NivaarExam PrepOfficial exam papers ↗

04-BS-7 · Undated paper

Question 2 of 13: Pitot-Static Tube in an Air Duct

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 2: Pitot-Static Tube in an Air Duct (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.

Part (a) — Sketch and explanation. A pitot-static tube carries two concentric passages: the central, forward-facing tip senses the stagnation (total) pressure where the flow is brought to rest, while a ring of small side ports further back on the probe senses the undisturbed static pressure of the duct flow. Each passage is piped to one leg of a U-tube manometer containing water; because the duct fluid (air) is far less dense than the manometer fluid (water), the manometer deflection is small but readable, and the deflection is driven purely by the dynamic pressure of the flow.

duct, air flow → stagnation tip static ports Δh = 24 mm water manometer
Stagnation port (high pressure, lower manometer leg) vs. static ports (lower pressure, higher manometer leg); the 24 mm difference is the dynamic pressure of the duct flow.

Approach for (b) and (c). The manometer reading converts to the true dynamic pressure via Δp = (ρwater − ρair)gΔh, and that same dynamic pressure equals ½ρairV² by the pitot-static (Bernoulli) relation — so a direct-reading pressure gauge would show exactly the same Δp, just without the density-difference amplification a liquid column provides.

  1. Part (b): dynamic pressure from the manometer reading. With Δh = 24 mm = 0.024 m: $$ \Delta p = (\rho_w-\rho_{air})g\,\Delta h = (1000-1.19)(9.81)(0.024) \approx 235.2\ \text{Pa} $$
  2. Solve for velocity. Setting Δp equal to the dynamic pressure of the duct air: $$ V = \sqrt{\frac{2\Delta p}{\rho_{air}}} = \sqrt{\frac{2(235.2)}{1.19}} $$ $$ \boxed{V \approx 19.9\ \text{m/s}} $$
  3. Part (c): the same velocity read on a differential pressure gauge. A gauge measures the stagnation–static difference directly, with no manometer-fluid amplification factor, so it reads the dynamic pressure itself: $$ \Delta p_{gauge} = \tfrac{1}{2}\rho_{air}V^2 = \tfrac12(1.19)(19.9)^2 $$ $$ \boxed{\Delta p_{gauge} \approx 0.235\ \text{kPa}} $$ This numerically equals the Step 1 result — the manometer and the gauge report the same physical dynamic pressure, the manometer simply displays it magnified into a visible liquid-column height.
QuantityResult
Air velocity in duct19.9 m/s
Differential gauge reading0.235 kPa