Question 4 of 7

Problem 4 (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 flow distribution in the pipe network (8 marks)

Given.

Node demands: inflow 1000 L/s at A; outflows 100 L/s at B, 300 L/s at C, 600 L/s at D (the outflows sum to the inflow, confirming the network is balanced).

Find. The steady-state flow in each of the five pipes.

Approach. No Hazen–Williams $C$ or Darcy $f$ is given, only lengths and diameters, so the standard exponent-2 Hardy-Cross form is used with relative resistance $K=L/D^5$ (the common friction-factor/constant term is identical for every pipe of the same material and cancels exactly in the loop-correction formula, so the converged flow split is fully determined even though absolute head loss is not). Two independent loops are needed for 5 pipes and 4 nodes: Loop I (A–B–C–A) and Loop II (B–D–C–B).

1. Assume an initial flow split satisfying continuity at every node. $Q_{AB}=600$, $Q_{AC}=400$ (sum to the 1000 L/s inflow at A); at B, $600=100+Q_{BC}+Q_{BD}$, take $Q_{BC}=250$, $Q_{BD}=250$; at C, $400+250=300+Q_{CD}$ gives $Q_{CD}=350$; at D, $250+350=600$ — continuity closes at every node with this starting split (all values L/s).
2. Compute relative resistance $K=L/D^5$ for each pipe and apply the Hardy-Cross loop correction $\Delta Q = -\dfrac{\sum K\,Q|Q|}{\sum 2K|Q|}$ to each loop in turn (Loop I: AB$+$, BC$+$, AC$-$; Loop II: BD$+$, CD$-$, BC$-$), updating the shared pipe BC by both corrections, and repeating until $\Delta Q$ is negligible (converges in under 10 iterations).
3. Converged flows. The iteration converges to the values below (continuity re-checked and satisfied at all four nodes to within 0.1 L/s): $$Q_{AB}=\boxed{453\ \text{L/s}}\quad Q_{AC}=\boxed{547\ \text{L/s}}\quad Q_{BC}=\boxed{52\ \text{L/s}}\quad Q_{BD}=\boxed{301\ \text{L/s}}\quad Q_{CD}=\boxed{299\ \text{L/s}}$$ Because $AC$ has lower resistance per unit flow ($K_{AC}=114{,}247$) than $AB$ ($K_{AB}=164{,}609$), the network naturally favours the AC branch over the initial 600/400 guess, ending up close to a 453/547 split instead.

(ii) Pressure head and cavitation-limited pump lift (4 marks)

Pressure head is the pressure at a point expressed as the height of a column of the fluid it would take to produce that pressure, $h=p/(\rho g)$, so it converts an energy-per-unit-weight quantity into a length that can be added directly to elevation and velocity head in the Bernoulli/energy equation. The theoretical maximum suction lift for a pump is limited by atmospheric pressure (about 10.3 m of water at sea level, since a perfect vacuum on the suction side could at most let atmospheric pressure push water up 10.3 m). In practice the achievable lift is always less than this because cavitation intervenes first: as the pump lowers the local pressure on the suction side to draw water upward, that pressure eventually falls to the liquid's vapour pressure at the operating temperature, at which point the water boils (forms vapour bubbles) even though it is at ambient temperature; those vapour bubbles collapse violently once they reach the higher-pressure region inside the pump, causing noise, vibration, loss of capacity and pitting damage to the impeller. The Net Positive Suction Head Available (NPSHA) must therefore stay above the pump's required NPSH, which is always well short of the full 10.3 m theoretical limit once vapour pressure, friction losses and velocity head in the suction line are subtracted.

(iii) Positive displacement pump principle (4 marks)

A positive displacement (PD) pump moves fluid by trapping a fixed volume of it in an expanding cavity (formed by a piston, gear teeth, a rotating lobe/vane, or a diaphragm) on the suction side, mechanically sealing that volume off from the discharge side, and then forcing the cavity to contract, physically displacing that same fixed volume out the discharge port. Because the volume delivered per cycle (or per revolution) is fixed by the geometry of the cavity, not by the pressure the pump is working against, a PD pump delivers a constant volumetric flow rate for a given speed almost independent of discharge pressure (unlike a centrifugal pump, whose flow falls as head rises along its curve) — which is also why a PD pump must never be run against a fully closed valve without a relief device, since it will keep trying to displace fluid into a fixed volume and pressure will rise until something fails.

(iv) System head curve for two pumps in series (4 marks)

Fig. 4 shows the standard construction: the pump head-capacity curve (blue) falls as flow increases; for two identical pumps piped in series, the combined curve is obtained by adding their heads at each common flow rate (so the series curve reaches roughly double the single-pump head at low flow, converging back toward the single curve only where flow becomes very large). The system curve (red) rises with flow, $H_{sys}=H_s+kQ^2$, combining the fixed static lift $H_s$ with velocity-squared friction losses. The operating point is where the series pump curve and the system curve intersect — the only flow/head combination at which the pumps' output exactly matches what the pipe system demands.