Question 1 of 6: Gravity PVC Main — Velocity Check and Re-sizing
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 1: Gravity PVC Main — Velocity Check and Re-sizing (20 marks)
Given. A single gravity main connects two fixed water surfaces whose difference drives the flow.
Given data — Question 1
Quantity
Symbol
Value
Upstream reservoir level
$z_1$
70 m
Downstream tank level
$z_2$
49 m
Hazen–Williams coefficient
$C$
132
Internal diameter
$D$
1057 mm = 1.057 m
Pipe length
$L$
1200 m
Find. (a) the flow velocity $v$ in the 1057 mm pipe; (b) whether $v \lt 3$ m/s and, if not, the diameter that limits the velocity to 3 m/s.
Approach. With minor and velocity-head losses negligible, the whole level difference is dissipated as pipe friction, so $S = (z_1-z_2)/L$; apply Hazen–Williams for the discharge, then $v = Q/A$. For (b), recognise that in a fixed-head gravity main the velocity is itself a function of diameter.
Available head and friction slope. All of the reservoir-to-tank drop is friction head:
$$h_f = z_1 - z_2 = 70 - 49 = 21\ \text{m}, \qquad S = \frac{h_f}{L} = \frac{21}{1200} = 0.0175.$$
Velocity in part (a). The cross-sectional area is $A = \tfrac{\pi}{4}D^2 = \tfrac{\pi}{4}(1.057)^2 = 0.877\ \text{m}^2$, so
$$v = \frac{Q}{A} = \frac{4.78}{0.877} = \boxed{5.44\ \text{m/s}}.$$
Compare with the 3 m/s limit. Since $5.44\ \text{m/s} \gt 3\ \text{m/s}$, the design guideline is not satisfied — the pipe must be re-sized.
How velocity depends on diameter (key insight). The head is fixed by the two water levels, so combining $v = Q/A$ with Hazen–Williams gives
$$v = \frac{0.278\,C\,D^{2.63}\,S^{0.54}}{\tfrac{\pi}{4}D^{2}} \;\propto\; D^{\,2.63-2} = D^{0.63}.$$
Velocity increases with diameter here: a larger pipe has a smaller friction gradient, so it flows faster under the same 21 m of head. To reduce the velocity the pipe must be made smaller.
Diameter for $v = 3$ m/s. Scaling from the known point $v = 5.44$ m/s at $D = 1.057$ m,
$$\left(\frac{D}{1.057}\right)^{0.63} = \frac{3}{5.44} \;\Rightarrow\; D = 1.057\left(\frac{3}{5.44}\right)^{1/0.63} = \boxed{0.410\ \text{m}\ (410\ \text{mm})}.$$
Check: with $D=0.410$ m, $Q = 0.278(132)(0.410)^{2.63}(0.0175)^{0.54}=0.397\ \text{m}^3/\text{s}$ and $v = 0.397/(\tfrac{\pi}{4}0.410^2) = 3.00\ \text{m/s}$ ✓.
Final results — Question 1
Quantity
Result
Discharge in the 1057 mm main
$Q = 4.78\ \text{m}^3/\text{s}$
Velocity in the 1057 mm main (a)
$v = 5.44\ \text{m/s}$ — exceeds 3 m/s
Diameter to reach $v = 3$ m/s (b)
$D \approx 0.410\ \text{m}$ (410 mm)
Delivered flow at that diameter
$Q \approx 0.40\ \text{m}^3/\text{s}$
Check / design note. Reducing the diameter to 410 mm satisfies the erosion-velocity limit but slashes the delivered flow from 4.78 to 0.40 m³/s — a factor of twelve. In practice one would not solve an over-velocity gravity main by shrinking the pipe; the realistic remedy is to keep the large diameter and dissipate surplus head with a control/throttling valve or an orifice plate, or to accept a lower flow. The 410 mm value answers the question exactly as posed (velocity-only criterion on a purely frictional gravity line).
