The Physics Philes, lesson 91: Go with the Flow
Last week we learned about the continuity equation in fluid mechanics. To refresh your memory, that equation says that the speed of the fluid flow can change along the path of the fluid. The pressure can vary, as well, depending on the height of the fluid and the speed of the flow. These are important relationships to know, because it “flows” (HA! See what I did there?) nicely into Bernoulli’s equation.
Bernoulli’s equation is a relationship between pressure, flow speed, and height of the flow of an ideal fluid. We know that the pressure of a fluid is dependent on the fluid’s speed from the continuity equation. We also know that when the cross sectional area of the flow tube changes, the speed of the fluid must change and thus have an acceleration. If the tube is horizontal, the force that causes the acceleration must come from the surrounding fluid, which in turn means that the pressure must be different in different regions of the fluid.
But we knew all of this already. What about this Bernoulli’s equation? What’s that about?
We can derive Bernoulli’s equation by applying the work-energy theorem to the fluid flowing through a cross section of a flow tube. Like last week, let’s think about a flow tube with fluid flowing through two different cross sections. It might look something like this, with the red and blue lines representing the cross sections.
The fluid at each cross section has a certain velocity. The distance the fluid moves in a tiny time interval is equal to the velocity times that tiny time interval. If the fluid is incompressible, we know from the continuity equation that the volume of the fluid passing through any cross section during a very short time interval is the same.
Again, that’s old news. But now let’s figure out the work done on the fluid during that tiny time interval. If we assume that there is no viscosity (as we do with ideal fluids), that means that the only force (besides gravity) that does work on the fluid is due to the pressure of the surrounding fluid. So the force at each cross section is equal to the cross sectional area times the pressure at the cross section. That means that the total work done during the displacement is
The second term is negative because the force opposes displacement.
Since the work done is due to nongravitational forces, it equals the change in total mechanical energy. (Which, if you’ll remember, is the sum of kinetic and gravitational potential energy.) At the beginning of our tiny time interval, the fluid at the cross section has a volume equal to the area times the distance the fluid travels, a mass equal to the density (ρ) times the volume, and kinetic energy equal to half of the product of the density, volume, and the square of the velocity. So the net change in kinetic energy (denoted dK) during our tiny time interval is
We can also find a change in gravitational potential energy (denoted dU). At the beginning of our tiny time interval, the potential energy for the mass of the fluid is the product of the mass, gravity, and vertical distance y, which is equal to the product of the density, volume, gravity, and vertical distance y. So our change in gravitational potential energy is
When we combine the change in kinetic energy with the change in gravitational potential energy in the energy equation dW = dK + dU, and fiddle around with it a bit, we get
This is Bernoulli’s equation. What it says in English is that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the changes in kinetic and potential energy per unit volume. When the fluid isn’t moving, the equation reduces down to an equation of pressure in a fluid of uniform density. Cool, right?
A word of caution: This only applies to fluids that are incompressible, in a steady flow, and with no viscosity. Make sure you don’t use the equation when it doesn’t apply!
Next week is the last post in fluid mechanics. We’ll learn a bit more about viscosity and turbulence.
Featured image credit: Michael Schack via Flickr