Radial Flow to a Drain Hole: 04-BS-7, May 2013, Q4

Question 4 of 13: Radial Flow to a Drain Hole

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

Notes on this paper

04-BS-7 Mechanics of Fluids — National Examination, 2013-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.

Reference texts: Crowe, C.T., Elger, D.F. & Roberson, J.A., Engineering Fluid Mechanics (the exam's own Moody chart and drag-coefficient chart are reproduced from this text); Douglas, J.F., Gasiorek, J.M., Swaffield, J.A. & Jack, L.B., Fluid Mechanics; White, F.M., Fluid Mechanics.

Check — assumptions used across this paper:

Where a question does not restate an ambient temperature, air is taken at the ISA sea-level standard of 15°C, giving ρ_air = 1.21 kg/m³ (the Constants table's 15°C value) — used in Q5, Q6, Q7 and Q9.
Q8's Moody chart and Q9's sphere drag-coefficient chart are supplied as attachments. Both are solved via the equations the charts themselves plot: the Colebrook–White equation for Q8 (the Moody chart is a graphical solution of Colebrook–White) and the Morrison (2013) curve-fit for sphere drag versus Reynolds number for Q9 (which reproduces the published "Sphere" curve to within a few percent over this Re range).
Q1, Q3 and Q11 depend on their figures; the relevant crop of the exam page is reproduced beside each, and every reading used is taken from it.

Question 4: Radial Flow to a Drain Hole (5 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. $Q=0.300\ \text{L/s}=3.00\times10^{-4}\ \text{m}^3/\text{s}$; radial distance from the hole $r=100\ \text{mm}=0.100\ \text{m}$; flow approaches the hole radially and symmetrically (hemispherical control surface, since the hole sits in a flat tank bottom).

Find. The velocity in the tank at $r=100$ mm from the hole.

Approach. Continuity through a hemispherical control surface of radius $r$ centred on the hole: the same flow $Q$ passes through every such surface as $r$ shrinks toward the hole, so $V(r) = Q/A(r)$ with $A(r)=2\pi r^2$ (a hemisphere, since the tank floor blocks the other half of a full sphere).

Hemispherical area at r = 100 mm. $A = 2\pi r^2 = 2\pi(0.100)^2 = 0.06283\ \text{m}^2$.
Velocity from continuity. $$V = \frac{Q}{A} = \frac{3.00\times10^{-4}}{0.06283} = \boxed{4.775\times10^{-3}\ \text{m/s} = 4.78\ \text{mm/s}}$$

Quantity	Value
Velocity at r = 100 mm	4.78 mm/s

Topic: Continuity for converging (sink) flow.

Key relations: continuity $Q=VA$; hemispherical control-surface area $A=2\pi r^2$.

Why this works: far from the hole, streamlines converge radially and symmetrically toward the opening; because the tank floor is a solid boundary, the flow occupies only the upper half-space, so the control surface at radius r is a hemisphere, not a full sphere. As $r\to0$ the same discharge is squeezed through a vanishing area, so velocity rises sharply near the hole — this is the classic "sink flow" idealization used before the flow accelerates into the orifice itself.

Common pitfall: using a full spherical area ($4\pi r^2$) instead of a hemisphere, which understates the velocity by a factor of two.

Source: Crowe, Elger & Roberson, Engineering Fluid Mechanics, Ch. 4 (continuity/control volumes).