22-Mec-A6 Fluid Machinery · December 2018

Question 2 of 6: Potential Flow — Discharge Pipe and Drain near a Bed

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Potential Flow — Discharge Pipe and Drain near a Bed (20 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. Take the bed as the wall $y=0$ (fluid in $y\gt 0$), with $x$ along the bed and the source at $(0,b)$, the drain (sink) at the origin. Source “strength” $m$ means volume flow per unit depth, so $v_r = m/(2\pi r)$ and $\psi = (m/2\pi)\theta$.

Find. (a) $\psi(x,y)$; (b) proof that $y=0$ is a streamline (bed simulated); (c) $u(x,0)$ along the bed; (d) the force on the source (pipe).

Approach. Represent the solid bed by the method of images: reflect the pipe source in $y=0$ to give an equal image source at $(0,-b)$. The drain lies on the symmetry axis, so it is its own image. Superpose the three line singularities for $\psi$, check the normal velocity on $y=0$, differentiate for the bed velocity, and use the Lagally theorem for the force on the pipe.

1. (a) Stream function. Summing a source $+m$ at $(0,b)$, its image $+m$ at $(0,-b)$, and the sink $-2m$ at the origin: $$\boxed{\psi(x,y)=\frac{m}{2\pi}\Big[\operatorname{atan2}(y-b,\,x)+\operatorname{atan2}(y+b,\,x)-2\,\operatorname{atan2}(y,\,x)\Big].}$$ (The strengths $+m,+m,-2m$ sum to zero, so the flow is closed: the drain swallows exactly the pipe discharge, which is why the sink is $-2m$ — only half of it, $-m$, acts in the physical half-plane $y\gt 0$.)
2. (b) Bed correctly simulated. The velocity normal to the bed is $v=\partial\psi/\partial x$-derived; evaluating the $y$-component of the superposed field on $y=0$, $$v(x,0)=\frac{m}{2\pi}\left[\frac{-b}{x^2+b^2}+\frac{+b}{x^2+b^2}-0\right]=0 \quad\text{for all }x\neq0 .$$ The source and its image contribute equal and opposite vertical velocities, and the drain (on the axis) contributes none; hence $y=0$ is a streamline — the bed is correctly represented. $\psi$ is constant along it (jumping by $m$ across the drain, the flux drawn out).
3. (c) Velocity along the bed. On $y=0$ the vertical velocity vanishes and the horizontal velocity is $$u(x,0)=\frac{m}{2\pi}\!\left[\frac{2x}{x^2+b^2}-\frac{2}{x}\right] =\boxed{-\,\frac{m\,b^{2}}{\pi\,x\,(x^{2}+b^{2})}.}$$ It is directed toward the drain everywhere (for $x\gt 0,\ u\lt 0$; for $x\lt 0,\ u\gt 0$), and vanishes far away as $1/x^3$.
4. (d) Force on the pipe. By the Lagally theorem the force on a source is $\mathbf{F}=-\rho\,m\,\mathbf{V}_{ind}$, where $\mathbf V_{ind}$ is the velocity at the source induced by all other singularities. At $(0,b)$ the image source contributes $+m/(4\pi b)$ upward and the drain contributes $-m/(\pi b)$: $$\mathbf V_{ind}=\Big(0,\ \tfrac{m}{4\pi b}-\tfrac{m}{\pi b}\Big)=\Big(0,\ -\tfrac{3m}{4\pi b}\Big).$$ $$\boxed{\mathbf F=-\rho m\,\mathbf V_{ind}=\Big(0,\ \dfrac{3\rho m^{2}}{4\pi b}\Big)\ \text{per unit length, directed vertically upward (away from the bed).}}$$ The strong drain repels the source more than the bed attracts it, so the net force lifts the pipe.