 In this segment we're going to take a look at one of the very important properties of a fluid and that is the viscosity of a fluid. So what viscosity does is it relates the local stresses within the fluid to the amount of strain or shear that is occurring within the fluid element itself. And what we'll begin with, and it's a very common approximation to be making with fluids, we will assume that the fluids that we're looking at are Newtonian and with that that means that the shear stress, we define shear stress with tau is proportional to and what we'll show in this segment, it will be proportional to the strain rate within the fluid and this typically holds for common fluids that we often deal with and examples of those would be water, oil, or air. There are other fluids that we'll talk about in a later segment that are non-Newtonian and for those this type of relationship would not hold. So when we look at this, if you recall back, if you've taken a course in solid mechanics, this relationship looks a little bit like Hooke's law. So it looks a little bit like Hooke's law that you would see in a course in solid mechanics. However, the difference here is that we're looking at the deformation rate and if you remember from one of the earliest segments in this course we talked about the difference between solids and fluids was that when you shear a solid it only deforms through an angle whereas when you deform or apply a shear to a fluid it will continue to deform and so that was the difference between a solid and a fluid and consequently we see it within the equation that we're going to derive here. So what we're going to do, we're going to start with a little element of fluid and so imagine we have a chunk of fluid look something like this and what we are going to do is we are going to assume that there is a shear stress. Well first of all let's define the axis or the size. So this chunk of fluid is delta y in the vertical and it is delta x in the horizontal that gives us the size of the chunk of fluid and then what we're going to assume is that up here we're applying a shear stress and that shear stress is going to be shown to be proportional to the deformation rate of the fluid itself and what we'll do we'll derive an equation for that so let's assume that we apply this shear stress. What will happen is the fluid element itself is going to deform and it will look something like that and this would be at a later point in time and assume that the velocity at which we're deforming at, let's say it's u and I'm going to denote that with a delta u and in a given amount of time the fluid element itself is going to deform through a certain angle and we'll call that angle delta theta and over here we will have a delta theta as well so delta thetas are the same and we know if the fluid element is moving at a speed of delta u and let's assume that this takes place over a period of time delta t we can then say that this distance of this deformation is going to be delta u that's the velocity that it's displacing multiplied by delta t or that time and down at the bottom what we're going to assume is that the lower chunk of the fluid element is fixed so it is attached to a wall and we'll call that the no slip condition we'll look at that in a later segment but for right now let's just say that it is attached to the wall and it's not moving so we can use geometry to write out an expression here so what we have we'll use trigonomic tree we have tan delta theta is equal to opposite over adjacent so that's delta u delta t divided by delta y so we come up with an expression or an equation and what we're going to do we're going to impose the small angle approximation and and so in doing this small angle approximation we're going to assume that all of the deltas go to differentials and so it will be a differential size the very very very small displacement and angle or in velocity so let's take a look at that if we make the small angle approximation the previous equation then becomes this and I've done a little bit of rearranging I've moved the delta t to the left hand side of the equation du dy so if you look back with the small angle approximation we had something like this delta u delta t over delta y and all I've done is I've taken this and I've moved it down there and I've converted the little differential elements into the the d for differential equation so with that if if we look back to what we were saying in the earlier slide that shear stress is proportional to delta theta by delta time what we have here is that the shear stress if you recall that we can say it was proportional to d theta by dt well we see from the equation from this little chunk of fluid what we have is shear stress is then equal to du by dy which is coming from this equation up here I'm bringing it in but there needs to be some sort of constant of proportionality and and that constant of proportionality that we use in fluid mechanics is called viscosity and so we put the viscosity in there that gives us the constant of proportionality for this relationship and what this does is it gives us the force actually it's it's stress but assuming it's over a unit area so the force due to shearing from velocity so if you shear a fluid this would be the force that resists that shearing motion and mu in here is a constant of proportionality and that is the viscosity and it actually turns out that this is what we call the absolute viscosity we'll look at another form of viscosity called the kinematic and it's basically this divided by the density but we'll look at that later so let's take a look at this now so we have this equation we have oops sorry we have tau tau is equal to mu du by dy let's take a look at the units in this equation now tau tau is a shear stress so the units of shear stress is going to be newtons per meter squared and on the right hand side of the equation we have the units of mu for right now i'm going to leave it unknown and then we have it multiplied by a velocity divided by a distance so velocity we know as meters per second and distance we know as meters so consequently what we can do is we can rearrange this and try to isolate and determine what the units of the absolute viscosity are so if we expand this the meters cancel out we have newtons seconds per meters squared newtons if you recall f equals ma a force is equal to mass times acceleration we can expand newtons as being kilograms times meters per second squared and then that's multiplied by seconds per meters squared we have some cancellations that are going to occur there that goes that goes so what we end up with for the viscosity is kilograms and it's divided by second meter per meter second and so essentially what we have are units of mass divided by length times time so those would be the units of viscosity and and you'll see different units in different books but really it could either be that or it could be that for the absolute viscosity now one last thing that i should say and i did mention this at the beginning this is assuming that we're dealing with a newtonian fluid and the characteristic of a newtonian fluid is that tau is a linear relation of du by dy now there are some fluids that are non-newtonian and for those fluids it is not a linear relationship so you would have a non-linear relationship here those are fairly complex fluids to analyze and we will not be analyzing them in this course but i will give you some examples of non-newtonian fluids in the next segment so what we've looked at here we've covered the viscosity of a fluid it's a very important parameter or property that we will be using in all of our analysis