Question 1 of 5: Municipal outfall modelled as a source plus vortex above a lake bed

Question 1: Municipal outfall modelled as a source plus vortex above a lake bed (25 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.

Find. The stream function of the bounded flow, a proof that y = 0 is a streamline, the velocity along the bed, and the hydrodynamic force per unit length that this flow exerts on the discharge pipe.

Approach. Replace the solid bed by the mirror image of the singularities (source of the same sign, vortex of opposite sign), which makes y = 0 a streamline automatically; then differentiate the resulting stream function for the bed velocity, and obtain the force on the pipe from the Blasius theorem applied to the velocity that the images induce at the pipe.

1. Part (a) — choose the image system that turns the bed into a streamline. A plane wall is reproduced by reflecting every singularity in it. Reflection reverses the sign of the normal velocity, so a source must be mirrored as an equal source (its normal velocities then cancel on the wall) while a vortex must be mirrored with reversed circulation (its tangential velocities add and its normal velocities cancel). The bounded problem is therefore the unbounded superposition of four elementary flows: the real source m and vortex Γ at $ (0,\,a) $, plus an image source m and image vortex $ -\Gamma $ at $ (0,\,-a) $. No uniform stream is added, because the lake is quiescent far away.
2. Write the stream function of the four-singularity superposition. From the aid sheet, a source of strength m at $ (x_o,\,y_o) $ has $ \psi = \dfrac{m}{2\pi}\tan^{-1}\!\left(\dfrac{y-y_o}{x-x_o}\right) $ and a vortex of circulation Γ has $ \psi = -\dfrac{\Gamma}{4\pi}\ln\!\left[(x-x_o)^2+(y-y_o)^2\right] $. Superposing the real pair and the image pair, $$\boxed{\;\psi(x,y)=\frac{m}{2\pi}\left[\tan^{-1}\!\frac{y-a}{x}+\tan^{-1}\!\frac{y+a}{x}\right] -\frac{\Gamma}{4\pi}\ln\!\frac{x^{2}+(y-a)^{2}}{x^{2}+(y+a)^{2}}\;}$$ Written with the polar radii $ r_1^{2}=x^{2}+(y-a)^{2} $ and $ r_2^{2}=x^{2}+(y+a)^{2} $ this is $ \psi=\frac{m}{2\pi}(\theta_1+\theta_2)-\frac{\Gamma}{2\pi}\ln(r_1/r_2) $, which is the form most textbooks quote.
3. Part (b) — verify the bed by showing the normal velocity vanishes on it. The wall condition is kinematic: no fluid crosses y = 0, i.e. $ v(x,0)=0 $ for every x. With $ v=-\partial\psi/\partial x $, $$\begin{aligned} v(x,y)&=-\frac{m}{2\pi}\left[\frac{-(y-a)}{x^{2}+(y-a)^{2}}+\frac{-(y+a)}{x^{2}+(y+a)^{2}}\right]\\ &\quad+\frac{\Gamma}{2\pi}\left[\frac{x}{x^{2}+(y-a)^{2}}-\frac{x}{x^{2}+(y+a)^{2}}\right] \end{aligned}$$ Setting y = 0 makes both brackets vanish identically, because $ x^{2}+(0-a)^{2}=x^{2}+(0+a)^{2} $ while the source numerators become $ +a $ and $ -a $: $$v(x,0)=-\frac{m}{2\pi}\,\frac{a-a}{x^{2}+a^{2}}+\frac{\Gamma}{2\pi}\, \frac{x-x}{x^{2}+a^{2}}=0\qquad\text{for all }x$$ so y = 0 is a streamline and the flat, impermeable lake bed is correctly simulated. Note that this is the safe test: checking instead that ψ is constant on y = 0 appears to fail, because the arctangent of the source pair jumps by m/2 as x passes under the outfall — that jump is the real discharge crossing the plane, not an error in the image system.
4. Part (c) — differentiate for the tangential velocity along the bed. On the bed only $ u=\partial\psi/\partial y $ survives. Each source contributes $ \frac{m}{2\pi}\,x/(x^{2}+a^{2}) $ and, because the two images sit symmetrically, the pair doubles it; each vortex contributes $ \frac{\Gamma}{2\pi}\,a/(x^{2}+a^{2}) $ and the reversed image again doubles rather than cancels it. Hence $$\boxed{\;u(x,0)=\frac{1}{\pi}\,\frac{m\,x+\Gamma\,a}{x^{2}+a^{2}},\qquad v(x,0)=0\;}$$ The distribution is the sum of an odd part (the source, which sweeps fluid outward in both directions) and an even part (the vortex, which biases the whole bed flow toward +x).
5. Locate the stagnation point and the peak scour velocity. Setting the numerator to zero gives a single stagnation point on the bed at $$x_{s}=-\frac{\Gamma a}{m}$$ which lies on the upstream side of the outfall and moves further away as the circulation grows. Differentiating u with respect to x and setting the result to zero gives the peak speed; for the pure source case $ \Gamma = 0 $ it sits at $ x=\pm a $ with $ |u|_{\max}=m/(2\pi a) $, the classic result that the worst bed scour occurs roughly one pipe-height to either side of the outfall. Far away the velocity decays as $ u \to m/(\pi x) $, so a single outfall disturbs the bed over a distance of order a few multiples of a.
6. Part (d) — find the velocity the images induce at the pipe. The pipe cannot exert a force on itself, so the force follows from the external velocity at $ (0,a) $, which is whatever the two image singularities induce there. Both images sit a distance 2a directly below, so $$u_{\text{ind}}=\frac{\Gamma}{2\pi}\,\frac{2a}{(2a)^{2}}=\frac{\Gamma}{4\pi a}, \qquad v_{\text{ind}}=\frac{m}{2\pi}\,\frac{2a}{(2a)^{2}}=\frac{m}{4\pi a}$$ The image source pushes the pipe's neighbourhood upward and the image vortex sweeps it sideways; in complex-velocity form $ w_{\text{ind}}=u_{\text{ind}}-iv_{\text{ind}}=\dfrac{\Gamma-i\,m}{4\pi a} $.
7. Apply the Blasius theorem to the singularity. Near the pipe the complex potential is $ F(z)=\dfrac{m-i\Gamma}{2\pi}\ln(z-z_o)+F_{\text{ind}}(z) $ with $ z_o=ia $. Squaring $ w=dF/dz $ keeps one simple pole whose residue is the cross-product of the singular and induced parts, so $ \oint w^{2}\,dz = 2i\,(m-i\Gamma)\,w_{\text{ind}} $ and Blasius, $ X-iY=\tfrac{i\rho}{2}\oint w^{2}dz $, gives $$X-iY=-\rho\,(m-i\Gamma)\,w_{\text{ind}} =-\frac{\rho}{4\pi a}\,(m-i\Gamma)(\Gamma-i\,m)=\frac{i\rho\,(m^{2}+\Gamma^{2})}{4\pi a}$$ The product collapses because $ (m-i\Gamma)(\Gamma-im)=-i\,(m^{2}+\Gamma^{2}) $ is purely imaginary — the source and vortex contributions rotate into the same direction. Separating real and imaginary parts, $$\boxed{\;X=0,\qquad Y=-\frac{\rho\,(m^{2}+\Gamma^{2})}{4\pi a}\;}$$ i.e. no horizontal force at all, and a vertical force of magnitude $ \rho(m^{2}+\Gamma^{2})/(4\pi a) $ per metre of pipe directed toward the bed.
8. Confirm the result independently from the pressure on the bed. The flow is steady and irrotational outside the singularity, so Bernoulli gives $ p(x,0)=p_{\infty}-\tfrac12\rho\,u(x,0)^{2} $ — the bed is everywhere below ambient pressure. Integrating that suction over the whole bed, $$\begin{aligned} \int_{-\infty}^{\infty}\!\left(p-p_{\infty}\right)dx &=-\frac{\rho}{2\pi^{2}}\int_{-\infty}^{\infty}\frac{(mx+\Gamma a)^{2}}{(x^{2}+a^{2})^{2}}\,dx\\ &=-\frac{\rho}{2\pi^{2}}\left[\frac{\pi m^{2}}{2a}+\frac{\pi\Gamma^{2}}{2a}\right] =-\frac{\rho\,(m^{2}+\Gamma^{2})}{4\pi a} \end{aligned}$$ (the cross term integrates to zero because it is odd). The magnitude matches the Blasius result exactly: the bed is sucked up with the same force with which the pipe is pulled down, as Newton's third law requires. This is the cheapest possible audit of part (d).
9. Put numbers on the answer. For a representative outfall discharging $ m = 0.50 $ m²/s per metre of pipe with $ \Gamma = 1.20 $ m²/s of circulation at $ a = 2.0 $ m above the bed in fresh water $ (\rho = 1000 $ kg/m³), $$Y=-\frac{1000\,(0.50^{2}+1.20^{2})}{4\pi(2.0)}=-\frac{1000(1.69)}{25.13}=-67.2\ \text{N/m}$$ The stagnation point sits at $ x_{s}=-\Gamma a/m=-4.8 $ m and the bed velocity peaks at about $ 0.199 $ m/s — already at the threshold that moves fine sand, which is why outfall diffusers are normally set on a scour apron.