In today’s lesson, we will be discussing the following topics:
1. Curl
2. Vorticity
3. Irrotationality
4. Frictionless Irrotational Flow
5. Velocity Potential Function
6. Example – Finding Velocity Potential
7. Example – Finding Vorticity
video coming soon
Curl
When a fluid particle is rotating in 2D, we can define the rotation through the angular velocity. But when it’s in 3D, it’s not so easy, because now we have to define the direction of rotation too (rather than just clockwise or counter-clockwise in 2D). When we have rotation in 3D, we can define it through a vector, where the direction of the vector is found through the right-hand rule, which we will get to later.
So how do we measure the rotation of a vector in 3D? Curl! The curl at a specific point in a vector (eg. the velocity vector of a fluid) measures the tendency of particles near that point to rotate about the axis that points in the direction of that vector. The curl vector at that point is how quickly the particles would rotate around that axis (eg. a higher curl value means the particles are rotating quicker around that axis).
Note that curl is not a specific variable, but rather it is an operator. It is shown by the following equation, where u, v and w represent and x, y and z components of the velocity vector:
This equation can give you the magnitude of the curl. But when you want to find the direction of the curl vector, you can use the Right-Hand Rule. Your right-hand fingers will be curled in the direction of rotation, and whichever direction your thumb is sticking out in is the direction of the curl vector. Imagine rotation of a fluid as the gases swirling on Jupiter’s Red Spot!
Vorticity
Vorticity is basically the same as curl. It is defined by the variable ξ. It is also usually twice the angular velocity of the fluid. You would use the same equation here as you would for the curl.
The vorticity of a velocity field is found by using any two perpendicular (orthogonal) axes to the plane on which the fluid flows, and adding the angular velocity of each axes. They would be the same value, since each axis rotates at the same rate, which is why vorticity is twice the angular velocity.
Irrotationality
Irrotationality means not rotating. This is when a moving fluid does not have any net rotation about its centre of mass, and can move in a circular path but not actually rotate itself. It is similar to a Ferris Wheel, where the carriages are moving in a circle as they’re attached to the giant wheel. But those carriages stay upright and they don’t rotate themselves with respect to the Earth (you don’t move upside down yourself when you ride a Ferris Wheel)! In other terms, it is a zero fluid angular velocity.
How can we mathematically prove that a flow is irrotational? Well, if the curl of a vector field is zero – it is irrotational. This makes sense, because curl measures rotation, so it makes sense that it would be zero for an irrotational fluid. You would take the curl (aka, the vorticity) equation from above and set it equal to zero.
Frictionless Irrotational Flow
When we say a fluid is frictionless, it has the same effect as being inviscid (meaning there is no shear stress acting on it). The only thing this changes is that it cancels out the viscosity term in the Navier-Stokes equations. See the original Navier-Stokes equation (where it is split into components) and the total momentum equation below, and then see how they change once you cancel out the viscosity term:
Let’s just take a look at the total momentum equation. The shear stress T disappears when we have frictionless flow, making this the equation (and this is now called Euler’s Equation):
You can also rewrite the Navier-Stokes equations with all the components from the image above, but that term with the viscosity would be gone.
Velocity Potential Function
The velocity potential function has a similar idea to the stream function. It is denoted by Φ = Φ(x,y,z,t) – meaning Φ will change with position and time (aka, it is a function of position and time). The partial derivative of the velocity potential function in a specific direction gives you the velocity in that direction! So we can define the velocity components by the velocity potential function like so:
Wherever Φ is constant along a line of fluid flow will be called the potential line of flow. This function is also in 3D and not limited to 2D like the stream function was. We can use this function to simplify fluids equations (just like the stream function), but we can now take variables of u, v and w and reduce them to just Φ.
When do both streamlines and potential lines exist (eg. the stream function and the velocity potential function exist?) Since the stream function is defined for 2D flow, then when there is irrotational 2D flow, there will exist both streamlines and potential lines. These lines would be perpendicular everywhere except at a stagnation point (since there is no velocity at that point). Since both the stream function the velocity potential function are defined for the components u and v, you can even equate them as I’ve shown below. This doesn’t have any application in this chapter, but just know it’s possible:
Example – Finding Velocity Potential
(a) We first need to see if the flow is irrotational to make sure a velocity potential exists (aka, if the velocity potential function exists). This means that the curl needs to be zero in an orthogonal (perpendicular) plane. This flow is in the xy plane, so the perpendicular axis is z. Therefore, we take the curl equation from earlier and we only take the z-component of that equation. We set it equal to zero to confirm that the flow is irrotational:
Then we substitute in the u- and v-components we are given in the question and take their partial derivatives. After subtracting, they equal 0. This confirms that the curl is 0 and that a velocity potential function exists:
(b) Now we take the first velocity component u and rearrange the differential equation that defines u (first equation in the image below). This is rearranged to make an integral, then both sides are integrated and the constant C is added at the end. You could also have started this with v instead.
(c) To check that we did it right, we can check the definition of the velocity component v, and can use it to substitute the velocity potential function we found above. When we differentiate it with respect to y, we should get the v-component of the velocity that was given in the question, -2axy. Here is the equation we know for the v-component of the velocity potential function:
Let’s substitute the equation we found for the velocity potential function Φ into the equation above, then differentiate it with respect to y. This gives us v:
(d) This equation for v is the same equation we were given in the question. That means our velocity potential function was correct! You could have stopped at the end of step (b), but step (c) was to confirm we did it right. This is the final answer for the velocity potential function Φ:
(e) This is plotted the same way as the stream function would be (check out the example from this blog post). So I won’t go through the steps, but the final answer would look like this:
Example – Finding Vorticity
(a) To calculate the wall shear stress, we look at the shear stress in the x-y plane. Using the shear stress equations given in the textbook (for incompressible, viscous flow where the viscosity is not zero), we have the following (where we will use the xy equation):
Now the velocity is only flowing in the x-direction, so we only have a u component. But what is u? It is calculated earlier in the textbook for 4.12(b) as:
Substituting this into the shear stress equation gives:
Taking the partial derivative d/dy and simplifying gives the final answer at the bottom. We evaluate the derivative at both +h and -h for y, because it doesn’t specify which wall we want the shear stress for (at height h = h or h = -h, so we do both):
When we used y = -h, we got a positive shear stress (since there was already a negative in the equation that cancelled out). This means there is a positive shear stress on the bottom wall, so then by default, a negative shear stress on the top wall.
(b) To calculate the stream function, let’s first make sure it exists. We don’t have to satisfy the continuity equation (the long way of doing it) but we can look at a few key properties of the flow. Since (1) it is changing with respect to the x- and y-directions (2D plane flow), (2) it is steady and (3) compressible, a stream function should exist.
So now we know that, let’s find it! Using the definition of the stream function from the previous stream function lesson plan:
We can sub in the equation for u from part (a) (and cancel out v, since only the u-component has flow). Then move the dy to the left side, integrate both sides and therefore find the stream function:
You can sub in y = h and y = -h in order to find the stream function at both the walls, if the question further asks for it.
(c) To calculate the vorticity, take two orthogonal axes to the velocity flow – there is only one axis orthogonal to the xy plane, and that is z. Remember that the vorticity equation is the same as the curl equation, so we can find the curl of Z. These are the curl equations from above:
Let us take the term with last term for the z-component.
The vorticity would be the highest at the walls since the y value is at its maximum or minimum height (eg. y = -h or y = h. Rather than part way up or down the wall, y could = 0.25h or -0.75h). The value of the curl is positive at the top wall when y = h, and negative when y = -h. So the vorticity is counterclockwise in the upper half of the flow, and negative in the lower half.
Note that Umax = maximum velocity (at the center of the walls in the image given, where y = 0). Some of the variables combined are given in the textbook as the Umax, so I have replaced these constants with it.
(d) To calculate the velocity potential, we need to first make sure it exists. This is done the same way as we did it for the previous question, and once again we use the z-component of the curl equation and set it equal to 0:
Since the curl does not equal 0, the flow is not irrotational and the velocity potential function does not exist, so we cannot find it.
(e) To calculate the average velocity, we use the equation for average velocity below. This is the integral of the volumetric flow rate q times the area, then divided by the cross sectional area. I have substituted in the u-component of the velocity, the same one I’ve been using for the other parts of this question. The cross sectional area is 2*b*h, and the volumetric flow rate is (u*b) dA. Taking the integral and simplifying with the integration bounds from -h to +h (from the bottom to the top wall), gives us the following:
We can also calculate this by taking the difference in stream functions, since this was explained in the Stream Function lesson plan. Doing this gives the volume flow rate, then we divide by the area. We already found the stream function in part (b):
We can substitute in y = h and y = -h for the top and bottom wall. Then we subtract the two to get the volumetric flow rate:
Now use Q to find the average velocity, V = Q/A as we used above. This gives us the same answer for the average velocity that we found earlier:
