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04-BS-7 · Undated paper

Question 10 of 13: Capillary Rise Between Two Square-Packed Rod Arrays

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

Notes on this paper

04-BS-7 Mechanics of Fluids — undated sitting, identified as May 2019 (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.” A complete paper is any 10 of the 13, each worth 5 marks. Every question is answered below (13 of 13) so the set can be used as a full study resource. Constants used throughout (from the paper's own Constants page, p.12): g = 9.81 m/s², patm = 100 kPa, ρwater = 1000 kg/m³, SGbenzene = 0.90, SGmercury = 13.56, SGcarbon tetrachloride = 1.59, ρair = 1.19 kg/m³ (20°C), μwater = 1.0×10⁻³ N·s/m², μair = 1.8×10⁻⁵ N·s/m².

Reference texts: F. M. White, Fluid Mechanics, 8th ed. (McGraw-Hill) — fluid statics and manometry (Ch. 2), hydrostatic forces on plane surfaces (Ch. 2), the Bernoulli/continuity pair and orifice flow (Ch. 3), the linear-momentum equation for moving vanes (Ch. 3), pipe friction and the Moody/Colebrook relation (Ch. 6), boundary layers and drag (Ch. 7), capillary rise (Ch. 1), high-lift devices and aircraft wing aerodynamics (J. D. Anderson, Fundamentals of Aerodynamics, Ch. 4–5).

Question 10: Capillary Rise Between Two Square-Packed Rod Arrays (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.

Approach. The general capillary-rise relation on the paper's own reference page expresses rise in terms of the wetted perimeter and cross-sectional area of the interstitial pore space, h = (σcosθ/ρg)×(perimeter/area); because Array B is simply Array A scaled up by a factor of 2 in every linear dimension, the perimeter-to-area ratio — and hence the rise — scales inversely with rod diameter.

Array A (small, d) Array B (large, 2d)
Array B repeats Array A's pattern at exactly twice the linear scale.
  1. Geometry of one square pore. For touching rods of diameter d in a square array, one unit cell (four quarter-rods bounding one pore) has: $$ \text{area} = d^2\left(1-\tfrac{\pi}{4}\right), \qquad \text{perimeter} = \pi d $$
  2. Perimeter-to-area ratio scales as 1/d. $$ \frac{\text{perimeter}}{\text{area}} = \frac{\pi d}{d^2(1-\pi/4)} = \frac{\pi}{d(1-\pi/4)} \propto \frac{1}{d} $$
  3. Apply the capillary-rise formula. With Array B's rods at 2d: $$ \frac{h_A}{h_B} = \frac{(\text{perimeter/area})_A}{(\text{perimeter/area})_B} = \frac{1/d}{1/(2d)} $$ $$ \boxed{h_A = 2\,h_B} $$
QuantityResult
Array with greater capillary riseArray A (small rods)
Ratio of riseshA = 2hB