Question 5 of 6: Question 5 (Part B, Question B2): Potential-Flow Model of a Pool Recirculating System

Question 5 (Part B, Question B2): Potential-Flow Model of a Pool Recirculating System (30 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. A stream function that represents the pool, proof that the floor is a streamline, the sink strength M, and the pressure difference between the two floor taps.

Approach. Build the flow by superposition of elementary irrotational solutions — a line source for the pipe, its mirror image to make the floor impermeable, and a line sink at the origin for the drain — then close the mass balance to size the sink and apply Bernoulli along the floor for the pressure difference.

1. Part (a) — choose the elementary singularities. With gravity neglected and the pool very large, the flow away from the pipe and the drain is inviscid and irrotational, so Laplace's equation governs and solutions may simply be added. The pipe is a line source of strength $m$ at $(0,a)$, whose stream function is $\psi = \tfrac{m}{2\pi}\theta$ measured about its own centre. The drain is a line sink of strength $M$ at the origin. Left alone these two do not satisfy the floor, so a third element is needed.
2. Part (a) — enforce the floor with an image. The floor $y = 0$ must be impermeable, and the method of images supplies exactly that: place an identical source of strength $m$ at the mirror point $(0,-a)$. The complete stream function in the pool is then $$\psi(x,y) = \frac{m}{2\pi}\left[\arctan\!\frac{y-a}{x} + \arctan\!\frac{y+a}{x}\right] - \frac{M}{2\pi}\arctan\!\frac{y}{x}$$ with the corresponding velocity potential $\phi = \tfrac{m}{2\pi}\left[\ln r_1 + \ln r_2\right] - \tfrac{M}{2\pi}\ln r_0$, where $r_1$, $r_2$ and $r_0$ are the distances from the pipe, its image and the drain.
3. Part (a) — verify that the floor is correctly modelled. The test is that the normal velocity vanishes on $y = 0$. The real source contributes a vertical velocity $-\tfrac{m}{2\pi}\tfrac{a}{x^{2}+a^{2}}$ there and the image contributes $+\tfrac{m}{2\pi}\tfrac{a}{x^{2}+a^{2}}$, which cancel exactly; the sink at the origin drives purely radial flow along the floor and so contributes no vertical velocity either. Hence $$v(x,0) = 0 \quad \text{for all } x \neq 0 \;\Longrightarrow\; \boxed{y = 0 \text{ is a streamline}}$$ Evaluating the stream function confirms it: $\psi = 0$ along the whole positive-$x$ floor and $\psi = -m$ along the negative-$x$ floor, each a constant, and the jump of exactly $m$ between them is the flow the drain removes — the drain is the singular point at which the two half-streamlines terminate.
4. Part (b) — size the sink by mass balance. The pipe discharges its full strength $m$ into the pool, since it sits inside the region $y>0$. The drain, however, lies on the boundary: a line sink of strength $M$ placed at the origin draws fluid from all directions, but only the half of its intake that lies in $y > 0$ is real water. Balancing what enters against what leaves, $$m = \tfrac{1}{2}M \;\Longrightarrow\; M = 2m = \boxed{2\ \text{m}^{3}\text{/s per metre}}$$ The factor of two is easy to lose and is the crux of this part: a drain in the floor must be given twice the nominal strength of the source it balances.
5. Part (c) — velocity along the floor. Adding the three horizontal contributions at a point $(x,0)$ — source, image and sink — gives $$u(x,0) = \frac{m}{2\pi}\frac{x}{x^{2}+a^{2}} + \frac{m}{2\pi}\frac{x}{x^{2}+a^{2}} - \frac{2m}{2\pi}\frac{1}{x} = -\,\frac{m\,a^{2}}{\pi\,x\,(x^{2}+a^{2})}$$ The result is negative for $x>0$, confirming that the floor flow runs back toward the drain, and it decays like $x^{-3}$ far away because the source and its image form a doublet-like pair when seen from a distance.
6. Part (c) — evaluate at the two taps. Substituting $m = 1$ and $a = 3$ m, $$\begin{aligned} u_b &= -\frac{(1)(9)}{\pi(1)(1+9)} = -0.2865\ \text{m/s} \\ u_c &= -\frac{(1)(9)}{\pi(2)(4+9)} = -0.1102\ \text{m/s} \end{aligned}$$ so the water is moving roughly two and a half times faster at the inner tap than at the outer one.
7. Part (c) — apply Bernoulli along the floor. The floor is a streamline of a steady irrotational flow with gravity neglected, so $p + \tfrac12\rho u^{2}$ is the same at both taps: $$p_b - p_c = \tfrac{1}{2}\rho\left(u_c^{2} - u_b^{2}\right) = \tfrac{1}{2}(1000)\left[(-0.1102)^{2} - (-0.2865)^{2}\right] = \boxed{-35.0\ \text{Pa}}$$ The negative sign says the inner tap reads the lower pressure, by 35.0 Pa, which is what one expects where the flow accelerates into the drain. A gauge connected between the two taps would read about 35 Pa, and because the reading scales as the square of the source strength it is a sensitive monitor of the recirculation rate.

Final Results.