 We're now going to solve an example problem for Newtonian fluid and what we're going to do is we will be given the velocity profile for laminar flow between two parallel plates and we're going to try to determine the shear stress on the upper plate and the direction of that shear. So we're told that the velocity profile, this is the given, it's going to be u as a function of some maximum velocity between the plates is this equation and the geometry that we're dealing with looks something like this. So we have an upper plate and both of the plates are fixed, they're not moving. We'll define a coordinate system x and y about the center line and we're told that the fluid flowing is water at 15 degrees Celsius and we're told that the channel height is h equals 0.5 millimeters. Okay, so we have water flowing and then looking at this profile that we have defined by the equation, this is a parabolic profile. On this flow we call it plane plus cell flow but we get something that looks like that and then here at the center line we would have u max which appears in the equation that we have. So that is the condition and what we're trying to find, oh and the other thing that I should say is u max equals 0.30 meters per second and fine. So what we're after, we want to determine the shear stress on the upper plate and we want to know the direction of the shear stress. Okay, so how do we go about solving this? Well, what we have in our back pocket thus far is we know that water is a Newtonian fluid and we've also seen that tau is equal to mu du by dy. So we have a velocity profile, a function of y, we'll have to look up the viscosity in order to get the proportionality constant in that equation so we can look it up for water and we can go about our solution. So let's work on the solution beginning with the Newtonian equation relating shear rate and shear stress. So we see we have the du by dy, let's evaluate the derivative of our velocity term and by taking the derivative remember I'll write out the velocity here, u over u max is 1 minus 2y over h squared. So we have to take the derivative with respect to y, we have u max, so we get that when we take the derivative of velocity with respect to y we can collapse that into minus 8 times u max times y divided by the denominator we have an h squared. Okay so we know from our Newtonian relationship between shear stress and shear rate that tau is equal to minus 8 mu u max y divided by h squared. So let's plug in the values and see what we get. Looking up the value for water, what did we say, a 15 degree C viscosity is equal to 1.3 times 10 to the minus 3 kilograms per meter second. So let's plug the numbers in and see what we compute. So you'll notice in the equation I've included all of the units just to be on the careful side you should do that although it's getting more practice you won't always need to do that but you can then check for the dimensions of your final result from your analysis but when we put in the numbers we calculate we get a negative 3.12 newtons per meter squared and so that comes out to be the final result that we get. Now if you recall with the question, let's take a look back, they were asking for the shear stress on the upper plate and give its direction. So when we went through this analysis and we determined this as being the shear stress is that shear stress on the upper plate or is that shear stress in the fluid? Well it turns out if you go back to the derivation that we used when we came up with tau equals mu du by dy we were considering a little chunk of fluid as it was deforming and we came up with a relationship from that so that is really the shear stress in the fluid not on the solid. So tau here as reflected in this result is actually the shear stress on the fluid and consequently the shear stress on the wall is going to be in the opposite direction. So if you remember our coordinate system, we had y in that direction, x in that direction that means that on the fluid the shear stress is in the negative x direction and on the wall it's actually in the positive x direction because it would be the equal and opposite. So if we were to sketch this out what we have is the shear stress on the fluid tau on fluid is in the negative x direction but the tau on the wall is in the positive x direction. So just like in statics if you draw a little picture and you say the direction that it's going in that would be good enough and then you would write 3.12 newtons per meters squared and what you could do is you could assign area as being unit area or one meter squared and with that you can then determine the stress that is on one meter squared of the upper wall. So your answer would be 3.12 newtons on the upper wall in the positive x direction. So that concludes the example of determining the shear stress on the upper wall and what we find that it is in this direction and it is 3.12 newtons assuming area equals one meter squared.