- #1

axim

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I have a question about aerodynamic drag. It sounds simple but when trying to understand why the relative velocity can be used in calculations I have some trouble. The formula is 0.5*rho*u

^{2}*cd*A where u is the relative velocity between the object and the fluid. The cd value depends on the shape. For sake of simplicity let it be 1 in the example.

For an object object moving trough a liquid (air) which is not moving, there is often a simplified explanation in textbooks.

If the object moves at a speed v1, it displaces air in front of it. In the time t1 a specific mass of air is pushed away. The mass can be easily calculated by v1*t1*A*density. v1*t1 is the distance passed, A the area and rho the density, so I know the displaced volume and with that the mass.

So for velocity v1 the air is accelerated and gains the kinetic energy of 0.5*m1*v1

^{2}. Of course this is simplified because the flow is more complicated.

Now let's say I triple the speed. Then the object will travel through 3 times the mass and it will be accelerated to 3 times the speed. According to the drag force formula I need a force 9 times higher due to the u

^{2}. This is also understandable and explainable with this model: I will accelerate 3*m1 to (3*v1)

^{2}so I have 27 times the energy of 0.5*m1*v1

^{2}: (0.5*3*m1*(3*v1)

^{2}) but I also have traveled 3 times the distance, so I can see the force required is 9 times higher (as expected for triple speed) and force*distance went into the kinetic energy of the air (and will finally end up as friction after later the movement has calmed down).

Everything fine so far. Now I have trouble to imagine how and where the energy is going when using a relative velocity.

Instead of the object moving at 3*v1, let's now say the object is moving at 2*v1 and there is a headwind with the speed -1*v1.

As the relative speed is used in the formula, the object will experience the same drag force: 0.5*rho*(2*v1-(-v1))

^{2}*A=0.5*rho*(3*v1)

^{2}*A.

Now, I try to find out how and where the energy "goes". And this is where I have trouble with this simplified model.

The object is moving at 2*v1, the air is approaching it with -1*v1. This means, the object will again displace three mass-units (3*m1) of air: It will move less distance 2*v1*t1, but meanwhile the air will travel -1*v1*t1 in direction of the object, so it was in contact with the same amount (3*m1) of air during t1. I created a small drawing so my explanation is not too confusing.

Then I tried to calculate it worst case: Every collision is inelastic, the air slows down the object. Then the object has to accelerate itself and the air to the speed of 2*v1 again. I always end up with the result that less work is required to travel the same distance and push away the air compared to the first example (I also considered that with v=2*v1 the object travels less distance during t1). If requested I can detail my calculations but I think it leads in the wrong direction.So this simplified Newtonian approach can't explain that I can use the relative speed in the formula.

I did also try to get the basics from transport phenomena lectures or explanations about the Navier Stokes equations but I end up with very complex mathematical derivations (I don't doubt these are correct), but still I miss the rough idea why I can just use the relative speed. The formulas with stress tensors just get too complex to grasp how the movement is going on. The transport phenomena explanations also usually touch the molecular level.

Is there any simplified model that can picture why the relative speed can be used if the object and fluid are moving?

Especially how the spent work relates to the kinetic energy changes and friction compared in the two examples: Relative speed is the same, but in one case only the object is moving, in the other case the object and fluid are moving in opposite direction.

Thanks in advance