Turbulent pipe flow, Moody chart and minor losses

Why it matters

Almost every practical pipe — cooling-water lines, fuel rails, exhaust ducts, city water mains — runs turbulent. The friction factor from the Moody chart and the loss coefficients of bends, valves and fittings set the pump head you must buy and the energy you pay for every year. Getting f or K wrong by a factor of four (Darcy vs Fanning) is one of the most common exam and design errors.

Key ideas

• Transition. In pipes, flow is laminar for Re below about 2000, transitional from about 2000 to 4000, and turbulent above about 4000. In turbulent flow the velocity at a point fluctuates randomly about a time-mean value; the eddies transport momentum far more effectively than molecular viscosity.
• Velocity profile. Turbulent profiles are much flatter than the laminar parabola: u_max/V is only about 1.2, often approximated by the 1/7 power law u/u_max = (y/R)^(1/7). Close to the wall there is a thin viscous sublayer where viscosity still dominates and the velocity gradient, and so the wall shear, is very high.
• Roughness and the sublayer.
  • Hydraulically smooth: roughness elements are buried inside the viscous sublayer; f depends on Re only.
  • Transitional: f depends on both Re and relative roughness ε/D.
  • Fully rough: roughness pokes through the sublayer; f depends on ε/D only and becomes constant with Re, so h_f ∝ V².
  • Typical ε: drawn tubing 0.0015 mm, commercial steel about 0.045 mm, cast iron about 0.26 mm — take the value for your material from a data book.
• Darcy–Weisbach gives the friction head loss in any fully developed pipe flow, laminar or turbulent. Only the friction factor changes.
• Moody chart plots Darcy f against Re with ε/D as a parameter. The laminar line f = 64/Re appears on the left. Colebrook's implicit equation reproduces the turbulent region; Haaland's explicit formula is within about 2 %.
• Darcy vs Fanning. Darcy f = 4 × Fanning f. Some books write h_f = 4fLV²/(2gD) using Fanning f. Check which one a formula or question uses.
• Head loss vs velocity. h_f ∝ V in laminar flow, about V^1.75 in smooth turbulent flow (Blasius), and V² in fully rough flow.
• Non-circular ducts. Use the hydraulic diameter D_h = 4A/P (area over wetted perimeter) in Re, ε/D and Darcy–Weisbach. It works well for turbulent flow.
• Minor (local) losses. Fittings, bends, valves, entrances, exits and area changes create separation and eddies. Each loss is written as K·V²/(2g), with V usually the velocity in the smaller pipe (or the upstream pipe for an expansion). In short systems with many fittings, minor losses can exceed pipe friction. An equivalent length L_e = K·D/f converts a fitting into extra pipe length.
• Typical K values (check a data book for a given fitting): sharp-edged entrance 0.5; well-rounded entrance about 0.04; exit into a large tank 1.0 (all kinetic energy is lost); 90° standard elbow about 0.9; fully open gate valve about 0.2; fully open globe valve about 10.
• Sudden expansion follows from momentum and energy: h_L = (V₁ − V₂)²/(2g). Sudden contraction: the loss happens as the vena contracta re-expands, h_L = (1/C_c − 1)²·V₂²/(2g), about 0.375·V₂²/(2g) for C_c = 0.62.

Formulas

h_f = f·(L/D)·V²/(2g) Darcy–Weisbach; h_f friction head loss (m), f Darcy friction factor (–), L length (m), D diameter (m), V mean velocity (m/s), g = 9.81 m/s².

h_f = 8·f·L·Q² / (π²·g·D⁵) The same written with flow rate Q (m³/s); shows h_f ∝ 1/D⁵ at fixed Q.

f = 64/Re (laminar) f = 0.316/Re^0.25 (Blasius, smooth, 4000 < Re < 10⁵)

1/√f = −2.0·log₁₀(ε/(3.7·D) + 2.51/(Re·√f)) Colebrook (turbulent, all roughness); ε absolute roughness (m).

1/√f = −1.8·log₁₀[(ε/(3.7·D))^1.11 + 6.9/Re] Haaland explicit approximation.

τ_w = f·ρ·V²/8 Wall shear stress (Pa) from Darcy f.

h_L = K·V²/(2g) L_e = K·D/f Minor loss (m); equivalent length (m).

h_L = (V₁ − V₂)²/(2g) sudden expansion (V₁ upstream, V₂ downstream) h_L = (1/C_c − 1)²·V₂²/(2g) sudden contraction (V₂ in the smaller pipe, C_c contraction coefficient)

D_h = 4·A/P Hydraulic diameter (m); A flow area (m²), P wetted perimeter (m).

Worked examples

Example 1 (standard) — friction loss with the Colebrook equation. Given: water (ν = 1.0 × 10⁻⁶ m²/s) flows at Q = 15 L/s through 200 m of commercial steel pipe, D = 100 mm, ε = 0.045 mm. Find f, h_f and the power lost.

1. A = π × 0.1²/4 = 7.854 × 10⁻³ m²; V = 0.015/7.854 × 10⁻³ = 1.910 m/s.
2. Re = VD/ν = 1.910 × 0.1/10⁻⁶ = 1.91 × 10⁵ → turbulent.
3. ε/D = 0.045/100 = 0.00045.
4. Colebrook, iterated from f = 0.02, converges to f = 0.0186 (Haaland gives 0.0184; the Moody chart reads about 0.019).
5. h_f = f·(L/D)·V²/(2g) = 0.0186 × 2000 × 1.910²/19.62 = 6.93 m.
6. Power lost = ρ·g·Q·h_f = 1000 × 9.81 × 0.015 × 6.93 = 1020 W. Answer: f ≈ 0.0186, h_f ≈ 6.9 m, about 1.0 kW dissipated.

Example 2 (GATE level) — two reservoirs with minor losses. Given: two reservoirs with a 12 m difference in water level are joined by a 300 m long, 150 mm pipe with a sharp-edged entrance (K = 0.5) and a submerged exit (K = 1.0). Take f = 0.02. Find the flow rate.

1. Energy equation between the two free surfaces (both at atmospheric pressure, negligible velocity): H = (K_entry + K_exit + f·L/D)·V²/(2g).
2. f·L/D = 0.02 × 300/0.15 = 40; total coefficient = 0.5 + 1.0 + 40 = 41.5.
3. V = √(2g·H/41.5) = √(2 × 9.81 × 12/41.5) = √5.673 = 2.382 m/s.
4. Q = (π × 0.15²/4) × 2.382 = 0.017671 × 2.382 = 0.0421 m³/s. Answer: Q ≈ 0.042 m³/s (42 L/s). The minor losses are 1.5/41.5 ≈ 3.6 % of the total here — small in a long pipe, but not in a short one.

Example 3 (sudden expansion). Q = 0.06 m³/s passes from a 150 mm to a 300 mm pipe. V₁ = 3.395 m/s, V₂ = 0.849 m/s; h_L = (3.395 − 0.849)²/19.62 = 0.331 m.

Common mistakes

• Mixing Darcy and Fanning friction factors (a factor of 4).
• Using f = 64/Re for turbulent flow, or reading the Moody chart with ε in mm and D in m.
• Forgetting the exit loss (K = 1) when the pipe discharges into a tank, or adding a velocity head at a free-jet outlet as well as an exit loss.
• Basing a minor-loss coefficient on the wrong velocity (an expansion uses the upstream velocity).
• Assuming h_f ∝ V² in all regimes.
• Using the geometric diameter instead of D_h for a rectangular duct.

For GATE ME

Expect: head loss or pressure drop from Darcy–Weisbach, the effect of doubling Q or halving D on h_f (∝ Q²/D⁵ at constant f), flow between two reservoirs including entrance and exit losses, sudden-expansion loss, wall shear from f, the regimes of the Moody chart, and power required to overcome friction. Practise converting between h_f, Δp and power.

Quick check

1. What is the relationship between Darcy and Fanning friction factors?
2. In the fully rough region, what does f depend on?
3. Water at 3 m/s flows through 50 m of 0.2 m pipe with f = 0.02. What is h_f?
4. What is the loss coefficient of a pipe exit into a large reservoir?
5. At constant f and Q, how does h_f change if D is halved? Answers: 1. f_Darcy = 4·f_Fanning. 2. Only on ε/D. 3. 0.02 × 250 × 9/19.62 = 2.29 m. 4. 1.0. 5. It rises 32 times (∝ 1/D⁵).