# Archived Explaining drag is proportional to speed squared with Bernouilli's Equation

1. Nov 17, 2013

### Jack.P

1. The problem statement, all variables and given/known data

How does Bernoulli's Equation for calculating the pressure in a moving fluid (as below) explain that drag increases proportionally to the square of the speed?

I have been dropping parachutes and recording the time to fall, so my question is in reference to free-fall motion.

2. Relevant equations

P1 + ρgy1 + ½ρv12 = P2 + ρgy2 + ½ρv12

3. The attempt at a solution

The third term is relevant because ½ρv12 = kinetic energy per unit volume, which is the kinetic or dynamic contribution to pressure. I assume the kinetic energy relates to drag.

Pressure as force per area is Drag= PA

Substituting P for Bernoulli's equation for the pressure in a moving fluid:
Drag=PA=(1/2 ρv^2 )A

Rearranging:

Drag = ½ρCAv^2

It now includes C, coefficient of drag.

I am unsure of a way to succinctly explain this, assuming it's correct! If you could help me to understand why Bernoulli's fluid equation is relevant to drag, I would be very appreciative!

Cheers

Last edited: Nov 17, 2013
2. Feb 5, 2016

### OrangeDog

Quite frankly, this is a bad question. The drag cannot be adequately explained using Bernoulli's equation because it assumes that energy is conserved along a streamline. With any body moving through a fluid this simply isn't true. You probably have not gotten this far in your fluids course, but there is something called "potential flow" which was used in modern times to do the first calculations on airplane wings. You cant calculate drag reliably using that method, which is based on Bernoulli's principle. However, you can use these equations to get a rough idea what the drag will be.

(P2-P1) + (1/2)*rho(V2^2-V1^2) = 0.

However, you can also use the conservation of mass and momentum (using the control volume method) to get a similar answer (this will help you understand your where your coefficient comes from):

Suppose you have an object in a wind tunnel (lets assume a square cross section) of height h (described by dimension y), with the object placed at h/2. You decide to measure the velocity profile before and after the object. A good assumption for a wind tunnel is that the velocity profile is uniform at the entrance to the test section. So, V1 = constant. When you measure the velocity behind the object, you'll see that the velocity should decrease as you get closer to the object and then return to normal as our measurement approaches the wall. So V2 = varies with the height, y.

You know mass is conserved so you know (I cant see symbols right now so if this looks bad please excuse me) ∫rhoV2 dA = ∫rhoV1 dA

Of course, analytically we have to make an assumption. Let's say as a first approximation that the velocity follows a cubic velocity profile behind the test object. Additionally, lets assume (ignoring the effects of the wall, if we don't this we will make the problem very difficult) that the velocity at y=0 and y=h equals the inlet velocity v1.

Since the velocity v2 is cubic we will have 4 unknowns, Ay^3 + By^2 + Cy^1 + D. We have three equations right now. What else can we assume? Well if momentum is conserved we can calculate that integral too. The equation will be ∫rhoV2 (V2⋅ n dA) = ∫rhoV1 (V1⋅ n dA)

Now you have 4 equations and 4 unknowns, you can solve the system. You can see from the final equation that the drag force is proportional to the difference in the square of the velocities. By assuming the velocity profile takes the form of a simple function, you can "guess" what it will look like by applying the conservation of mass, momentum, and energy. I picked a polynomial because it is the most simple to illustrate. Maybe the velocity profile is better approximated by a cosine curve? Maybe a higher order polynomial will be sufficient? Try some stuff out.

The point is: if you want to get a feeling for why that coefficient is there, do not view the velocity strictly as that, but rather a function over a line or an area. The purpose of the drag coefficient is to quantify changes in drag as the orientation of the body changes in time or space (or maybe the fluid itself changes!). This coefficient is almost always determined via experiment because aerodynamics gets pretty complicated in the real world. So your simple formula could have very easily been viewed using bernoullis as:

D = 1/2 rho A (ay^3 + by^2 +cy +d)^2
or
D = 1/2 rho A (a cos(by+c))^2
or
any function you think is cool.

The question is really silly for someone who is new to fluids. Im sorry your prof made you answer it. I hope this helped.