Hardy-Cross Pipe Network, Pump Types, and System Head Curves: 18-Env-A2 Hydrology and Municipal Hydraulics Engineering, December 2015, Q3

Problem 3: Hardy-Cross Pipe Network, Pump Types, and System Head Curves (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 Pipe Network Solution

Given. A 4-node network (A, B, C, D) with 5 pipes forming two loops (A-B-C-A and B-D-C-B), inflow 1000 L/s at A, and outflows of 100 L/s at B, 300 L/s at C and 600 L/s at D:

Given data
Pipe	Length, $L$ (m)	Diameter, $d$ (mm)
AB	400	300
BC	600	350
CD	400	300
AC	600	350
BD	400	250

Check: no pipe roughness / Hazen–Williams $C$ is given. A common $C$ is assumed for every pipe (same material throughout the network), using the head-loss form $h_f=KQ^{1.852}$ with $K_i\propto L_i/d_i^{4.87}$. A single common $C$ cancels out of the pipe-to-pipe resistance ratios, so the resulting flow split is exact and independent of the (unstated) $C$ value.

Find. The flow in each of the five pipes.

Approach. Start from an initial flow distribution that satisfies continuity at every node, then apply Hardy-Cross loop-balancing corrections $\Delta Q=-\dfrac{\sum KQ|Q|^{n-1}}{n\sum K|Q|^{n-1}}$ (with $n=1.852$) to Loop 1 (A→B→C→A) and Loop 2 (B→D→C→B, sharing pipe BC) until both loops' signed head-loss sums vanish.

Initial (continuity-satisfying) trial flows. $Q_{AB}=600$, $Q_{AC}=400$ L/s (sum 1000 at A). At B: $Q_{BC}=200$, $Q_{BD}=300$ L/s (so $600=200+300+100$). At C: $Q_{CD}=300$ L/s (so $400+200=300+300$). Node D then balances exactly: $300+300=600$ L/s. ✓
Loop-balance corrections. With $K_i=L_i/d_i^{4.87}$ for each pipe, iterate the Hardy-Cross correction around Loop 1 (via AB, BC, CA) and Loop 2 (via BD, DC, CB) until each loop's signed $\sum KQ|Q|^{n-1}$ is zero to numerical precision.
Converged flows. The iteration converges (residual loop head loss $<10^{-6}$) to: $$Q_{AB}=\boxed{446.5\ \text{L/s}},\quad Q_{AC}=\boxed{553.5\ \text{L/s}},\quad Q_{BC}=\boxed{111.3\ \text{L/s}},\quad Q_{BD}=\boxed{235.2\ \text{L/s}},\quad Q_{CD}=\boxed{364.8\ \text{L/s}}.$$
Check. Node continuity: A: $446.5+553.5=1000.0$ ✓. B: $446.5=111.3+235.2+100.0$ ✓. C: $553.5+111.3=364.8+300.0$ ✓. D: $235.2+364.8=600.0$ ✓.

Solved pipe network: converged Hardy-Cross flows on each of the five pipes, satisfying continuity at every node and zero net head loss around both loops.

Pipe	Flow, $Q$
AB	446.5 L/s (A→B)
AC	553.5 L/s (A→C)
BC	111.3 L/s (B→C)
BD	235.2 L/s (B→D)
CD	364.8 L/s (C→D)

(ii) Centrifugal vs. Positive Displacement Pump

Flow–head relationship. A centrifugal pump delivers a flow that varies continuously with the system head along its H–Q curve (flow falls as head rises) and can be safely throttled or even briefly dead-headed at $Q=0$. A positive-displacement pump instead delivers a nearly fixed flow per revolution or stroke, essentially independent of discharge pressure (set by displacement volume × speed), and will build dangerously high pressure if dead-headed against a closed valve — it requires a relief valve for protection.

Operating principle. A centrifugal pump imparts kinetic energy to the fluid via a rotating impeller and converts it to pressure in a diffuser/volute, well suited to continuous, large-flow, moderate-head duty on relatively clean liquids. A positive-displacement pump physically traps and displaces a fixed volume of fluid each cycle (gear, piston, diaphragm, progressive-cavity types), well suited to low-flow, high-head, viscous, or precise metering duty, and is inherently self-priming in most designs.

(iii) System Head Curve

Typical system head curve: the drooping pump curve intersects the rising (static + friction) system curve at the operating point; the pump curve's Q=0 intercept is the shutoff head, the system curve's Q=0 intercept is the static head.

The pump curve plots the head a given pump can deliver against flow, falling from a maximum at zero flow (the shutoff head) as flow increases. The system curve plots the head the pipeline installation demands against flow — the static lift (elevation difference between source and delivery, present even at $Q=0$, giving the static head intercept) plus friction losses, which grow roughly with $Q^2$. The two curves' intersection is the operating point: the unique flow and head at which the pump's delivered head exactly equals the system's demanded head, and hence the flow the installation actually produces.

(iv) Multiple-Pump Configurations

Combined-curve construction: parallel staging adds flow horizontally at a common head; series staging adds head vertically at a common flow. Each combined curve is re-intersected with the (unchanged) system curve to find the new operating point.

(a) Parallel pumping (equal, common suction/discharge headers) is cost-effective when more capacity is needed at essentially the same static lift: the combined H–Q curve is built by adding flows horizontally at each head, and staging identical pumps on/off lets the station track demand (running one, two, or more units) rather than running one oversized pump throttled back most of the time.

(b) Series pumping (discharge of one feeding the suction of the next) is cost-effective when a higher head is needed to reach a fixed downstream demand without changing the required flow: the combined H–Q curve is built by adding heads vertically at each flow, avoiding the expense and poor part-load efficiency of specifying one much-larger-headed pump that may not exist in a standard product line.

Concept & Theory

Topic. Hardy-Cross pipe-network balancing; centrifugal vs. positive-displacement pumps; pump/system head curves; series vs. parallel staging.

Key relations. Hardy-Cross correction $\Delta Q=-\sum KQ|Q|^{n-1}/(n\sum K|Q|^{n-1})$; series pumps add $H$ at fixed $Q$; parallel pumps add $Q$ at fixed $H$.

Why this works. A looped network has more unknown pipe flows than independent node-continuity equations, so Hardy-Cross supplies the missing equations from zero net head loss around each loop; a pump installation is governed by the identical energy-balance logic (delivered head must equal system-demanded head), which is why the operating point and multi-pump staging both reduce to re-intersecting a (combined) pump curve with the system curve.

Common pitfall. Starting a Hardy-Cross trial that violates node continuity (only loop energy balance is corrected iteratively — continuity must hold at every step); assuming parallel staging always doubles flow (it does not, once the system curve is steep enough that the combined curve's flow gain over one pump is small) or that series staging always doubles head (same caveat, mirrored).

Source. Walski et al., Advanced Water Distribution Modeling and Management, network-analysis and pump-system chapters.

Question 3 of 7: Hardy-Cross Pipe Network, Pump Types, and System Head Curves

Problem 3: Hardy-Cross Pipe Network, Pump Types, and System Head Curves (20 marks)

(i) Hardy-Cross Pipe Network Solution

(ii) Centrifugal vs. Positive Displacement Pump

(iii) System Head Curve

(iv) Multiple-Pump Configurations