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04-BS-7 · May 2016

Question 5 of 13: Total Head Loss from a Reservoir-Fed Pipe Test

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

Notes on this paper

04-BS-7 Mechanics of Fluids — May 2016 (National Examinations, three hours, closed book). Section A (Calculative) offers 9 questions and instructs "do seven"; Section B (Analytical/Graphical) 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. Constants used throughout (from the paper's own Constants page): g = 9.81 m/s², patm = 100 kPa (an atmospheric head of 10 m of water is specified separately for Question 1), ρwater = 1000 kg/m³, SGglycerine = 1.26, SGmercury = 13.56, ρconcrete = 2400 kg/m³, ρair = 1.19 kg/m³ (20°C) / 1.21 kg/m³ (15°C), μwater = 1.0×10⁻³ N·s/m², Rair = 287 J/kg·K.

Reference texts: F. M. White, Fluid Mechanics, 8th ed. (McGraw-Hill) — fluid statics and manometry (Ch. 2), hydrostatic forces and the middle-third rule (Ch. 2), dimensional analysis and drag (Ch. 5, 7), pipe friction and the Moody/Colebrook relation (Ch. 6), control-volume momentum (Ch. 3); B. R. Munson et al., Fundamentals of Fluid Mechanics — jets, orifices, and streamline patterns (Ch. 5, 8); J. D. Anderson, Fundamentals of Aerodynamics — wave/compressibility drag divergence (Ch. 5) for the Boeing 747 wind-tunnel chart used in Question 9.

Question 5: Total Head Loss from a Reservoir-Fed Pipe Test (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
Pipe inlet elevation z16 m
Reservoir surface above the inlet3 m (⇒ surface at z = 9 m)
Pipe outlet elevation z2 (open to atmosphere)4 m
Pipe diameter D50 mm
Pipe length L120 m
Flow rate Q0.006 m³/s

Find. The total head loss hL in the system.

datumWL +3 mtotal head loss h_L = 4.52 mopen end, z=4 mz=6 mD, L pipe
Fig. Q5 — reservoir at +9 m feeds a 50 mm, 120 m pipe discharging at +4 m; the HGL (red) drops 4.52 m of total head loss below the reservoir level.

Approach. Apply the steady-flow energy equation between the reservoir free surface (V ≈ 0, p = 0 gauge) and the open pipe outlet (p = 0 gauge, V = Q/A), with the smooth entrance meaning all losses are lumped into a single hL term.

  1. Exit velocity. $$A = \frac{\pi}{4}(0.050)^2 = 1.9635\times10^{-3}\ \text{m}^2, \qquad V = \frac{Q}{A} = \frac{0.006}{1.9635\times10^{-3}} = 3.056\ \text{m/s}$$
  2. Energy equation, reservoir surface (z=9 m) to pipe outlet (z=4 m). $$0 + 9 + 0 = 0 + 4 + \frac{V^2}{2g} + h_L$$
  3. Solve for hL. $$h_L = (9-4) - \frac{V^2}{2g} = 5 - \frac{3.056^2}{2(9.81)} = 5 - 0.476 = \boxed{4.52\ \text{m}}$$
QuantityResult
Exit velocity V3.06 m/s
Total head loss hL4.52 m