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16-Civ-A5 Hydraulic Engineering · December 2016

Question 2 of 6: Twin Pumps Delivering Over a High Point

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

Notes on this paper

Paper format. National Exams, December 2016 — 98-Civ-A5 Hydraulic Engineering. Closed book, 3 hours, one aid sheet permitted. Six questions of equal value (20 marks each); candidates answer any five. All six are solved here as a study resource. Take water density ρ = 1000 kg/m³, kinematic viscosity ν = 1.31 × 10−6 m²/s; local losses and velocity head are negligible unless stated.

Reference texts. Mays, Water Resources Engineering (Wiley) — pipe/pump systems & distribution networks; Chow, Open-Channel Hydraulics (McGraw-Hill) — normal/critical depth, specific energy, unsteady flow; Transportation Association of Canada, Geometric Design Guide for Canadian Roads — gutter/roadway drainage. Permitted equation set (from the exam cover): Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$; Manning $Q = \tfrac{1}{n}A R^{2/3} S^{1/2}$; total dynamic head $\text{TDH} = H_s + H_f$.

Question 2: Twin Pumps Delivering Over a High Point (20 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. Water is pumped uphill from a low reservoir to a higher one, across an intermediate ridge A.

Given data — Question 2
QuantitySymbolValue
Upstream (suction) reservoir level$z_s$13 m
Downstream (discharge) reservoir level$z_d$68 m
High point A: distance / ground elevation$L_A$ / $z_A$1000 m / 55 m
Pipe length, diameter, $C$$L,D,C$3000 m, 0.450 m, 115
Each pump curve (parallel × 2)$\text{TDH}$$80 - 10Q^2$ (m)
Minimum required pressure head at A—28 m

Find. (a) total pumped discharge; (b) pressure head at A and whether it clears the 28 m minimum; (c) if deficient, two remedies.

Suction res. 13 m A (55 m) Discharge res. 68 m P 2 pumps ‖ (parallel) HGL p/γ ≈ 29.5 m at A
Longitudinal profile: two pumps lift water 55 m from the suction reservoir over ridge A to the discharge reservoir. The hydraulic grade line (HGL) clearance above the ridge is the pressure head at A.

Approach. Two identical pumps in parallel share the head and add their flows, giving a combined curve $H = 80 - 2.5Q^2$. Intersect it with the system curve (static lift + friction) to find $Q$; then trace the HGL from the pump to A and subtract the ground elevation.

  1. Combined pump curve (parallel). Parallel pumps deliver the same head, so at head $H$ each passes $q=\sqrt{(80-H)/10}$ and the total is $Q=2q$. Inverting, $H = 80 - 10\,(Q/2)^2 = 80 - 2.5\,Q^2$.
  2. System curve. The head the pipe demands is the static lift plus friction over the full 3 km: $$H_{\text{sys}}(Q) = (z_d - z_s) + h_f = 55 + 0.278^{-1.852}\!\left(\frac{Q}{C D^{2.63}}\right)^{1.852}\!\!L .$$ Numerically $h_f = 3000\,(Q/3.915)^{1.852}$ with $0.278\,C\,D^{2.63}=3.915$.
  3. Operating point. Setting pump = system, $80 - 2.5Q^2 = 55 + h_f(Q)$, and solving iteratively: $$\boxed{Q \approx 0.294\ \text{m}^3/\text{s} = 294\ \text{L/s}}, \qquad \text{TDH} = 80 - 2.5(0.294)^2 = 79.8\ \text{m}.$$ The line is friction-controlled: $h_f \approx 24.8$ m over the 3 km.
  4. HGL at the pump and at A. Immediately downstream of the pumps the HGL sits at $z_s+\text{TDH}=13+79.8 = 92.8$ m. Friction to A (first 1000 m of 3000 m) is one-third of the total: $$h_{f,A} = \tfrac{1}{3}(24.8) = 8.3\ \text{m}, \qquad \text{HGL}_A = 92.8 - 8.3 = 84.5\ \text{m}.$$
  5. Pressure head at A. Subtract the ridge ground elevation: $$\frac{p_A}{\gamma} = \text{HGL}_A - z_A = 84.5 - 55 = \boxed{29.5\ \text{m}}.$$
  6. Compare with the minimum. Since $29.5\ \text{m} \gt 28\ \text{m}$, the pressure head at A is above the 28 m minimum — but only by ~1.5 m of margin.
Final results — Question 2
QuantityResult
Total pumped flow (a)$Q \approx 0.294\ \text{m}^3/\text{s}$ (294 L/s)
Total dynamic head at duty point$\text{TDH} \approx 79.8\ \text{m}$
Pressure head at A (b)$p_A/\gamma \approx 29.5\ \text{m}$ — above 28 m minimum

Part (c). The condition in (b) is satisfied, so no corrective action is strictly required. Because the margin is small (~1.5 m), the following measures would raise the pressure head at A if a larger safety margin were wanted (the same list applies whenever such a ridge point is deficient): (1) increase the pipe diameter between the pumps and A, cutting friction loss and lifting the HGL at A; (2) add a booster pump (or a larger/higher-head pump) to raise the HGL everywhere upstream of A; other options include re-routing the alignment to lower the ridge crossing, or reducing the delivered flow (friction $\propto Q^{1.85}$, so a modest flow cut recovers head quickly).