Forces on a Discharge Pipe near a Tank Bed (Potential Flow): 22-Mec-A6 Fluid Machinery, May 2017, Q2

Question 2: Forces on a Discharge Pipe near a Tank Bed (Potential Flow) (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. Two-dimensional incompressible potential flow: a source of strength $m$ at $(0,b)$ and a sink of strength $-2m$ at the origin $(0,0)$, which lies on the flat tank bed $y=0$; fluid occupies $y\gt0$.

Find. (a) the stream function $\psi$; (b) proof that $y=0$ is a streamline (no normal velocity); (c) $u(x)$ along the bed; (d) the resultant force per unit length on the pipe (source).

Figure 2: Source at $(0,b)$, sink $-2m$ at the origin, and the image source $+m$ at $(0,-b)$ that renders the bed $y=0$ a streamline.

Approach. Model the flat wall by the method of images: reflect the source across $y=0$ with an image source of equal sign and strength. Superpose stream functions, verify the wall, differentiate for the bed velocity, and apply the Lagally theorem for the force on the source.

(a) Build the image system and superpose. A wall at $y=0$ requires an image source $+m$ at $(0,-b)$; the sink sits on the wall and is its own image. With $\theta_i=\operatorname{atan2}(y-y_i,\,x-x_i)$ for each singularity, $$\boxed{\psi=\frac{m}{2\pi}\Big[\operatorname{atan2}(y-b,\,x)+\operatorname{atan2}(y+b,\,x)\Big]-\frac{2m}{2\pi}\operatorname{atan2}(y,\,x).}$$
(b) Verify the bed is a streamline. The normal (vertical) velocity on $y=0$ is $v=\partial\psi/\partial x$ evaluated there; more directly, superpose the radial fields. At a bed point $(x,0)$ the source at $(0,b)$ contributes $v_y=\dfrac{m}{2\pi}\dfrac{-b}{x^2+b^2}$ and the image at $(0,-b)$ contributes $v_y=\dfrac{m}{2\pi}\dfrac{+b}{x^2+b^2}$; these cancel, and the sink on the axis gives $v_y=0$. Thus $v_y\equiv0$ along the entire bed, so $y=0$ is a streamline — the wall is correctly simulated.
(c) Velocity along the bed. Only the horizontal component survives. Summing the three contributions at $(x,0)$, $$u(x)=\frac{m}{2\pi}\frac{2x}{x^2+b^2}-\frac{2m}{2\pi}\frac{1}{x}=\frac{m}{\pi}\left[\frac{x}{x^2+b^2}-\frac1x\right],$$ $$\boxed{u(x)=-\frac{m\,b^{2}}{\pi\,x\,(x^{2}+b^{2})}.}$$ The flow along the bed is everywhere drawn toward the drain at the origin, as expected.
(d) Force on the pipe. By the Lagally theorem the force per unit length on a source equals $\rho m$ times the velocity induced at its location by all other singularities. At $(0,b)$ the image source contributes $v_y=+\dfrac{m}{4\pi b}$ and the sink contributes $v_y=-\dfrac{m}{\pi b}$, so the induced velocity is $\left(0,\,-\dfrac{3m}{4\pi b}\right)$. Hence $F_x=0$ and $$\boxed{F_y=-\frac{3\rho m^{2}}{4\pi b}\quad(\text{directed toward the bed}).}$$ The strong drain ($-2m$, nearer at $b$) overpowers the repulsion of the image source (at $2b$), so the net force pulls the pipe down toward the tank bed.

Question 2 — results
Quantity	Result
Stream function	$\psi=\dfrac{m}{2\pi}[\theta_{\text{src}}+\theta_{\text{img}}]-\dfrac{m}{\pi}\theta_{\text{sink}}$
Bed normal velocity	$v_y=0$ (wall verified)
Bed velocity	$u(x)=-\dfrac{m b^{2}}{\pi x (x^{2}+b^{2})}$
Force on pipe	$F_x=0,\quad F_y=-\dfrac{3\rho m^{2}}{4\pi b}$ (toward bed)

Topic: Two-dimensional potential flow — superposition of sources/sinks, the method of images for a plane wall, and forces on singularities.

Key relations: source stream function $\psi=\tfrac{m}{2\pi}\theta$ and radial velocity $v_r=\tfrac{m}{2\pi r}$; image of a source in a wall is a like source; Lagally theorem $\mathbf F=\rho m\,\mathbf u_{\text{induced}}$.

Why this works: potential flows superpose linearly, so a boundary is enforced by adding image singularities that make the boundary a streamline. Reflecting a source with a source of the same sign makes the wall-normal velocities cancel in pairs, leaving $v_y=0$ on the wall. Once the field is known, the Lagally theorem gives the force on each singularity from the regular part of the velocity there: like singularities induce a mutually repulsive velocity, opposite ones attractive, so the sign of each contribution follows physical intuition (the source is pushed from its image but pulled toward the sink).

Common pitfall: using an image of the wrong sign (a sink) — that would make the wall a stagnation streamline of the wrong type and give zero tangential velocity. Also, the singularity sitting on the wall (the drain) must not be doubled; being on the plane of symmetry it is already self-imaged.

Source: Currie, Fundamental Mechanics of Fluids, Ch. 4–5; White, Fluid Mechanics, 8th ed., §8.3–8.4 (Lagally / image methods).

Question 2 of 6: Forces on a Discharge Pipe near a Tank Bed (Potential Flow)

Question 2: Forces on a Discharge Pipe near a Tank Bed (Potential Flow) (20 marks)