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Aerospace Difference between normal viscous stress and pressure

  1. Oct 17, 2016 #1
    Hello,

    I came to know that a normal component of the viscous stress can exist at a solid surface for unsteady, compressible flows(though even in that case, the normal component of the viscous stress is typically much smaller than the tangential component). I tried googling it myself but could not find any relevant information.
    To me, it seems that in calculating the aerodynamic force on an airfoil, the normal viscous stress contribution is completely ignored.

    Can someone explain the difference between this normal viscous stress and a pressure?
     
  2. jcsd
  3. Oct 18, 2016 #2

    boneh3ad

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    Perhaps the simplest answer is that pressure is entirely kinetic and exists whether the fluid is moving or not, let alone whether or not it is compressible. Viscous normal stresses are viscous in nature and only occur in flows in which there is dilatation.
     
  4. Oct 18, 2016 #3
    A lot of intro aerodynamics books seemed to suggest that the shear stress acts only tangential to the surface. Then, are we technically ignoring the contribution of the normal viscous stress in calculating the aerodynamic force on a body? Even when we have dilation of fluid particles, I believe there has to be a change in momentum of the fluid parcels through some mechanism for it to develop a resisting force. And if this physical term is named "normal" viscous stress, I suppose there has to be a chance in momentum in the perpendicular direction. Perhaps, the magnitude of this normal viscous stress is so small that it's neglected. Would you be able to explain what is actually happening to fluid parcels? So essentially I'm curious to know in which mechanism the normal viscous stress arises and how is it different from that of the pressure in its molecular nature.

    Thank you...
     
  5. Oct 19, 2016 #4

    K41

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    The Cauchy equation of motion relates the acceleration of a control volume of fluid to the sum of all the forces acting on the control volume. One of the forces are body forces. These include gravitational and electromagnetic forces. The other forces are called surface forces.

    In continuum mechanics (which we are using to model the fluid flow), the surface forces arise from two sources:
    - intermolecular forces of attraction
    - fluxes across a control volume due to fluctuating velocities*.

    *The velocity you see in the Cauchy equation of motion (or the Navier-Stokes Eq.) is a "spatial instantaneous average" of the molecular velocities of the fluid. So if you are looking at a single point in the fluid, and assuming its continuous (from the continuum approximation), then at that single point we have averaged all the molecular velocities of the discrete fluid molecules, (so in water they would be H20 molecules). There would be a fluctuation about that mean too. Now, say the fluid is moving from left to right on the computer, we would say U1, the instantaneous spatial average velocity = 1m/s. Lets say velocity up and down is U2. We would say, U2 = 0. So momentum cannot be transferred due to the "mean" (or "average" if you want to call it that) velocity. Momentum can however be transferred due to fluctuations and these appear as surface force contributions

    Now moving on, we define the surface force as a stress vector, which is the summation of all the forces acting per differential unit of area. Using some derivation, the stress tensor can then be equivalent to a stress tensor dot product with a unit normal on the surface through that point. It is this stress tensor which appears in the Cauchy equation.

    Next, you need constitutive equations to expand on what the stress tensor is, because there are more unknowns than equations. How you define the constitutive equations will determine what exactly what pressure is in the Navier-Stokes equations. I haven't reached that part of the book though so I cannot comment further. What I do believe is that the determination of the pressure will have a direction influence on what the the viscous stress is.

    You will have to read transport phenomena books for more details:
    Advanced Transport Phenomena - Gary Leal
    Transport Phenomena - Bird, Stewart and Lightfoot

    Boneh3ad is also correct. Pressure is always kinetic. This means that, regardless of whether the fluid is in motion or it isn't, there will always be some form of pressure. Bird et. al explicitly state that the viscous stresses require a velocity gradient.
     
  6. Oct 19, 2016 #5

    boneh3ad

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    Shear stress does act only tangential to a surface; that's the definition of "shear". Shear stresses are captured in the off-diagonal terms of the Cauchy stress tensor for a Newtonian fluid. Both the pressure and the normal viscous stresses show up in the diagonal components of the stress tensor. Generally, the overall stress tensor is made up of two components: pressure (which is a diagonal tensor where all three elements are identical, i.e. ##-p\delta_{ij}##) and the viscous stress tensor (which is, in general, full). The viscous stress tensor's components include terms proportional to ##u_{i,j} + u_{j,i}##, which are the shear stress terms and are nonzero when the velocity gradients are nonzero. It also includes terms proportional to ##u_{k,k}##, which are the normal viscous stresses. Those normal viscous stresses are nonzero only if the divergence of the velocity field is nonzero, and therefore only if the fluid is incompressible. Usually the effect is quite small, and it has no effect on viscous drag since that phenomenon is due to shear stresses.

    Pressure is due to particle collisions with surfaces or each other. Really, it is just due to the fact that fluid molecules are always moving about randomly and exert a force on whatever they hit, even if the bulk fluid motion is zero. Viscous normal forces are intermolecular in nature and only matter when compressibility is important.
     
  7. Oct 19, 2016 #6
    Thank you so much for the inputs djpailo and boneh3ad... There are quite a number of new terms that I have not come across before. I think it will take a while for me to go through the books that djpailo recommended. Would you guys say a Mechanical Engineering student who has not yet taken the fluid mechanics 1 can easily grasp the concepts contained in the book? I have been just reading intro aerodynamics books so I don't have a strong mathematical background yet. Would you be able to perhaps suggest another reading prior to reading these books?
     
  8. Oct 20, 2016 #7

    boneh3ad

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    It depends on which book. Bird, Stewart, and Lightfoot is often used to teach introductory fluid courses, especially in chemical engineering programs. I am not familiar with the other book he mentioned.
     
  9. Oct 23, 2016 #8
    Thank you!
     
  10. Jan 13, 2017 #9

    This is what I have always thought, and I guess just more or less accepted. However, I am thinking about it now and have become confused. Say that you have a velocity gradient ∂u/∂x, which doesn't imply incompressibility. This will result in a normal viscous force of 2μ∂u/∂x according to stress tensor. Sure there will be the other gradients to satisfy to incompressibility condition; however, we still have a normal viscous force in the x direction. So, what is it that separates the normal viscous forces from the pressure forces?

    Given your questions, it is very impressive that you have not taken a fluid dynamics course yet. You seem very much ahead.
     
  11. Jan 14, 2017 #10
    I totally disagree with this last statement. The viscous normal stress in the x direction is given by ##2\eta \frac{du_x}{dx}##, which can certainly be non-zero even if the divergence of the velocity field is zero. Certainly, fluid impingement on a surface includes such a term. The total normal stress in the x direction is equal to ##-p+2\eta \frac{du_x}{dx}##. So viscous normal stresses can contribute to the total normal stress even if the fluid is incompressible.
     
  12. Jan 14, 2017 #11

    boneh3ad

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    Right. I was going off of the definition of the mean mechanical pressure and then overgeneralizing. What I had been saying really only holds for the mean normal stress, not the individual normal stresses. Call it a case of me not thinking, then overthinking.
     
    Last edited: Jan 14, 2017
  13. May 30, 2017 #12
    But in incompressible flow normal viscous stress on solid surface should be zero? If solid surface is assumed vertical then on wall there is dv/dy=0 and dw/dz=0 which implies (by incompressible continuity equation) that du/dx=0, so normal viscous stress on surface is zero. Am I wright?
     
  14. May 30, 2017 #13
    No. Who says that in incompressivle flow, the normal viscous stress on a solid surface is zero. That would require that du/dx=0, which is not in general true.
     
  15. May 30, 2017 #14
    From incompressible continuity equation we have: du/dx+dv/dy+dw/dz=0. If dv/dy=0 and dw/dz=0 which is thru if there is no-slip condition then du/dx must be zero to fulfill continuity constrain?
     
  16. May 30, 2017 #15
    Oops. Of course you're right. Don't know how I missed that. Sorry for any confusion I might have caused.

    Chet
     
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