Question 2 of 6: Discharge pipe above a tank bed with a drain

Question 2: Discharge pipe above a tank bed with a drain (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. A two-dimensional line source of strength m representing the discharge pipe, sitting a height b above a flat, impermeable tank bed; a line sink of strength −2m representing the drain, located on the bed directly beneath the pipe; fluid density ρ; quiescent fluid far away; free-surface effects neglected. Coordinates are taken from the figure: x along the bed, y vertically upward, origin at the drain, so the pipe sits at (0, b).

Find. A stream function for the flow, a demonstration that it really does reproduce the flat bed, the velocity distribution along that bed, and the resultant force per unit length that the flow exerts on the discharge pipe.

[Figure not reproduced: Figure 2 (redrawn) with the image system added. Mirroring the source in the bed makes y = 0 a streamline; the drain sits on the plane of symmetry, so its printed strength −2 m already includes its own image and exactly absorbs the 2 m the source pair emits. See the official exam paper.]

Approach. Build the flow from three elementary singularities — the real source, its image in the bed, and the drain — then verify the wall by showing the normal velocity vanishes, differentiate for the bed velocity, and evaluate the force with the Blasius/Lagally theorem, whose one non-obvious ingredient is that the singularity does not act on itself.

Throughout, a line source of strength m is taken in the standard convention $$\psi=m\theta,\qquad \phi=m\ln r,\qquad u_r=\frac{m}{r},$$ so the volume discharged per unit length of pipe is Q = 2πm. (Papers that define the strength as the discharge itself simply replace m by Q/2π everywhere below; every formula that follows is homogeneous of degree two in m, so nothing else changes.)

1. Part (a) — place the image and write the stream function. A plane wall is reproduced by reflecting each singularity in it with the same sign for a source. Mirroring the pipe at (0, b) puts an image source of strength m at (0, −b). The drain already lies on the wall, so its image coincides with itself — which is exactly why the paper prints its strength as −2m. Superposing the three, $$\boxed{\;\psi(x,y)=m\arctan\!\frac{y-b}{x}+m\arctan\!\frac{y+b}{x}-2m\arctan\!\frac{y}{x}\;}$$ with each arctangent understood as the two-argument form measured from the positive x axis.
2. Check the mass budget before going further. In the full-plane model the source pair emits 2π(2m) and the drain absorbs 2π(2m): the system is closed, which is the physical statement that the drain removes exactly what the pipe discharges. Had the drain been given strength −m, half the discharge would have had to escape to infinity and the far-field “quiescent” assumption in the question would be violated.
3. Part (b) — verify the bed. The bed is correctly simulated if the normal velocity vanishes on it. Differentiating the superposition, the vertical component at a general point is $$v=\frac{m(y-b)}{x^{2}+(y-b)^{2}}+\frac{m(y+b)}{x^{2}+(y+b)^{2}}-\frac{2my}{x^{2}+y^{2}},$$ and setting y = 0 the first two terms cancel identically while the third vanishes: $$v(x,0)=\frac{-mb}{x^{2}+b^{2}}+\frac{mb}{x^{2}+b^{2}}-0=0\quad\text{for all }x\neq 0 .$$ So y = 0 is a streamline and the bed is exactly reproduced.
4. A warning about testing the wall with ψ instead of v. Evaluating the stream function on the bed gives ψ = 0 for x > 0 but ψ = −2πm for x < 0. That jump is not a failure: it equals the volume flux the drain removes, and any streamline function must be multivalued across a sink placed on the boundary. The physically meaningful test is the vanishing normal velocity in Step 3, and it is the test to present.
5. Part (c) — the velocity along the bed. The horizontal component is $$u=\frac{mx}{x^{2}+(y-b)^{2}}+\frac{mx}{x^{2}+(y+b)^{2}}-\frac{2mx}{x^{2}+y^{2}},$$ which on y = 0 collapses neatly: $$u(x,0)=\frac{2mx}{x^{2}+b^{2}}-\frac{2m}{x}=2m\,\frac{x^{2}-(x^{2}+b^{2})}{x\,(x^{2}+b^{2})}$$ $$\boxed{\;u(x,0)=-\frac{2mb^{2}}{x\,(x^{2}+b^{2})},\qquad v(x,0)=0\;}$$ The sign is the physically satisfying part: u is negative for x > 0 and positive for x < 0, so the bed flow converges on the drain from both directions, as it must when the drain swallows the entire discharge. The speed decays like 2mb²/x³ far away and grows without bound at the drain itself, where the point-sink idealisation loses meaning within a radius of order the real drain opening.
6. Part (d) — set up the force calculation. Work with the complex potential of the whole system, $$w(z)=m\ln(z-ib)+m\ln(z+ib)-2m\ln z,\qquad \frac{dw}{dz}=u-iv .$$ The Blasius theorem applied to a small contour around the pipe, combined with the fact that a singularity exerts no force on itself, gives the Lagally result $$X-iY=-2\pi\rho\,m\,W_{0},$$ where W₀ is the complex velocity of everything except the pipe itself, evaluated at the pipe’s own position.
7. Evaluate the induced velocity at the pipe. At z₀ = ib the image and the drain contribute $$W_{0}=\frac{m}{ib+ib}+\frac{-2m}{ib}=-\frac{im}{2b}+\frac{2im}{b}=\frac{3im}{2b}.$$ The image alone would push the pipe upward at m/2b; the drain, being twice as strong and only half as far in this sense, reverses and triples the result.
8. Collect the force. Substituting, $$X-iY=-2\pi\rho m\left(\frac{3im}{2b}\right)=-\frac{3i\pi\rho m^{2}}{b},$$ so that $$\boxed{\;X=0,\qquad Y=+\frac{3\pi\rho m^{2}}{b}\ \ \text{per unit length (directed away from the bed)}\;}$$ The horizontal force vanishes because the image and the drain are both on the vertical axis through the pipe, leaving nothing to break the left–right symmetry.
9. Sanity-check against the classical wall result. Switch the drain off and the same formula returns Y = −πρm²/b — the textbook result that a source is attracted to a nearby wall, which can be confirmed independently by integrating the bed pressure. Turning the drain on flips the sign and multiplies the magnitude by three: the pipe is pushed away from the bed. That reversal is the engineering message for the supervisor — opening the drain does not merely change the load on the pipe supports, it changes its direction.
10. Put numbers on it. For a discharge of Q = 0.40 m³/s per metre of pipe (so m = Q/2π = 0.0637 m²/s), a standoff b = 1.5 m and ρ = 998 kg/m³, $$Y=\frac{3\pi(998)(0.0637)^{2}}{1.5}=25.4\ \text{N/m (upward)},$$ and the bed speed one metre from the drain is 2(0.0637)(1.5)²/[1.0(1.0 + 2.25)] = 0.088 m/s — slow enough that scour is not a concern, while the 25 N per metre of uplift is a real design load on a long pipe run.