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Hydrodynamic Drag in two Directions

Hydrodynamic Drag in two Directions

Hydrodynamic Drag in two Directions

(OP)
Hello everyone,

I have just started my engineering career in the offshore industry in Houston and this is my first post on this website. I have browsed here before looking for answers to engineering and CAD questions and have always been impressed with the level of knowledge on the site. I am happy to join you all as a fellow member.

I have a pretty simple theoretical question that is causing me a headache.

If two perpendicular (separated by pi/2) currents with velocity, v, push a sphere, what is the resultant force on the sphere?

I see two ways to solve this problem that unfortunately do not yield the same results.

1st Method: Resolve the velocities and calculate the drag.

vnet = sqrt(v^2 + v^2) = v*sqrt(2)

Drag = 1/2*Cd*A*rho*(v*sqrt(2))^2 = Cd*A*rho*v^2

2nd Method: Resolve the constituent forces.

Net Drag = sqrt( (1/2*Cd*A*rho*v^2)^2 + (1/2*Cd*A*rho*v^2)^2 ) = (Cd*A*rho*v^2)/sqrt(2)

Why do these answers not match?

If I can get a handle on this problem, some other project tasks will be easier for me.

Thanks in advance,    

RE: Hydrodynamic Drag in two Directions

Congradulations.

Roughly
In the first, you're multiplying the sqrt of (V^2+V^2)^1/2 by 1/2 Cd A ρ
In the second, you're multiplying by the sqrt of 1/2 Cd A ρ

Anyway, I don't think the sphere will ever see two Cds in two directions.  Using a Cd of 0.6 is only valid when the sphere is placed in a unidirectional flow only.  I believe, if your 2 current flow was possible, your 2 current flow would require a different Cd.  I think one Cd = 0.6 would be in the direction of the resultant current.  But you might want to post that question in the Aerodynamics Engineering Forum.

 

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RE: Hydrodynamic Drag in two Directions

Think about the ball for a minute, and let's suppose you have two fire hoses, squirting streams of water of diameter equal to the ball's diameter, and the two streams are impinging at 90 degrees seperation on the ball.

What happens about 1 second after the flow starts?  The ball moves away, and the two streams impinge.  Even before that happens, there will be flow reversal along a meridian line 45 deg. between the two streams, and not the smooth flow that a Cd drag calc. is based upon.

Going further, the impact of two streams does not resolve into a single stream at 45 deg., but more typically into a spray fan of droplets, spreading at some angle, with the plane of the fan at 45 deg. and perpendicular to the plane of the original streams.  Useless fact, unless you are designing spray bars or rocket engine injectors.

BigInch is right, you can't get two Cd's in two directions.  The closest I've gotten to a 2-D flow problem is flow past rotating circular discs (pumps, propellers, jet engine hubs...).  You have the tangential velocity from the disc rotation inducing drag, and the slowed fluid also undergoing centripetal forces causing radial flow (pumping).  Fun stuff.  Von Karman wrote a lovely NACA Tech Note on the subject.

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