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Laminar/turbulent flow and friction factor

Short answer

Laminar flow: smooth parallel layers, Re<2000. Turbulent flow: random eddies, Re>4000. For D=0.7m, Q=1.2m³/s, L=50m, hf=0.6m: V=3.118m/s, V²/2g=0.4956m. Darcy-Weisbach gives f = hf D / (L V²/2g) = 0.017. Re≈2.2×10⁶, turbulent.

Define laminar and turbulent flows. Water flows through a 700 mm diameter pipe with flow rate 1200 lps. Calculate the value of friction factor if the head loss in 50 m of pipe length is 0.6 m.

Laminar and turbulent flow

  • Laminar flow: fluid particles move in smooth, parallel layers without mixing; viscous forces dominate; Re<2000Re < 2000 in pipes. Example: thick oil in a thin tube.
  • Turbulent flow: particles move randomly with eddies and mixing; inertia forces dominate; Re>4000Re > 4000 in pipes. Example: water in a penstock.

Numerical

Given: D=0.7D = 0.7 m, Q=1200Q = 1200 L/s =1.2= 1.2 m³/s, L=50L = 50 m, hf=0.6h_f = 0.6 m.

Velocity:

A=π4(0.7)2=0.3848 m2V=QA=1.20.3848=3.118 m/sV22g=3.11822×9.81=0.4956 m\begin{aligned} A &= \frac{\pi}{4}(0.7)^2 = 0.3848\ \text{m}^2 \\ V &= \frac{Q}{A} = \frac{1.2}{0.3848} = 3.118\ \text{m/s} \\ \frac{V^2}{2g} &= \frac{3.118^2}{2 \times 9.81} = 0.4956\ \text{m} \end{aligned}

Darcy–Weisbach equation:

hf=fLDV22gh_f = f\frac{L}{D}\frac{V^2}{2g} f=hf DL⋅2gV2=0.6×0.750×0.4956=0.01695\begin{aligned} f &= \frac{h_f\, D}{L}\cdot\frac{2g}{V^2} \\ &= \frac{0.6 \times 0.7}{50 \times 0.4956} = 0.01695 \end{aligned}

Answer: Darcy friction factor f≈0.017f \approx 0.017. (If the form hf=4fLV22gDh_f = \frac{4fLV^2}{2gD} is used, the coefficient of friction is f/4≈0.0042f/4 \approx 0.0042.) At V=3.12V = 3.12 m/s in a 0.7 m pipe, Re≈2.2×106Re \approx 2.2\times10^6, so the flow is turbulent.

✓ Answer checked by Claude Opus 5.5 (high effort)

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