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Velocity for fully turbulent flow in rough pipe

Short answer

The fully rough criterion is u₊k/ν ≥ 100. Using ν = 10⁻⁶ m²/s, k = 0.001 m, D = 1.2 m, the rough-pipe friction factor f = 0.0188, then u₊ = 0.1 m/s and V = u₊/√(f/8) ≈ 2.06 m/s.

A 120 cm diameter pipe has equivalent roughness height of 1 mm. What will be the velocity at which the roughness causes the flow to become fully turbulent?

Criterion. Flow is fully turbulent (hydrodynamically rough) when the roughness Reynolds number (Nikuradse) is

u∗kν≥100\frac{u_* k}{\nu} \ge 100

Take ν=1.0×10−6\nu = 1.0\times10^{-6} m²/s (water at about 20 °C), k=1k = 1 mm =0.001= 0.001 m, D=1.2D = 1.2 m.

Step 1: rough-pipe friction factor (von Karman-Prandtl, D/k=1200D/k = 1200):

1f=2log⁡10Dk+1.14  ⇒  f=0.01877,f/8=0.04844\frac{1}{\sqrt f} = 2\log_{10}\frac{D}{k} + 1.14 \;\Rightarrow\; f = 0.01877, \qquad \sqrt{f/8} = 0.04844

Step 2: shear velocity at the limit

u∗=100 νk=100(1.0×10−6)0.001=0.100 m/su_* = \frac{100\,\nu}{k} = \frac{100(1.0\times10^{-6})}{0.001} = 0.100\ \text{m/s}

Step 3: mean velocity

V=u∗f/8=0.1000.04844=2.06 m/sV = \frac{u_*}{\sqrt{f/8}} = \frac{0.100}{0.04844} = 2.06\ \text{m/s}

Answer: The flow becomes fully turbulent (rough) for V≥2.06V \ge 2.06 m/s, about 2 m/s. (If the limit u∗k/ν=70u_*k/\nu = 70 is used, the velocity is 1.45 m/s.)

✓ Answer checked by Claude Sonnet 5.5

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