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04-BS-7 · May 2015

Question 5 of 13: Pitot-Static Tube Air-Velocity Measurement

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

Notes on this paper

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 5: Pitot-Static Tube Air-Velocity Measurement (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.

QuantityValue
Water manometer reading, Δh24 mm
Manometer fluid density1000 kg/m³
Air density (20°C, Constants page)1.19 kg/m³

Find. (a) the air velocity, and (b) the pressure-gauge reading a differential pressure gauge would show for the same flow.

airstagnation tapstatic tapΔh = 24 mmV = 19.89 m/s
Fig. Q5 — the stagnation tap reads p0, the static tap reads p; the manometer height is just one way of displaying the same Δp a pressure gauge would read directly.

Approach. Convert the manometer reading to a dynamic pressure using the water-manometer relation, then invert the stagnation-pressure formula to get velocity; the "differential gauge" reading is simply that same Δp expressed in kPa.

  1. Dynamic pressure from the manometer (water column, air above it is negligible in the manometer leg): $$\Delta p = \rho_{water}\,g\,\Delta h = 1000(9.81)(0.024) = 235.4\ \text{Pa}$$
  2. Pitot-static relation — the stagnation-to-static pressure difference equals the dynamic pressure of the oncoming air: $$\Delta p = \tfrac{1}{2}\rho_{air}V^2 \ \Rightarrow\ V=\sqrt{\frac{2\Delta p}{\rho_{air}}}$$
  3. Solve for V: $$V=\sqrt{\frac{2(235.4)}{1.19}}=\boxed{19.89\ \text{m/s}}$$
  4. Differential pressure gauge reading. A pressure gauge connected across the same two taps measures the identical physical quantity — the same 235.4 Pa the water column supported — just displayed directly instead of as a column height: $$\Delta p_{gauge} = \boxed{0.235\ \text{kPa}}$$
QuantityResult
Dynamic pressure235.4 Pa
Air velocity19.89 m/s
Equivalent gauge reading0.235 kPa