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Torque of rotating coaxial cylinders with gas between them.

  1. Dec 17, 2015 #1
    I was going through a worked example in book "Concepts in Thermal Physics" by S.J. Blundell and K.M.Blundell. The example talks about measuring viscosity of a gas between two coaxial cylinders.
    1. The problem statement, all variables and given/known data
    Two vertical coaxial cylinders. Outer cylinders is rotated by a motor at constant angular speed, while the inner one is suspended by a torsion wire. The torque is transmitted from rotating cylinder to suspended one via the gas. Find torque on the inner cylinder (i.e. couple on inner cylinder required to keep it from rotating) and show how can you measure viscosity using this set up.
    2. Relevant equations
    The momentum flux of x momentum in z direction is Πz = -η duz/dz
    Momentum flux is equal to - shear stress
    Force = stress * area
    Linear velocity us related to angular velocity as u(r) = r*ω(r)
    3. Solution
    To find the force on the on inner cylinder we just need to find the flux of momentum in radial direction to find shear stress. Then should be easy to find torque and so on and so forth. So authors solve the problem by first considering the velocity gradient. They say that that molecules of gas will be travelling in a circular motion and consider the velocity as u(r) = r*ω(r) . The do not actually find the velocity profile of the flow, they just consider the velocity gradient of the flow. So they find:
    du/dr = ω + r dω/dr (1)​
    But then the next part of their solution has me completely stuck. They completely ignore the first term. They state (referring to equation (1) ): "but the first term on the right hand side simply corresponds to the velocity gradient due to rigid rotation and does not contribute to the viscous shearing stress which thus is just ηrdω/dr".
    This does not make sense to me. ω is a function of r, I really struggle to understand the reasoning to ignore the first term. Shearing stress is related to velocity gradient. du/dr is the velocity gradient in radial direction. Why would you completely a term in it? Does anyone have a nice explanation? I found an article which considers a similar problem, but goes to find the actual velocity profile of the flow: http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid301.html
    There again, it just says:

    From the velocity profile, the momentum flux (shear stress) is determined as:s
    img19.gif
    I imagine the reason it uses such a weird term r d/dr(vθ/r) is because of the same logic of ignoring the "velocity gradient due to rigid rotation", but is there a "mathematical" explanation? Is my understanding of shear stress and momentum flux too simplistic? Can this be shown mathematically by properly considering the stress tensor?
    The rest of the example makes sense - they find force from stress. the torque from force, integrate between two cylinders and integrate between ω0 and 0.
     
  2. jcsd
  3. Dec 17, 2015 #2
    The book you are using did you a disservice by giving you this problem without your having had significant fluid mechanics background. I'm guessing that you haven't had much experience with tensor analysis and with deformational kinematics. When you are working in cylindrical coordinates like this, you need to be extra attentive to handling gradients in vector- and tensor quantities, because the unit vectors change direction, and this needs to be accounted for. There are two parts to this. 1. Expressing the differential equation of motion in cylindrical coordinates (involving the components of the stress tensor). 2. Determining the (a) velocity gradient tensor and, from that, (b) the rate of deformation tensor. The stress tensor is directly proportional to the rate of deformation tensor. The rate of deformation tensor is obtained from the velocity gradient tensor by summing the velocity gradient tensor with its transpose, and dividing by 2. This has the effect of subtracting out any local rigid body rotation, which does not contribute to shear rate. So this is what they were alluding to when they talked about subtracting the rigid body rotation.

    I don't know how to get you to the point where you can understand all this, because the development is lengthy. The only thing I can suggest is for you to consider the case in which the gap between the cylinders is very small compared to their radii. Under these circumstances, the curvature of the system can be neglected, and you can approximate the shear between the cylinders as just shear between two infinite parallel plates in Cartesian geometry. Then you can use your usual equations to get the shear rate and shear stress.

    I apologize for not being able to help you more.

    Chet
     
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