Equivalent-pipe network flow: 16-Civ-A5 Hydraulic Engineering, May 2015, Q1

Question 1: Equivalent-pipe network flow (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. Seven identical pipes connect reservoir A (head 100 m) to reservoir B (head 70 m); each pipe has $L=1{,}000\text{ m}$, $C=120$, $D=250\text{ mm}=0.25\text{ m}$. Available head $\Delta H = 100-70 = 30\text{ m}$.

Figure 1. Pipe network: A—P1—J1, then two parallel routes (P3+P4) and (P2+P5) to J2, then P6—P7—B.

Find. The total discharge $Q$ delivered A→B, and the discharge carried by pipe 5.

Approach. Because every pipe is identical, write each pipe loss as $h_f=kQ^{1.852}$, reduce the network by series–parallel rules to a single equivalent line, and set the summed loss equal to the 30 m of available head.

Per-pipe resistance. Inverting Hazen–Williams, $h_f = L\left(\dfrac{Q}{0.278\,C\,D^{2.63}}\right)^{1/0.54}=kQ^{1.852}$, with $$k=\frac{L}{(0.278\,C\,D^{2.63})^{1.852}}=\frac{1000}{(0.278\cdot120\cdot0.25^{2.63})^{1.852}}=1293\ \text{s}^{1.852}\text{m}^{-4.556}.$$
Read the topology. Pipe 1 is in series from A to junction J1. Between J1 and J2 there are two routes in parallel — the upper route P3+P4 and the lower route P2+P5 — each route being two identical pipes in series. Pipes 6 and 7 are again in series from J2 to B.
Split the parallel block. The two routes are identical, so they share the flow equally: each carries $Q/2$. The head loss across the parallel block (two pipes in series, each passing $Q/2$) is $$\Delta H_{\text{block}} = 2k\left(\tfrac{Q}{2}\right)^{1.852}.$$
Assemble the equivalent line. Pipes 1, 6 and 7 each pass the full $Q$, so $$\underbrace{3\,kQ^{1.852}}_{\text{P1, P6, P7}}+\underbrace{2k\left(\tfrac{Q}{2}\right)^{1.852}}_{\text{parallel block}} = kQ^{1.852}\big[\,3+2\cdot2^{-1.852}\,\big]=3.554\,kQ^{1.852}=30\text{ m}.$$
Solve for total flow. $Q^{1.852}=\dfrac{30}{3.554\cdot1293}=6.53\times10^{-3}$, hence $$\boxed{Q = 0.0661\ \text{m}^3/\text{s} \approx 66.1\ \text{L/s}.}$$
Flow in pipe 5. Pipe 5 lies on the lower parallel route, which carries half the total: $$\boxed{Q_5 = \tfrac{Q}{2}=0.0330\ \text{m}^3/\text{s} \approx 33.0\ \text{L/s}.}$$

A direct nodal (Hardy–Cross-free) solve of all seven pipes reproduces these values to four significant figures and gives junction heads $H_{J1}=91.6\text{ m}$, $H_{J2}=86.9\text{ m}$, confirming the reduction.

Final results — Question 1
Quantity	Value
Per-pipe resistance $k$	1293 (SI)
Total system flow $Q$	0.0661 m³/s (66.1 L/s)
Flow in pipe 5	0.0330 m³/s (33.0 L/s)

Topic. Equivalent-pipe reduction of a series–parallel pipe network.

Key relations. Hazen–Williams $h_f=kQ^{1.852}$; parallel branches share a common head loss; series pipes add their losses at a common flow.

Why this works. Identical parallel routes must dissipate the same head between the same two junctions, which forces an equal flow split; that turns an apparently looped network into a single equivalent resistance in the flow variable $Q$, so no iterative Hardy–Cross balancing is required. The non-linear exponent 1.852 is handled by collecting all full-flow pipes together and treating the half-flow block with the factor $2^{-1.852}$.

Common pitfall. Treating pipes 3, 4, 2, 5 as four independent parallel pipes (they are two series pairs in parallel), or forgetting that each parallel route carries only $Q/2$ when computing pipe 5.

Source. Mays, Water Resources Engineering, Ch. 3 (pipe networks & equivalent pipes).

Question 1 of 6: Equivalent-pipe network flow

Question 1: Equivalent-pipe network flow (20 marks)