Question 7 of 7: Question 7 (Part B, Question B3): Source and Sink Above a Plane Wall — Method of Images

Question 7 (Part B, Question B3): Source and Sink Above a Plane Wall — Method of Images (29 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A plane solid wall taken as \(y=0\), with the fluid occupying \(y>0\). A two-dimensional sink of strength \(m\) sits at \((0,\,b)\) and a source of strength \(2.5m\) sits at \((0,\,2b)\), both on the same vertical line above the wall. The fluid is ideal (inviscid, incompressible) of density \(\rho\), the flow is irrotational, gravity is neglected, and \(P_o\) is the pressure far from the wall where the fluid is at rest. The aid sheet supplies \(\Phi_{\text{source}}=\frac{m}{4\pi}\ln\!\left[(x-x_o)^2+(y-y_o)^2\right]\) and \(\Psi_{\text{source}}=\frac{m}{2\pi}\tan^{-1}\!\left(\frac{y-y_o}{x-x_o}\right)\).

Find. (a) a potential (or stream function) that satisfies the wall condition, with proof; (b) the velocity along the wall and its stagnation points, plus a sketch; (c) the wall pressure distribution and the location of its maxima.

Approach. Impose the wall by the method of images: reflect every singularity in the plane \(y=0\), keeping its sign and strength, so that the normal velocities cancel on the wall by symmetry. Then differentiate the resulting potential to obtain the wall velocity, set it to zero for the stagnation points, and apply Bernoulli to convert the speed distribution into a pressure distribution.

1. Part (a) — construct the image system. A solid wall requires zero normal velocity, \(v=0\) on \(y=0\). Placing an identical image of each singularity at its mirror point below the wall makes the vertical velocity of each pair cancel there. The image system is therefore a sink \(-m\) at \((0,-b)\) and a source \(+2.5m\) at \((0,-2b)\), and the complete potential for the half-plane \(y\ge0\) is \[\boxed{\begin{aligned}\Phi(x,y)=&\ \frac{2.5m}{4\pi}\ln\!\left[x^{2}+(y-2b)^{2}\right]\\&+\frac{2.5m}{4\pi}\ln\!\left[x^{2}+(y+2b)^{2}\right]\\&-\frac{m}{4\pi}\ln\!\left[x^{2}+(y-b)^{2}\right]\\&-\frac{m}{4\pi}\ln\!\left[x^{2}+(y+b)^{2}\right]\end{aligned}}\] with the companion stream function \[\begin{aligned}\Psi(x,y)=&\ \frac{2.5m}{2\pi}\left[\tan^{-1}\frac{y-2b}{x}+\tan^{-1}\frac{y+2b}{x}\right]\\&-\frac{m}{2\pi}\left[\tan^{-1}\frac{y-b}{x}+\tan^{-1}\frac{y+b}{x}\right].\end{aligned}\] Each term is a solution of Laplace's equation away from its own singular point, so the sum is harmonic in the fluid and the flow is irrotational everywhere the fluid actually is.
2. Show explicitly that the wall is represented. The vertical velocity of a single source of strength \(q\) at \((0,c)\) is \(v=\partial\Phi/\partial y=\frac{q}{2\pi}\frac{y-c}{x^2+(y-c)^2}\). Evaluating the four contributions at \(y=0\), \[\begin{aligned}v(x,0)=&\ \frac{2.5m}{2\pi}\left[\frac{-2b}{x^{2}+4b^{2}}+\frac{+2b}{x^{2}+4b^{2}}\right]\\&-\frac{m}{2\pi}\left[\frac{-b}{x^{2}+b^{2}}+\frac{+b}{x^{2}+b^{2}}\right]=0 .\end{aligned}\] Each real singularity is cancelled term-by-term by its own image, so \(v\equiv0\) along the whole wall and \(y=0\) is a streamline: the wall is exactly reproduced. (Equivalently, substituting \(y=0\) into \(\Psi\) gives \(\Psi=0\) for all \(x\ne0\), since \(\tan^{-1}(-c/x)+\tan^{-1}(c/x)=0\) — but the \(v=0\) test is the safer one, because the arctangent branch cut can make \(\Psi\) appear to jump.)
3. Part (b) — velocity distribution along the wall. The horizontal velocity of a source \(q\) at \((0,c)\) is \(u=\partial\Phi/\partial x=\frac{q}{2\pi}\frac{x}{x^2+(y-c)^2}\); on the wall each real singularity and its image contribute equally, so their sum is simply doubled: \[\boxed{u(x,0)=\frac{m\,x}{\pi}\left[\frac{2.5}{x^{2}+4b^{2}}-\frac{1}{x^{2}+b^{2}}\right],\qquad v(x,0)=0}\] The two terms compete: the closer sink dominates near the origin and pulls fluid inward, while the stronger source dominates far away and pushes fluid outward.
4. Locate the stagnation points. Setting \(u(x,0)=0\), either \(x=0\) or the bracket vanishes: \[\begin{aligned}\frac{2.5}{x^{2}+4b^{2}}&=\frac{1}{x^{2}+b^{2}}\\\Longrightarrow\quad 2.5\left(x^{2}+b^{2}\right)&=x^{2}+4b^{2}\\\Longrightarrow\quad 1.5x^{2}&=1.5b^{2}.\end{aligned}\] \[\boxed{x=0,\qquad x=+b,\qquad x=-b}\] Three stagnation points lie on the wall. The one at the origin is the point directly beneath the singularity pair; the pair at \(x=\pm b\) marks the dividing streamline that separates fluid drawn into the sink from fluid emitted by the source that escapes to infinity.
5. Describe and sketch the wall velocity. \(u\) is odd in \(x\), so the picture is antisymmetric. For \(0<x<b\) the bracket is negative and \(u<0\): fluid slides along the wall toward the origin, on its way up into the sink. For \(x>b\) the bracket is positive and \(u>0\): fluid runs outward, fed by the net source strength \(2.5m-m=1.5m\). The extrema are found from \(du/dx=0\) and sit at \(x=0.463b\) (inward maximum, \(u=-0.0339\,m/b\)) and \(x=3.116b\) (outward maximum, \(u=+0.0883\,m/b\)); far away \(u\to 1.5m/(\pi x)\), the signature of a net source of strength \(1.5m\) bounded by a wall.

6. Part (c) — apply Bernoulli along the wall. The flow is ideal, irrotational and steady, so Bernoulli's constant is the same throughout the field and may be evaluated far away, where the fluid is at rest at pressure \(P_o\). Along the wall \(v=0\), so the local speed is \(|u|\) and \[P(x,0)+\tfrac{1}{2}\rho u^{2}(x,0)=P_o .\]
7. Write the wall pressure distribution. Substituting the velocity from step 3, \[\boxed{P(x,0)=P_o-\frac{\rho m^{2}x^{2}}{2\pi^{2}}\left[\frac{2.5}{x^{2}+4b^{2}}-\frac{1}{x^{2}+b^{2}}\right]^{2}}\] Because the bracket is squared, the pressure deficit is even in \(x\): the wall pressure distribution is symmetric about the origin even though the velocity is antisymmetric.
8. Identify the pressure maxima and minima. The pressure is greatest wherever the speed is zero, which is precisely at the three stagnation points: \[\boxed{P_{\max}=P_o\quad\text{at }x=0,\ x=+b,\ x=-b}\] The pressure never exceeds \(P_o\) anywhere on the wall, since the far field is at rest and Bernoulli only allows a deficit. The two minima occur at the speed maxima, \(x=\pm3.116b\), where \[P_{\min}=P_o-\tfrac{1}{2}\rho(0.0883\,m/b)^{2}=P_o-0.00389\,\frac{\rho m^{2}}{b^{2}},\] with a weaker local minimum at \(x=\pm0.463b\) of \(P_o-0.000576\,\rho m^{2}/b^{2}\). Physically the wall is sucked hardest just outboard of the outer stagnation points, and the net upward suction on the wall is what a real installation would have to resist.