The definition of a fluid can be slightly different depending on the source and application for which you’re looking it up. However, for a fluid mechanics course, a fluid is a substance or collection of particles that continually move and change shape as a result of subjecting it to an external shearing stress. In general, think of a fluid as a gas or a liquid. To solve any of the problems we’re gonna be solving in this fluid mechanics course, we’ll be using what is usually called conservation of mass, momentum, and energy. But in a more accurate way, we will in fact use conservation of mass, technically not conservation of momentum, but Newton’s second law (more on that later), and again, more specifically, the first and second laws of thermodynamics. To use and meet all of these conditions we’ll use common fluid properties, like density, specific volume (which is the inverse of density), specific weight (or the density times the gravity constant), and specific gravity (which is the ratio between the density of the fluid and the density of water). And for many of our applications, not all though, we will use the ideal gas equation: P/ρ=RT The viscosity is one of the most important properties in fluid mechanics, so let’s look into that one in more detail. When you picture a fluid flowing over a surface, for example the wind, you can see how the speed of the fluid (in this case the moving air) is increasing as you go higher and higher. The wind that you can feel on the rooftop of a 3-story high building is not much compared to the wind that you would feel on a 15th floor rooftop pool, for example. That is the main reason why a wind turbine is not located and ground level, but as high as it’s reasonably possible. And it works both ways. The lower or closer you are to the surface where the fluid is flowing over, the slower it travels. So much so, that exactly on the surface, the speed of the fluid is exactly zero. This is what’s known as the No-slip condition. In general, a fluid that is in contact with a surface, will have the velocity of that surface, whether that is zero or not. For example, if a fluid is trapped between two horizontal surfaces, placed at a distance h apart, where the bottom surface is static, and the top surface is moving with velocity V, from the no-slip condition, we know that the fluid particles at the very bottom are not moving, and that the fluid particles at the top are moving with the same velocity, V, of the top surface. After a very small time, dt, the top plate will have moved dt times its velocity, and we can call this dx; or that dx over dt is the velocity of the plate. From the triangle that we see here, we can state that the tangent of the bottom angle is equal to the opposite side, dx, over the adjacent side, h. Since this angle, dθ, is very small, and if you remember from calc that the limit of tangent of x when x approaches zero is just x, then we can write this as dθ equals dx over h. From your mechanics of materials course, you probably remember a similar derivation. The small angle change is what we then defined as shear strain (in that case, we used gamma). If we now divide both sides by dt, the left hand side is what we can therefore define as the shear strain rate, and we see that the right-hand side is the velocity V, over h. The units of the shear strain rate are therefore radians over second, or just 1 over second, since radians is a dimensionless unit. Now, we’re not only interested in finding the shear strain rate at the top, where the top plate is located. If we define the velocity of the fluid at any vertical location, dy, as the velocity du, we know that V over h is proportional to du over dy. Now, the connection between the shear stress, that term that we mentioned in our definition of a fluid, and the shear strain rate, is very similar to what the relationship between shear stress and strain is in solid mechanics. In that case, we defined that shear stress over shear strain is the property of the material that we call shear modulus or modulus of rigidity, G. Of course, this linear relationship between the two is ONLY true, AS LONG AS the material is being deformed elastically, not plastically. It is only true AS LONG AS we are looking at the linear deformation in a stress-strain diagram. Very similarly here, the relationship between the shear stress and the shear strain rate will be linear, AS LONG AS our fluid is what we call, a Newtonian Fluid. If we’re pulling the top plate, that has an area A, with a force, F, then the shear stress can be calculated as τ equal to F over A. And just like in a solid material, that is being elastically deformed, the shear stress over the shear strain is a constant called the shear modulus, here, in fluids, but specifically to Newtonian Fluids only, the shear stress over the shear strain RATE, is also a constant that we call Viscosity. Now, this viscosity is specifically the dynamic viscosity, and we use the µ Greek letter as its variable. The units for dynamic viscosity, in metric, can be found by substituting newtons per meter squared for tau, one over second for the shear strain rate, and rearranging to either Pa∙s, or substituting Newton by kg∙m/s2 to write the units as kg over m∙s. Going back to our analogy, just like we could plot the elastic stages of the stress strain relationship between some metals, for example aluminum 69, a titanium alloy 110, and steel 180 (meaning that steel has a higher elastic modulus and is therefore stiffer than aluminum), we could plot air, water and oil on a shear stress vs strain rate diagram to say that oil is more viscous than water. Brief side note: non-newtonian. Now, since we will be using the density of the fluid for many of our calculations (like it was stated a few minutes ago), and that density is usually next to our dynamic viscosity, µ, we also define what the kinematic viscosity is with the letter ν (nu), as the dynamic viscosity over the density of the fluid. By substituting the units of each, we see that the kinematic viscosity, ν, has units of m2 over s. An easy way to remember this is that “kinematic”, which means motion, should be similar to velocity, which has units of m/s. In this case there’s just an extra meter on top. Let’s look at a simple example where we make use of what we’ve learned today, and if you want to check out more examples on this topic (some of them using English units too), or any of the other lectures of this fluid mechanics course (like Bernoulli’s equation or head loss calculations), make sure to check out the links in the description below. Let’s say there’s a snowmobile that reached an area that is covered in ice, instead of snow. Between the sleds and the ice, there’s a very thin film of water, 14 microns in thickness, and the total contact area between the sleds and the water is 960 squared centimeters. What is the thrust force that the snowmobile should generate to maintain a constant speed of 40km/h. A simplified schematic of the sleds on the ice over a thin film of water would show a rectangular plate, the sleds, of area 960 cm2, 14 microns above the static ice surface, with water between them. The shearing stress would therefore be the viscosity of the water times the shearing strain rate, du/dy. Since V over h is the same as du/dy (or the slope of the diagonal), and the shearing stress tau, can be written as the force that is being overcome by snowmobile over the area, we can solve for the force, and calculate its value, if we find the viscosity of the water first. To do that, and since we know that the viscosity if a function of the temperature, we can assume that the temperature of that water is near zero degrees Celsius. We look up the dynamic viscosity for water at zero degrees, and we find it to be 1.787 mPa∙s. We substitute the values, make sure that the units are correct, and find the necessary force to be 136.2 N, for the snowmobile to maintain its 40 km/h speed. The links to other examples, other lectures of Fluid Mechanics, and other engineering courses, are found in the description of this video; so make sure to check them out! Thanks for watching!