Bernoulli's Equation

Energy per unit volume

Bernoulli's equation is the work–energy idea applied to a fluid. We use the ideal fluid model: the fluid is incompressible (constant density $\rho$), has no viscosity (no internal friction), and flows steadily (the flow pattern does not change in time). Along a flow path:

Each term has units of $\text{J/m}^3$, which is the same as $\text{Pa}$ (energy per unit volume):

• $P$ is the pressure (the work done by the surrounding fluid per unit volume).

• $\rho g y$ is the gravitational potential energy per unit volume.

• $\tfrac{1}{2}\rho v^2$ is the kinetic energy per unit volume (compare $K=\tfrac12 mv^2$ with $m=\rho V$).

Why does it work? When a small volume of fluid moves along the pipe, the pressure force from the fluid behind it does positive work and the pressure force from the fluid ahead does negative work. By the work–energy theorem, the net work equals the change in $K+U_g$. Dividing everything by the volume gives the equation above.

Fast flow, low pressure

Take a horizontal pipe, so $y_1 = y_2$ and the $\rho g y$ terms cancel:

If the speed goes up, the pressure must go down. Think of it as energy bookkeeping: the fluid gains kinetic energy, and that energy comes from the pressure difference pushing it along. A fluid speeds up when it moves from a region of higher pressure to a region of lower pressure.

Combine this with the continuity equation $A_1 v_1 = A_2 v_2$: a narrow section means faster flow, which means lower pressure.

Height changes

When the fluid also changes height, keep the full equation. A pipe that rises gains gravitational potential energy per unit volume, so something has to give: pressure, speed, or both. If the pipe has the same cross-section, the speed is the same (continuity), and the pressure alone drops:

This is exactly the hydrostatic result. Because the same equation works with absolute pressure or gauge pressure (the $P_0$ cancels if both points use the same one), you can use gauge pressure whenever both pressures are gauge.

Speed of water leaving a tank

A large open tank in Houston has a small hole a depth $h$ below the water surface. Apply Bernoulli between point 1 (the surface) and point 2 (the hole). Both are open to the atmosphere, so $P_1 = P_2 = P_0$. The surface is so wide that its downward speed is negligible, $v_1 \approx 0$. Take $y_2 = 0$ and $y_1 = h$:

This is Torricelli's law. It is the speed of an object falling freely through height $h$: the potential energy per unit volume at the top turns completely into kinetic energy per unit volume. Notice that $\rho$ cancels and the hole's size does not appear.

Combining continuity and Bernoulli

Harder problems use both equations. The standard recipe:

1. Pick two points on the flow path and list $P$, $y$, $v$ at each.

2. Use continuity, $A_1v_1 = A_2v_2$ (or $Q = Av$), to find the speeds.

3. Use Bernoulli to find the unknown pressure.

4. Check that the answer makes sense (faster means lower pressure, higher means lower pressure).

Back to the start: a common explanation for the shower curtain is that the falling water drags the air inside the shower into motion. By Bernoulli's equation, this faster-moving air has lower pressure than the still air outside the curtain, so the higher outside pressure pushes the curtain inward. Physicists still debate the full story, since convection from the warm water and the Coanda effect (a moving stream clinging to a surface) also play a role, so treat this as a likely explanation, not the only one. A hurricane roof lifts because the fast wind above it lowers the pressure there.

Summary