Shear Stress between Parallel Plates

Shear Stress between Parallel Plates

So in this particular problem what we have
is laminar flow between parallel plates, and the velocity distribution is given by u over
big U equals one minus two, y over h, that whole quantity squared. This big U generally
we us it big U when we are talking about a fixed plate and a moving plate, and that big
U is the velocity of the moving plate. However when you are talking about a flow, laminar
flow between plates. Generally what is causing the flow is not a moving plate, but its pressure
driven. So instead of saying big U we actually say umax, which is the maximum velocity of
the fluid, and that is going to be what we normalize our velocity distribution by, and
so what we are asked to do here is consider water at 15 degree C. Umax equals 0.10 m/s,
and h, which is the distance separating the plate is 0.25 mm, and we are going to set
the origin midway between the plates. The reason for that is if you actually drew the
velocity distribution it out look like this, and what you should hopefully be able to see
from this. Is that it is symmetrical with a max being at the origin, right here, and
what we are asked to find is the shear stress on the upper plate, and its direction. So
our governing equation is that the shear stress our tau equals mu, which is the viscosity
times du/dy. So were do we find the viscosity, well the reason that they tell us its water
at 15 degree C is that we go to a table look up the viscosity of water at 15 degree C,
and we find that it equals 1.14×10^-3 N*s/m^2. The next thing we do is take the derivative
of u with respect to y, and we have to rewrite it as u equals Umax, one minus two y, over
h, the quantity squared. So when we take this derivative, it is again a relatively simple
derivative we find that du/dy equals Umax, minus 4 y, over h, times 2. Or minus h, Umax,
y, over h^2. Sorry I forgot that h^2 down there. So we are trying to find the shear
stress on the upper plate. So if you recall our origin is in the center, and our y in
the plus direction is like that. So what we are really trying to find is du/dy at y=positive
h/s. So when we put that in our equation we find that our tau equals mu, then minus 8,
Umax over h^2, times h/2, which is minus 4, mu,Umax, over h, and so when we plug our numbers
in. Including what our h is, and our viscosity that we looked up. We find that our tau is
minus 1.83 N/m^2. Notice that is a force divided by an area. Since our tau is a negative number,
what direction does it act in? It has to act in the negative direction or against the flow.
So what if we were asked to find it instead on the lower plate. Now instead of having
y be h/2, y would be negative h/2, and so what would the difference be. The difference
would be that the magnitude would be the same. However the sign would change. So we would
have the exact same magnitude, but in a different direction, and finally lets check to make
sure that max is actually at the center. So how do we do that? Well what we do is we find
du/dy, and again that is negative 8, Umax, y, over h^2, and how do we find what the maximum
is? You take the derivative, and set it equal to 0. So we set this equal to 0. The only
way that could be is if y equal 0. They way we set up our coordinate system. For fun what
you might want to do is change the coordinate system, put it on the very bottom, and see
how that changes your answer.

14 comments on “Shear Stress between Parallel Plates

  1. dsf sadf Post author

    how did you get tow to equal 1.83 when i am doing the problem i keep getting .001824, I am not sure if I am plugging a number in wrong but is the equation( 4 (1.14*10^-3)(0.1))/.25

  2. jamilbigj Post author

    If y = h then shear stress would -3.66 N/m^2 and y = 0, shear stress would be 0. But, I am having trouble finding for what value of y is umax, maximum.

  3. LearnChemE Post author

    We are looking at the top with the origin in the middle, so we are evaluating at y = h/2. When we plug in our values (make sure to use correct units), h is still the distance between plates. y = 0 is the middle between the two plates, where the velocity, U, is a maximum.

  4. Alex Zenteno Post author

    In the lower plate shouldn´t the shear stress be in the same direction that in the upper one? Cause, I think that always the shear stress is in oposite direction to the velocity and decreases while the velocity grows so it would be zero at the origin and have both maximum values (negative to the velocity) near the plates. I would realy appreciate that you answer my question. Greetings

  5. Safuan Sabri Post author

    if the question ask about the shears stress for both plates. are we have to sum the answer or leave it for two answer?

  6. runinguy Post author

    Thanks for the great video (along with the others in the Fluid Mechanics playlist). Do you have citable sources for the the two equations presented in this video? 

  7. Loay Al Ahmad Post author

    Since the final value is negative (-1.83), the direction in the plus x direction. Therefore you made a mistake in the end, please correct that.


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