Laminar and turbulent pipe flow, Moody chart and losses

Why it matters

Fuel feed lines, hydraulic actuators, environmental-control ducting, wind-tunnel return circuits and rocket propellant feed systems all lose pressure to friction and fittings. Pump and fan sizing, line diameters and the pressure available at an injector depend on predicting these losses. Pipe flow is also the cleanest example of laminar versus turbulent behaviour, which reappears in boundary layers on wings.

Key ideas

Reynolds' experiment. Dye injected into pipe flow stays a straight thread at low speed (laminar), wavers in a transitional range, and mixes across the pipe at high speed (turbulent). The controlling parameter is Re = ρV̄D/μ = V̄D/ν. In ordinary pipes:

• Re < about 2300: laminar.
• about 2300–4000: transitional (intermittent).
• Re > about 4000: turbulent. With very smooth inlets and no disturbances, laminar flow can persist to much higher Re, so the lower limit is the robust one.

Entrance length. Flow enters with a nearly flat profile; boundary layers grow from the walls until they meet. Beyond the entrance length the profile no longer changes (fully developed). Laminar: L_e ≈ 0.06·Re·D. Turbulent: L_e ≈ 4.4·D·Re^(1/6), typically 20–40 diameters.

Laminar flow has the parabolic Hagen–Poiseuille profile (previous topic), with f = 64/Re, independent of wall roughness.

Turbulent flow. Velocity fluctuates randomly about a mean; turbulent eddies transfer momentum across the pipe much more effectively than molecular viscosity. The mean profile is much flatter in the core (roughly a 1/7-power law, u/u_max = (y/R)^(1/7)), with a thin viscous sublayer at the wall and steep wall gradients, so wall shear and losses are much higher than in laminar flow at the same Re. Friction depends on Re and on relative roughness ε/D:

• Hydraulically smooth: roughness elements are buried in the viscous sublayer; f depends on Re only (Blasius f = 0.316·Re^(−1/4) for Re up to about 10⁵).
• Fully rough: roughness protrudes through the sublayer; f depends on ε/D only (flat lines at the right of the Moody chart).
• Transitional roughness: both matter (Colebrook equation).

Moody chart. A log–log plot of Darcy f against Re with curves of ε/D. It is a graphical form of the Colebrook equation. Haaland's explicit formula is within about 2 % of Colebrook and is handy in exams. Typical ε: drawn tubing 0.0015 mm, commercial steel 0.045 mm, cast iron 0.26 mm (take values from your data book).

Head loss. Major (friction) loss follows Darcy–Weisbach. Minor losses from entrances, exits, bends, valves and sudden area changes are written as K·V²/(2g); K comes from tables. The sudden-expansion loss can be derived from momentum: h = (V₁ − V₂)²/(2g); the exit into a large reservoir has K = 1. Energy equation between two points: p₁/(ρg) + V₁²/(2g) + z₁ + h_pump = p₂/(ρg) + V₂²/(2g) + z₂ + h_turbine + Σh_L.

Non-circular ducts use the hydraulic diameter D_h = 4A/P (A area, P wetted perimeter) in Re and in Darcy–Weisbach for turbulent flow.

Formulas

Re = ρ·V̄·D / μ = V̄·D / ν

• V̄: mean velocity (m/s); D: diameter (m); μ (Pa·s); ν (m²/s).

h_f = f·(L/D)·V̄²/(2g) and Δp = f·(L/D)·ρ·V̄²/2

• h_f: head loss (m); f: Darcy friction factor (dimensionless); L: length (m); Δp (Pa).

f = 64 / Re (laminar) and f = 0.316·Re^(−1/4) (smooth, 4000 < Re < 10⁵) 1/√f = −2.0·log₁₀(ε/(3.7D) + 2.51/(Re·√f)) (Colebrook) 1/√f = −1.8·log₁₀((ε/(3.7D))^1.11 + 6.9/Re) (Haaland)

• ε: absolute roughness (m).

h_m = K·V̄²/(2g), sudden expansion h = (V₁ − V₂)²/(2g)

• K: loss coefficient (from tables; ≈ 0.5 sharp entrance, 1.0 exit).

τ_w = f·ρ·V̄²/8 and D_h = 4A/P

• τ_w: wall shear stress (Pa); D_h: hydraulic diameter (m).

Worked examples

Example 1 (standard): head loss in a steel pipe. Given: water (ν = 1.0×10⁻⁶ m²/s), D = 0.1 m, V̄ = 2 m/s, L = 100 m, ε = 0.045 mm.

1. Re = V̄·D/ν = 2 × 0.1/10⁻⁶ = 2.0×10⁵, turbulent. ε/D = 0.00045.
2. Colebrook (iterated) gives f = 0.0186 (Haaland 0.0184; Moody chart ≈ 0.019).
3. h_f = f·(L/D)·V̄²/(2g) = 0.0186 × 1000 × 4/19.62 = 3.78 m.
4. Δp = ρg·h_f = 1000 × 9.81 × 3.78 = 37.1 kPa. Answer: h_f ≈ 3.8 m (Δp ≈ 37 kPa).

Example 2 (GATE level): gravity feed between two tanks. Given: water flows from an upper to a lower open tank through 200 m of 0.15 m pipe (ε = 0.15 mm) with a sharp entrance (K = 0.5), two bends (K = 0.9 each) and an exit (K = 1.0). Required Q = 0.04 m³/s. Find the necessary difference in water levels.

1. A = π × 0.15²/4 = 0.017671 m², V̄ = 0.04/0.017671 = 2.264 m/s, V̄²/(2g) = 0.2611 m.
2. Re = 2.264 × 0.15/10⁻⁶ = 3.40×10⁵; ε/D = 0.001; Colebrook f = 0.0205.
3. Friction: h_f = 0.0205 × (200/0.15) × 0.2611 = 7.14 m.
4. Minor: ΣK = 0.5 + 1.8 + 1.0 = 3.3, so h_m = 3.3 × 0.2611 = 0.86 m.
5. Energy equation between the free surfaces (p and V equal): Δz = h_f + h_m. Answer: Δz ≈ 8.0 m.

Common mistakes

• Mixing the Darcy factor f with the Fanning factor (f_Fanning = f_Darcy/4). Check which one a chart or formula uses.
• Using 64/Re for turbulent flow, or a Moody curve for laminar flow.
• Using roughness in mm with D in m: ε/D must be dimensionless.
• Forgetting the exit loss (K = 1) or counting it twice with the outlet kinetic energy.
• Using the diameter instead of the hydraulic diameter for rectangular ducts.
• Treating Re = 2300 as a sharp switch. Transition is a range.

For GATE AE

Expect: flow regime from Re; laminar friction factor and pressure drop; Darcy–Weisbach head loss with a given f; reading f from Colebrook, Haaland or Blasius; minor losses and the energy equation between reservoirs; pumping power ρgQH/η; and hydraulic diameter of ducts. Practise one complete reservoir-to-reservoir problem and keep the laminar and smooth-turbulent friction laws in memory.

Quick check

1. Laminar f at Re = 1600?
2. Blasius f at Re = 10⁴?
3. Head loss for f = 0.02, L/D = 2000, V̄ = 1.5 m/s?
4. Hydraulic diameter of a 0.2 m × 0.1 m duct?

Answers: 1. 0.04 2. 0.0316 3. 4.59 m 4. 4 × 0.02/0.6 = 0.133 m