04-BS-7 · May 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
04-BS-7 Mechanics of Fluids — May 2015 (National Examinations, three hours, closed book). Section A (Calculative) offers 9 questions and instructs "do seven"; Section B (Analytical) offers 4 questions and instructs "do three." Every question is answered below (13 of 13), so students can use the full paper as a study resource.
Reference texts: F. M. White, Fluid Mechanics, 8th ed. (McGraw-Hill) — fluid statics (Ch. 2), the linear-momentum and energy equations (Ch. 3), pipe friction and the Moody chart (Ch. 6), and drag on immersed bodies (Ch. 7).
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.
Draining through a small orifice follows Torricelli's law, exit velocity V = √(2gh), where h is the instantaneous depth of water above the hole. The instantaneous drain rate is then Q = CdAhole√(2gh), and by mass conservation this must equal −Asurface(h)(dh/dt), where Asurface(h) is the free-surface area at that instant. Combining and separating variables, the total drain time is
$$t_{drain} = \int_0^{h_{max}} \frac{A_{surface}(h)}{C_d A_{hole}\sqrt{2gh}}\,dh$$
(a) Which orientation drains faster. With axis vertical, the free surface is the drum's full circular end (constant area, A = πD²/4, for the whole drain). With axis horizontal, the free surface is a chord of the circular cross-section that SHRINKS as the water level falls, and for most of the drain (away from the very top and bottom) that chord width is smaller than the vertical orientation's constant circular area of the same nominal size. With height equal to diameter, the horizontal orientation drains faster overall, because the reduced surface area for most of the draining history more than compensates for the brief periods (near-full and near-empty) where the horizontal chord is momentarily wide.
(b) Why. The integral above shows drain time is directly proportional to the free-surface area at every instant — a smaller Asurface(h) at a given h means less volume must pass through the hole to drop the level by dh, so the level (and hence the driving head h) falls faster. The vertical orientation keeps the largest possible surface area (a full circle) present for the entire drain, which is the slowest possible arrangement for a given hole and total volume.
(c) A faster alternative. Tilting the drum so its axis is neither vertical nor horizontal, with the drain hole at the lowest corner and the drum resting so that the water surface area shrinks monotonically and rapidly as it drains (approaching a cone/wedge-like emptying shape rather than a cylinder draining side-on) reduces the average free-surface area further still. In the practical limit, standing the drum on a corner/edge so it drains through a steadily narrowing wedge-shaped water volume gives the shortest time of the three, because Asurface(h) can be made to shrink toward zero well before the volume is exhausted, in contrast to both the vertical case (constant full-circle area throughout) and the horizontal case (which still passes through its widest chord at mid-depth).
