage design: Directly used to estimate the drain time of storage tanks, swimming pools, and the sedimentation basins of water treatment plants. The time to empty a tank for maintenance or cleaning is predicted, and the size of drain valves and pipes is determined.

Level measurement and flow meters: The relationship between flow through an orifice or nozzle and the head (differential pressure) is the very principle of orifice flow meters and Venturi meters. Torricelli's law provides the theoretical foundation of these flow-measurement instruments.

Batch operations in chemical plants: Used to calculate the time to transfer liquid from a reactor or storage tank to the next process. Understanding the characteristics of unsteady outflow is important for batch process scheduling and piping design.

Hydraulic model experiments and education: Torricelli's law is one of the first unsteady problems learned in an introduction to fluid mechanics. As teaching material that lets you experience Bernoulli's equation, the continuity equation, and solving differential equations all at once, it is covered in the hydraulics labs of many universities.

Common Misconceptions and Cautions

The most common misconception is to assume the outflow velocity depends on the tank cross-section A. The cross-section A does not appear in Torricelli's law $v = C_d\sqrt{2gh}$. The outflow velocity is determined by the water level h alone; whether the tank is wide or narrow, it flows out at the same velocity at the same level. A affects the "drain time" instead — a larger cross-section takes longer to drain. Vary only A in the simulator and confirm that the outflow velocity v does not change.

Next is the mistake of trying to get away with the simple calculation of "dividing by the average flow rate." Because the flow rate varies moment by moment as the level changes, dividing the total volume by the full-tank flow rate badly underestimates the drain time. The correct formula is $t_{empty} = (A/C_d a)\sqrt{2h_0/g}$, derived by solving the differential equation. In fact, you can confirm from the equation that the correct drain time is exactly twice the value obtained by dividing by the full-tank flow rate.

Finally, the pitfall of treating the discharge coefficient Cd as 1.0. Using the theoretical formula $v = \sqrt{2gh}$ as-is overestimates the flow rate by 30–40 percent compared to reality. For a sharp-edged orifice, contraction and viscosity give Cd ≈ 0.62, and in real piping the losses of fittings and valves are added as well. In design, always use a Cd that is measured or taken from reliable literature, and allow a conservative margin.