Sanity check on falling steel ball in water

[Sherwood Botsford]

Homework Statement

What is the speed of a 5 mm steel ball falling through water at 10 C[/B]

Viscosity of water: 1.3059 * 10^-3 Pa*s (Various online sources)

Homework Equations

Laminar flow:

Vt = gd² (ρp - ρm)/18μ

Vt = terminal velocity
g = gravity (10m/s²)
d = diameter of ball
ρ_p = density of ball
ρ_m = density of medium
μ = viscosity of medium

Turbulent drag

F_d=12ρC_dAv²

F_d = drag
C_d = drag coefficient = 1
A = cross section area
v = velocity.

Force of gravity on ball.
F_g = mg

Mass of ball
m = ρ*4πr³/3
I used 8000kg/m³ for density.

At this point I'm ignoring buoyancy effects.

The Attempt at a Solution

I initially did this using the stokes law calculator here:

http://www.meracalculator.com/physics/fluid-mechanics/stokes-law.php

But this gives me an answer of 75 m/sec. Which seems to me to be absurd, even for laminar flow.

This equation is for stokes law laminar flow. Units error for viscosity?

So try it for turbulent flow.

Mass of ball = .0005 kg = .5 gm this seems reasonable.
Fg = .005 N

Equating the above two formulas F_g = F_d and solve for v

v = √(12ρC_dA/F_d)

Plugging in numbers I get 6.7 m/s While closer this still seems high. I would expect something on the order of a meter/s

I've been unable to find an online calculator to check on this[/B]

 

[mfb]

v = √(12ρC_dA/F_d)

That would mean a steel ball with a larger surface area and a smaller mass would fall faster. Is that plausible?
The square root doesn't have units of speed either.

 

[haruspex]

F_d=12ρC_dAv²

Define ρ here.

 

[Sherwood Botsford]

Crap. Comes from doing things in my head.

Cross section of ball = πd²/4
= 3.14 * 0.005²/4
= 20e-6 m²v² = mg/(12ρC_dA)

v=√(mg/(12ρC_dA))

=√(.005/12*1000kg/m³* 1* 20e-6 m²))

=0.14 m/sec

Algebra mistake. This looks more reasonable. I think.

 

[Sherwood Botsford]

Define ρ here.

Density of the medium being displaced by the falling sphere -- in this case water.

Yes?

 

[mfb]

That looks good.