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18-Env-A2 Hydrology and Municipal Hydraulics Engineering · May 2014

Question 5 of 7: Pipe Network Analysis and Sanitary Sewer Design

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

Notes on this paper

National Exams — May 2014 — 04-Env-A2 / Hydrology and Municipal Hydraulics Engineering. 3 hours duration; closed book with an 8.5×11 in double-sided aid sheet; Casio or Sharp approved calculator only. Any five questions constitute a complete paper (first five answers marked); all seven are solved below for completeness. Each question is worth 20 marks.

Reference texts. Chow, Open-Channel Hydraulics; Linsley, Kohler & Paulhus, Hydrology for Engineers (3rd ed.); Walski et al., Advanced Water Distribution Modeling and Management; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.).

Question 5: Pipe Network Analysis and Sanitary Sewer Design (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.

(i) Hardy-Cross Pipe Network Solution

Given. A 4-node network (A, B, C, D) with 5 pipes forming two loops (A-B-C-A and B-D-C-B), inflow 1000 L/s at A, and outflows of 100 L/s at B, 300 L/s at C and 600 L/s at D:

Given data
PipeLength, $L$ (m)Diameter, $d$ (mm)
AB500400
BC700250
CD500200
AC700350
BD500300
Check: no pipe roughness / Hazen–Williams $C$ is given. A common $C$ is assumed for every pipe (same material throughout the network) using a Hazen–Williams head-loss form $h_f = K Q^{1.852}$ with $K_i \propto L_i/d_i^{4.87}$. Because a single common $C$ cancels out of the *relative* pipe-to-pipe resistance ratios, the resulting flow split is exact and independent of the (unstated) $C$ value — only the absolute head losses (not requested) would depend on it.

Find. The flow in each of the five pipes.

Approach. Assume an initial flow distribution that satisfies continuity at every node, then apply Hardy-Cross loop-balancing corrections $\Delta Q = -\dfrac{\sum K Q|Q|^{n-1}}{n\sum K|Q|^{n-1}}$ (with $n=1.852$) to each of the two independent loops until both loops' signed head-loss sums vanish.

  1. Initial (continuity-satisfying) trial flows. Split A's 1000 L/s in proportion to the larger-diameter pipe carrying more: $Q_{AB}=600$, $Q_{AC}=400$ L/s. At B: $Q_{AB}=Q_{BC}+Q_{BD}+100 \Rightarrow$ try $Q_{BC}=200$, $Q_{BD}=300$ L/s. At C: $Q_{AC}+Q_{BC}=Q_{CD}+300 \Rightarrow Q_{CD}=300$ L/s. Node D then balances exactly: $Q_{BD}+Q_{CD}=300+300=600$ L/s. ✓
  2. Loop-balance corrections. With $K_i = L_i/d_i^{4.87}$ for each pipe, iterate the Hardy-Cross correction around Loop 1 (A→B→C→A, via AB, BC, CA) and Loop 2 (B→D→C→B, via BD, DC, CB, sharing pipe BC with Loop 1) until each loop's signed $\sum KQ|Q|^{n-1}$ is zero to numerical precision.
  3. Converged flows. The iteration converges (residual loop head-loss $<10^{-4}$) to: $$Q_{AB}=\boxed{609.3\ \text{L/s}}, \quad Q_{AC}=\boxed{390.7\ \text{L/s}}, \quad Q_{BC}=\boxed{58.1\ \text{L/s}}, \quad Q_{BD}=\boxed{451.2\ \text{L/s}}, \quad Q_{CD}=\boxed{148.8\ \text{L/s}}.$$
  4. Check. Node continuity: A: $609.3+390.7=1000.0$ ✓. B: $609.3=58.1+451.2+100.0$ ✓. C: $390.7+58.1=148.8+300.0$ ✓. D: $451.2+148.8=600.0$ ✓.
AB: L=500 m, d=400 mm Q = 609.3 L/s BC: L=700 m, d=250 mm Q = 58.1 L/s CD: L=500 m, d=200 mm Q = 148.8 L/s AC: L=700 m, d=350 mm Q = 390.7 L/s BD: L=500 m, d=300 mm Q = 451.2 L/s 1000 L/s in 100 L/s out 300 L/s out 600 L/s out A B C D
Solved pipe network: converged Hardy-Cross flows on each of the five pipes, satisfying continuity at every node and zero net head loss around both loops.
PipeFlow, $Q$
AB609.3 L/s (A→B)
AC390.7 L/s (A→C)
BC58.1 L/s (B→C)
BD451.2 L/s (B→D)
CD148.8 L/s (C→D)

(ii) Gravity, Pressure and Vacuum Sanitary Sewers

Conditions and advantages
SystemCondition favouring its useAdvantage
Gravity sewerTerrain has consistent, adequate downhill fall toward the treatment plant/pump station without excessive trench depthSimplest and most reliable; no per-connection mechanical/electrical components; lowest operating cost; easiest to inspect and maintain (CCTV, rodding)
Pressure sewer (grinder-pump)Flat or undulating terrain, or rock/high groundwater that makes deep continuous gravity trenching impracticalPipe can follow the ground surface contour at shallow, uniform burial depth; smaller-diameter pipe suffices since flow is pumped, reducing excavation in difficult terrain
Vacuum sewerFlat, low-lying terrain with a high water table (e.g. coastal/waterfront development) served by a central vacuum stationPipe network operates under vacuum, so any leak draws groundwater IN rather than sewage OUT, virtually eliminating exfiltration; shallow, slope-independent installation

(iii) The Harmon Formula — $M$ and $p$

$M$ is the peaking factor: the dimensionless ratio of peak-hour to average dry-weather sanitary flow, used to convert an average design flow into the peak flow a sewer must be sized to carry. $p$ is the tributary population served, expressed in thousands of persons. As $p$ grows large, $\sqrt{p}$ dominates the denominator and $M\to1$: a large sewershed averages out individual households' peak-use variability, so its peak-to-average ratio approaches unity. Conversely, for a small $p$, $M$ is large, reflecting that a handful of connections' simultaneous peak usage can dominate a small collection system's instantaneous flow — exactly the physical behaviour the formula is calibrated to reproduce.