In previous tutorials, we learned about two new flows: uniform and source flows. Now we will look at what happens if they are superimposed with one another.
Let's picture a polar coordinate system with a flow source strength of $ \Lambda $ located at the origin. If we were to superimpose this with a uniform flow with velocity $ v_{\infty} $ moving to the right, we would get a stream function of
[ 1 ]
$$ \psi = v_{\infty} r \sin \theta + \frac{\Lambda}{2 \pi} \theta $$
Eq. (1) is simply addition of the two stream functions for a uniform flow and a source; an example of this can be seen below.
Fig. 1 - Image of uniform and source flow superimposed.
Notice how the superimposed flow looks when the source point is added in the uniform flow. I like to think of it as a rock obstructing the natural flow of a water stream. There is one point that is of particular interest in the superimposed graphic, and that is point B. Notice how the streamline going from D to B to moving to the left, until is stops at point B. Hence, B is a stagnation point in our flow! We will touch on this more later.
Since we know that the uniform stream function and source stream function are solutions of Laplace's Equation, then we know that Eq. (1) satisfies Laplace's Equation as well. Hence, we could set Eq. (1) equal to a constant value.
[ 2 ]
$$ \psi = v_{\infty} r \sin \theta + \frac{\Lambda}{2 \pi} \theta = const $$
We can obtain the velocity field by differentiating Eq. (1)
[ 3 ]
$$ v_r = \frac{1}{r} \frac{\partial \psi}{\partial \theta} = v_{\infty} \cos \theta + \frac{\Lambda}{2 \pi r} $$ $$ v_{\theta} = -\frac{\partial \psi}{\partial r} = -v_{\infty} \sin \theta $$
I want to point out, the radial velocity from a source is $ \frac{\Lambda}{2 \pi r} $ and the freestream velocity in the radial direction is $ v_{\infty} \cos \theta $, so $ v_r $ is simply the sum of the two radial velocities.
Before, we had seen that point B in Figure 1 is a stagnation point because the flow stops moving to the left and starts to split and move up as well as down and to the right. Hence, we can solve for the distance point B is away from D, or $ r $, and the angle at which point B is in our system, or $ \theta $. Now, off the bat, we can deduct what $ \theta $ is. Since the positive x direction is pointing horizontally to the right, and since point B is directly to the left of point D, point B will be 180° from the positive x axis, or $ \theta = \pi $, but we will solve for this to make sure.
[ 4 ]
$$ v_{\infty} \cos \theta + \frac{\Lambda}{2 \pi r} = 0 $$ $$ v_{\infty} \sin \theta = 0 $$
We know that $ \sin \theta $ is only zero when $ \theta $ is equal to $ ..., -2 \pi, -\pi, 0, \pi, 2 \pi... $, but which do we select? Since point B is directly to the left of point D, and the positive x axis is directly to the right of point D, we can say point B is 180°, or $ \pi $ radians away from point D, the origin. We can then plug this into Eq. (4) to get
[ 5 ]
$$ -v_{\infty} + \frac{\Lambda}{2 \pi r} = 0 $$
$$ v_{\infty} • 0 = 0 $$
$$ \Rightarrow \theta = \pi $$
$$ \Rightarrow r = \frac{\Lambda}{2 \pi v_{\infty}} $$
Hence, we can say that $ (r, \theta) = (\Lambda/2 \pi v_{\infty}, \pi) $ at point B. Now, if we substituted these values in Eq. (2), we would get
[ 6 ]
$$ \psi = v_{\infty} \frac{\Lambda}{2 \pi v_{\infty}} \sin \pi + \frac{\Lambda}{2 \pi} \pi = const $$ $$ \psi = \frac{\Lambda}{2} = const $$
So, now we know the streamline that goes through ABC in Figure 1! What is interesting is since we are working with an inviscid flow, the velocity of the flow would be tangential to a solid surface. Imagine if we had a solid body thats edge was inset to the ABC streamline. Since we are in an inviscid scenario, the flow would not feel any difference. Hence, we can use this superimposed model for basic approximations of flow and then scale up to more complex models afterwards!
Consider if we had a polar coordinate system with a uniform flow going from left to right at a freestream velocity of $ v_{\infty} $. Now try to picture if we put in a source and a sink in the uniform flow with strengths $ \Lambda $ and $ -\Lambda $, respectively. Figure 2 depicts this below.
Fig. 2 - A source and sink in a uniform flow.
Now we can say for any point P in the flow, it will have a stream function of
[ 7 ]
$$ \psi = v_{\infty} r \sin \theta + \frac{\Lambda}{2 \pi} \theta_1 - \frac{\Lambda}{2 \pi} \theta_2 $$ $$ \Rightarrow \psi = v_{\infty} r \sin \theta + \frac{\Lambda}{2 \pi} (\theta_1 - \theta_2) $$
Notice we have a positive $ \frac{\Lambda}{2 \pi} \theta $ term as well as a negative $ \frac{\Lambda}{2 \pi} \theta $ term. The positive term is for the source flow's contribution to the flow and the negative term is for the sink's contribution. In Figure 2, it can be see that there are in fact 2 stagnation points, point A and point B. Using geometric relations, we get that
[ 8 ]
$$ OA = OB = \sqrt{b^2 + \frac{\Lambda b}{\pi v_{\infty}}} $$
The streamline equations are given by Eq. (7) as
[ 9 ]
$$ \psi = v_{\infty} r \sin \theta + \frac{\Lambda}{2 \pi} (\theta_1 - \theta_2) = const $$
However, because point A is at a $ \theta = \theta_1 = \theta_2 = \pi $ and point B is at a $ \theta = \theta_1 - \theta_2 = 0 $, Eq. (9) can be set equal to zero.
[ 10 ]
$$ \psi = v_{\infty} r \sin \theta + \frac{\Lambda}{2 \pi} (\theta_1 - \theta_2) = 0 $$
Hence, the stagnation streamline is zero for source and sink flows superimposed in a uniform flow!