18-Env-A2 Hydrology and Municipal Hydraulics Engineering, December 2016, Q5

Problem 5 (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) Hardy-Cross solution for the pipe network (10 marks)

Fig. 5 — Network topology: inflow 1000 L/s at A; outflows 100 L/s at B, 300 L/s at C, 600 L/s at D. Two independent loops share edge BC: Loop I (A–B–C–A) and Loop II (B–D–C–B).

Given. Pipe lengths/diameters per the table above; nodal continuity: inflow 1000 L/s at A, outflows 100/300/600 L/s at B/C/D (balances: $100+300+600=1000$). Five pipes and four nodes give $5-4+1=2$ independent loops.

Find. The flow (magnitude and direction) in each of the five pipes.

Check: no Hazen–Williams C or Darcy friction factor is given for the pipes, so a Darcy–Weisbach exponent of $n=2$ is used with a common friction factor assumed across all (same-material) pipes; the relative pipe resistance is then $K_i \propto L_i/D_i^{5}$, and the common friction-factor constant cancels exactly in the Hardy-Cross flow-correction formula, so it does not affect the resulting flow split — only the (unreported) head losses would need the actual $f$.

Approach. Assume an initial flow distribution satisfying continuity at every node, then apply the Hardy-Cross loop-balancing correction $$\Delta Q = -\frac{\sum K Q|Q|}{\sum 2K|Q|}$$ to each loop repeatedly (with the shared pipe BC receiving both loops' corrections, opposite in sign) until the corrections become negligible.

Relative pipe resistance $K = L/D^5$.

Pipe $L$ (m) $D$ (m) $K=L/D^5$

AB 600 0.300 246,914

BC 700 0.350 133,270

CD 600 0.300 246,914

AC 500 0.250 512,000

BD 500 0.200 1,562,500
Initial assumed flows (satisfying continuity at every node). $Q_{AB}=600$, $Q_{AC}=400$ (sum 1000 at A); $Q_{BC}=250$, $Q_{BD}=250$ (so B balances: $600=100+250+250$); $Q_{CD}=350$ (C balances: $400+250=300+350$; D balances: $350+250=600$). All in L/s.

Pipe	$L$ (m)	$D$ (m)	$K=L/D^5$
AB	600	0.300	246,914
BC	700	0.350	133,270
CD	600	0.300	246,914
AC	500	0.250	512,000
BD	500	0.200	1,562,500

Iterate the loop correction for Loop I (A–B–C–A: AB and BC taken positive in the loop-traversal direction, AC negative) and Loop II (B–D–C–B: BD positive, CD and BC negative), updating the shared pipe BC by both loops' corrections each pass:

Iteration	$\Delta Q_{I}$ (L/s)	$\Delta Q_{II}$ (L/s)	Q_AB	Q_AC	Q_BC	Q_BD	Q_CD
1	−19.80	−57.88	580.2	419.8	288.1	192.1	407.9
2	−4.98	−6.30	575.2	424.8	289.4	185.8	414.2
3	−0.60	−0.50	574.6	425.4	289.3	185.3	414.7
4	−0.05	−0.05	574.6	425.4	289.3	185.3	414.7

Corrections fall below 0.1 L/s by the fourth pass, so the solution has converged.

Pipe	Converged flow	Direction
AB	574.6 L/s	A → B
AC	425.4 L/s	A → C
BC	289.3 L/s	B → C
BD	185.3 L/s	B → D
CD	414.7 L/s	C → D

Continuity check: A: $574.6+425.4=1000$; B: $574.6=100+289.3+185.3$; C: $425.4+289.3=300+414.7$; D: $414.7+185.3=600$ — all balance.

(ii) Cavitation in a water distribution system (5 marks)

Cavitation is the formation and violent collapse of vapour bubbles in a liquid when the local absolute pressure falls below the fluid's vapour pressure — typically at a pump impeller eye, a partly closed valve throat, or any high-velocity/low-pressure point — followed by implosion of those bubbles microseconds later when they are swept into a region of higher pressure. Potential problems: (1) pitting and erosion of impeller vanes, casings and valve/fitting surfaces from the repeated micro-jet impacts of imploding bubbles, shortening equipment life; (2) loss of pump head/efficiency, along with characteristic noise, vibration and unstable flow that can fatigue shafts, bearings and seals. Solutions: (1) ensure the available NPSH exceeds the pump's required NPSH with an adequate margin — raise the suction water level, lower the pump elevation relative to the source, or shorten/enlarge the suction piping to cut suction losses; (2) select and operate the pump to avoid its low-pressure zones — avoid running an oversized pump far right of its best-efficiency point, use a lower-speed or double-suction impeller, or add an inducer/booster pump ahead of the main pump.

(iii) Net positive suction head (NPSH) (5 marks)

Fig. 6 — Pump suction schematic: lift $h_{lift}$ and suction-side head $h_{suction}$ relative to the free water surface.

Definition. NPSH is the margin, expressed as a head of liquid (m), by which the total absolute pressure head at the pump suction/impeller eye exceeds the liquid's vapour pressure head at the operating temperature: $$\text{NPSH}_{available} = \frac{P_{atm}}{\gamma} \pm h_{static} - h_{f,suction} - \frac{P_{vapour}}{\gamma}$$ For safe operation, $\text{NPSH}_{available}$ (a property of the installation/system) must exceed $\text{NPSH}_{required}$ (a property of the specific pump, given on its performance curve), normally with a safety margin.

Significance. NPSH is the direct design check against cavitation: whenever the available margin drops below what the pump requires, the local pressure at the impeller eye falls to the vapour pressure and cavitation begins, so verifying $\text{NPSH}_{available} > \text{NPSH}_{required}$ at the design flow (and at any off-design condition the pump will actually see) is mandatory in any pump/suction-piping layout, particularly where the pump sits above the source (a suction lift $h_{suction}$, as in Fig. 6) rather than in a flooded-suction arrangement.

Two key terms. Vapour pressure is the pressure at which the liquid, at its operating temperature, begins to flash into vapour; it rises sharply with temperature, which is why hot-water or high-elevation (lower atmospheric pressure) installations are more cavitation-prone. $\text{NPSH}_{required}$ is the minimum suction head margin the particular pump's internal geometry (impeller eye design, inlet losses, speed) needs to keep the lowest-pressure point inside the pump above vapour pressure; it is a manufacturer-tested characteristic of the pump itself (increasing with flow rate), distinct from $\text{NPSH}_{available}$, which depends only on the suction-side installation.

Topic: Hardy-Cross pipe network balancing; cavitation and net positive suction head.

Key relations: loop correction $\Delta Q = -\dfrac{\sum KQ|Q|}{\sum 2K|Q|}$ with $K\propto L/D^5$; $\text{NPSH}_{available} = P_{atm}/\gamma \pm h_{static} - h_{f} - P_{vapour}/\gamma \ \geq\ \text{NPSH}_{required}$.

Why this works: Hardy-Cross exploits the fact that a correctly balanced network satisfies both continuity (built into the initial assumed flows and preserved by every loop correction) and energy conservation (zero net head loss around every closed loop); iterating loop-by-loop corrections drives the head-loss imbalance in each loop toward zero without ever violating nodal continuity. Cavitation is fundamentally a local pressure problem, so NPSH re-expresses "is the pressure at the impeller eye above vapour pressure" as a head-margin bookkeeping exercise that can be checked at the design stage before a pump is ever installed.

Common pitfall: forgetting that the shared pipe between two adjacent loops (BC here) receives corrections from both loops each iteration, with opposite sign because the two loops traverse it in opposite directions; and, for NPSH, confusing suction lift (pump above source, subtracts from available head) with a flooded suction (pump below source, adds to it).

Source: Walski et al., Advanced Water Distribution Modeling and Management, pipe network analysis (Hardy-Cross) and pump cavitation/NPSH chapters.

Question 5 of 7

Problem 5 (20 marks)

(i) Hardy-Cross solution for the pipe network (10 marks)

(ii) Cavitation in a water distribution system (5 marks)

(iii) Net positive suction head (NPSH) (5 marks)