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Section 1: Molecular Momentum Transport

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Section 1: Molecular Momentum Transport Section 1: Molecular Momentum Transport Newton’s Law of Viscosity What is viscosity? Viscosity is a physical property of fluids (liquids and gasses) that pretty much measures their resistance to flow. On a more fundamental level, the term “flow” refers to molecules moving along due to some net force. The more a fluid is able to withstand that force, the greater its viscosity. Using a more anecdotal example, if you’re not someone who normally “goes with the flow”, you might consider yourself viscous. As we’ll talk about in just a bit, viscosity is very much tied in with velocity of a fluid. If you were to pour honey and water out of a cup, obviously the honey would flow slower and would be deemed more viscous than water. While this is true, I challenge you to think of it more like “under the same amount of force (i.e gravity), honey will flow much slower than water due to its viscosity”. Parallel Plate Example Let’s consider a pair of large parallel plates. They each have a surface area of A and are separated by a distance of Y. In between, there is some fluid (which again, could be a liquid or a gas) that reside in layers between the plates. A Y y x Now consider what happens at four different time intervals: a) b) c) d) Some additional context: - Stationary plate begin to move at our relative t = 0 - We assumed the “no-slip” condition, where the layer of molecules in direct contact with the bottom plate move together with the same velocity in the x-direction, vx - The first layer of molecules “drag” the subsequent layers, each which has a slightly lower velocity – this creates a velocity gradient (further explained in a bit). By logic, you can reason the following proportionalities: - If you had larger plates (A), you would need a greater force (F) to maintain the same velocity (v) - If you wanted a greater velocity (v), you would have to apply more force (F) - Full disclosure: to this day I still don’t have a good grasp of why F is inversely proportional to Y, so here’s an example of some hand-waviness This yields the following symbolic relationship, where μ is the viscosity of the fluid. 퐹 푣 = 휇 퐴 푌 If you are familiar with material sciences, you may notice that F/A is essentially the term for stress. There are in fact two types of stresses; tensile and shear. A simple way to distinguish the two is that tensile stress acts perpendicularly on the surface of an object/molecule, whereas shear stress acts tangentially to a surface. As we are dealing with molecules “dragging” and “sliding” past each other to drive flow, we are more interested in shear forces acting on the molecule, generally denoted by τ. = = In the example for the parallel plate, we can further understand the behaviour of shear stress and how it acts as a means of momentum transport. As the first plate moves with some momentum (which is just velocity when you include considerations of mass), molecules on the first layer shears against the next layer and begins to drag them in the same direction. Albeit, it will have a slightly lower momentum due to the viscosity of the fluid. And then the second layer drags the third, and so on, with each layer gaining some amount of momentum like a decreasing domino effect. Overall, momentum of fluid changes from high to low as you move in the positive y- direction. In other words, there is a transfer of x-momentum as you move in the y- direction. As such, shear stress here can be further denoted as τyx, which has the net effect of causing flux of x-momentum in the positive y-direction, where flux means “flow per unit area”. It’s important to note that flux is always perpendicular to the flow direction. F/A is thus replaced by τyx. Finally, we replace V/Y with -휕vx/휕y. Essentially this explicitly represents the rate at which momentum in the x-direction decreases as you move in the y-direction. We then arrive at Newton’s Law of Viscosity: 휕푣 휏 = −휇 푥 푦푥 휕푦 As an exercise, could you determine units of viscosity from this equation? (Hint: stress is measured in ‘N/m2’, velocity is measured in ‘m/s’, and distance is measured in ‘m’) This may seem trivial, but in all honestly unit analysis is a skill that I don’t believe is emphasized enough for engineers. It serves as a means of ensuring your models are set up properly and that you didn’t accidently forget to include a key term. If it’s not arrantly obvious, try to prove on a piece of paper or something that viscosity has units of N•s/m2, or Pa•s. Not all fluids follow this law exactly. Usually gasses or liquids with molecular weights < 5000 g/mol behave in this manner, and are called “Newtonian Fluids”. Newtonian fluids are your friend. While they are still somewhat of a simplification, they are still surprisingly apt at representing most common fluids. Furthermore, we can actually prove that stress and momentum flux are one in the same through unit analysis. Let’s first consider the units of stress, where [=] denotes “has units of”. 휕푣 푚 1 휏 = −휇 푥 [=] 푃푎 ∙ 푠 ( ∙ ) 푦푥 휕푦 푠 푚 Right away, we can cancel out the distance and time units, which leaves us with just Pa. Then expand Pa -> N/m2. Since N is the unit of force, and F = ma, we can then replace N with the units of mass and acceleration. 푚 푚 1 푁 (푘푔 ∙ 2) 푃푎 ∙ 푠 ( ∙ ) = ( ) = 푠 푠 푚 푚2 푚2 Momentum, p, is mass times velocity, so we can collect the “kg” and an (m/s), then clean up the rest. 푚 1 (푘푔 ∙ 2) 푝 ∙ 푠 = 푠 푚2 푚2 푝 ∴ 휏 [=] 푚2푠 Momentum per unit area is defined as momentum flux. It is also per unit time, which is because it’s a time-dependent process that changes until it reaches steady-state. We’re going to use a similar approach to relate future terms to momentum flux. Generalization of Newton’s Law of Viscosity In the parallel plate example, we only needed to consider vx (velocity of fluid moving in the x-direction), however there exists other situations where vy and vz come into play for momentum transport. In order to apply Newton’s Law of Viscosity to all scenarios, we need to take into account the transport of momentum from all forces in all directions Assume you have an infinitesimally small cubic element in a fluid with axes in the x, y, and z direction. We can use this as a model to describe pressure and viscous forces. y-plane z-plane x-plane Pressure forces act perpendicularly to a plane or surface. In relation to fluids, this is also known as hydrostatic pressure, which acts on a fluid whether it is stationary or moving. It can also be used to drive fluid flow, as pressure exerts a stress in the direction of high-pressure to low-pressure (think about how a straw works). Pressure forces are denoted by P. Viscous forces can act on a plane in any direction, except perpendicularly. Fortunately, as with all vectors in Cartesian coordinates, you can decompose viscous forces at any angle into terms of x-, y- and z-components. Be careful here – although we’re saying that viscous forces do not directly act perpendicularly to a plane, there can be components that do (but we will group these with pressure forces later on). As you may have figured out by now, viscous forces are actually the same thing as shear stress described in the last section as τ. Here’s a picture distinguishing between pressure forces (left) and the components of viscous forces (right) relative to a typical xyz axis. Going one step further, you can imagine how both pressure and viscous forces can move fluids. If you kick a ball square-on, it will move. If you skim the top of the ball with your heel, it will still move. Taking a holistic view, you can combine the two forces, and write out the components that will act on each plane: X-Plane Y-Plane Z-Plane τxx + P τyx τzx τxy τyy + P τzy τxz τyz τzz + P *Don’t forget that τij means “motion in the j-direction with flux in the i-direction”. You can refer back to the parallel plate example to make sure this resonates. This can be written as a matrix and is denoted by Пij, or otherwise known as a “molecular stress tensor”. As the name kind of insinuates, Пij captures both of the molecular-driven forces (viscous and pressure) in all directions. Here is the full form: One last concluding notes regarding Newton’s Law of Viscosity. The full form of the equation is actually as follows: 휕푣 휕푣 2 휕푣 휕푣 휕푣 휏 = −휇 ( 푗 + 푖) + ( 휇 − 푘) ( 푥 + 푦 + 푧) 훿 , where 훿 = 0 for i ≠ j 푖푗 휕푖 휕푗 3 휕푥 휕푦 휕푧 푖푗 푖푗 Under the assumption that the fluids (or the molecules) we’re working with are non- compressible, the whole second term of the sum equals zero. Furthermore, you might notice that there are two velcoities derivatives that are considered in the first term now. Fortunately, you’ll often find that one or both of those terms will equal zero as well.
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