When a severe cyclone strikes the coast of Odisha, the wind can blow at \(50\,\text{m/s}\) or more. After the storm, whole villages have lost the tin sheets of their roofs. Many walls still stand, and the sheets lie scattered far away. The wind did not push the roofs sideways off the walls. It lifted them.
Measuring a flow with two tubes
A venturimeter is a short length of pipe with a narrow part, the throat, put into a water main. An open tube stands on the inlet and another on the throat. The water is faster in the throat, so its pressure is lower, and its column stands lower. The difference of levels \(\Delta h\) grows with the flow, and it can be turned into a speed.
Here is how the levels give the speed. Continuity gives \(v_2 = (A_1/A_2)\,v_1\). Bernoulli for the level pipe gives \(P_1 - P_2 = \tfrac12\rho(v_2^2 - v_1^2)\). The two columns give \(P_1 - P_2 = \rho g\,\Delta h\), as in the chapter on Pressure in Fluids. Put them together, and \(\rho\) cancels.
\[ \rho g\,\Delta h = \tfrac12\rho v_1^2\left(\dfrac{A_1^2}{A_2^2} - 1\right), \qquad v_1 = \sqrt{\dfrac{2g\,\Delta h}{(A_1/A_2)^2 - 1}} \]
The flow rate is then \(Q = A_1 v_1\). Because \(\Delta h\) grows as \(v_1^2\), doubling the flow makes the difference of levels four times as large.
A venturimeter has an inlet of \(10\,\text{cm}^2\) and a throat of \(5\,\text{cm}^2\). The water in its two tubes differs in level by \(15\,\text{cm}\). Take \(g = 10\,\text{m/s}^2\). Find the pressure difference, the speeds at the inlet and the throat, and the flow rate.
\(A_1/A_2 = 2\), so \(v_2 = 2v_1\). The pipe is level.
\(P_1 - P_2 = \rho g\,\Delta h\), and \(P_1 - P_2 = \tfrac12\rho(v_2^2 - v_1^2) = \tfrac12\rho \times 3v_1^2\).
\(P_1 - P_2 = 1000 \times 10 \times 0.15 = 1500\,\text{Pa}\). Then \(1500 v_1^2 = 1500\), so \(v_1 = 1\,\text{m/s}\) and \(v_2 = 2\,\text{m/s}\).
The flow rate is \(Q = 10 \times 10^{-4} \times 1 = 10^{-3}\,\text{m}^3/\text{s}\), one litre per second. Set the slider of the sim above to \(1.0\,\text{L/s}\) to see these numbers.
A venturimeter has an inlet of \(8\,\text{cm}^2\) and a throat of \(4\,\text{cm}^2\). The levels in its two tubes differ by \(60\,\text{cm}\). Take \(g = 10\,\text{m/s}^2\). Find the flow rate, in L/s.
A tube that faces the flow
Bend a tube at a right angle and point its open mouth into the flow. The water that meets the mouth has nowhere to go, so it comes to rest there. That point is a stagnation point. Bernoulli along the streamline that ends there gives \(P_s = P + \tfrac12\rho v^2\). The kinetic energy per volume of the stream has become extra pressure. A straight tube on the side of the pipe feels only the pressure \(P\) of the moving water.
This is the Pitot tube. The extra height of its column is \(h = (P_s - P)/\rho g = v^2/2g\), the velocity head of the last lesson. So \(v = \sqrt{2gh}\). In Fig 6.1, \(h = 45\,\text{cm}\) gives \(v = \sqrt{2 \times 10 \times 0.45} = 3\,\text{m/s}\). An aeroplane carries a Pitot tube on its nose or wing, and its airspeed indicator reads \(\tfrac12\rho v^2\) of the air. At \(60\,\text{m/s}\) that is \(\tfrac12 \times 1.2 \times 3600 = 2160\,\text{Pa}\).
A Pitot tube in a stream of water shows a level \(80\,\text{cm}\) higher than a straight side tube. Take \(g = 10\,\text{m/s}^2\). Find the speed of the stream, in m/s.
Henri Pitot, a French engineer, wanted to know how fast the Seine flowed in Paris at different depths. In 1732 he lowered into the river a glass tube bent at a right angle, its mouth facing the current, beside a straight tube. The water in the bent tube stood higher, and the difference gave the speed. His measurements showed, against the belief of the time, that the water flowed more slowly near the river bed than near the surface.
Lift on a wing
The wing of an aeroplane is shaped and tilted so that the air flowing over its top moves faster than the air flowing under it. By Bernoulli's theorem, the faster air above has the lower pressure. The pressure below the wing then exceeds the pressure above, and the difference pushes the wing up. This upward push is the dynamic lift.
Air of density \(1.2\,\text{kg/m}^3\) flows over the top of a wing at \(60\,\text{m/s}\) and under it at \(50\,\text{m/s}\). The wing has an area of \(20\,\text{m}^2\). Take \(g = 10\,\text{m/s}^2\). Find the pressure difference, the lift, and the mass the wing can support.
The wing is thin, so the air above and below is at nearly the same height. Bernoulli along the flow gives \(P + \tfrac12\rho v^2\) the same above and below.
\(P_{\text{bottom}} - P_{\text{top}} = \tfrac12\rho(v_{\text{top}}^2 - v_{\text{bottom}}^2)\), and \(F = \Delta P\,A\).
\(\Delta P = \tfrac12 \times 1.2 \times (3600 - 2500) = 660\,\text{Pa}\). \(F = 660 \times 20 = 13\,200\,\text{N}\).
The lift of \(13\,200\,\text{N}\) can hold up a mass of \(13\,200/10 = 1320\,\text{kg}\).
A popular story says that the air over the top must go faster "to meet the air from below at the trailing edge". That is not true. Air over a real wing arrives at the trailing edge well before the air underneath. The shape and the tilt of the wing turn the flow, and the air above speeds up. Bernoulli's theorem then links that speed to the lower pressure.
The lift on a wing of fixed shape and tilt grows as the square of the speed of the air past it. At half its take-off speed, an aeroplane gets only a quarter of the lift it needs. So it must race along the runway until the air moves fast enough over its wings. At take-off the pilot also tilts the wings up a little and extends flaps. These turn the flow more and give lift at a lower speed.
Roofs in a cyclone
Now the cyclone. The air inside a closed house is still, at atmospheric pressure. Outside, a \(50\,\text{m/s}\) wind sweeps over the roof. Both the still air and the wind came from air at rest far away, so they share one Bernoulli constant. The fast air over the roof therefore has a pressure lower by \(\tfrac12\rho v^2 = \tfrac12 \times 1.2 \times 2500 = 1500\,\text{Pa}\). A tin sheet of \(3\,\text{m}^2\) feels a net upward push of \(1500 \times 3 = 4500\,\text{N}\), the weight of \(450\,\text{kg}\).
A wind of \(40\,\text{m/s}\) blows over a flat roof of area \(10\,\text{m}^2\). The air inside the house is still. Take the density of air as \(1.2\,\text{kg/m}^3\). Find the net upward force on the roof, in N.
Sprayers
A spray pump on a farm, a perfume atomiser and the old flit gun all work the same way. A fast stream of air is blown across the top of a thin vertical tube. The lower end of the tube dips into the liquid. The fast air lowers the pressure at the top of the tube by \(\tfrac12\rho_{\text{air}}v^2\). The atmosphere pushes on the liquid surface below, and it pushes liquid up the tube. At the top, the fast air tears the liquid into fine droplets.
The liquid reaches the top when the drop in pressure can hold up the column below it: \(\tfrac12\rho_{\text{air}}v^2 = \rho_{\text{liquid}}\,g h\). For a column of water \(5\,\text{cm}\) tall, \(\tfrac12 \times 1.2 \times v^2 = 1000 \times 10 \times 0.05 = 500\,\text{Pa}\). So \(v = \sqrt{500/0.6} \approx 29\,\text{m/s}\). The carburettor of a scooter and the Bunsen burner of a school laboratory use the same effect.
In a sprayer, air of density \(1.2\,\text{kg/m}^3\) is blown at \(30\,\text{m/s}\) across the top of a vertical tube dipping into water. Take \(g = 10\,\text{m/s}^2\). To what height above the water surface can the water rise in the tube, in cm?
Hang two sheets of paper side by side, a few centimetres apart, and blow into the gap between them. They do not fly apart. They swing together. The air moving fast between the sheets has a lower pressure than the still air outside them. The still air pushes each sheet inward, towards the other.
Two sheets of paper hang side by side, a few centimetres apart. What happens when you blow a stream of air between them, and why?
The spinning ball
A ball flying through still air can be thought of the other way round: the ball at rest, with the air flowing past it. If the ball does not spin, the air goes round it equally on both sides, and there is no sideways force. A spinning ball drags a thin layer of air round with it. On the side where the surface moves with the airflow, the air is sped up. On the other side, where the surface moves against it, the air is slowed down.
In the spinning chip of Fig 6.2, the streamlines crowd together above the ball, so the air there is faster. Its pressure is lower than the pressure below the ball, and the ball is pushed towards the fast side. This sideways push on a spinning ball is the Magnus effect. A spin bowler's ball drifts in the air before it pitches, and a cut shot in table tennis swerves, for this reason.
A ball flies to the left through still air, spinning so that its top surface moves to the right. Which way does the ball curve, and why?
The flow of air round a real ball is not ideal. A thin layer next to the surface feels viscosity, and behind the ball the flow becomes a turbulent wake. The spin shifts the point where the flow leaves the ball, so the wake is thrown to one side. The air is pushed one way, and the ball the other. The ideal-flow picture of Fig 6.2 gives the right direction, but not the full size of the force.
Now return to the roofs on the Odisha coast. The walls feel the wind sideways, but they are heavy and set in the ground. The tin sheets of the roof are light and wide. A \(50\,\text{m/s}\) wind over them lowers the pressure above by \(1500\,\text{Pa}\), while the still air inside pushes up at the full atmospheric pressure. A sheet of \(3\,\text{m}^2\) is pushed up with \(4500\,\text{N}\), far more than its own weight. So the roof lifts off while the walls stand.
- Using the difference of levels of a venturimeter as if \(v_1 = \sqrt{2g\,\Delta h}\).Both ends move. Use continuity too: \(v_1 = \sqrt{2g\,\Delta h/((A_1/A_2)^2 - 1)}\).
- Using the density of water for the air over a wing or a roof.\(\tfrac12\rho v^2\) of the moving fluid is the air's: \(\rho = 1.2\,\text{kg/m}^3\).
- Saying that a spinning ball curves towards the side where the air is slower.The ball moves towards the fast side, where the pressure is lower.