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04-BS-7 · May 2016

Question 8 of 13: Centrifugal Pump – System Operating Point

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 8: Centrifugal Pump – System Operating Point (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
Pump curve constantsA = 0.000060, B = 1.6
System curve constantsC = 20 m (static head), D = 10
Pump speed N900 rev/min

Find. The operating point (Q, H) where the pump and system curves intersect.

Approach. The operating point is where the pump's H–Q curve and the system's H–Q curve intersect, i.e. where both equations give the same H for the same Q — set them equal and solve for Q, then back-substitute for H.

  1. Set pump head equal to system head. $$AN^2 - BQ^2 = C + DQ^2 \quad\Rightarrow\quad Q^2 = \frac{AN^2-C}{B+D}$$
  2. Substitute the given values. $$AN^2 = 0.000060(900^2) = 48.6\ \text{m}, \qquad Q^2 = \frac{48.6-20}{1.6+10} = \frac{28.6}{11.6} = 2.4655$$ $$Q = \boxed{1.570\ \text{m}^3/\text{s}}$$
  3. Back-substitute for H (system curve; check against the pump curve). $$H = C+DQ^2 = 20+10(2.4655) = \boxed{44.66\ \text{m}}$$ Check: $AN^2-BQ^2 = 48.6-1.6(2.4655)=48.6-3.94=44.66$ m — matches.
Q (m3/s)H (m)pump: H=AN2-BQ2system: H=C+DQ2operating point (1.57, 44.7)0.01.02.0
Fig. Q8 — pump characteristic (falling) and system characteristic (rising) intersect at the operating point Q=1.57 m³/s, H=44.66 m.
QuantityResult
Operating flow Q1.570 m³/s
Operating head H44.66 m