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Bernoulli's equation

Bernoulli’s equation says that along a streamline in a flowing fluid, this sum does not change:

$$p + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$$

$p$ is the pressure, $\rho$ the density, $v$ the speed and $h$ the height. It is the work-energy theorem with every term divided by volume, which the units confirm: pressure in Pa is N/m$^2$ = J/m$^3$, and $\tfrac{1}{2}\rho v^2$ in (kg/m$^3$)(m$^2$/s$^2$) is also J/m$^3$. Kinetic energy per unit volume, gravitational potential energy per unit volume, and pressure, which is the capacity to do work per unit volume. Their total is conserved for the same reason mechanical energy is.

Four hypotheses go into that derivation, and naming them first is not pedantry - nearly every misuse of the equation is one of them being dropped. The flow must be steady, so the velocity at a given point does not change with time. It must be incompressible, so $\rho$ is the same everywhere. It must be non-viscous, so no energy leaks into heat against the walls or against itself. And the equation compares two points on the same streamline; the constant belongs to that streamline, and the derivation says nothing about a point on a different one.

Emptying a tank

Torricelli’s result drops out of Bernoulli in three lines, which makes it the cleanest demonstration that the equation is doing real work.

Take a wide open tank with a small hole punched in the side, a depth $h$ below the surface. Point 1 is the surface, point 2 is the hole. Both are open to the atmosphere, so $p_1 = p_2$ and the pressure terms cancel. The tank is wide and the hole is small, so the surface drops far more slowly than the jet moves, and $v_1 \approx 0$. Measuring height from the hole, $h_1 = h$ and $h_2 = 0$. What is left is

$$\rho g h = \tfrac{1}{2}\rho v_2^2 \quad\Longrightarrow\quad v = \sqrt{2gh}$$

Density cancelled. Mercury and water leave the hole at the same speed, which is not obvious until the algebra says so. With $h = 2.50$ m, $v = \sqrt{2 \times 9.8 \times 2.50} = \sqrt{49} = 7.00$ m/s.

$\sqrt{2gh}$ is also the speed of an object dropped from rest through a height $h$, and it is the same expression for the same reason: a parcel of fluid at the surface has converted $\rho g h$ of potential energy per unit volume into kinetic energy per unit volume, with nothing taken out along the way. The pressure term played no part because it was equal at both ends.

A Venturi, where continuity does half the work

Bernoulli alone cannot solve a pipe problem, because it has two unknowns at the narrow section. It needs continuity: for an incompressible fluid in steady flow, the volume passing every cross-section per second is the same, so $A_1v_1 = A_2v_2$.

Water flows at $1.50$ m/s through a horizontal pipe of inside diameter $6.00$ cm, which necks down to $3.00$ cm. Use $\rho_{\text{water}} = 1000$ kg/m$^3$.

Areas first. $A_1 = \pi(0.0300)^2 = 2.827\times10^{-3}$ m$^2$ and $A_2 = \pi(0.0150)^2 = 7.069\times10^{-4}$ m$^2$. Halving the diameter quarters the area, so $A_1/A_2 = 4.00$ exactly, and continuity gives

$$v_2 = v_1\frac{A_1}{A_2} = 1.50 \times 4.00 = 6.00\ \text{m/s}$$

The pipe is horizontal, so the $\rho g h$ terms are equal and drop out, leaving pressure to pay for the extra kinetic energy:

$$p_1 - p_2 = \tfrac{1}{2}\rho\left(v_2^2 - v_1^2\right) = 500 \times (36.0 - 2.25) = 1.69\times10^{4}\ \text{Pa}$$

That is $16.9$ kPa, about a sixth of an atmosphere, from nothing but a pipe getting narrower.

Check it as a height of water, since that is what a manometer across the throat would actually show: $\Delta p/(\rho g) = 16875/(1000 \times 9.8) = 1.72$ m. Check the flow rate too, which continuity says must match at both ends: $Q = A_1v_1 = 4.24\times10^{-3}$ m$^3$/s and $A_2v_2 = 7.069\times10^{-4} \times 6.00 = 4.24\times10^{-3}$ m$^3$/s, about $4.24$ litres per second. The energy bookkeeping closes as well - kinetic energy density rose by $16875$ J/m$^3$ and the height term did not move, so the pressure had to fall by exactly that much and did.

This is how a Venturi meter works backwards: measure $\Delta p$, know the two areas, and solve for the flow rate without putting anything in the stream.

What it does not explain

The viscosity hypothesis is the one that fails silently, because nothing in the algebra complains. Attach a garden hose to a tap at $300$ kPa gauge pressure and apply Bernoulli between the tap and the open end, both at the same height. The pressure difference converts entirely to speed:

$$v = \sqrt{\frac{2\Delta p}{\rho}} = \sqrt{\frac{2 \times 3.00\times10^{5}}{1000}} = \sqrt{600} = 24.5\ \text{m/s}$$

Through a $15$ mm bore that is $A = 1.77\times10^{-4}$ m$^2$ and $Q = 4.33\times10^{-3}$ m$^3$/s, which is $260$ litres a minute. A hose would fill a ten-litre bucket in $2.3$ seconds. No hose does this. Most of that $300$ kPa is spent dragging water along twenty metres of rubber, and viscous loss is precisely what hypothesis three assumed away. Bernoulli is for short runs, sudden constrictions and free jets, where the fluid has no time to lose much to friction. Over a long narrow pipe you want Poiseuille’s law instead.

Which brings us to aeroplane wings. The story that gets told is that the air travelling over the curved upper surface has further to go and must arrive at the trailing edge at the same time as the air underneath, so it goes faster, so by Bernoulli its pressure is lower, so the wing is pushed up. The Bernoulli step in that chain is fine. The step before it is not. Nothing requires two air parcels split at the leading edge to meet again at the back, and measurements of real aerofoils find the upper parcel arrives first, by a wide margin. The equal-transit-time premise is simply false, so the argument gets the right conclusion from a wrong reason.

What Bernoulli genuinely tells you is that where the flow is faster, the pressure is lower. It does not tell you why the flow over the wing is faster, and that is the part the explanation needs. The honest account involves the circulation the wing establishes around itself and the resulting downward deflection of a large mass of air, which by Newton’s third law pushes the wing up. A symmetric wing at an angle of attack flies perfectly well, and so does an aircraft inverted, neither of which the curved-path story accommodates.

Every hypothesis in the list is a place a problem can be set. The Bernoulli practice set pairs the equation with continuity the way real pipe problems do, and the simulations let you change an area ratio and watch the pressure drop follow the square.

References