Viscometer Question - calculating viscosity?

  • #1

Homework Statement


The velocity distribution of a fluid in two 0.3m long rotating cylinders is given by u=0.4/r-1000r (m/s). The diameter of the cylinders are 2cm and 4cm. If the inner cylinder rotates at 500rpm the power on the inner cylinder is measured to be 1Watt. Find the viscosity of the fluid?

Homework Equations



P=τ . ω
Torque = μ x ((ω . R1)/(R1-R0))

The Attempt at a Solution


So far my only attempt has been to converting the rpm and power output to find a torque value and then use this to find a viscosity, however, this brings me to a negative value?

torque = 1/52.36 = 0.0191

viscosity = (0.0191)(1-2)/1 = -0.0191

I also understand that the velocity distribution is not linear, however, do not understand how this works into finding the answer. Any help would be great! Thanks!
 

Answers and Replies

  • #2
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4,672
This is a very confusing problem statement. What does u represent? Is it an axial flow? What are the units of r in your equation for u? Your equation for the torque is incorrect. The equation you give for torque is really an approximation to the shear stress. You are correct in saying that the circumferential velocity distribution is not linear in the radius. To solve this problem, you first need to solve for the circumferential velocity distribution.
 
  • #3
Yes, that's exactly what I thought. They haven't given any further information about the equation given only that it is for velocity distribution. However, from some research I believe that "u" is the tangential velocity and "r" is the location of the fluid within the gap between the two cylinders. What would my next step be?
 
  • #4
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4,672
Yes, that's exactly what I thought. They haven't given any further information about the equation given only that it is for velocity distribution. However, from some research I believe that "u" is the tangential velocity and "r" is the location of the fluid within the gap between the two cylinders. What would my next step be?

If that is the case, then it doesn't seem consistent with the rotational speed of the inner cylinder of 500 rpm = 500 x 2π/60 radians/sec. If r is in meters, then I get zero velocity at the inner cylinder, and - 30 m/s at the outer cylinder (from the velocity equation). The actual velocity at the inner cylinder should be the radians/sec x 0.02 m. The velocity at the outer cylinder should be zero. The general form of the velocity equation seems correct, but the constants in the equation don't seem right. What do you think?

Chet
 
  • #5
Yes, I agree.. The values given out by the velocity distribution do not seem to match with the specific situation of this question. The equation suggests that the distribution would be of parabolic nature with the velocity getting higher as it gets closer to the inner cylinder? Officially confused.
 
  • #6
21,158
4,672
Yes, I agree.. The values given out by the velocity distribution do not seem to match with the specific situation of this question. The equation suggests that the distribution would be of parabolic nature with the velocity getting higher as it gets closer to the inner cylinder? Officially confused.

If you want, we can derive the correct equation for the velocity that is consistent with the stated boundary conditions. From the solution to the navier stokes equations, the form of the circumferential velocity as a function of radial position is going to be:
[tex]u=C_1r+\frac{C_2}{r}[/tex]
Use the stated boundary conditions to determine the constants C1 and C2.

The shear rate γ as a function of radial position is, for this situation, given by:
[tex]γ=r\frac{d\left(\frac{u}{r}\right)}{dr}[/tex]
and the shear stress is given by:

σ=ηγ

Once you know the shear stress at the inner cylinder, you can get the torque on the cylinder.

Chet
 
  • #7
I have talked to my lecturer and he say's we shouldn't be changing the distribution formula.. Therefore, I was wondering if there is any way of doing this without using this equation?
 
  • #8
21,158
4,672
I have talked to my lecturer and he say's we shouldn't be changing the distribution formula.. Therefore, I was wondering if there is any way of doing this without using this equation?
Sure, by solving for the correct velocity distribution that satisfies the stated boundary conditions. This would be done starting with the circumferential component of the Navier-Stokes equations. This leads to a simple ODE for the velocity distribution. If you want to see the whole thing worked out in detail, see Transport Phenomena by Bird, Stewart, and Lightfoot, Example 3.6-3. See Eqn. 3.6-32. Also see Eqn. 3.6-31 for the torque in the case in which the outer cylinder is rotating, and the inner cylinder is fixed.

Chet
 

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