At a water treatment plant, a level pipe narrows where it passes under a road, and a pressure gauge sits on each part. Most people expect the narrow part, where the water is squeezed, to show the highest pressure. The gauges say the opposite. Where the pipe is narrowest and the water fastest, the gauge reads lowest.
Energy along a tube of flow
The equation of continuity came from counting mass. Bernoulli's theorem comes from counting energy. Take a tube of flow of an ideal fluid in steady flow. At its lower end, section 1 has area \(A_1\), height \(h_1\), pressure \(P_1\) and speed \(v_1\). At its upper end, section 2 has \(A_2\), \(h_2\), \(P_2\) and \(v_2\). Apply the work-energy theorem to the fluid between the two sections.
Watch the slices in Fig 3.1. In a short time \(\Delta t\), the fluid at section 1 moves forward by \(v_1\,\Delta t\), and the fluid at section 2 by \(v_2\,\Delta t\). By continuity the two slabs have the same volume, \(\Delta V = A_1v_1\,\Delta t = A_2v_2\,\Delta t\), and the same mass \(\rho\,\Delta V\). The fluid in the middle is replaced by identical fluid, since the flow is steady. So the net change is as if one slab had been carried from 1 to 2.
\[ W_P = P_1A_1\,v_1\Delta t - P_2A_2\,v_2\Delta t = (P_1 - P_2)\,\Delta V \]
Why. the force \(P_1A_1\) pushes forward through \(v_1\Delta t\), the force \(P_2A_2\) pushes back through \(v_2\Delta t\), and both products of area and distance equal \(\Delta V\).
\[ W_g = -\rho\,\Delta V\,g\,(h_2 - h_1) \]
Why. gravity does negative work as a slab of mass \(\rho\,\Delta V\) is raised from \(h_1\) to \(h_2\).
\[ (P_1 - P_2)\,\Delta V - \rho\,\Delta V\,g\,(h_2 - h_1) = \tfrac12\rho\,\Delta V\,(v_2^2 - v_1^2) \]
Why. the work-energy theorem: the total work equals the gain in kinetic energy of the slab. An ideal fluid has no friction to do other work.
\[ P_1 + \tfrac12\rho v_1^2 + \rho g h_1 = P_2 + \tfrac12\rho v_2^2 + \rho g h_2 \]
Read it. divide by \(\Delta V\) and gather the terms of each section on its own side.
In the steady flow of an ideal fluid, \(P + \tfrac12\rho v^2 + \rho g h\) is the same at every point of a streamline. The three terms are the pressure, the kinetic energy per unit volume and the potential energy per unit volume.
In a level pipe, faster means lower pressure
In a level pipe \(h_1 = h_2\), and the height terms cancel. Then \(P + \tfrac12\rho v^2\) is the same everywhere along a streamline. Where the water moves faster, the pressure must be lower. The cause is Newton's second law. Water entering a narrow part speeds up, so it needs a net forward force. That force can only come from a higher pressure behind than ahead.
Water flows along a level pipe whose area halves twice, so the speed is \(1\), \(2\) and \(4\,\text{m/s}\) in the three parts. An open tube stands on each part. In which part does the water rise lowest in its tube?
In the figure above, the pressure in the wide part is \(1.2 \times 10^5\,\text{Pa}\), absolute. Choose each chip. In the middle part \(P = 1.2 \times 10^5 - \tfrac12 \times 1000 \times (2^2 - 1^2) = 1.185 \times 10^5\,\text{Pa}\). In the narrow part \(P = 1.2 \times 10^5 - 500 \times (4^2 - 1^2) = 1.125 \times 10^5\,\text{Pa}\). Each tube is open to the air, so its column shows the gauge pressure, \(P - P_0 = \rho g h\). That puts the three columns at \(2.00\), \(1.85\) and \(1.25\,\text{m}\) above the axis.
Water flows at \(1\,\text{m/s}\) along a level pipe of area \(4\,\text{cm}^2\), where the pressure is \(1.2 \times 10^5\,\text{Pa}\). The pipe narrows to \(2\,\text{cm}^2\). Find the speed and the pressure in the narrow part.
Level pipe, so the height terms cancel. The areas are in the ratio \(2 : 1\).
\(A_1v_1 = A_2v_2\), and \(P_1 + \tfrac12\rho v_1^2 = P_2 + \tfrac12\rho v_2^2\).
\(v_2 = 4 \times 1/2 = 2\,\text{m/s}\). \(P_2 = 1.2 \times 10^5 - \tfrac12 \times 1000 \times (4 - 1) = 1.2 \times 10^5 - 1500 = 1.185 \times 10^5\,\text{Pa}\).
The water runs at \(2\,\text{m/s}\) in the narrow part, and its pressure there is \(1500\,\text{Pa}\) lower, \(1.185 \times 10^5\,\text{Pa}\).
Water flows at \(2\,\text{m/s}\) along a level pipe at a pressure of \(1.5 \times 10^5\,\text{Pa}\). The pipe narrows to half its area. Find the pressure in the narrow part, in Pa.
Three energies per volume
Read the three terms as energies carried by each cubic metre of fluid. \(\tfrac12\rho v^2\) is the kinetic energy per unit volume, since \(\rho\) is the mass per unit volume. \(\rho g h\) is the potential energy per unit volume. \(P\) is the pressure energy per unit volume. Step 1 of the derivation showed why: pushing a volume \(\Delta V\) against a pressure \(P\) takes work \(P\,\Delta V\). Bernoulli's theorem says that the total energy per unit volume stays the same along a streamline.
The units agree. A pascal is \(\text{N/m}^2\), and \(\text{N/m}^2 = \text{N·m/m}^3 = \text{J/m}^3\). For the second term, \(\rho v^2\) has the unit \(\text{kg/m}^3 \times \text{m}^2/\text{s}^2 = \text{kg/(m·s}^2) = \text{N/m}^2\).
Heads in metres
Divide every term by \(\rho g\). Each term becomes a length, called a head. \(P/\rho g\) is the pressure head, \(v^2/2g\) is the velocity head, and \(h\) is the height, or elevation head. Their sum is the total head, and along a streamline it stays the same.
\[ \dfrac{P}{\rho g} + \dfrac{v^2}{2g} + h = \text{constant} \]
Heads can be seen. In the figure of the predict above, each open tube shows the gauge pressure head of its part. The blue bar above each column is the velocity head, \(0.05\), \(0.20\) and \(0.80\,\text{m}\). Column and bar together always reach the dashed line, the total head of \(2.05\,\text{m}\). What the water gains in speed it loses in pressure, metre for metre.
A pump at the ground floor sends water up a vertical pipe of uniform bore to a flat \(12\,\text{m}\) higher. At the bottom of the pipe the pressure is \(2.5 \times 10^5\,\text{Pa}\). Treat the water as an ideal fluid and take \(g = 10\,\text{m/s}^2\). Find the pressure at the top of the pipe.
The bore is uniform, so by continuity the speed is the same at both ends, and the velocity terms cancel.
Bernoulli then gives \(P_{\text{top}} = P_{\text{bottom}} - \rho g h = 2.5 \times 10^5 - 1000 \times 10 \times 12 = 1.3 \times 10^5\,\text{Pa}\).
Every metre of rise costs \(10^4\,\text{Pa}\) of pressure, exactly as in water at rest.
What does the term \(\tfrac12\rho v^2\) in Bernoulli's equation measure?
Water flows at \(2\,\text{m/s}\) through a pipe of uniform area that rises \(3\,\text{m}\). Take \(g = 10\,\text{m/s}^2\). By how much does the pressure fall between the lower and the upper ends, in Pa?
Where the theorem holds
Bernoulli's theorem is the work-energy theorem with no losses. It holds when the flow is steady, the fluid is incompressible, and no friction turns energy into heat. It links two points on one streamline. For an irrotational flow it links any two points, and this chapter uses it that way for the smooth flows of tanks and pipes.
A pump adds energy per volume
A pump or a fan does work on the fluid. Between its two sides \(P + \tfrac12\rho v^2 + \rho g h\) is not constant: it rises by the energy the pump gives each cubic metre. The power of the pump is that energy per cubic metre times the volume per second. A building's pump that lifts \(1\,\text{L/s}\) through \(20\,\text{m}\) and delivers it at \(10\,\text{m/s}\) adds \(\rho g h + \tfrac12\rho v^2 = 2 \times 10^5 + 5 \times 10^4 = 2.5 \times 10^5\,\text{J/m}^3\). Its power is \(10^{-3} \times 2.5 \times 10^5 = 250\,\text{W}\).
A pump draws water from a still sump and sends \(0.5\,\text{L/s}\) out of a pipe \(25\,\text{m}\) higher, at \(10\,\text{m/s}\) and at atmospheric pressure. Treat the flow as ideal and take \(g = 10\,\text{m/s}^2\). Find the power the pump gives the water, in watts.
Real fluids lose head
A real fluid has viscosity. Along a level pipe of uniform bore, Bernoulli's theorem would keep the pressure the same. In fact the pressure falls slowly along the pipe, because friction between the layers turns some energy into heat. The chapter on Viscosity measures this loss. Turbulent flow loses even more.
A limit to the speed
Bernoulli's theorem also sets a limit. In a level pipe, the faster the water, the lower its pressure. But the pressure in a liquid cannot fall below zero. Water that enters a throat at \(1 \times 10^5\,\text{Pa}\), moving slowly, can reach at most \(\sqrt{2P_1/\rho} = \sqrt{200} \approx 14\,\text{m/s}\) there. Make the throat narrower still, and the water near zero pressure boils into bubbles of vapour. This is cavitation. The bubbles collapse violently further on, and they pit the blades of pumps and the propellers of ships.
Water flows slowly along a level pipe at an absolute pressure of \(2 \times 10^5\,\text{Pa}\) and then through a narrow throat. Neglect the speed in the wide pipe. At what speed in the throat would the pressure there fall to zero, in m/s?
"Fast fluid means low pressure" is true only along one streamline, with no energy added or lost. The jet from a hose is fast, yet its pressure is atmospheric, the same as the still air around it. The water got its speed from the pressure inside the hose, which was higher. Compare points in one flow, never a fast jet with still air elsewhere.
Between which pairs of points can Bernoulli's theorem be used directly, with the same constant on both sides?
Water flows from a wide part of a level pipe into a narrow part. Explain, using forces, why the pressure in the narrow part must be lower than in the wide part.
By continuity the water moves faster in the narrow part, so each bit of water speeds up as it passes from the wide part into the narrow part. To speed up it needs a net forward force. In a level pipe with no friction, the only forces along the pipe come from the pressure of the water behind and in front. So the pressure behind, in the wide part, must be greater than the pressure ahead, in the narrow part.
Daniel Bernoulli was a Swiss physician and mathematician who worked at St Petersburg. In 1738 he published Hydrodynamica, the book that gave the science of moving fluids its name. He measured the pressure on the wall of a pipe by piercing it with a small open tube and watching how high the water rose. He found that the faster the water, the lower the column. He explained it by the conservation of what he called living force, our kinetic energy.
Now return to the gauges on the pipe under the road. The water is squeezed into the narrow part, but squeezing does not raise its pressure. To pass the same volume each second, the water must move faster there. To speed up, it must be pushed from behind by a higher pressure. So the gauge on the narrow part reads lowest, by \(\tfrac12\rho(v_2^2 - v_1^2)\).
- Expecting the pressure to rise where a pipe narrows, because the water is squeezed.The narrow part has the faster water, and faster water along a streamline has the lower pressure.
- Using Bernoulli's theorem across a pump or a fan.A pump adds energy per volume. Write the energy it adds as an extra term, or apply the theorem on each side separately.
- Comparing a fast jet in the open air with still air elsewhere.Use two points of the same flow. The pressure of a free jet is atmospheric.
- Forgetting that the height term cancels only for a level pipe.Keep \(\rho g h\) whenever the two points are at different heights.