- #1

Sherwood Botsford

- 91

- 22

## 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 = gd

^{2}(ρp - ρm)/18μ

Vt = terminal velocity

g = gravity (10m/s

^{2})

d = diameter of ball

ρ

_{p}= density of ball

ρ

_{m}= density of medium

μ = viscosity of medium

Turbulent drag

F

_{d}=12ρC

_{d}Av

^{2}

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*4πr

^{3}/3

I used 8000kg/m

^{3}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

_{d}A/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]