Ten-pipe network — flow split and pressures: 16-Civ-A5 Hydraulic Engineering, December 2019, Q4

Question 4: Ten-pipe network — flow split and pressures (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. Ten identical pipes ($D=0.250\ \text{m}$, $L=200\ \text{m}$, $C=130$) between reservoir A (95 m) and B (70 m), arranged as two parallel branches (Figure 2). Node ground elevations are read from the dashed contours: the three first-tier nodes lie on the 90 m contour, the two second-tier junctions on the 80 m contour, and the collector node before P10 on the 75 m contour.

Find. (a) total discharge; (b) maximum and minimum pressure head; (c) which branch carries more flow and why.

Figure 4-1. Network topology (Figure 2 on the exam). Upper branch: two parallel two-pipe routes (P1+P4)∥(P2+P5), then P8. Lower branch: P3, then the parallel pair P6∥P7, then P9. Both branches meet before P10, which delivers the whole flow to B. Dashed lines are ground contours (90, 80, 75 m).

Approach. Identical pipes let each carry a friction head $h_f=k\,Q^{1.852}$ with the same $k$, so the network reduces by series–parallel algebra — no Hardy–Cross needed. Equate the head across the two parallel branches, add the P10 loss to reach the 25 m reservoir difference, solve for the branch flows, then trace the hydraulic grade line (HGL) to get nodal pressure heads $p/\gamma = \text{HGL}-z$.

Per-pipe resistance. Invert Hazen–Williams to $h_f=k\,Q^{1.852}$ with $k = L\big/(0.278\,C\,D^{2.63})^{1/0.54}$. For $D=0.25$, $L=200$, $C=130$: $k = 223$ ($Q$ in m³/s, $h_f$ in m).
Reduce each branch. Two identical pipes in series carry the same flow, so their resistance adds; two identical routes in parallel each take half the branch flow. Writing branch losses in $Q^{1.852}$: $$h_{upper}=\Big[\underbrace{2^{\,1-1.852}}_{(P1+P4)\|(P2+P5)}+\underbrace{1}_{P8}\Big]k\,Q_u^{1.852}=1.554\,k\,Q_u^{1.852},$$ $$h_{lower}=\Big[\underbrace{1}_{P3}+\underbrace{2^{-1.852}}_{P6\|P7}+\underbrace{1}_{P9}\Big]k\,Q_\ell^{1.852}=2.277\,k\,Q_\ell^{1.852}.$$
Solve the flow split. The two branches share the same head, and their combined flow $Q_t=Q_u+Q_\ell$ then drops through P10 to reach B: $$1.554\,k\,Q_u^{1.852}=2.277\,k\,Q_\ell^{1.852},\qquad 1.554\,k\,Q_u^{1.852}+k\,Q_t^{1.852}=95-70=25\ \text{m}.$$ Solving simultaneously, $$\boxed{Q_u = 0.135\ \text{m}^3/\text{s},\quad Q_\ell = 0.110\ \text{m}^3/\text{s},\quad Q_t = 0.245\ \text{m}^3/\text{s}\ (245\ \text{L/s}).}$$ Check: each branch drops 8.5 m, P10 drops 16.5 m, total 25 m, and the HGL lands exactly on B at 70 m.

Trace the HGL and pressure heads (part b). Starting from HGL $=95$ at A and subtracting each pipe loss (P1–P5 carry a half-branch flow, P3 the full lower flow, P8 the full upper flow):

Nodal HGL and pressure head $p/\gamma=\text{HGL}-z$
Node	Ground $z$ (m)	HGL (m)	$p/\gamma$ (m)
First upper (P1/P4)	90	93.5	3.5
Middle (P2/P5)	90	93.5	3.5
First lower (P3)	90	91.3	1.26 (min)
Upper junction (before P8)	80	92.0	11.97 (max)
Lower junction (before P9)	80	90.2	10.2
Collector (before P10)	75	86.5	11.5

$$\boxed{p/\gamma_{max}=11.97\ \text{m (upper junction)},\qquad p/\gamma_{min}=1.26\ \text{m (first lower node)}.}$$ The minimum sits at the first lower node: it is on high ground (90 m) yet fed by the single pipe P3 that carries the whole lower flow, so its HGL is drawn down the most. The maximum sits at the upper junction, where the HGL is still high but the ground has dropped to 80 m.

Which branch carries more (part c). The upper branch offers two complete parallel two-pipe routes, giving an equivalent resistance $1.554\,k$, whereas the lower branch is mostly single pipes in series ($2.277\,k$). At equal head the lower-resistance branch carries the larger flow: $$\boxed{\text{upper branch} \;(Q_u=135\ \text{L/s}) \gt \text{lower branch}\;(Q_\ell=110\ \text{L/s}).}$$

Question 4 — results
Quantity	Value
(a) Total flow $Q_t$	0.245 m³/s (245 L/s)
(a) Upper / lower branch flow	135 L/s / 110 L/s
(b) Maximum pressure head	11.97 m (upper junction, $z=80$)
(b) Minimum pressure head	1.26 m (first lower node, $z=90$)
(c) Higher-flow branch	upper (lower equivalent resistance)

Topic. Series–parallel pipe-network reduction; hydraulic grade line and pressure head.

Key relations. $h_f=k\,Q^{1.852}$ with $k=L/(0.278CD^{2.63})^{1.852}$; parallel of $m$ identical routes divides flow equally so equivalent $k_{eq}=k_{route}\,m^{-0.852}$; series adds $k$; pressure head $p/\gamma=\text{HGL}-z$.

Why this works. Identical pipes share one resistance constant, so the whole network collapses by inspection instead of iteration: parallel routes take equal flow and reach the same node head, series pipes stack their losses. The exponent 1.852 (not 2) is the only wrinkle — it changes the parallel-division factor from $2^{-1}$ to $2^{-0.852}$. The HGL is then a bookkeeping walk from the source, and pressure head is what is left after subtracting ground elevation, which is why a high-ground node near the source can be the low-pressure point.

Common pitfall. Splitting flow by ratios of diameter or length instead of recognising identical pipes carry equal flow, and forgetting that pressure head is HGL minus ground — a node with a high HGL can still be the pressure minimum if it sits high, and vice-versa.

Source. Chin, Water-Resources Engineering, §2.4 (pipe networks); Roberson, Cassidy & Chaudhry, Hydraulic Engineering, ch. 4.

Question 4 of 6: Ten-pipe network — flow split and pressures

Question 4: Ten-pipe network — flow split and pressures (20 marks)