Pipe Network Analysis, Water Hammer, and Pump System Curves: 18-Env-A2 Hydrology and Municipal Hydraulics Engineering, December 2014, Q3

Problem 3: Pipe Network Analysis, Water Hammer, and Pump System 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	600	250
BC	800	300
CD	600	250
AC	800	300
BD	600	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 Hazen–Williams head-loss form $h_f = K Q^{1.852}$ 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 $\Delta Q = -\dfrac{\sum K Q|Q|^{n-1}}{n\sum K|Q|^{n-1}}$ (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 $<10^{-4}$) to: $$Q_{AB}=\boxed{419.4\ \text{L/s}}, \quad Q_{AC}=\boxed{580.6\ \text{L/s}}, \quad Q_{BC}=\boxed{19.2\ \text{L/s}}, \quad Q_{BD}=\boxed{300.3\ \text{L/s}}, \quad Q_{CD}=\boxed{299.7\ \text{L/s}}.$$
Check. Node continuity: A: $419.4+580.6=1000.0$ ✓. B: $419.4=19.2+300.3+100.0$ ✓. C: $580.6+19.2=299.7+300.0$ ✓. D: $300.3+299.7=600.0$ ✓.

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	419.4 L/s (A→B)
AC	580.6 L/s (A→C)
BC	19.2 L/s (B→C)
BD	300.3 L/s (B→D)
CD	299.7 L/s (C→D)

(ii) Water Hammer

Water hammer is the pressure transient produced when the velocity of a moving water column is changed abruptly: the column's kinetic energy converts into a pressure pulse that propagates back and forth along the pipe at the pressure-wave celerity (typically 1000–1400 m/s for water in a rigid steel/ductile-iron main). A pump system causes water hammer whenever the change in flow is rapid relative to the pipe's characteristic reflection time $2L/a$ (the round-trip travel time of the pressure wave along the pipe of length $L$) — classically, sudden pump trip (power failure), fast valve closure, or a pump starting against a closed or partially-open discharge valve. If the closure/stoppage time is shorter than $2L/a$, the full Joukowsky pressure rise $\Delta p = \rho a \Delta V$ is realized; if it is slower, the surge is attenuated.

One effective way to reduce or eliminate the effect is to slow the rate of flow change so that it exceeds the critical time $2L/a$ — for example, a slow-closing check valve or a controlled valve-closure schedule at the pump discharge (or, on the pump-trip side, a flywheel that extends the pump's coast-down time). This converts a "rapid" closure into a "slow" one and greatly reduces the peak transient pressure without the added infrastructure of a surge tank or air chamber.

(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 the pump can deliver against flow, falling from its maximum at zero flow (the shutoff head) as flow increases. The system curve plots the head the pipeline demands against flow — the static lift (elevation difference between source and delivery, present even at $Q=0$) plus friction losses, which grow 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 will actually deliver.

(iv) Multiple-Pump Configurations

(a) Increase discharge head, same flow — pumps in series. Staging two (or more) identical pumps in series (the discharge of one feeding the suction of the next) adds their heads at a common flow: the combined H–Q curve is obtained by adding heads vertically at each Q. This is the cost-effective route when a higher head is needed to reach a fixed downstream demand without changing the flow requirement, since a single much-larger-headed pump may not exist in a standard line or may be far less efficient at the required duty point.

(b) Increase discharge rate, same static head — pumps in parallel. Staging identical pumps in parallel (common suction and discharge headers) adds their flows at a common head: the combined H–Q curve is obtained by adding flows horizontally at each H. This is the cost-effective route when more capacity is needed at essentially the same static lift, and it has the added operational benefit that pumps can be staged on/off to track demand (running one, two, or three units) rather than running one oversized pump throttled back for most of its duty cycle.

Concept & Theory

Topic. Hardy-Cross pipe-network balancing; water-hammer transients; pump/system head curves; series vs. parallel pump staging.

Key relations. Hardy-Cross correction $\Delta Q = -\sum KQ|Q|^{n-1}/(n\sum K|Q|^{n-1})$; Joukowsky surge $\Delta p = \rho a \Delta V$; 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 exactly the same energy-balance logic (pump head must equal system head), which is why the operating point, water hammer (a transient violation of that balance) and multi-pump staging are all naturally discussed together.

Common pitfall. Starting a Hardy-Cross trial that violates node continuity (only loop energy balance is corrected iteratively, so continuity must hold at every step); confusing "static head" (system curve's $Q=0$ intercept) with "shutoff head" (pump curve's $Q=0$ intercept) — they are numerically different unless the pump is dead-headed exactly at the static lift.

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

Question 3 of 7: Pipe Network Analysis, Water Hammer, and Pump System Curves

Problem 3: Pipe Network Analysis, Water Hammer, and Pump System Curves (20 marks)

(i) Hardy-Cross Pipe Network Solution

(ii) Water Hammer

(iii) System Head Curve

(iv) Multiple-Pump Configurations