# How to estimate a drag of flow reversing cylinder? I am assuming a turbulent flow and a full reversal of the flow (in the picture the arrows are at non-straight angle but please consider the skew as negligible).

Assuming we are talking about regular earth atmosphere and small velocities, can I estimate the drag using Newton's law and simply say that:

$$Drag \approx 2 V^2A\rho\qquad?$$ V is air velocity, A is cross section of the cylinder, $\rho$ is air density.

If this approximation is valid under given assumptions, what would be the limiting velocity of its validity?

• I answered, with the assumption that this is something that you stick to the end of a pipe. If you really mean 'drag' as in an object that you put in a windy place, then the answer would be different. – Han-Kwang Nienhuys May 31 '16 at 19:49
• How different would it be? – user68634 May 31 '16 at 19:52
• If the object is placed in an open place with a flow past it, then the arrows don't represent the flow. – Joce Jun 1 '16 at 15:12

Pressure drop in duct flow

For fluid flow through a duct, we normally talk about pressure drop, not drag. Pressure drop is usually expressed as $\Delta p=\xi\rho V^2/2$, where $\xi$ is the minor loss coefficient (minor as opposed to major, which refers to pressure drop over a long straight pipe). In a case like this, you will have an additional loss due the Bernoulli pressure drop $\rho (V_{\mathrm{out}}^2-V_{\mathrm{in}}^2)/2$.

The minor loss coefficient in a case like this is highly dependent on the exact shape of the place where the fluid makes the turn and to some extent the Reynolds number of your flow, but likely, you will find $0.5<\xi<2$, if you evaluate $V$ at the point of the highest flow velocity. Tabulated values for many types of flow restrictions are in Idelchik, Handbook of Hydraulic Resistance (warning: 21 MB download). I'm not sure whether this particular configuration is there, but you could look in section 6 (PDF page 195) or section 9 (PDF page 359).

Drag on an object in a free stream

Apparently, I misunderstood the question when I wrote the answer above. If you place this object out in the wind, the drag force would be $$F = \frac12 C_d A \rho V^2,$$ where $C_d$ is the drag coefficient. For this object, the drag coefficient would probably about the same as for a cylinder. You will never get more than the free-stream velocity $V$ at the throat of the funnel; since the throat is much smaller than the cross section of the cylinder, the mass flow rate through the funnel will be very small compared tho $\rho V A$ with $A$ the cross section of the mouth of the funnel.

Another reason why you should expect that flow rate to be small is that the reversed stream, if it exists, must push against the free-stream flow. Think of it: if you hold a U-shaped pipe in a downward stream, you won't get a spontaneous flow from the left leg of the U to the right leg.

For a cylinder, Wikipedia lists a drag coefficient $C_d=0.82$. So, that would be a bit over a factor 2 less than what you assumed at first.

• Does the fact that I am reversing the flow has no influence over the drag? Won't the momentum of the flow entering the cylinder be reversed? – user68634 May 31 '16 at 20:27
• Answer updated. – Han-Kwang Nienhuys May 31 '16 at 20:52