Question 7 of 7

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.

Darcy–Weisbach friction factor $f=0.05$ (same for all three pipes); total head loss $h_L=30$ 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.

1. Equivalent length and equivalent diameter. $$L_e=\sum L_i=600+400+300=1300\ \text{m}$$ $$\sum\frac{L_i}{D_i^5}=\frac{600}{(0.300)^5}+\frac{400}{(0.450)^5}+\frac{300}{(0.250)^5}=246914+21701+307200=575{,}790\ \text{m}^{-4}$$ $$D_e=\left(\frac{L_e}{\sum L_i/D_i^5}\right)^{0.2}=\left(\frac{1300}{575{,}790}\right)^{0.2}=\boxed{0.2956\ \text{m}=296\ \text{mm}}$$
2. Equivalent resistance $K_e$. $$K_e=\frac{8fL_e}{\pi^2gD_e^5}=\frac{8(0.05)(1300)}{\pi^2(9.81)(0.2956)^5}=\boxed{2379\ \text{s}^2/\text{m}^5}$$
3. Flow $Q$, from $h_L=K_eQ^2$ (Darcy–Weisbach, the relation the $K_e$ formula itself is built on). $$Q=\sqrt{\frac{h_L}{K_e}}=\sqrt{\frac{30}{2379}}=\boxed{0.1123\ \text{m}^3/\text{s}=112.3\ \text{L/s}}$$ Check: $V=Q/(\pi D_e^2/4)=0.1123/0.0686=1.64$ m/s through the equivalent pipe — a physically reasonable gravity-main velocity, confirming the exponent-0.5 solution.

(ii) Hardy-Cross Method: three main assumptions/limitations (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 assumptions/limitations designers must be aware of:

(1) Continuity must be satisfied at every node from the very first trial flow onward. The method only ever corrects loop head-loss imbalance ($\sum K Q|Q|^{n-1}=0$ around each loop); it never re-balances nodal inflow=outflow, so if the very first guessed flow pattern does not already satisfy continuity at every junction, the iteration will converge to a self-consistent but WRONG flow pattern (or fail to converge at all). In practice this means the initial trial flows are always assigned by hand-balancing each node before the first loop correction is computed.

(2) Pipe roughness/resistance coefficients ($C$ for Hazen–Williams or $f$ for Darcy–Weisbach) are assumed known and constant for every pipe. Real pipe roughness increases with age (tuberculation, biofilm), and $f$ is itself Reynolds-number (hence flow-rate) dependent for Darcy–Weisbach — so the "solved" network is only as accurate as these assumed, typically-fixed input coefficients, and a network's real performance should be periodically re-calibrated against field-measured pressures/flows rather than trusted indefinitely from design-stage assumptions.

(3) The method solves a single steady-state demand pattern at a time. Each Hardy-Cross solution represents the network's flow distribution under one fixed, static set of nodal demands (e.g., peak-hour demand); it does not capture transient behaviour (water hammer, pump start/stop surges) or the network's response as demands shift through the day, so a full design check requires re-running the method (or an equivalent network solver) for multiple demand scenarios (average day, peak hour, fire flow) rather than relying on a single balanced solution.

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

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-series curve, obtained by adding the single pump's head to itself at every value of $Q$ (series pumps pass the same flow but each adds its own head, so $H_{series}(Q)=2H_{single}(Q)$); (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-series curve) with the system curve — this is where the pumps will actually run, since it is the only flow at which the head the pumps can deliver exactly equals the head the system demands. The shutoff head (each curve's $y$-intercept at $Q=0$, the head developed at zero flow, e.g. valve fully closed) is marked for both the single-pump and the series configuration; for identical pumps in series the shutoff head doubles relative to a single unit, which is the entire point of choosing a series arrangement when the system's static lift/friction exceeds what one pump alone can develop.