Question 3 of 7: Pipe Network Analysis, Head Concepts, and Pump System Curves

Problem 3: Pipe Network Analysis, Head Concepts, 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:

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.

1. 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. ✓
2. 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 $\sum KQ|Q|^{0.852}$ is zero to numerical precision (21 iterations to $10^{-6}$ convergence).
3. Converged flows. $$Q_{AB}=\boxed{672.3\ \text{L/s}}, \quad Q_{AC}=\boxed{327.8\ \text{L/s}}, \quad Q_{BC}=\boxed{117.4\ \text{L/s}}, \quad Q_{BD}=\boxed{454.9\ \text{L/s}}, \quad Q_{CD}=\boxed{145.1\ \text{L/s}}.$$
4. Check. Node continuity: A: $672.3+327.8=1000.0$ ✓. B: $672.3=117.4+454.9+100.0$ ✓. C: $327.8+117.4=145.1+300.0$ ✓. D: $454.9+145.1=600.0$ ✓.

(ii) Static Head vs. Friction Head

Static head $H_s$ is the elevation difference alone between the supply (suction) water surface and the discharge water surface — the head a pump must overcome even at zero flow, entirely independent of pipe size, roughness or flow rate: $H_s = z_{discharge} - z_{suction}$. Friction head $H_f$ is the additional head consumed by fluid friction along the pipe as flow actually moves through it, and it grows with the square of velocity (Darcy–Weisbach $H_f = f(L/d)(V^2/2g)$, or equivalently $H_f = KQ^2$ for a fixed pipe): it is zero at zero flow and rises steeply as $Q$ increases. The total dynamic head the pump must supply at any operating flow is $H = H_s + H_f(Q) = H_s + KQ^2$, which is exactly the system curve plotted in part (iii).

(iii) System-Pump Curve for a Low Static Head System

(iv) Pump Efficiency and Hydraulic Power

(a) Pump efficiency is the ratio of useful hydraulic power delivered to the fluid to the shaft (brake) power supplied to the pump: $$\eta_{pump} = \frac{P_{hydraulic}}{P_{shaft}} \times 100\%,$$ where $P_{shaft}$ is read from the motor/driver curve (or measured by a torque-and-speed test) at the operating point. Manufacturer pump curves typically plot $\eta$ directly against $Q$ alongside the head curve, so $P_{shaft}=P_{hydraulic}/\eta$ can be read off without a separate torque measurement.

(b) Hydraulic (water) power is the rate of useful work done raising and moving the fluid against the total dynamic head at the operating point: $$P_{hydraulic} = \rho g Q H = \gamma Q H,$$ with $\rho$ the fluid density (kg/m³), $g$ gravitational acceleration, $Q$ the discharge (m³/s) and $H$ the total head (m) delivered by the pump at that $Q$ — i.e. $Q$ and $H$ are read at the operating point identified in part (iii), not at shutoff or free discharge.