Question 5 of 7: Question 5 (Part B, Question B1): Municipal Discharge Pipe Modelled as a Source and Vortex above a Lake Bed

Question 5 (Part B, Question B1): Municipal Discharge Pipe Modelled as a Source and Vortex above a Lake Bed (26 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 including whatever image system makes the bed a streamline; a proof that the bed is a streamline; the bed velocity distribution in terms of m and Γ; the numerical circulation; and the bed pressure distribution.

Approach. Use the method of images. Reflecting the singularities in the plane y = 0 — with the source kept the same sign and the vortex reversed — cancels the normal velocity on the bed, turning it into a streamline without disturbing the flow far away. All five parts then follow from differentiating the resulting stream function and applying Bernoulli.

1. Part (a) — assemble the stream function from the aid-sheet building blocks. The aid sheet gives, for a singularity at $(x_o,y_o)$, $$\begin{aligned} \Psi_{\text{source}}&=\frac{m}{2\pi}\tan^{-1}\!\left(\frac{y-y_o}{x-x_o}\right)\\ \Psi_{\text{vortex}}&=-\frac{\Gamma}{4\pi}\ln\!\left[(x-x_o)^{2}+(y-y_o)^{2}\right] \end{aligned}$$ Place the real pipe at $(0,a)$ and its image at $(0,-a)$. A wall requires zero normal velocity, which is achieved by an image source of the same strength (its outflow toward the wall cancels the real source's) and an image vortex of opposite circulation (a mirrored rotation is reversed). Superposing all four, $$\boxed{\begin{aligned} \Psi(x,y)=&\ \frac{m}{2\pi}\left[\tan^{-1}\!\frac{y-a}{x}+\tan^{-1}\!\frac{y+a}{x}\right]\\ &-\frac{\Gamma}{4\pi}\ln\!\left[x^{2}+(y-a)^{2}\right] +\frac{\Gamma}{4\pi}\ln\!\left[x^{2}+(y+a)^{2}\right] \end{aligned}}$$ Equivalently the last two terms combine to $\dfrac{\Gamma}{4\pi}\ln\dfrac{x^{2}+(y+a)^{2}}{x^{2}+(y-a)^{2}}$, which makes the antisymmetry of the vortex pair explicit.
2. Part (b) — show the vertical velocity vanishes on the bed. The aid-sheet relation is $v=-\partial\Psi/\partial x$. Differentiating a source term, $$v_{\text{source}}=-\frac{m}{2\pi}\frac{\partial}{\partial x}\tan^{-1}\!\frac{y-y_o}{x} =\frac{m}{2\pi}\frac{y-y_o}{x^{2}+(y-y_o)^{2}} .$$ Setting $y=0$, the real source at $y_o=a$ contributes $-\dfrac{m}{2\pi}\dfrac{a}{x^{2}+a^{2}}$ and the image at $y_o=-a$ contributes $+\dfrac{m}{2\pi}\dfrac{a}{x^{2}+a^{2}}$; they cancel identically. Similarly for the vortex pair, $v_{\text{vortex}}=\dfrac{\Gamma}{2\pi}\dfrac{x}{x^{2}+(y-y_o)^{2}}$, which on $y=0$ gives $+\dfrac{\Gamma}{2\pi}\dfrac{x}{x^{2}+a^{2}}$ from the real vortex and $-\dfrac{\Gamma}{2\pi}\dfrac{x}{x^{2}+a^{2}}$ from its reversed image. Hence $$\boxed{\ v(x,0)=0\quad\text{for every }x\ }$$ and with no flow through it, the line $y=0$ is a streamline — the bed is reproduced exactly.
3. Note why the velocity test is the right one to use here. The question offers the alternative of showing $\Psi$ is constant along the bed. That route is a trap with this particular superposition: on $y=0$ the two arctangent terms give $\Psi_{\text{source}}=\tfrac{m}{2\pi}\left[\tan^{-1}(-a/x)+\tan^{-1}(a/x)\right]$, and the two logarithms cancel exactly, so $\Psi$ is indeed piecewise constant — but the principal branch of the inverse tangent jumps by exactly m as x passes through zero beneath the pipe. That jump is physically correct (it is the discharged flux crossing the plane through the pipe), yet it looks like a failure of the test. Demonstrating $v\equiv0$ avoids the branch-cut bookkeeping entirely, which is why it is the demonstration given above.
4. Part (c) — differentiate to get the horizontal velocity on the bed. With $u=\partial\Psi/\partial y$, the source terms give $\dfrac{m}{2\pi}\dfrac{x}{x^{2}+(y-y_o)^{2}}$ each, and on $y=0$ the real and image sources contribute equally, for a total of $\dfrac{m}{\pi}\dfrac{x}{x^{2}+a^{2}}$. The vortex terms give $-\dfrac{\Gamma}{2\pi}\dfrac{y-y_o}{x^{2}+(y-y_o)^{2}}$ for the real vortex and the sign-reversed image expression; on $y=0$ these also add, for a total of $\dfrac{\Gamma}{\pi}\dfrac{a}{x^{2}+a^{2}}$. Combining, $$\boxed{\ u(x,0)=\frac{m\,x+\Gamma\,a}{\pi\left(x^{2}+a^{2}\right)},\qquad v(x,0)=0\ }$$ The image system has doubled both contributions relative to the free-space values — the wall reflects the flow back into the fluid. The source part is odd in x (fluid spreads symmetrically both ways from beneath the pipe) while the vortex part is even (a uniform sweep in one direction), and a single stagnation point sits where they cancel, at $x_{\text{stag}}=-\Gamma a/m$.
5. Part (d) — identify the source strength, then solve for the circulation. The pipe sits wholly inside the water at $y=a>0$, so its entire free-space output enters the fluid and $m$ equals the stated discharge directly: $m=0.5\ \text{m}^{3}/\text{s}$ per metre of pipe. (Had the pipe discharged flush with the bed, only half of a free-space source would be real fluid and $m$ would be twice the discharge; that is not the case here.) Evaluating the part-(c) result at the origin, where the source term vanishes, $$u(0,0)=\frac{\Gamma a}{\pi a^{2}}=\frac{\Gamma}{\pi a} \;\Rightarrow\;\Gamma=\pi a\,u(0,0)=\pi\times2\times0.07958,$$ so $$\boxed{\ \Gamma=0.500\ \text{m}^{2}/\text{s}\ }$$ The stated 0.07958 m/s is $1/(4\pi)$ to four figures, which is the signature of $\Gamma=m=0.5$ — the paper has set the two strengths equal. With those values the bed velocity is $u(x,0)=(0.5x+1.0)/\left[\pi(x^{2}+4)\right]$ m/s, plotted in Figure B1(b). Differentiating, $mx^{2}+2\Gamma ax-ma^{2}=0$ locates the peak at $x=a\left(-\Gamma+\sqrt{\Gamma^{2}+m^{2}}\right)/m=2(\sqrt2-1)=0.828\ \text{m}$, where $u=0.09606\ \text{m/s}$; the stagnation point is at $x=-\Gamma a/m=-2.00\ \text{m}$.
6. Part (e) — apply Bernoulli along the bed. The flow is steady, incompressible and irrotational everywhere outside the singularities, so Bernoulli's constant is the same throughout the field. Comparing a general bed point with the far field, where the velocity vanishes and the pressure is $P_o$, and noting that the bed is horizontal so the elevation term cancels, $$P(x,0)+\tfrac{1}{2}\rho\,u^{2}(x,0)=P_o+0 ,$$ hence $$\boxed{\ P(x,0)=P_o-\frac{\rho}{2}\left[\frac{m\,x+\Gamma\,a} {\pi\left(x^{2}+a^{2}\right)}\right]^{2}\ }$$ The bed pressure is everywhere at or below the far-field value, and it is a suction that is largest where the flow runs fastest. With $\rho=1000\ \text{kg/m}^{3}$ the numbers are small: the deficit is 3.17 Pa directly beneath the pipe, peaks at 4.61 Pa at $x=0.828\ \text{m}$, and returns exactly to $P_o$ at the stagnation point $x=-2.00\ \text{m}$. Scour is therefore not a concern — a pressure difference of a few pascals over the bed is three orders of magnitude below what mobilises sediment.