Bernoulli's Equation

Bernoulli's principle

On observing a river, the speed of the water decreases in a wide region and increases where the river is narrow. One might think that the pressure in the narrower region would be greater, but in fact the pressure within the fluid in the narrower part is less and that in the wider part is more. Daniel Bernoulli found that the speed of a fluid in a narrow region increases while its internal pressure in that region decreases. This is Bernoulli's principle.

Bernoulli's equation relates the speed of a fluid at a point, the pressure at that point, and the height of that point above a reference level. It is an application of the work–energy theorem to a fluid in flow, and is consistent with the principle of conservation of energy.

Diagram to draw

An ideal fluid flowing through a tube of varying cross section and varying height. Mark the lower end P and the upper end R. At P: area $A_{1}$, speed $v_{1}$, pressure $P_{1}$, height $h_{1}$, displacement $d_{1}$ in time $dt$. At R: area $A_{2}$, speed $v_{2}$, pressure $P_{2}$, height $h_{2}$, displacement $d_{2}$ in the same time. Show the forces $P_{1}A_{1}$ pushing the element forward at P and $P_{2}A_{2}$ opposing it at R. Q and S mark the intermediate cross sections after the displacements.

Derivation

The volume $dV$ of fluid passing through any cross section during the time interval $dt$ is the same:

$$dV = A_{1}d_{1} = A_{2}d_{2}$$

The fluid is ideal, so there is no internal friction, and the only non-gravitational force doing work on the fluid element is the pressure of the surrounding fluid. The net work done on the element from P to R is

$$W = P_{1}A_{1}d_{1} - P_{2}A_{2}d_{2}$$

The second term is negative because the force at R opposes the displacement of the fluid. Using $dV = A_{1}d_{1} = A_{2}d_{2}$,

$$W = (P_{1} - P_{2})\,dV$$

Since $W$ is due to forces other than the conservative force of gravity, it equals the change in the total mechanical energy of the fluid element:

$$W = \Delta \text{K.E.} + \Delta \text{P.E.}$$

The mechanical energy of the fluid between the sections Q and R does not change, so only the end portions matter.

Kinetic energy. At the beginning of $dt$ the mass between P and Q is $\rho A_{1}d_{1}$ and its kinetic energy is $\dfrac{1}{2}\rho A_{1}d_{1}v_{1}^{2}$. At the end of $dt$ the kinetic energy of the fluid between R and S is $\dfrac{1}{2}\rho A_{2}d_{2}v_{2}^{2}$. Hence

$$\Delta \text{K.E.} = \dfrac{1}{2}\rho\,dV\,v_{2}^{2} - \dfrac{1}{2}\rho\,dV\,v_{1}^{2} = \dfrac{1}{2}\rho\,dV\left(v_{2}^{2} - v_{1}^{2}\right)$$

Potential energy. At the beginning of $dt$ the gravitational potential energy of the mass between P and Q is $\rho\,dV g h_{1}$, and at the end that of the mass between R and S is $\rho\,dV g h_{2}$. Hence

$$\Delta \text{P.E.} = \rho\,dV g\,(h_{2} - h_{1})$$

Substituting the three results,

$$(P_{1} - P_{2})\,dV = \dfrac{1}{2}\rho\,dV\left(v_{2}^{2} - v_{1}^{2}\right) + \rho\,dV g\,(h_{2} - h_{1})$$

$$P_{1} - P_{2} = \dfrac{1}{2}\rho\left(v_{2}^{2} - v_{1}^{2}\right) + \rho g\,(h_{2} - h_{1})$$

This is Bernoulli's equation. It states that the work done per unit volume of a fluid by the surrounding fluid is equal to the sum of the changes in kinetic and potential energies per unit volume that occur during the flow. Rearranged,

$$P_{1} + \dfrac{1}{2}\rho v_{1}^{2} + \rho g h_{1} = P_{2} + \dfrac{1}{2}\rho v_{2}^{2} + \rho g h_{2}$$

$$P + \dfrac{1}{2}\rho v^{2} + \rho g h = \text{constant}$$

Reading the equation

Dimensionally, pressure is energy per unit volume, and each of the three terms has that dimension. The term $P$ is called the pressure head, $\dfrac{1}{2}\rho v^{2}$ the velocity head, and $\rho g h$ the potential head. Bernoulli's principle is thus consistent with the principle of conservation of energy. For a fluid at rest, $v_{1} = v_{2} = 0$ and the equation reduces to $P_{1} - P_{2} = \rho g (h_{2} - h_{1})$, the hydrostatic result.

Applications

a) Speed of efflux — Torricelli's law

Consider a liquid of density $\rho$ in a tank of large cross section $A_{1}$ with an orifice of cross section $A_{2}$ at the bottom, $A_{2} \ll A_{1}$, the free surface standing at a height $h$ above the orifice. Both the free surface and the orifice are exposed to the atmosphere, so the pressure at both is $p_{0}$. Bernoulli's equation gives

$$p_{0} + \dfrac{1}{2}\rho v_{1}^{2} + \rho g h = p_{0} + \dfrac{1}{2}\rho v_{2}^{2}$$

From the equation of continuity, $v_{1} = \dfrac{A_{2}}{A_{1}}v_{2}$, so

$$v_{2}^{2}\left[1 - \left(\dfrac{A_{2}}{A_{1}}\right)^{2}\right] = 2gh \qquad\Longrightarrow\qquad v_{2} = \sqrt{\dfrac{2gh}{1 - \left(A_{2}/A_{1}\right)^{2}}}$$

If $A_{2} \ll A_{1}$ this reduces to

$$v_{2} = \sqrt{2gh}$$

The speed of efflux is the same as that of a particle falling freely through the height $h$ under gravity. When the liquid is instead driven out by a gauge pressure $p$ inside a closed tank, the same argument gives $v = \sqrt{2p/\rho}$.

b) Venturi tube

A venturi tube measures the speed of flow of a fluid in a tube. It has a constriction: as the fluid passes through it, the speed increases by the equation of continuity, and the pressure therefore decreases as required by Bernoulli's equation. Two vertical tubes are connected at the wide part (area $A_{1}$, speed $v_{1}$, pressure $p_{1}$) and at the constriction (area $A_{2}$, speed $v_{2}$, pressure $p_{2}$). For a horizontal tube,

$$p_{1} + \dfrac{1}{2}\rho v_{1}^{2} = p_{2} + \dfrac{1}{2}\rho v_{2}^{2} \qquad\Longrightarrow\qquad p_{1} - p_{2} = \dfrac{1}{2}\rho\left(v_{2}^{2} - v_{1}^{2}\right)$$

If the difference in the liquid levels in the two vertical tubes is $h$, then $p_{1} - p_{2} = h\rho g$, and combining this with $A_{1}v_{1} = A_{2}v_{2}$ lets the rate of flow be calculated from the areas $A_{1}$ and $A_{2}$. With a mercury manometer measuring the pressure difference in a water pipe, $\dfrac{1}{2}\rho_{w}\left(v_{2}^{2} - v_{1}^{2}\right) = h\rho_{\text{Hg}}g$.

c) Lifting up of an aeroplane

Because of the aerodynamic shape of the cross section of the wings, when the aeroplane runs on the runway the streamlines of air are crowded above the wings compared with those below. The air above the wings therefore moves faster than that below. By Bernoulli's principle the pressure above the wings decreases while that below does not change. This pressure difference produces an upward force on the bottom of the wings, called the dynamic lift; when it exceeds the weight of the aeroplane, the aeroplane takes off.

d) Working of an atomizer

The carburettor of an automobile engine, a paint gun, a scent spray and an insect sprayer all work on Bernoulli's principle. A tube T is dipped in the liquid, and air is blown at high speed over the tip of this tube by a piston P in a cylinder C. The high speed air creates a low pressure over the tube, so the liquid rises in it and is then blown off as very small droplets with the expelled air.

e) Blowing off of roofs by stormy wind

When a high speed stormy wind blows over a roof top it causes a low pressure $p$ above the roof, in accordance with Bernoulli's principle. The air below the roof, inside the room, is still at the atmospheric pressure $p_{0}$. Because of this pressure difference the roof is lifted up and is then blown off by the wind.

Standard results

• $P + \dfrac{1}{2}\rho v^{2} + \rho g h =$ constant, for an ideal fluid in steady flow.
• Horizontal pipe: $p_{1} - p_{2} = \dfrac{1}{2}\rho\left(v_{2}^{2} - v_{1}^{2}\right)$.
• Torricelli: $v = \sqrt{2gh}$; under a gauge pressure $p$, $v = \sqrt{2p/\rho}$.
• Basis of Bernoulli's principle: conservation of energy (work–energy theorem applied to a fluid element).