# Concerning drag on a flow past a cylinder

I understand why the drag is high to begin with (point 2), when the boundary layer separates and the recirculations appear, the pressure behind the cylinder (where the wakes are located) will drop, creating a drag. This is similar to what happens when vortices appear at the end of an airplane wing, the result is a drag force.

However, I am wondering about the following:
1) Why does the drag decrease as the Re goes up?
2) Why does the drag coefficient take a dive at point 5
3) Why does it rise after point 5?

The text doesn't really explain why the drag coefficient is increasing or decreasing.

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I can't give you an exact answer for this geometry, but I can give you some general principles. There are really 2 contributions to the drag force. One comes from the inertia of the fluid (the object actually collides with molecules) and the other comes from the viscosity. As Re increases, the size of the viscosity, relative to the other parameters, is decreasing. This means that the viscosity term in Navier-Stokes equation is relatively less important. Since the viscosity plays a smaller role, we would expect the contribution to drag due to viscosity to decrease. –  Kevin Driscoll Apr 11 '13 at 4:01
As for point 5, I believe that around Re of $10^5$ the effects of compressibility of the fluid become important. I can't explain exactly what is going on, but I think this is why there is a qualitative change. –  Kevin Driscoll Apr 11 '13 at 4:05
This effect is not due to compressibility. I could achieve $Re = \frac{Ud}{\nu} = 10^5$ with a very low freestream velocity and very large diameter. The large dive in the drag coefficient is called the drag crisis. This happens when the boundary layer over the cylinder transitions from laminar to turbulent. Turbulent boundary layers are more resistant to separation, so the flow stays attached longer resulting in decreased form drag. As the Reynolds number increases further the higher skin friction from the turbulent boundary layer increases the drag. –  OSE Apr 11 '13 at 15:19
Hmmm... doesn't really make intuitive sense to me that the form/pressure drag decreases as a result of a more resistant boundary layer as in the turbulent case. Since the boundary layer now hugs a larger part of the cylinder, and all that fluid that hugs that cylinder is pushing on it (causing a drag), how can the drag decrease? –  l3win Apr 11 '13 at 15:59
The pressure in the wake region behind the cylinder is very low. The pressure on the front of the cylinder is near the stagnation pressure which is much higher. In an approximate sense, the pressure drag is $D = A(P_F-P_B)$, where $A$ is the frontal area of the cylinder, $P_F$ is the pressure on the front of the cylinder, and $P_B$ is the pressure on the back of the cylinder. To reduce drag it is desirable to reduce the width of this wake region---turbulent boundary layers do an excellent job at this. –  OSE Apr 11 '13 at 16:20

1. The drag coefficient decreases at low Reynolds numbers because the flow is in a regime called creeping flow. That is, the inertial forces are negligible. In this region, the Reynolds number isn't all that great of a parameter for the drag coefficient because the governing equations are independent of the Reynolds number. In spite of this, it is possible to analytically find the drag coefficient at very low Reynolds number to be $C_d \propto \frac{1}{Re}$. Once the drag coefficient levels off, the flow is now in the laminar region with significant inertial effects.