Laminar Flow: Finding Viscosity of Oil

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
6 replies · 2K views
Panphobia
Messages
435
Reaction score
13

Homework Statement



Finding the viscosity of oil...

Homework Equations



η = b/(6∏r)
Fr = -bv

The Attempt at a Solution


The question only gives a radius, mass, density of oil, and terminal velocity. Is it possible to get the viscosity with the given information?
 
Physics news on Phys.org
Panphobia said:

Homework Statement



Finding the viscosity of oil...

Homework Equations



η = b/(6∏r)
Fr = -bv

The Attempt at a Solution


The question only gives a radius, mass, density of oil, and terminal velocity. Is it possible to get the viscosity with the given information?
Yes. This sounds like a falling ball viscometer test. It is often used to measure the viscosity of highly viscous fluids. You do a force balance on the ball, taking into account the buoyant force on the ball, the weight, and the drag force. The drag force of the fluid on the ball is given by the equations you wrote down. This is called Stokes' Law.
 
Hmmm we haven't learned this. I will have to look it up on my own.
 
Ok so I just saw that the velocity given was the terminal velocity so then Fr = mg, and you can figure out b and η pretty easily. What I do not understand is this, if you wanted to get the time it takes for this object to go from 0 to half the terminal velocity, how would you get the it if the acceleration of the object is not constant?
 
Panphobia said:
Ok so I just saw that the velocity given was the terminal velocity so then Fr = mg, and you can figure out b and η pretty easily. What I do not understand is this, if you wanted to get the time it takes for this object to go from 0 to half the terminal velocity, how would you get the it if the acceleration of the object is not constant?

[tex]ma = mg-6πrη v-\frac{mgρ_F}{ρ_B}[/tex]

The last term on the right is the buoyant force. ρF is the density of the fluid, and ρB is the density of the ball material.

Chet
 
but this has nothing to do with getting the time for the ball to get to half of terminal velocity right? the terminal velocity = 4 cm/s, so half = 2 cm/s, even if I got the acceleration of the ball at that instant, that wouldn't bring me any closer to finding the time, right? Can v(t) = -(mg*e^((-b/m)t))/b + mg/b give me the right time?
 
Panphobia said:
but this has nothing to do with getting the time for the ball to get to half of terminal velocity right? the terminal velocity = 4 cm/s, so half = 2 cm/s, even if I got the acceleration of the ball at that instant, that wouldn't bring me any closer to finding the time, right? Can v(t) = -(mg*e^((-b/m)t))/b + mg/b give me the right time?
That's the solution to the equation I gave you with the buoyant force neglected and v(0) = 0. You did know that, in the equation I gave you, you were supposed to substitute a = dv/dt, right?

Chet