Pipe network with two reservoirs (Hardy Cross): 16-Civ-A3 Elementary Environmental Engineering, December 2019, Q4

Question 4: Pipe network with two reservoirs (Hardy Cross) (25 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. Two reservoirs A (85 m) and G (102 m); demands at C, F, E; head loss $h=KQ^2$ with the $K$ values and assumed initial flows below.

Given — pipe resistances and assumed flows
Pipe	AB	BC	BF	CF	DC	DE	EF	GD
K (s²/m⁵)	194	1900	1900	678	1630	678	1630	423
Q₀ (m³/s)	0.200	0.100	0.100	0.050	0.050	0.200	0.100	0.250

Find. The method, the loop-correction equations, and the pipe flows after one Hardy Cross iteration.

Approach. Use the Hardy Cross head-balance method: assume flows that satisfy continuity (given), then correct each loop by $\Delta Q=-\dfrac{\sum KQ|Q|}{\sum 2K|Q|}$ until every loop’s head loss closes. Two real loops plus one pseudo-loop through the two fixed-head reservoirs are needed.

Pipe network: reservoirs A (85 m) and G (102 m); demands at C, F, E. Loop I = B-C-F; Loop II = C-D-E-F; pseudo-loop A-B-C-D-G closes on the reservoir head difference.

Part 1 — Method. Because the network is looped with two fixed-grade sources, the flows cannot be found by continuity alone; the energy (head-loss) balance around every independent loop must also hold. The Hardy Cross method does this iteratively: (i) express each pipe head loss as $h=KQ^{2}$; (ii) assume a set of pipe flows that satisfy junction continuity ($\sum Q_{in}=\sum Q_{out}$, given here); (iii) identify the independent loops — two real loops plus one pseudo-loop linking the two reservoirs, whose head loss must equal $H_G-H_A$; (iv) compute a circulating correction $\Delta Q$ for each loop; (v) apply the corrections (a pipe shared by two loops receives both) and repeat until each $\Delta Q\to0$.

Part 2 — Correction equations. Summing $h=KQ^2$ around a loop and linearizing ($h=KQ|Q|$, $dh/dQ=2K|Q|$) gives, for each loop $k$,

\Delta Q_k = -\frac{\sum_k K\,Q\,|Q|}{\sum_k 2K\,|Q|}\,,\qquad \text{(pseudo-loop numerator adds } H_G-H_A).

With clockwise-positive sign convention and traversals Loop I: B→C→F→B, Loop II: C→D→E→F→C, pseudo-loop III: A→B→C→D→G:

\Delta Q_{I}=-\frac{K_{BC}Q_{BC}^2+K_{CF}Q_{CF}^2-K_{BF}Q_{BF}^2}{2\left(K_{BC}|Q_{BC}|+K_{CF}|Q_{CF}|+K_{BF}|Q_{BF}|\right)}

\Delta Q_{II}=-\frac{-K_{DC}Q_{DC}^2+K_{DE}Q_{DE}^2+K_{EF}Q_{EF}^2-K_{CF}Q_{CF}^2}{2\left(K_{DC}|Q_{DC}|+K_{DE}|Q_{DE}|+K_{EF}|Q_{EF}|+K_{CF}|Q_{CF}|\right)}

\Delta Q_{III}=-\frac{\left(H_G-H_A\right)+K_{AB}Q_{AB}^2+K_{BC}Q_{BC}^2-K_{DC}Q_{DC}^2-K_{GD}Q_{GD}^2}{2\left(K_{AB}|Q_{AB}|+K_{BC}|Q_{BC}|+K_{DC}|Q_{DC}|+K_{GD}|Q_{GD}|\right)}

Part 3 — One iteration. Substituting the assumed flows:

Loop I correction. Numerator $=1900(0.1)^2+678(0.05)^2-1900(0.1)^2 = 1.695$; denominator $=2[1900(0.1)+678(0.05)+1900(0.1)]=827.8$. $\Delta Q_I=-1.695/827.8=\boxed{-0.0020\ \text{m}^3/\text{s}}$.
Loop II correction. Numerator $=-1630(0.05)^2+678(0.2)^2+1630(0.1)^2-678(0.05)^2 = 37.65$; denominator $=2[1630(0.05)+678(0.2)+1630(0.1)+678(0.05)]=828.0$. $\Delta Q_{II}=-37.65/828.0=\boxed{-0.0455\ \text{m}^3/\text{s}}$.
Pseudo-loop III correction. Numerator $=(102-85)+194(0.2)^2+1900(0.1)^2-1630(0.05)^2-423(0.25)^2 = 13.25$; denominator $=2[194(0.2)+1900(0.1)+1630(0.05)+423(0.25)]=832.1$. $\Delta Q_{III}=-13.25/832.1=\boxed{-0.0159\ \text{m}^3/\text{s}}$.
Apply corrections. Add each $\Delta Q$ to the pipes of its loop with the loop sign (shared pipes receive both). For example $Q_{CF}$ (in Loop I as $+$, Loop II as $-$) becomes $0.050+\Delta Q_I-\Delta Q_{II}=0.050-0.0020+0.0455=0.093$; $Q_{BC}$ (Loops I and III, both $+$) becomes $0.100-0.0020-0.0159=0.082$. The updated flows are tabulated below; junction continuity is preserved exactly.

Pipe flows after one Hardy Cross iteration (m³/s)
Pipe	AB	BC	BF	CF	DC	DE	EF	GD
Q₀	0.200	0.100	0.100	0.050	0.050	0.200	0.100	0.250
Q₁	0.184	0.082	0.102	0.093	0.111	0.155	0.055	0.266

Check: the resistance table is used as printed — each $K$ was confirmed against $K=fL/(2gDA^{2})$ (e.g. BC: $0.021\times350/(0.20\cdot19.62\cdot0.0314^2)=1.90\times10^{3}$). Continuing the iteration to convergence gives $Q_{AB}=0.201$, $Q_{GD}=0.249\ \text{m}^3/\text{s}$ and reproduces the reservoir constraint (HGL at G $=102.0$ m along A-B-C-D-G), confirming the loop set-up and signs.

Topic: Hardy Cross analysis of a looped network with fixed-grade reservoirs.

Key relations: $h=KQ^2$; loop correction $\Delta Q=-\dfrac{\sum KQ|Q|}{\sum 2K|Q|}$; number of independent loops $=$ real loops $+$ (reservoirs $-1$); pseudo-loop numerator carries $(H_A-H_G)$.

Why this works: the assumed flows already satisfy mass balance; each circulating $\Delta Q$ leaves continuity untouched while driving the loop head-loss residual to zero. The two-reservoir pseudo-loop enforces that the head loss between the fixed grades equals their elevation difference, so the source split (A vs. G) is solved too.

Common pitfall: omitting the reservoir pseudo-loop (which leaves $Q_{AB},Q_{GD}$ frozen at their assumed values and ignores the given heads), and dropping the sign on shared pipes when applying two loop corrections.

Source: Mays, Water Resources Engineering, Ch. 4 (pipe networks); Davis & Cornwell, Ch. 10.

Question 4 of 5: Pipe network with two reservoirs (Hardy Cross)

Question 4: Pipe network with two reservoirs (Hardy Cross) (25 marks)