# Calculating the drag force provided by a laminar flow [duplicate]

If I drop a ball into viscous fluid (e.g. oil), by calculation I know that the fluid flow is laminar. So which equation do I need to calculate the drag force provided by fluid?

1. The drag equation, $$F=\frac12\rho CAv^2$$ where $ρ$ is the density of fluid, $C$ is drag coefficient, $A$ is cross-sectional area of ball, and $v$ is the velocity of ball.

2. Or the Stoke's law, which is $$F=6\pi\eta rv$$ where $\eta$ is the viscosity of fluid, $r$ is radius of ball, and $v$ is velocity of ball.

## marked as duplicate by John Rennie, Kyle Kanos, HDE 226868, Qmechanic♦Jul 29 '15 at 19:54

Depends on the Reynolds number. Stokes can be used for purely laminar flow. Complete answer to be found here.

I would use the traditional drag equation:

F = 0.5 * p * v^2 * Cd * A

Reasoning: Stokes' law is for very small Reynolds number flows only (much less that 1), and I don't believe the more common drag model has any such restriction. The problem is that you will need to know the drag coefficient of the ball in question to use this force model.

Sources: I have taken two university-level fluids courses, and these wikipedia articles:

• The drag model does have such restrictions. It will actually under-estimate the drag for low Reynolds numbers, particularly if it is much less than 1. If you're calculating some transit time for some very small thing, the drag model will give answers which are way off. – Alan Rominger Jul 29 '15 at 15:08
• @AlanRominger Yes, that makes sense. Thanks Alan. Does it also have a restriction for high Reynolds numbers? For example, supersonic flow? – hakzatchel Jul 29 '15 at 15:42
• Restrictions for high numbers? Not likely. Consider atmospheric reentry - projectiles will vaporize before the drag equation loses its validity. This is because the equation is valid for totally ballistic particle collisions. The drag coefficient will change, but that's always dependent on the specific flow pattern anyway. Even when the mean free path is larger than the projectile, it works. Just assume there is just a column of particles in front that bounce off in some constant angle, you get the v^2 form. That's different from the normal application, but it's also strangely the same... – Alan Rominger Jul 29 '15 at 19:54