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Pipe flow and head losses

Guided step by step from velocity to total head loss: Reynolds number, friction factor on the Moody chart, Darcy–Weisbach friction loss and fitting losses, drawn as energy and hydraulic grade lines.

Guided · step 1 of 10

Know the pipe

Water flows from the upper tank to the lower tank through the pipe. We want the head lost on the way: friction along the length (major loss) plus losses at fittings (minor losses). That loss is the level difference the tanks need, or the extra head a pump must give.

Q = 15 L/s = 0.015 m³/s, D = 100 mm = 0.1 m, L = 250 m

Galvanised iron (GI): ε = 0.15 mm; water at 20 °C: ν = 1.004 × 10⁻⁶ m²/s

Pipe and flow

Fittings

Entrance (square-edged) K = 0.51
Entrance (bell-mouth) K = 0.040
90° elbow (standard) K = 0.94
45° elbow K = 0.40
90° long-radius bend K = 0.30
Tee (flow through branch) K = 1.80
Gate valve (fully open) K = 0.22
Globe valve (fully open) K = 100
Non-return (check) valve K = 2.51
Exit into tank K = 11

Energy and hydraulic grade lines

90°90°90°90°GateGateNon-returnH = 12.109 mEGLL = 250 m. Fittings are drawn evenly spaced; heights are not to scale with length.
Total loss H
12.109m
hf (friction)
10.622m
hm (fittings)
1.487m
V
1.91m/s
Re
1.9 × 10⁵
turbulent
f
0.0229

Moody chart

10³10⁴10⁵10⁶10⁷10⁸0.0080.010.0150.020.030.040.050.070.164/Re0.050.030.020.010.0050.0020.0010.00050.00020.00010.00005Reynolds number Refε/D

Your pipe: Re = 1.9 × 10⁵, ε/D = 1.5 × 10⁻³, so f = 0.0229 (orange curve and point). Left of Re = 2000 the flow is laminar and f = 64/Re.

Full solution, all steps at once
  1. Know the pipe

    Q = 15 L/s = 0.015 m³/s, D = 100 mm = 0.1 m, L = 250 m

    Galvanised iron (GI): ε = 0.15 mm; water at 20 °C: ν = 1.004 × 10⁻⁶ m²/s

  2. Area and mean velocityA = πD²4V = QA

    A = π × 0.1²4 = 7.854 × 10⁻³ m²

    V = 0.0157.854 × 10⁻³ = 1.91 m/s

  3. Velocity headhᵥ = V²2g

    V²2g = 1.91²2 × 9.81 = 0.1859 m

  4. Reynolds number: laminar or turbulent?Re = VDν

    Re = 1.91 × 0.11.004 × 10⁻⁶ = 1.902 × 10⁵

    The flow is turbulent (Re > 4000)

  5. Relative roughnessεD

    ε/D = 0.15100 = 1.5 × 10⁻³

  6. Friction factor (Colebrook–White)1/√f = −2 log₁₀(ε3.7D + 2.51Re·√f)f₀ = 0.25[log₁₀(ε3.7D + 5.74/Re0.9)]²

    Swamee–Jain: f₀ = 0.25[log₁₀(1.5 × 10⁻³3.7 + 5.741.902 × 10⁵0.9)]² = 0.02303

    Colebrook iteration 1: f = 0.02285

    Colebrook iteration 2: f = 0.022854

    Colebrook iteration 3: f = 0.022854

    f = 0.02285

  7. Major loss: friction along the pipe (Darcy–Weisbach)hf = f·(L/D)·V²2g

    hf = 0.02285 × (2500.1) × 0.1859

    hf = 0.02285 × 2500 × 0.1859 = 10.622 m

    Friction slope = hfL = 0.04249 (42.49 m per km)

  8. Minor losses: entrance, bends, valves, exithm = ΣK·V²2g

    ΣK = 0.5 + 4 × 0.9 + 2 × 0.2 + 2.5 + 1 = 8

    hm = 8 × 0.1859 = 1.487 m

  9. Total head loss, EGL and HGLH = hf + hm

    H = 10.622 + 1.487 = 12.109 m

    So the upper tank must stand 12.109 m above the lower one (or a pump must add 12.109 m of head). Minor losses are 12.3% of the total.

  10. Cross-check with Hazen–Williamshf = 10.67·L·Q1.852C1.852·D4.87

    C = 120 for galvanised iron (gi)

    hf = 10.67 × 250 × 0.0151.8521201.852 × 0.14.87 = 11.695 m

    Darcy–Weisbach gave 10.622 m: a difference of 10.1%.