Viscous Flow and Flow Through Pipes
Reynolds number, laminar flow in pipes and between plates (Hagen-Poiseuille), Darcy-Weisbach friction, Moody chart, minor losses, and pipes in series and parallel.
Real fluids lose energy to friction. In pipes, this loss decides pump sizes and flow rates. The first question is always: is the flow laminar or turbulent?
Reynolds number
Re = ρVD/μ = VD/ν (inertia forces / viscous forces)
- In a pipe: laminar below about 2000, turbulent above about 4000, transitional in between.
Laminar flow in a pipe (Hagen–Poiseuille)
- Velocity profile is a parabola: u = u_max (1 − r²/R²).
- u_max = 2 V_avg.
- Pressure drop: Δp = 32μVL/D².
- Friction factor f = 64/Re.
- Shear stress varies linearly from zero at the centre to its maximum at the wall.
Between fixed parallel plates: u_max = 1.5 V_avg and Δp = 12μVL/h².
Head loss (Darcy–Weisbach)
h_f = f L V²/(2 g D)
- Laminar: f = 64/Re.
- Turbulent: f depends on Re and relative roughness ε/D (Moody chart / Colebrook equation). For smooth pipes, Blasius: f = 0.316/Re^0.25 (Re < 10⁵).
- In terms of flow: h_f = f L Q²/(12.1 D⁵), so h_f ∝ 1/D⁵ at fixed Q.
Minor losses
h = K V²/(2g):
- Sudden expansion: h = (V₁ − V₂)²/(2g).
- Pipe entrance: K ≈ 0.5. Exit into a tank: K = 1.
- Bends, valves and fittings each have their own K.
Pipes in series and parallel
- Series: same Q; head losses add.
- Parallel: same head loss across each branch; flows add.
- Equivalent pipe: a single pipe with the same loss as a series combination: L/D⁵ = Σ Lᵢ/Dᵢ⁵.
Worked example
Oil (μ = 0.1 Pa·s, ρ = 900 kg/m³) flows at 1 m/s in a 50 mm pipe 100 m long. Find Re and the head loss.
- Re = 900 × 1 × 0.05/0.1 = 450 → laminar.
- f = 64/450 = 0.1422; h_f = 0.1422 × 100 × 1/(2 × 9.81 × 0.05) = 14.5 m of oil.
Practice
Water (ν = 10⁻⁶ m²/s) flows at 0.5 m/s in a 20 mm pipe. What is the Reynolds number?
Show answer and explanation
Answer: 10000Re = VD/ν = 0.5 × 0.02/10⁻⁶ = 10,000 (turbulent).
In laminar pipe flow, the maximum velocity is how many times the average velocity?
Show answer and explanation
Answer: 2The parabolic profile gives u_max = 2V_avg.
At constant flow rate and friction factor, if the pipe diameter is halved, the head loss becomes how many times larger?
Show answer and explanation
Answer: 32h_f ∝ Q²/D⁵, so halving D multiplies h_f by 2⁵ = 32.
A pipe of 0.1 m diameter has f = 0.02, length 200 m and V = 2 m/s. What is the head loss in m? (g = 9.81 m/s², give 2 decimal places.)
Show answer and explanation
Answer: 8.15h_f = fLV²/(2gD) = 0.02 × 200 × 4/1.962 = 8.15 m.