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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).

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: v263

HGE - Hydraulics / Pipe Network / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A sealed vertical cylinder of diameter 1.1 m is completely filled with water and rotates at 1200 rpm. Its height is 2.4 m.

Find the pressure increase at the wall relative to the axis.

  1. 2388.44 kPa
  2. 1865.37 kPa
  3. 2715.66 kPa
  4. 3011.83 kPa

Using the axis as the zero-pressure reference, state the center pressure.

  1. 0 kPa
  2. 10 kPa
  3. 20 kPa
  4. 30 kPa

Find the resultant force on the top cover.

  1. 1134.91 kN
  2. 886.36 kN
  3. 1290.39 kN
  4. 1431.12 kN
Rigid-body rotation gives a parabolic pressure head; integrate that radial pressure over the circular cover.
Computed answers: 2388.44426506, 0, 1134.90724219.