18-Env-A2 Hydrology and Municipal Hydraulics Engineering, Undated paper, Q7

Problem 7 (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) Equivalent pipe diameter and flow, three pipes in series (6 marks)

Given.

Pipe	Length	Diameter
1 (A side)	$L_1=500$ m	$D_1=200$ mm
2 (middle)	$L_2=600$ m	$D_2=400$ mm
3 (B side)	$L_3=400$ m	$D_3=150$ mm

Darcy–Weisbach friction factor $f=0.03$ (same for all three pipes); total head loss $h_L=20$ m; minor losses ignored.

[Figure not reproduced: Fig. 5 — Three pipes in series between tank A and tank B, as printed on the exam. See the official exam paper.]

Find. Equivalent diameter $D_e$; corresponding flow $Q$.

Approach. Use the supplied formula for $D_e$ (a single pipe of length $L_e=\sum L_i$ that reproduces the same $\sum L_i/D_i^5$), compute the equivalent resistance $K_e$, then solve $h_L=K_eQ^2$ for $Q$ — note the printed exponent on the $Q$ formula (0.2) does not match the Darcy–Weisbach relationship implied by the $K_e$ formula itself (which gives $h_L=K_eQ^2$, i.e. $Q=(h_L/K_e)^{0.5}$); the physically consistent exponent 0.5 is used, and the discrepancy is flagged below.

Equivalent length and equivalent diameter.$$L_e=\sum L_i=500+600+400=1500\ \text{m}$$$$\sum\frac{L_i}{D_i^5}=\frac{500}{(0.200)^5}+\frac{600}{(0.400)^5}+\frac{400}{(0.150)^5}=156{,}250+58{,}594+6{,}673{,}740=6{,}888{,}583\ \text{m}^{-4}$$$$D_e=\left(\frac{L_e}{\sum L_i/D_i^5}\right)^{0.2}=\left(\frac{1500}{6{,}888{,}583}\right)^{0.2}=\boxed{0.1852\ \text{m}=185\ \text{mm}}$$
Equivalent resistance $K_e$.$$K_e=\frac{8fL_e}{\pi^2gD_e^5}=\frac{8(0.03)(1500)}{\pi^2(9.81)(0.1852)^5}=\boxed{17{,}075\ \text{s}^2/\text{m}^5}$$
Flow $Q$, from $h_L=K_eQ^2$ (Darcy–Weisbach, the relation the $K_e$ formula itself is built on).$$Q=\sqrt{\frac{h_L}{K_e}}=\sqrt{\frac{20}{17{,}075}}=\boxed{0.0342\ \text{m}^3/\text{s}=34.2\ \text{L/s}}$$Check: $V=Q/(\pi D_e^2/4)=1.27$ m/s through the equivalent pipe — a physically reasonable gravity-main velocity, confirming the exponent-0.5 solution.

Quantity	Value
Equivalent diameter, $D_e$	185 mm
Equivalent resistance, $K_e$	$17{,}075$ s²/m&sup5;
Flow, $Q$	0.0342 m³/s (34.2 L/s)

Check: the exam's printed formula shows $Q=(h_L/K_e)^{0.2}$, but $K_e$ is defined exactly so that $h_L=K_eQ^2$ (this falls straight out of Darcy–Weisbach, $h_f=fLV^2/(2gD)=8fLQ^2/(\pi^2gD^5)=K_eQ^2$) — so the algebraically consistent exponent is 0.5, not 0.2, and the printed "0.2" is treated as a misprint (it also matches the $D_e$ formula's own exponent, suggesting the two may have been mis-copied together). Using the printed 0.2 literally gives $Q=(20/17{,}075)^{0.2}=0.2593\ \text{m}^3/\text{s}$, shown here for transparency but not carried forward as the answer, since it is inconsistent with the very $K_e$ expression the question supplies and implies an unrealistic $V\approx9.6$ m/s through the equivalent pipe.

(ii) Hardy-Cross Method: three main steps (7 marks)

The Hardy-Cross method balances a looped pipe network iteratively, correcting an initial trial flow pattern loop-by-loop using $\Delta Q=-\sum KQ|Q|^{n-1}/\sum nK|Q|^{n-1}$ ($n=2$ for Darcy–Weisbach, $n=1.85$ for Hazen–Williams) until every loop's head-loss imbalance is negligible. Three main steps:

(1) Assign an initial trial flow distribution that satisfies continuity at every node. Before any loop correction is computed, flows must be assigned by hand-balancing each junction so that inflow equals outflow everywhere (Kirchhoff's node law); the Hardy-Cross correction step never re-checks or re-balances continuity, so an inconsistent starting guess propagates as a wrong — though internally self-consistent — final answer.

(2) Compute and apply the loop head-loss correction. For each closed loop, sum the signed head losses $\sum KQ|Q|^{n-1}$ around the loop (clockwise positive) and the loop's "loop conductance" $\sum nK|Q|^{n-1}$, then apply the correction $\Delta Q=-\sum KQ|Q|^{n-1}/\sum nK|Q|^{n-1}$ to every pipe in that loop (added to clockwise flows, subtracted from counter-clockwise flows) — this correction drives that loop's own head-loss imbalance toward zero while automatically preserving the nodal continuity established in Step 1, since equal and opposite $\Delta Q$ is applied to shared pipes between adjacent loops.

(3) Iterate to convergence. Steps (2) is repeated, loop by loop, recomputing $K$-weighted sums with the updated flows each pass, until the largest $|\Delta Q|$ across all loops falls below an acceptable tolerance (e.g., <1% of the smallest pipe flow, or a fixed small flow such as 0.001 m³/s) — at convergence, every loop simultaneously satisfies both continuity (built in from Step 1 and preserved by Step 2's balanced correction) and the energy equation (zero net head-loss imbalance around every loop), which together define the unique steady-state flow distribution in the network.

Note: strictly, Hardy-Cross balances LOOPED networks; a purely branched (tree) network has no closed loops at all, so its flows are already fully determined by continuity alone (working from the dead-end branches back to the source) and requires no iterative loop-balancing step — Hardy-Cross becomes necessary specifically once the network contains at least one loop, creating more unknown pipe flows than independent continuity equations.

(iii) System-pump curve for two pumps in parallel (7 marks)

Fig. 6 — Example system-pump curve for two identical pumps in parallel: the single-pump characteristic curve, the two-pumps-in-parallel curve (flow doubled at equal head), the system (demand) curve, the resulting operating point, and the shared shutoff head.

The figure identifies the four key points requested: (1) the single-pump characteristic curve (head $H$ falling with increasing flow $Q$, the pump's own $H$-$Q$ performance); (2) the two-pumps-in-parallel curve, obtained by adding the single pump's flow to itself at every value of $H$ (parallel pumps see the same head but each contributes its own flow, so $Q_{parallel}(H)=2Q_{single}(H)$ — the curve is "stretched" horizontally rather than vertically, the opposite of the series-pump case); (3) the system (demand) curve, $H_{sys}=H_{static}+kQ^2$, representing the static lift plus friction/minor losses the network imposes as flow increases; and (4) the operating point, the intersection of the pump curve actually in service (here, the two-pumps-in-parallel curve) with the system curve — the only flow/head combination at which the pumps' delivered head exactly equals the system's demanded head. The shutoff head (each curve's $y$-intercept at $Q=0$) is IDENTICAL for the single-pump and parallel configurations, since at zero flow neither the second pump nor the piping between them changes the head a closed valve sees — unlike series pumps, parallel pumps do not increase shutoff head, they only extend the curve's reach in $Q$.

This is also why the actual flow gain from adding a second pump in parallel is normally much less than double: because the system curve rises with $Q^2$, the operating point on the flatter parallel curve is pulled up to a higher $H$ (and correspondingly lower $Q$ per pump) than either pump would deliver alone at low system resistance — the true gain in delivered flow depends entirely on where the combined curve intersects the system curve, exactly analogous to (and the mirror image of) how series pumps deliver less than double the head once the system curve is taken into account.

Topic: Equivalent-pipe representation of pipes in series; Hardy-Cross network-balancing steps; pump combination curves for two pumps in parallel.

Key relations: $D_e=(\sum L_i/\sum L_i/D_i^5)^{0.2}$; $K_e=8fL_e/(\pi^2gD_e^5)$ with $h_L=K_eQ^2$; Hardy-Cross $\Delta Q=-\sum KQ|Q|^{n-1}/\sum nK|Q|^{n-1}$; parallel pumps add flow at equal head, $Q_{parallel}(H)=2Q_{single}(H)$.

Why this works: the equivalent-diameter formula works because it preserves $\sum L_i/D_i^5$, the exact quantity that sets total friction resistance for a fixed $f$; Hardy-Cross converges because each loop correction drives that loop's own head-loss imbalance toward zero while never disturbing the nodal continuity fixed in the starting guess; and pumps in parallel add flow (not head) at each common head because each pump draws from and discharges to the same shared header, so the combined flow is simply the sum of what each pump alone would deliver at that head — which is why the operating point must be read off the INTERSECTION of the combined pump curve with the system curve, not off either curve alone.

Common pitfall: applying a printed formula's exponent literally without checking it against the physics the rest of the question already established (as in part (i)); starting a Hardy-Cross iteration from a trial flow pattern that does not already balance every node, or applying the method to a branched network that needs no iteration at all; and assuming pumps in parallel simply double the delivered flow (they double the flow ONLY at a fixed head; the actual flow gain depends on where the combined curve crosses the system curve, and is normally much less than double once system friction is included).

Source: Walski et al., Advanced Water Distribution Modeling and Management, equivalent-pipe and pump-combination chapters.

Question 7 of 7

Problem 7 (20 marks)

(i) Equivalent pipe diameter and flow, three pipes in series (6 marks)

(ii) Hardy-Cross Method: three main steps (7 marks)

(iii) System-pump curve for two pumps in parallel (7 marks)