Before the monsoon, the overhead tank on the roof is emptied for cleaning. The caretaker opens the drain at the bottom of the full tank. At first the water gushes out in a strong, far-reaching jet. As the level falls, the jet weakens and droops. The last few centimetres of water dribble out slowly, and they seem to take for ever.
Bernoulli from the surface to the hole
Take a wide tank open to the air, with a small hole in its side at a depth \(h\) below the free surface. Pick a streamline that starts at the surface and ends in the jet just outside the hole. At the surface the pressure is \(P_0\), the atmospheric pressure. In the jet the pressure is also \(P_0\), since the jet is open to the air. The tank is wide and the hole small, so by continuity the surface hardly moves.
\[ P_0 + \tfrac12\rho\,(0)^2 + \rho g h = P_0 + \tfrac12\rho v^2 + 0 \]
Why. Bernoulli between the surface, at height \(h\) above the hole and nearly at rest, and the jet at the level of the hole.
\[ \rho g h = \tfrac12\rho v^2 \]
Why. the atmospheric pressure appears on both sides and cancels.
\[ v = \sqrt{2gh} \]
Read it. the density cancels too. What is left is the speed of a body that has fallen freely through \(h\).
A liquid flows out of a small hole in a wide open tank with the speed \(v = \sqrt{2gh}\). Here \(h\) is the depth of the hole below the free surface. It is the speed a body gains in falling freely through \(h\).
A hole \(1.8\,\text{m}\) below the surface gives \(v = \sqrt{2 \times 10 \times 1.8} = 6\,\text{m/s}\). A stone dropped from the level of the surface would pass the hole at the same \(6\,\text{m/s}\). The speed of efflux is set by the depth alone. The pressure of the water above the hole does the work. Each cubic metre that leaves carries off the potential energy of a cubic metre at the surface.
Two things that do not matter
The density cancelled in step 2. Mercury and water leave holes at the same depth with the same speed. A denser liquid has a greater pressure at the hole, but it also has more mass to speed up, and the two effects match exactly. This is the same reason that heavy and light stones fall together.
The size of the hole does not enter either, as long as the hole is small compared with the tank. A larger hole passes more water each second, \(Q = av\), where \(a\) is its area. But the water leaves at the same speed. A hole of \(1\,\text{cm}^2\) at a depth of \(5\,\text{m}\) gives \(10\,\text{m/s}\) and passes \(10^{-4} \times 10 = 10^{-3}\,\text{m}^3\), one litre, each second.
A jet pointed upward
Fit a short nozzle to the hole and point it straight up. The water leaves at \(\sqrt{2gh}\), and it rises until its speed is zero, a height \(v^2/2g = h\). So the jet climbs back exactly to the level of the water in the tank. In practice it stops a little lower, because viscosity and the air take some of its energy. Torricelli used this very fact to argue that the jet has the speed of a falling body.
A wide open tank has a small hole \(3.2\,\text{m}\) below the surface of the water. Take \(g = 10\,\text{m/s}^2\). With what speed does the water leave the hole, in m/s?
Two open tanks, one of water and one of mercury, each have a small hole at a depth of \(1\,\text{m}\). The hole in the mercury tank is twice as large. How do the speeds of the two jets compare?
A nozzle \(1.8\,\text{m}\) below the surface of the water in a wide tank points straight up. Ignoring friction and air resistance, how high does the jet rise above the nozzle?
The jet pushes back
A jet carries momentum away from the tank each second. The chapter on Momentum and Its Conservation found the force of a steady stream. It equals the momentum the stream carries off each second. The hole of area \(a\) passes a mass \(\rho a v\) each second at speed \(v\). So the thrust on the tank is \(\rho a v^2 = \rho a (2gh) = 2\rho g h a\).
A tank has a hole of area \(2\,\text{cm}^2\) in its side, \(5\,\text{m}\) below the surface. Take \(g = 10\,\text{m/s}^2\). Find the mass of water leaving each second and the thrust of the jet on the tank. The jet then strikes a wall and stops. What force does it exert on the wall?
The speed is \(v = \sqrt{2 \times 10 \times 5} = 10\,\text{m/s}\), so the mass flow is \(\rho a v = 1000 \times 2 \times 10^{-4} \times 10 = 2\,\text{kg/s}\).
The momentum carried off each second is \(2 \times 10 = 20\,\text{N}\). This equals \(2\rho g h a = 2 \times 1000 \times 10 \times 5 \times 2 \times 10^{-4} = 20\,\text{N}\), pushing the tank backward.
The thrust is twice \(\rho g h a = 10\,\text{N}\), the push of the still water on a patch of wall of the same area. The wall stops \(20\,\text{kg·m/s}\) of momentum each second, so it feels \(20\,\text{N}\).
A closed tank with gas above
In the chapter on Pressure in Fluids, gauge pressure was the pressure above atmospheric, \(P - P_0\). Now close the tank and pump air into the space above the water, to a pressure \(P\). The surface and the jet are no longer at the same pressure. Bernoulli from the surface to the jet gives
\[ P + \rho g h = P_0 + \tfrac12\rho v^2, \qquad v = \sqrt{\dfrac{2(P - P_0)}{\rho} + 2gh} \]
The gauge pressure of the gas acts like extra depth. In Fig 4.1 the gas is at \(0.32 \times 10^5\,\text{Pa}\) gauge, worth \(0.32 \times 10^5/(1000 \times 10) = 3.2\,\text{m}\) of water. The jet leaves as if the hole were \(1.8 + 3.2 = 5\,\text{m}\) deep: \(\sqrt{2 \times 10 \times 5} = 10\,\text{m/s}\).
A knapsack sprayer on a farm has a closed tank of pesticide solution, density \(1000\,\text{kg/m}^3\). The farmer pumps air into it until the air is at a gauge pressure of \(2 \times 10^5\,\text{Pa}\). The nozzle is held level with the surface of the liquid in the tank. Treat the flow as ideal. Find the speed of the spray at the nozzle.
Take a streamline from the liquid surface in the tank to the nozzle, at the same height. At the surface the pressure is \(P\) and the speed nearly zero. At the nozzle the pressure is \(P_0\).
\(P + 0 + 0 = P_0 + \tfrac12\rho v^2 + 0\), so \(v = \sqrt{2(P - P_0)/\rho}\).
\(v = \sqrt{2 \times 2 \times 10^5/1000} = \sqrt{400} = 20\,\text{m/s}\).
The spray leaves at \(20\,\text{m/s}\). As the tank empties, the trapped air expands, its pressure falls, and the spray weakens until the farmer pumps again.
A sealed tank can also do the opposite. Suppose the air above the water is trapped and nothing lets more air in. As water flows out, the trapped air expands and its pressure falls below \(P_0\). The flow stops when \(P + \rho g h = P_0\). That is why a sealed can with one small hole soon stops pouring. A second hole to let air in keeps it flowing.
A closed tank has air above the water at a gauge pressure of \(0.6 \times 10^5\,\text{Pa}\). A small hole is \(1.2\,\text{m}\) below the surface and opens to the atmosphere. Take \(g = 10\,\text{m/s}^2\). Find the speed of efflux, in m/s.
How long a tank takes to empty
A jet from a full tank is fast, and one from a nearly empty tank is slow. So the level cannot fall at a steady rate. Let the tank have a cross-section \(A\) and a hole of area \(a\) at its base, with \(a\) much smaller than \(A\). When the depth is \(h\), the hole passes \(a\sqrt{2gh}\) each second. By continuity, the surface falls at the rate \(-dh/dt = (a/A)\sqrt{2gh}\).
\[ A\,\dfrac{dh}{dt} = -a\sqrt{2gh} \]
Why. the volume lost from the tank each second equals the volume leaving through the hole.
\[ dt = -\dfrac{A}{a\sqrt{2g}}\,\dfrac{dh}{\sqrt{h}} \]
Why. separate the variables: all the \(h\) on one side, \(t\) on the other.
?
Your turn. integrate from the full depth \(H\) down to zero. Which line is it?
\[ t = \dfrac{A}{a}\sqrt{\dfrac{2H}{g}} \]
Read it. \(2\sqrt{H}/\sqrt{2g} = \sqrt{2H/g}\).
For the tank of Fig 4.2, \(A/a = 2/(2 \times 10^{-4}) = 10^4\). The time to empty is \(t = 10^4 \times \sqrt{2 \times 1.25/10} = 10^4 \times 0.5 = 5000\,\text{s}\), about \(83\) minutes. At the start the jet leaves at \(5\,\text{m/s}\) and the level falls by \(0.5\,\text{mm}\) each second. The graph is a parabola, \(h = H(1 - t/T)^2\), steep at first and flat at the end.
Look at the dashed lines of Fig 4.2. After half the time, \(42\) minutes, the depth is not half but a quarter, \(0.31\,\text{m}\). The top three quarters of the water leave in the first half of the time. The last quarter takes the other half, because its jet is slow. The time to fall from \(H\) to \(h\) grows with \(\sqrt{H} - \sqrt{h}\), and from \(H\) to \(H/4\) that is exactly half of \(\sqrt{H}\).
A tank of cross-section \(0.5\,\text{m}^2\) holds water \(0.8\,\text{m}\) deep. A hole of area \(1\,\text{cm}^2\) is opened in its base. Take \(g = 10\,\text{m/s}^2\). How long does the tank take to empty, in seconds?
A tank empties through a hole in its base in a time \(T\). How long does the level take to fall from the full depth \(H\) to \(H/4\)?
Evangelista Torricelli, who succeeded Galileo at Florence, studied water spouting from holes in vessels. In a book on the motion of heavy bodies, published in 1644, he gave the answer. The water at the hole moves as fast as a body that has fallen from the surface. His evidence was a jet turned upward, which rises almost to the level of the water in the vessel. The same Torricelli made the first mercury barometer.
Now return to the overhead tank. When it is full, the water leaves the drain at \(\sqrt{2gH}\), a strong jet. As the level falls, the depth above the drain shrinks and so does the speed. The level falls quickly at first and then more and more slowly. Half the time goes on the last quarter of the water, which is why the end seems to take for ever.
- Measuring \(h\) in \(v = \sqrt{2gh}\) from the base of the tank.\(h\) is the depth of the hole below the free surface.
- Adding the atmospheric pressure to the head of an open tank.The surface and the jet are both at \(P_0\), so it cancels. Only a gauge pressure above the liquid adds head.
- Assuming a tank drains at a steady rate.The speed falls as the depth falls. Use \(t = (A/a)\sqrt{2/g}\,(\sqrt{H} - \sqrt{h})\).
- Taking the thrust of a jet as \(\rho g h a\).The jet carries off \(\rho a v^2\) of momentum each second, which is \(2\rho g h a\).