In today’s lesson, we will be discussing the following topics:
1. Couette Flow
2. Poiseuille Flow
3. Example with Couette Flow
4. Example with Poiseuille Flow
5. Velocity Profile
6. Example with Fully Developed Laminar Pipe Flow
Couette Flow
Couette flow is a type of fluid flow in which one plate is moving with respect to the other. This motion of one plate relative to the other creates a shearing action in the fluid between the plates. Fully Developed Couette (or Poiseuille) Flow is one where the velocity profile does not change with downstream distance.
Poiseuille Flow
Poiseuille flow is a type of flow in which both the plates are not moving, and the fluid flow is cause by an external pressure gradient/pressure change. We can have fully developed flow here, just like with Couette flow as mentioned above.
Example with Couette Flow
Consider the following example. The velocity profile shown in this example is one of fully developed pipe flow:
This is 1D flow. To calculate the velocity field, we will first have to figure out which variables we are given and what equation we can use.
(a) Let us assume steady, incompressible flow. In incompressible flow, the density is constant over time (dp/dt term = 0).
(b) We are told to neglect the effects of pressure or gravity, so if we use any equations with those variables, we can cancel them out as well.
(c) To get the change in velocity over time, we will use the continuity equation that uses velocity components of u, v and w (the x, y and z components of velocity). Note that these are all partial derivatives (with the curly d). This equation was discussed in this lesson here, so check it out if you don’t know where I got it from. I have taken out the term with density and then factored out the remaining densities and divided them on the right side because it is incompressible flow, so that leaves one side of the equation equal to zero:
We can cancel out the last two terms because the flow is only moving in the x-direction horizontally, not in the y- or z-directions:
This simplifies to:
(d) Because we are using just the x-component of velocity, we will look at just the x-component of momentum in order to solve for the velocity profile. The Navier-Stokes equation gives us the momentum conservation equations, so taking the x-component here and then canceling out certain terms. Note that the density cancelled out because it was multiplied by the brackets that end up being zero:
(e) Rewriting the remaining terms, then writing out the boundary conditions below gives the following. This is the velocity field u!
(f) But we don’t have the C1 and C2 constant, which need to be filled in with actual numbers rather than leaving them as unknowns. Let’s find C1 and C2 first, then plug them back into this above equation. We need to then figure out what the boundary conditions are, so we can plug them into this equation. This velocity field equation u = C1*y + C2 contains u (velocity) and y (height) aside from the C1 and C2 which we need to find. So that means our boundary conditions need to contain something that relates the velocity and the height!
We can see at a height of zero and then at the full height of h, the velocity of the fluid right along the bottom plate is zero because the plate is not moving – this is one of the boundary conditions (no-slip condition) we learned in the Boundary Conditions lesson plan (click here if you missed that). The idea for this condition is that the velocity of a fluid along a wall/plate is zero because that wall/plate is not moving, so the fluid right along it is considered to be at zero velocity as well.
Now the velocity at the top is equal to V, which is the velocity of the top plate (we don’t need to use a number, we can leave it as V).
(g) We have two boundary conditions, so we can substitute these into the velocity field equation for u and h and make two equations with it. Then we have two equations, and two unknowns C1 and C2 (recall from math – you need as many equations as you have unknowns in order to solve for any of the unknowns. So if we had two unknowns, we would need at least 2 equations for solve for C1 and C2).
(h) Now substituting C1 and C2 into the equation for the velocity field u gives the final answer for the velocity field:
(i) To calculate the shear force per unit area acting on the bottom plate, we need to know which equation to use. Let us first use Newton’s Law, F = m*a. We need to figure out the m*a first, then from there we can find the shear force since it is a part of the net force F. We can use this equation for the acceleration from the textbook and figure out whether the fluid is accelerating first. If you missed the Acceleration Field of a Fluid lesson, check it out here.
If a = 0, then we don’t need the mass because then the net force will also be zero, since F = m* a and if a = 0, then m * 0 = 0 anyways.
(j) All the terms in the acceleration equation cancel out because we have fully developed, steady flow. This means acceleration is zero, so the net force is zero! Let us now set up the net force equation, since F = m*a = 0. How do we find the force balance on a fluid? Let’s set up an image showing the forces acting on a fluid particle in all directions. This takes into account the force due to pressure and the force due to shear stress. The directions of the pressure and shear stress acting on the fluid are given by red arrows:
(k) Shear stress and pressure both have units of pressure, we need to multiply these by the area to get the forces and calculate the net force. This is because Pressure = Force/Area (this is a general engineering equation that can be applied here). Then we add up the force due to both shear stress and pressure to get the following net force equation The positive and negative signs correspond to the arrows drawn in the image above, where the positive direction is up and to the right:
Writing this equation into mathematical terms and simplifying:
(l) Substitute the shear stress equation into the shear force (in the picture right above, that we ended up with from the force balance). For incompressible and viscous flow, these following equations are given for shear stress in the textbook (greek letter tau, the curly T, is shear stress. The subscripts on the T tell you which plane the shear stress is in). The one term that I cancelled out is the change in the y-component of velocity (v) in the x-direction, which is zero since the velocity does not move in that direction:
(m) This is the final answer for the shear force per unit area along the bottom wall. Shear stress is already in units of force per area, so finding the shear force per unit area refers to finding the shear stress tau, which we just found above. You can put the i component next to it to indicate it is acting horizontally (in the x-direction) along the bottom plate:
Example with Poiseuille Flow
(a) This is 1D flow since it is only in the x-direction. We are told it is: steady, incompressible viscous, fully developed flow. We know the type of fully developed flow is Poiseuille Flow because both plates are not moving. We also know that the flow is only in the x-direction, and there is no gravity effect.
(b) To calculate the velocity field, I am going to first start with the continuity equation, and simplify the terms according to the information we have been given:
(c) Since only the x-direction component of velocity (u) is applicable here, we will take the x-component of the Navier Stokes equation first.
(d) Simplifying the equation:
(e) Now we can use the y- and z-direction Navier Stokes momentum equations too, even though the flow is not in those directions. These equations will still help us later! Here are the general equations, then in the second picture I cancel out the terms on the same basis I used to cancel out the terms in the x-direction Navier Stokes equation above. Only the pressure terms remain in both equations, because all other terms are not applicable:
(f) We already used the Continuity equation from the first step to plug that information in for one of the terms in the x-direction Navier Stokes equation in step (c). But we still have all three Navier Stokes equations that we haven’t used anywhere else! What can we do with them to get to the velocity field that we were asked to find?
Usually, the next step when we have equations with derivatives (which we always end up with in these fluids equations) is integration in order to remove the derivative and isolate the variable we need. We need to integrate the equation twice to get the squared term off the u to make it just a u instead of a d^2(u). This is how it goes:
(g) So this is the final equation for velocity! But we don’t know C1 and C2 yet. So let’s make some boundary conditions to evaluate C1 and C2, then plug them back into this equation to get the final answer. Aside from the constant values (dP/dx, mew (curly u)) and the unknowns C1 and C2, we have y and u which are the variables we need to use in the boundary conditions.
The boundary conditions are based on the image that was given (shown below). We look at a height of zero, and a height of 2h. Since both plates are stationary, the velocity at these plates are also zero. This is based on the same boundary condition we used for the previous example (no-slip condition) – the velocity right at a stationary wall/plate is zero, since the wall/plate is not moving.
Then using the velocity field equation we had found earlier, we can use it twice and substitute in each boundary condition (equations 1 and 2). We made two equations, and we had two unknowns C1 and C2. This makes the systems of equations solvable (you need at least as many equations as you have unknowns – that is a rule in math). Then we simplify it to solve for C1 and C2:
(h) Lastly, we plug C1 and C2 back into the velocity field equation to obtain the final answer. u and y (dependent and independent variables) here are like y and x in an equation like y = x^2.
(i) Now let’s find the volumetric flow rate per unit width. The equation for volumetric flow rate passing through unit width is the following:
(j) We have the velocity field equation from earlier, and we will integrate this with respect to A. Using our integration rules, we do the following steps to get the final answer:
(k) Now let’s plot the velocity distribution (velocity profile). It is a parabolic distribution because the independent variable (in this case, y) has a squared on it.
Whenever you have an equation in math (eg. when you take functions in high school), we learn some of the basic parents functions such as linear, quadratic, square root, exponential, trigonometric, etc. So this should be recognizable as a parabolic graph since the independent variable is squared.
Velocity Profile
A velocity profile, if you haven’t seen this term before this lesson, is a graph of the speed of fluid flow (velocity) as a function of distance perpendicular to the direction of flow. It shows the velocity distribution. So for example, let’s say we have flow through two fixed plates here. The vertical distance goes from 1 to 5 meters (the height of the flow). The flow is going horizontally, so this is the perpendicular distance.
The velocity profile graphs the velocity of the fluid (u) as a function of this perpendicular distance. Then this is what the velocity profile could look like, if the flow is slower as it gets closer to the fixed plates (at 1 meter and 4 meters), and fastest in the middle (at 2 meters). Sometimes you would see the velocity profile drawn directly onto the flow above, but I will draw a separate graph here:
Here is a picture of the velocity profile of both types of flow. Couette flow is in (a) without a pressure gradient and one moving plate, which has a linear velocity profile. Poiseuille flow has a parabolic velocity profile and has a pressure gradient with two fixed plates in (b). (b) is similar to the example velocity profile I drew above.
Example with Fully Developed Laminar Pipe Flow
Let’s do an example of laminar flow through a pipe, in cylindrical coordinates. I have written out the question, then started steps for the answer and assumptions we can make:
Below, I have cancelled out terms in the equation above, with an explanation of why I did so. Note that I mentioned above there is no swirl in this question – that means no change with respect to theta. Parallel flow means that the flow is parallel to the pipe (only horizontal), so there is no change with respect to that direction.
Even without labelling these terms as parallel flow or no swirl, the terms that were differentiated with respect to theta or z below would have been cancelled anyways because the flow is only in the z-direction and changing with respect to the r-direction (the height).
Lastly, I crossed off some terms with the explanation that the continuity equation makes it zero. I did not show the steps of finding out theses values are zero through the continuity equation, but let’s do that here. The continuity equation (and these Navier-Stokes equations we just used) were given in a previous lesson plan (linked here) in cylindrical coordinates. Looking at the continuity equation in cylindrical coordinates for incompressible flow:
Crossing off the first two terms (because velocity is only changing in the z-direction) gives the following:
Because this term dVz/dz = 0, it was substituted into the Navier-Stokes equation above which ended up cancelling out the term.
Now we had the Navier-Stokes equation above with all the terms cancelled. We can rewrite it and integrate it twice to solve for u, and find the constants C1 and C2 which arise after the two integrations.
