Torricelli’s law gives the speed of a liquid leaving a small hole in a tank.
It is a direct application of Bernoulli’s equation between the free surface of the liquid and the hole.
1. Physical situation
A tank contains a liquid of density \(\rho\). A small hole is located a vertical distance \(h\) below the free surface.
The liquid leaves the hole as a jet.
The usual ideal assumptions are:
- The liquid is incompressible.
- The flow is steady.
- The hole is small compared with the tank cross-section.
- The liquid speed at the free surface is approximately zero.
- Viscous losses and contraction losses are ignored unless a discharge coefficient is used.
2. Bernoulli equation
Apply Bernoulli’s equation between the free surface and the hole:
\[
\begin{aligned}
P+\frac{1}{2}\rho v_{\mathrm{surface}}^2+\rho gy_{\mathrm{surface}}
&=
P_0+\frac{1}{2}\rho v^2+\rho gy_{\mathrm{hole}}.
\end{aligned}
\]
Because the tank is large compared with the hole, the free-surface speed is approximately zero:
\[
\begin{aligned}
v_{\mathrm{surface}} &\approx 0.
\end{aligned}
\]
Let
\[
\begin{aligned}
h &= y_{\mathrm{surface}}-y_{\mathrm{hole}}.
\end{aligned}
\]
Then Bernoulli’s equation becomes:
\[
\begin{aligned}
P+\rho gh
&=
P_0+\frac{1}{2}\rho v^2.
\end{aligned}
\]
3. General pressurized-tank formula
Rearranging for \(v\):
\[
\begin{aligned}
\frac{1}{2}\rho v^2
&=
P-P_0+\rho gh.
\end{aligned}
\]
Multiply by \(2/\rho\):
\[
\begin{aligned}
v^2
&=
\frac{2(P-P_0)}{\rho}+2gh.
\end{aligned}
\]
Therefore:
\[
\begin{aligned}
v
&=
\sqrt{\frac{2(P-P_0)}{\rho}+2gh}.
\end{aligned}
\]
4. Open tank formula
If the tank is open to the atmosphere, then the free surface pressure and outside pressure are equal:
\[
\begin{aligned}
P &= P_0.
\end{aligned}
\]
The pressure term cancels:
\[
\begin{aligned}
v
&=
\sqrt{2gh}.
\end{aligned}
\]
This is the standard form of Torricelli’s law.
5. Meaning of the formula
The formula
\[
\begin{aligned}
v &= \sqrt{2gh}
\end{aligned}
\]
is the same speed an object would gain by falling freely through height \(h\), ignoring air resistance.
The gravitational potential energy per unit mass \(gh\) becomes kinetic energy per unit mass \(v^2/2\).
6. Flow rate through the hole
If the hole has area \(A_{\mathrm{hole}}\), the ideal volume flow rate is
\[
\begin{aligned}
Q_{\mathrm{ideal}}
&=
A_{\mathrm{hole}}v.
\end{aligned}
\]
For a circular hole of diameter \(d\):
\[
\begin{aligned}
A_{\mathrm{hole}}
&=
\frac{\pi d^2}{4}.
\end{aligned}
\]
A real hole usually discharges less than the ideal prediction. A discharge coefficient \(C_d\) corrects this:
\[
\begin{aligned}
Q_{\mathrm{actual}}
&=
C_dA_{\mathrm{hole}}v.
\end{aligned}
\]
7. Draining time
As a tank drains, the water level falls, so the depth \(h\) decreases and the jet slows down.
If the tank cross-section is \(A_{\mathrm{tank}}\), then
\[
\begin{aligned}
A_{\mathrm{tank}}\frac{dh}{dt}
&=
-C_dA_{\mathrm{hole}}
\sqrt{\frac{2(P-P_0)}{\rho}+2gh}.
\end{aligned}
\]
Integrating from initial depth \(h_0\) to final depth \(h_f\) gives:
\[
\begin{aligned}
t
&=
\frac{A_{\mathrm{tank}}}{C_dA_{\mathrm{hole}}g}
\left[
\sqrt{\frac{2(P-P_0)}{\rho}+2gh_0}
-
\sqrt{\frac{2(P-P_0)}{\rho}+2gh_f}
\right].
\end{aligned}
\]
For an open tank, \(P=P_0\), so this becomes:
\[
\begin{aligned}
t
&=
\frac{A_{\mathrm{tank}}}{C_dA_{\mathrm{hole}}}
\sqrt{\frac{2}{g}}
\left(\sqrt{h_0}-\sqrt{h_f}\right).
\end{aligned}
\]
8. Jet as a horizontal projectile
If the jet leaves horizontally from height \(y_{\mathrm{hole}}\) above the ground, its flight time is
\[
\begin{aligned}
t_{\mathrm{flight}}
&=
\sqrt{\frac{2y_{\mathrm{hole}}}{g}}.
\end{aligned}
\]
The horizontal range is
\[
\begin{aligned}
x
&=
vt_{\mathrm{flight}}.
\end{aligned}
\]
For an open tank with total surface height \(H\) above the ground:
\[
\begin{aligned}
H &= h+y_{\mathrm{hole}}.
\end{aligned}
\]
The range becomes
\[
\begin{aligned}
x
&=
2\sqrt{hy_{\mathrm{hole}}}.
\end{aligned}
\]
This range is maximum when
\[
\begin{aligned}
h &= y_{\mathrm{hole}}=\frac{H}{2}.
\end{aligned}
\]
9. Worked example
A hole is \(5\ \mathrm{m}\) below the water surface in an open tank.
Use \(g=9.81\ \mathrm{m/s^2}\).
\[
\begin{aligned}
h &= 5\ \mathrm{m}.
\end{aligned}
\]
Since the tank is open, use
\[
\begin{aligned}
v &= \sqrt{2gh}.
\end{aligned}
\]
Substitute:
\[
\begin{aligned}
v
&=
\sqrt{2(9.81)(5)}\\
&=
\sqrt{98.1}\\
&\approx 9.90\ \mathrm{m/s}.
\end{aligned}
\]
So the efflux speed is about \(9.9\ \mathrm{m/s}\).
10. Formula summary
| Goal |
Formula |
Use |
| Open-tank efflux speed |
\(v=\sqrt{2gh}\) |
Find jet speed from depth below surface |
| Pressurized-tank efflux speed |
\(v=\sqrt{2(P-P_0)/\rho+2gh}\) |
Include extra pressure at liquid surface |
| Hole area |
\(A_{\mathrm{hole}}=\pi d^2/4\) |
Convert hole diameter to area |
| Actual flow rate |
\(Q=C_dA_{\mathrm{hole}}v\) |
Estimate real discharge rate |
| Open-tank draining time |
\(t=\frac{A_{\mathrm{tank}}}{C_dA_{\mathrm{hole}}}\sqrt{2/g}(\sqrt{h_0}-\sqrt{h_f})\) |
Estimate time for level to fall |
| Projectile flight time |
\(t=\sqrt{2y_{\mathrm{hole}}/g}\) |
Find time for jet to reach ground |
| Jet range |
\(x=vt\) |
Find horizontal distance reached by the jet |
11. Assumptions and limitations
- The hole is small compared with the tank cross-section.
- The liquid is incompressible.
- The free-surface speed is approximately zero.
- The flow is steady at each instant.
- Air resistance on the jet is ignored.
- The jet leaves horizontally.
- The discharge coefficient is only a simple correction for real losses.
- Very viscous liquids, very large openings, turbulence, and splashing can make real results differ from the ideal model.
Key idea: a deeper hole gives a faster jet because gravitational potential energy is converted into kinetic energy.
