Pipe Flow (Darcy-Weisbach)
Pressure loss · Friction factor λ · Reynolds number · Flow velocity · Pipe sizing
Pipe Flow Calculator
Formulas & Fundamentals
Δp = λ · (L/D) · (ρ·v²)/2 [Pa]
λ = friction factor | L = length | D = diameter
ρ = density | v = velocity
h_f = λ · (L/D) · v²/(2·g) [m]
g = 9.81 m/s² | Δp = ρ·g·h_f
Re = v·D / ν
ν = kinematic viscosity [m²/s]
Re < 2300: laminar | Re > 4000: turbulent
laminar: λ = 64 / Re
turbulent (Colebrook-White):
1/√λ = −2·log₁₀( ε/(3.7·D) + 2.51/(Re·√λ) )
implicit – solved iteratively (Swamee-Jain initial value)
Q = v · A = v · π·D²/4 [m³/s]
v = 4·Q / (π·D²) [m/s]
D = √(4·Q / (π·v_allow)) [m]
from flow rate Q and target velocity
Pipe roughness ε (typical values)
| Material | ε [mm] |
|---|---|
| Copper, glass (smooth) | 0.0015 |
| PVC, plastic | 0.007 |
| Steel new / seamless | 0.05 |
| Galvanized steel | 0.15 |
| Cast iron | 0.25 |
| Concrete smooth | 0.3–3.0 |
Pipe Flow & Pressure Loss – Fundamentals of Hydraulics
What does the Darcy-Weisbach equation describe?
The Darcy-Weisbach equation is the fundamental relationship for calculating the pressure loss due to pipe friction in a flowing line. It is universally valid for laminar and turbulent flow as well as for all Newtonian fluids (water, oil, air), making it the standard in process engineering, water supply, heating and ventilation systems.
The term ρ·v²/2 is the dynamic (velocity) pressure; λ·(L/D) is the dimensionless resistance coefficient of the line.
Flow regimes
| Reynolds number | Flow |
|---|---|
| Re < 2300 | laminar (layered flow) |
| 2300 – 4000 | transition region |
| Re > 4000 | turbulent |
Recommended velocities
| Application | v [m/s] |
|---|---|
| Drinking water service | 1.0–2.0 |
| Pressure line (pump) | 1.5–3.0 |
| Suction line | 0.7–1.5 |
| Heating (circulation) | 0.5–1.5 |
| Compressed air | 10–20 |
The friction factor λ – core of the calculation
The pipe friction factor λ (dimensionless) depends on the Reynolds number and the relative roughness ε/D. The Moody diagram presents this relationship graphically. Three regions:
λ = 64 / Re
Independent of roughness – only viscosity determines the resistance.
λ = 0.3164 / Re0.25
1/√λ = −2·log₁₀( ε/(3.7·D) + 2.51/(Re·√λ) )
Implicit equation – solved iteratively. Explicit approximation by Swamee-Jain:
λ = 0.25 / [ log₁₀( ε/(3.7·D) + 5.74/Re0.9 ) ]²
Worked example: drinking water pressure line
Problem:
Steel pipe DN100 (D = 100 mm), length L = 250 m, flow rate Q = 15 l/s, water 20 °C (ρ = 998 kg/m³, ν = 1.004·10⁻⁶ m²/s), roughness ε = 0.05 mm. Find: pressure loss Δp.
Solution:
- Cross-section: A = π·0.1²/4 = 7.854·10⁻³ m²
- Velocity: v = Q/A = 0.015/7.854·10⁻³ = 1.91 m/s
- Reynolds: Re = v·D/ν = 1.91·0.1/1.004·10⁻⁶ = 190,200 → turbulent
- Relative roughness: ε/D = 0.05/100 = 0.0005
- Colebrook-White (iterative): λ ≈ 0.0193
- Δp = 0.0193·(250/0.1)·(998·1.91²)/2 = 87,900 Pa ≈ 0.88 bar
- Head loss: h_f = Δp/(ρ·g) = 87900/(998·9.81) = 8.98 m
Frequently Asked Questions
Summary
Pressure loss
Δp = λ·(L/D)·ρv²/2
h_f = Δp/(ρg)
Friction factor λ
laminar: 64/Re
turbulent: Colebrook-White
Reynolds number
Re = v·D/ν
limit: 2300
Typical applications
- Water supply: sizing of drinking water and pressure lines, pump head
- Heating & ventilation: pipe network calculation, hydraulic balancing
- Process engineering: pipe sizing in plants, pump selection
- Fire protection: sprinkler and hydrant lines, minimum pressure
- Civil engineering: pressure pipelines, sewage pressure mains
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