Library / Fluid Mechanics / FLU-05
Fluid Mechanics
Laminar flow in a pipe (Hagen–Poiseuille)
Flow rate through a pipe from the pressure drop for slow, viscous flow.
- Ref
- FLU-05
- Rev
- v1
- Field
- Fluid Mechanics
- Result
- Q [mL/s]
Steady, incompressible Newtonian fluid in a straight circular pipe with no slip. The pressure-drop expression assumes a horizontal pipe. Use the diameter-based Reynolds number; the flow is laminar below about Re = 2300.
Thick liquids in thin pipes flow smoothly, and the flow rate rises very steeply with pipe diameter.
Calculator
- Mean velocity0.1563 m/s
- Wall shear stress Δp D/(4L)12.5 Pa
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Doubling the pipe diameter raises the flow sixteenfold for the same pressure drop.
- The friction factor 64/Re shows the pressure drop is proportional to speed (not speed squared) in laminar flow.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mL/s) | Local step | Change in result |
|---|---|---|---|---|---|---|
| D Diameter | 0.5 to 50 mm | 3.068e-4 | 30,680 | 30,680 | +10% | +46.4 % |
| Δp Pressure drop | 0.1 to 1,000 kPa | 0.01534 | 153.4 | 153.4 | +10% | +10 % |
| μ Dynamic viscosity | 0.001 to 1 Pa·s | 0.1534 | 153.4 | 153.2 | +10% | −9.09 % |
| L Length | 0.1 to 100 m | 0.06136 | 61.36 | 61.3 | +10% | −9.09 % |
Sweep of Diameter
| Diameter (mm) | Flow rate (mL/s) |
|---|---|
| 0.5 | 3.068e-4 |
| 0.8891 | 0.003068 |
| 1.581 | 0.03068 |
| 2.812 | 0.3068 |
| 5 | 3.068 |
| 8.891 | 30.68 |
| 15.81 | 306.8 |
| 28.12 | 3,068 |
| 50 | 30,680 |
Sweep one variable
Variables and how each one contributes
Flow rate
What the equation gives you. Everything else feeds it.
Diameter
Pipe internal diameter. Fourth power: the strongest driver.
Pressure drop
Pressure drop along the pipe. Directly proportional.
Dynamic viscosity
Fluid viscosity (water 0.001, light oil about 0.05). Inversely proportional.
Length
Pipe length. Inversely proportional.
Worked example with the values above
- D⁴ = (0.005 m)⁴ = 6.25×10−10 m⁴.
- Q = π D⁴ Δp / (128 μ L) = π × 6.25×10−10 × 20,000 / (128 × 0.05 × 2) = 3.068×10−6 m³/s.
- Q = 3.068 mL/s.
Assumptions and limits
- Laminar, steady, fully developed flow.
- Horizontal circular pipe.
Common mistakes
- Using the formula for turbulent flow (Re above about 2300).
- Using radius where diameter is meant.
Statistics and uncertainty
Because D appears to the fourth power, a 2% error in diameter gives about 8% error in flow.
Where it is used
- Hydraulic lines and capillary flow.
- Viscometer and filter design.
Maths you need
- Viscosity: μ
- Reynolds number: Re