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Transport phenomena -- Direction of shear stress

  1. May 29, 2015 #1
    Greetings, PF! I have some questions regarding the problem I attached below. It is some sort of Couette flow variation. It's not homework, I'm just learning the basics of TP on my own. I fully solved the problem with the Navier-Stokes and continuity equations, using some assumptions in order to simplify the models. Here are the solutions
    1. [itex]v_x = \left( \frac{U_1 - U_2}{b} \right)y + U_2[/itex]
    2. [itex]Q = \int_0^W \int_0^b v_x \ dy \ dz = Wb \left( \left( \frac{U_1 - U_2}{2} \right) + U_2 \right)[/itex], where W is the width of the flow in the z direction
    3. [itex]\tau_{yx} = - \mu \frac{d v_x}{dy} = -\mu\left( \frac{U_1 - U_2}{b} \right)[/itex]
    Now, what I'm trying to understand is the meaning of the sign of the shear stress, and how it affects its direction. The velocity and stress profiles in the diagram were added by me. According to my understanding so far, and to the math also, if the velocity profile of the fluid is linear, then the shear stress is constant; now, [itex]\left( \frac{U_1 - U_2}{b} \right)[/itex] is always positive, so the shear stress component is always oriented in the negative x direction, even when the velocity of the fluid is oriented in the positive x direction. What does this result actually mean? How is this explained physically?

    Thanks in advance for any input!
     

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    Last edited: May 30, 2015
  2. jcsd
  3. May 31, 2015 #2
    To do this without making any mistakes, you need to learn a little about how to work with tensors. The stress tensor has a bi-directional nature. You are going to have to learn how to express 2nd order tensors (like the stress tensor) in terms of so-called dyadic tensor notation, as sums of components times juxtaposed pairs of unit vectors. After that, you are going to need to learn about how to apply the Cauchy Stress Relation, which enables you to use the stress tensor to confidently determine the stress on a surface of specified orientation.

    Step 1 is to learn about dyadic representation of a tensor. To learn this, see Section A.3 (Appendix A) of BSL. Equation A.3-7 shows you the dyadic representation, and Eqn. A.3-18 shows you how to take the dot product of a tensor with a vector, to yield a second vector. Learning the dot product operation is necessary to know in preparation for learning how to apply the Cauchy Stress Relation. Look over these equations, and see if you have any questions. Then I'll tell you how to apply the Cauchy Stress Relationship.

    Chet
     
  4. May 31, 2015 #3
    All right. The stress tensor is a second-order tensor, which is the product of the scalar [itex]\mu[/itex] and the dyadic product [itex]\nabla \mathbf{v}[/itex], which is also a second-order tensor. The negative sign in the expression [itex]\boldsymbol{\tau} = - \mu \nabla \mathbf{v}[/itex] just indicates that momentum flows from regions of high velocity to regions of low velocity. It can be represented by a jacobian matrix.

    I also got a grasp of the following operations
    [itex](\boldsymbol{\sigma} : \boldsymbol{\tau})[/itex] yields a scalar
    [itex]\{ \boldsymbol{\sigma} \cdot \boldsymbol{\tau} \}[/itex] yields a second-order tensor
    [itex][\boldsymbol{\tau} \cdot \mathbf{v}][/itex] yields a vector
    Where [itex]\boldsymbol{\sigma}[/itex] and [itex]\boldsymbol{\tau}[/itex] are second-order tensors and [itex]\mathbf{v}[/itex] is a vector. I understand how to work with these three operations in terms of their capital-sigma notation.

    [itex]\tau_{yx}[/itex] is a bi-directional vector component of the stress tensor [itex]\boldsymbol{\tau}[/itex]. It is the only significant component in this problem, because all the other velocity derivatives are zero.
     
    Last edited: May 31, 2015
  5. May 31, 2015 #4
    This equation is not correct. It should read:
    [tex]\boldsymbol{\tau} = - \mu (\nabla \mathbf{v}+ (\nabla \mathbf{v})^T)[/tex]
    where the superscript T denotes the transpose of a tensor. The transpose of a tensor is obtained by interchanging the two unit vectors of each term of the dyadic summation.
    This is not correct. Please write out the dyadic summation for the stress tensor [itex]\boldsymbol{\tau}[/itex] in your problem in terms of the velocity derivatives and the unit vectors ##\boldsymbol{i}_x## and ##\boldsymbol{i}_y##. You will find that there are two non-zero terms in the summation.

    After you get back with me, I will tell you about the Cauchy Stress Relationship and help you to apply it to your specific problem to get the stress exerted by the fluid on the upper boundary (with confidence about the sign).

    Chet
     
  6. Jun 1, 2015 #5
    Oops. You're right, I will re-read its derivation in BSL carefully.
    And you're right again. Using the equation you provided me with, here's the correct stress tensor for this problem
    [tex]\boldsymbol{\tau} = - \mu \left( \frac{dv_x}{dy} \mathbf{ji} + \frac{dv_x}{dy} \mathbf{ij} \right)[/tex]
     
  7. Jun 1, 2015 #6
    Nicely done. Now for the Cauchy Stress Relationship:

    If you have a small (fictitious) element of surface dA situated between two adjacent portions of a fluid, and you want to find the viscous force (vector) per unit area t exerted by the fluid on one side of the surface on the fluid on the other side of the surface, you draw a unit normal n from the "by" side of the surface to the "on" side of the surface and take the dot product of the viscous stress tensor τ with the unit normal:
    $$\vec{t}=\vec{τ}\centerdot \vec{n}$$
    This automatically gives you the viscous force (vector) per unit area ##\vec{t}##, and with the correct sign. This method also works for solid surfaces, solid materials (Hooke's law in 3D), and for the pressure portion of the overall stress tensor. Now, how would this apply to your problem?

    Chet
     
  8. Jun 1, 2015 #7
    Here we go
    [tex]\mathbf{t} = [\boldsymbol{\tau} \cdot \mathbf{n}] = -\mu \left( \frac{dv_x}{dy} [\mathbf{ij} \cdot \mathbf{j}] + \frac{dv_x}{dy} [\mathbf{ji} \cdot \mathbf{j}] \right)[/tex]
    [tex]\mathbf{t} = -\mu \left( \frac{dv_x}{dy} \mathbf{i} \delta_{jj} + \frac{dv_x}{dy} \mathbf{j} \delta_{ij} \right)[/tex]
    But δjj = 1 and δij = 0, therefore the viscous force exerted by the fluid on the upper plate is given by
    [tex]\mathbf{t} = -\mu \frac{dv_x}{dy} \mathbf{i} = -\mu \left( \frac{U_1 - U_2}{b} \right) \mathbf{i}[/tex]
    I assume I performed the dot product correctly, as I arrived at the same result than before. However, now I know how to obtain viscous forces in surfaces which I guess is a useful skill in transport phenomena/fluid mechanics, and I've never seen before this calculation in BSL. I'm also now more confident about why the shear stress is oriented in that direction.

    I also just noticed, given the correct stress tensor expression and the Cauchy stress relationship, that while the layers of fluid exert a viscous force in the negative x direction in the above layer, they also exert a positive viscous force in the layer below.

    Do you have any other basic problems, besides the ones included in BSL, where I can get useful insights while solving them?
    Thanks so far for the input, Chet!
     
    Last edited: Jun 1, 2015
  9. Jun 2, 2015 #8
    Very nicely done!!
    Excellent!!!!
    Well, here are some additional ideas on how you can get some practice using the Cauchy Stress Relationship:

    1. The contribution of the pressure to the overall stress tensor is given by: p(ii + jj + kk). Show that, whenever you dot this with a unit normal n to any surface (real or fictitious), you get the vector pn, irrespective of the direction of n.

    2. Use the Cauchy Stress relationship to unambiguously set up the shell force balance for viscous flow in a tube.

    Chet
     
  10. Jun 2, 2015 #9
    Alright, this is an easy one. Letting
    [tex]\boldsymbol{\sigma} = p(\mathbf{ii} + \mathbf{jj} + \mathbf{kk})[/tex]
    [tex]\mathbf{n} = \mathbf{i} + \mathbf{j} + \mathbf{k}[/tex]
    We take the following dot product
    [tex]\mathbf{p} = [\boldsymbol{\sigma} \cdot \mathbf{n}] = p(\mathbf{i} [\mathbf{i} \cdot \mathbf{i}] + \mathbf{j} [\mathbf{j} \cdot \mathbf{j}] + \mathbf{k} [\mathbf{k} \cdot \mathbf{k}])[/tex]
    [tex]\mathbf{p} = p(\mathbf{i} \delta_{ii} + \mathbf{j} \delta_{jj} + \mathbf{k} \delta_{kk}) = p(\mathbf{i} + \mathbf{j} + \mathbf{k})[/tex]
    [tex]\mathbf{p} = p \mathbf{n}[/tex]
    The meaning of this result is that the stress being exerted due to pressure is always normal to the surface it is being exerted on. That's why it is called normal stress, right?
    This one seems fun. I will take my time with this one as I haven't set up a shell momentum balance "manually" before. I'll keep posting my progress.
     
  11. Jun 2, 2015 #10
    This is not a unit vector. You need to pre-multiply each unit vector by its component in the arbitrary direction.
    Yes. Well said. This is one of the things that an enormous number of members posting on Physics Forums struggle with.
    Excellent.
     
  12. Jun 2, 2015 #11
    Right, I didn't realize that wasn't a unit vector. For any arbitrary surface s, can I use [itex]\mathbf{n} = \frac{\nabla s}{\| \nabla s \|}[/itex] to define its unit normal vector?
     
  13. Jun 3, 2015 #12
    Usually, the unit normal is obvious, as determined by the orientation of an actual solid surface, or for simple straight streamlines. The equation you wrote (assuming s is the stream function) only applies to 2D flow. Normals to more intricate surfaces are more complicated to establish.

    Chet
     
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