Closed Conduit Energy Losses
Closed conduit flow fills the pipe completely, so pressure may be above or below atmospheric. Head loss includes friction and minor losses.
$$h_L=h_f+\sum h_m$$$$h_f=f\frac{L}{D}\frac{V^2}{2g}$$$$h_m=K\frac{V^2}{2g}$$
Flow Regime by Reynolds Number
$$Re=\frac{VD}{\nu}$$
Typical classification: laminar for $Re<2000$, transition near 2000 to 4000, and turbulent for larger values.
Pipe Pressure Required
Water flows at 0.040 m3/s in a 150 mm pipe, 120 m long. If $f=0.020$, find the friction head loss.
$$A=\frac{\pi(0.15)^2}{4}=0.01767\text{ m}^2$$$$V=\frac{0.040}{0.01767}=2.26\text{ m/s}$$$$h_f=0.020\frac{120}{0.15}\frac{(2.26)^2}{2(9.81)}=8.33\text{ m}$$
Answer: The friction head loss is about 8.33 m.
Problem: Reynolds Number and Flow Regime
Water at 20°C (kinematic viscosity ν = 1.004 × 10⁻⁶ m²/s) flows through a 200 mm pipe. Determine the velocity at which the flow transitions from laminar to turbulent. Also find the velocity above which fully turbulent flow is assured (Re > 4000). What discharge corresponds to Re = 4000?
$$Re = \frac{VD}{\nu}$$
$$V_{laminar} = \frac{Re_{cr} \cdot \nu}{D} = \frac{2000(1.004 \times 10^{-6})}{0.200} = 0.01004 \text{ m/s}$$
$$V_{turbulent} = \frac{4000(1.004 \times 10^{-6})}{0.200} = 0.02008 \text{ m/s}$$
$$Q = VA = 0.02008\frac{\pi(0.200)^2}{4} = 0.02008(0.03142) = 6.31 \times 10^{-4} \text{ m}^3/\text{s}$$
Answer: Laminar flow transitions at V = 0.010 m/s. Fully turbulent flow begins above V = 0.020 m/s (Q = 0.63 L/s). In practice, water pipelines always operate at velocities orders of magnitude higher — turbulent flow is the norm in hydraulics.
Problem: Pipeline with Minor Losses — Total Head Loss
Water flows at Q = 0.060 m³/s through a 200 mm diameter pipeline 150 m long (f = 0.020). The system includes: a sharp-edged entrance (K = 0.5), two 90° elbows (K = 0.9 each), and a fully open gate valve (K = 0.2). Find the total head loss in the pipeline.
$$A = \frac{\pi(0.20)^2}{4} = 0.03142 \text{ m}^2$$
$$V = \frac{Q}{A} = \frac{0.060}{0.03142} = 1.910 \text{ m/s}$$
$$\frac{V^2}{2g} = \frac{(1.910)^2}{2(9.81)} = 0.1862 \text{ m}$$
$$h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.020\frac{150}{0.20}(0.1862) = 2.793 \text{ m}$$
$$h_m = \sum K \frac{V^2}{2g} = (0.5 + 0.9 + 0.9 + 0.2)(0.1862)$$
$$h_m = 2.5(0.1862) = 0.466 \text{ m}$$
$$h_{L,total} = 2.793 + 0.466 = 3.259 \text{ m}$$
Answer: Total head loss is 3.259 m (friction: 2.793 m + minor losses: 0.466 m). In short pipes with many fittings, minor losses become significant; in long pipelines they are typically neglected.
Problem: Pressure Distribution Along a Pipeline — HGL
Water flows from reservoir A (surface elevation 50 m) through a 250 mm pipe (f = 0.022, L = 600 m) to reservoir B (surface elevation 25 m). All elevations are above a common datum. Find: (a) the discharge, (b) the pressure at the midpoint of the pipe if the pipe centerline is at elevation 35 m throughout, and (c) state whether the pipe is under positive or negative pressure at midpoint.
$$\Delta z = 50 - 25 = 25 \text{ m}$$
$$h_f = 25 \text{ m}, \quad h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.022\frac{600}{0.25}\frac{V^2}{2g} = 52.8\frac{V^2}{2g}$$
$$\frac{V^2}{2g} = \frac{25}{52.8} = 0.4735 \text{ m} \Rightarrow V = 3.047 \text{ m/s}$$
$$Q = \frac{\pi(0.25)^2}{4}(3.047) = 0.04909(3.047) = 0.1496 \text{ m}^3/\text{s}$$
At midpoint (300 m from A, z = 35 m), friction loss consumed = half total = 12.5 m:
$$50 + 0 + 0 = 35 + \frac{p_{mid}}{\gamma} + \frac{V^2}{2g} + 12.5$$
$$\frac{p_{mid}}{\gamma} = 50 - 35 - 0.4735 - 12.5 = 2.027 \text{ m}$$
$$p_{mid} = 9.81(2.027) = 19.88 \text{ kPa (positive)}$$
Answer: Q = 0.150 m³/s. Midpoint pressure is 19.88 kPa positive gage — the pipe is under pressure at this point. The hydraulic grade line (HGL) drops linearly from 50 m to 25 m along the pipe, and since the pipe is at 35 m, the HGL is above the pipe centerline everywhere (positive pressure throughout).
Problem: Finding Pipe Diameter for a Given Head Loss Constraint
A pump delivers 0.10 m³/s of water through a straight steel pipe (f = 0.018) that is 500 m long. The maximum allowable head loss due to pipe friction is 15 m. Determine the minimum pipe diameter required. Also find the pump power if the elevation difference is 10 m (pump at the lower end).
$$h_f = f\frac{L}{D}\frac{V^2}{2g} = f\frac{L}{D}\frac{Q^2}{2g(A)^2} = \frac{8fLQ^2}{\pi^2 g D^5}$$
$$15 = \frac{8(0.018)(500)(0.10)^2}{\pi^2(9.81)D^5}$$
$$D^5 = \frac{8(0.018)(500)(0.01)}{9.81\pi^2(15)} = \frac{0.72}{1452.0} = 4.959 \times 10^{-4}$$
$$D = (4.959 \times 10^{-4})^{1/5} = 0.2185 \text{ m} = 219 \text{ mm}$$
Use next standard size: D = 250 mm. Pump head = elevation + head loss = 10 + 15 = 25 m:
$$P = \gamma Q h_p = 9.81(0.10)(25) = 24.53 \text{ kW (hydraulic power)}$$
Answer: Minimum diameter = 219 mm; use 250 mm standard pipe. Pump hydraulic power = 24.53 kW (shaft power depends on pump efficiency, typically divide by 0.75–0.85).