Question 1 of 13: Capillary Rise in a Rectangular Channel
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2017-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Graphical & 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: White, F.M., Fluid Mechanics (8th ed.) — fluid statics and capillarity (Ch. 2), integral analysis and buoyancy (Ch. 3), viscous flow in ducts, pipe friction and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), open-channel flow and the hydraulic jump (Ch. 10).
Check — assumptions used across this paper:
Q2's figure shows the vertical leaf of the gate rising above the free surface, so water acts over the full depth x = 2.0 m; both "0.3 m" dimensions locate the centre of gravity (0.3 m right of the vertical leaf, 0.3 m above the arm). Reservoir water also fills the space beneath the 1.2 m arm (the vertical lines below O are dimension extension lines, not a wall), so the arm carries uplift at pressure ρgx, and the arm tip bears up against a lip on the spillway crest — the gate can only open by rotating clockwise, vertical leaf toward the spillway.
Q4's flow coefficient K ≈ 0.69 is read from the attached VDI chart on the Do/D1 = 0.50 curve (third from the top, the one that levels off at 0.62) at the approach Reynolds number R ≈ 2550. The printed curve gives 0.695 at R = 2000 and 0.684 at R = 3000; a reading of 0.68–0.70 moves the answer by only ±0.3 kPa.
Q7 and Q8 friction factors are obtained from the Colebrook–White/Haaland equation the Moody chart itself plots (smooth-wall case, as both problems specify or imply smooth surfaces).
Q9's sphere drag coefficient vs. Reynolds number is obtained from the Schiller–Naumann correlation $C_D=\tfrac{24}{Re}\left(1+0.15\,Re^{0.687}\right)$, which reproduces the published sphere-drag curve (the same curve reprinted in the attachment) to within a few percent for $Re<1000$ — used here because the resulting Reynolds number (≈19–20) is well above the range where the simple Stokes'-law formula ($C_D=24/Re$) alone is valid.
Question 1: Capillary Rise in a Rectangular Channel (5 marks)
Given. Rectangular channel cross-section $a=1\ \text{mm}$, $b=2\ \text{mm}$; water surface tension $\sigma=0.073\ \text{N/m}$; contact angle $\theta=0^\circ$; water density $\rho=1000\ \text{kg/m}^3$ (Constants table); $g=9.81\ \text{m/s}^2$.
Water rises inside the narrow rectangular channel; the meniscus wets the glass fully at θ=0°.
Find. The equilibrium capillary rise height $h$; explain the role of the contact angle.
Approach. Apply the reference-sheet capillary-rise relation for a non-circular tube, $h=(\sigma\cos\theta/\rho g)\times(\text{perimeter/area})$, using the wetted perimeter and cross-sectional area of the rectangular channel.
Perimeter and area of the channel. $P=2(a+b)=2(0.001+0.002)=0.006\ \text{m}$; $A=ab=0.001\times0.002=2\times10^{-6}\ \text{m}^2$, so $P/A=3000\ \text{m}^{-1}$.
Capillary rise. Substituting into the reference equation,
$$h=\frac{\sigma\cos\theta}{\rho g}\times\frac{P}{A}=\frac{0.073\times\cos0^\circ}{1000\times9.81}\times3000=\boxed{0.02232\ \text{m} = 22.3\ \text{mm}}$$
Quantity
Value
Wetted perimeter / area, P/A
3000 m⁻¹
Capillary rise h
22.3 mm
The contact angle enters the rise formula purely through the factor $\cos\theta$. At $\theta=0^\circ$ (water fully wetting clean glass) $\cos\theta=1$ and the rise is maximum. As $\theta$ increases toward $90^\circ$, $\cos\theta\to0$ and the rise vanishes — the liquid surface inside the tube becomes flat, level with the free reservoir surface. For $\theta>90^\circ$ (a non-wetting liquid, e.g. mercury against glass), $\cos\theta$ is negative and the formula predicts a capillary depression: the liquid inside the narrow passage sits below the free surface, with a convex (bulging) meniscus instead of the concave one sketched above. Physically, $\theta$ measures the balance between the liquid's cohesive forces and its adhesive attraction to the solid wall — a small $\theta$ means adhesion dominates and the liquid climbs the wall, pulling the whole meniscus up with it.
