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Topic: Couette Viscometer Problem  (Read 9219 times)

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Offline EngineerMom

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Couette Viscometer Problem
« on: December 02, 2014, 11:22:55 PM »
I have to complete this problem to revalidate a class I took 6 years ago.  I've gone through my books, notes, old tests - and it's not coming back to me as fast as I would like!

Can someone please help me with the following problem:

1.   Consider a Newtonian fluid of constant density flowing in the gap between two long vertical concentric cylinders when the outer one (of radius R2) is stationary (at r = R2, vθ = 0) and the inner one (of radius R1) is turning at a constant angular velocity ω (at r = R1, vθ = ωR1). Assume that the flow is steady, that the fluid moves only in the θ direction (vr = 0 and vz = 0), that derivates of velocity and pressure with respect to q are zero (due to symmetry), and that end effects are negligible so that vθ does not vary with z.
 

(a)   In the q component of the equation of motion, shown below, cross out the terms that are zero, explaining under each crossed out term why it is zero. Then rewrite the equation with only the remaining terms. Note that since vq is a function of r only, partial derivatives of vq with respect to r can be changed to ordinary derivatives.



I know that anything with vr and vz is going to be zero.  I'm stuck after that part.

(b)   Show that the solution to the equation remaining in part (a) for vθ, the velocity in the θ direction in cylindrical coordinates, as a function of r (the radial position) is given by


 
where C1 and C2 are constants.

(c)   Use the boundary conditions (at r = R1, vθ = ωR1; at r = R2, vθ = 0) to find the constants C1 and C2 and the final expression for vq as a function of r, R1, R2, and ω.


Any help would be greatly, greatly appreciated!  This is the last thing holding me up from being cleared for graduation!!

Offline curiouscat

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Re: Couette Viscometer Problem
« Reply #1 on: December 03, 2014, 12:14:49 AM »
When you write q do you mean θ?

Offline curiouscat

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Re: Couette Viscometer Problem
« Reply #2 on: December 03, 2014, 12:16:15 AM »
Quote
I know that anything with vr and vz is going to be zero.  I'm stuck after that part.

What about derivatives wrt. θ? Those go to zero too.  No?

Also, part (c) seems straightforward. Please substitute & show what two equations you get.

Offline EngineerMom

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Re: Couette Viscometer Problem
« Reply #3 on: December 03, 2014, 07:00:50 PM »
When you write q do you mean θ?

Yes, sorry.  That was a typo on my part!

Offline EngineerMom

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Re: Couette Viscometer Problem
« Reply #4 on: December 03, 2014, 07:12:14 PM »
Quote
I know that anything with vr and vz is going to be zero.  I'm stuck after that part.

What about derivatives wrt. θ? Those go to zero too.  No?

Also, part (c) seems straightforward. Please substitute & show what two equations you get.

OK - so if the derivative of velocity wrt θ is zero, how can you solve for this derivative (as asked in part B) if there are none left in the equation?  Wouldn't they all have gone to zero?

Offline EngineerMom

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Re: Couette Viscometer Problem
« Reply #5 on: December 03, 2014, 07:53:32 PM »
OK - does this look correct for Part A?


Offline curiouscat

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Re: Couette Viscometer Problem
« Reply #6 on: December 03, 2014, 09:52:20 PM »
They say flow is steady. So time derivative can also go.

Also they say no end effects. So z derivative can also go away.

Offline EngineerMom

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Re: Couette Viscometer Problem
« Reply #7 on: December 03, 2014, 10:01:02 PM »
They say flow is steady. So time derivative can also go.

Also they say no end effects. So z derivative can also go away.

So I am left with the following:



Since the value for gravity is given wrt θ, can I still cancel it out?  I know for horizontal flow you would cancel it out, but this is vertical flow. 

Offline curiouscat

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Re: Couette Viscometer Problem
« Reply #8 on: December 03, 2014, 10:17:21 PM »

Since the value for gravity is given wrt θ, can I still cancel it out?  I know for horizontal flow you would cancel it out, but this is vertical flow.

I think so, yes. Cancel it. So long as the axis of rotation is vertical.

How is this vertical flow? Is at all this is tangential / angular / rotational flow.

Offline EngineerMom

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Re: Couette Viscometer Problem
« Reply #9 on: December 03, 2014, 10:20:21 PM »

Since the value for gravity is given wrt θ, can I still cancel it out?  I know for horizontal flow you would cancel it out, but this is vertical flow.

I think so, yes. Cancel it. So long as the axis of rotation is vertical.

How is this vertical flow? Is at all this is tangential / angular / rotational flow.

Sorry, I meant to say rotational, not vertical.  OK - so the right side of the equation is zero, since everything on that side canceled out.  So now I integrate the equation and I should come up with the equation in B, correct?

Offline curiouscat

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Re: Couette Viscometer Problem
« Reply #10 on: December 03, 2014, 11:01:04 PM »
Sorry, I meant to say rotational, not vertical.  OK - so the right side of the equation is zero, since everything on that side canceled out.  So now I integrate the equation and I should come up with the equation in B, correct?

I think so, yes.

Offline EngineerMom

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Re: Couette Viscometer Problem
« Reply #11 on: December 03, 2014, 11:27:23 PM »
I can't for the life of me figure out Part C.  All my examples I'm working off of are assuming an open ended boundary, whereas this one is closed on both ends.  Any help on how to start this?  (And thank you so much for all of your assistance!!)

Offline curiouscat

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Re: Couette Viscometer Problem
« Reply #12 on: December 04, 2014, 02:29:09 AM »
I can't for the life of me figure out Part C.  All my examples I'm working off of are assuming an open ended boundary, whereas this one is closed on both ends.  Any help on how to start this?  (And thank you so much for all of your assistance!!)

Just substitute the two boundary conditions & show what two equations you get.

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