Fluid mechanicsMVP
Pressure drop
Pressure needed to drive a flow through a circular channel (Hagen–Poiseuille).
Inputs
Result
Enter values and calculate to see the result.
Interpretation
Enter the fluid, channel, and flow rate to get the pressure drop. The strong 1/D⁴ dependence means small diameter changes matter a lot.
Formula
ΔP = 128 · μ · L · Q / (π · D⁴)
Exact for fully developed laminar flow in a circular channel.
Variables
| Symbol | Variable | Unit |
|---|---|---|
| ΔP | Pressure drop | Pa |
| μ | Dynamic viscosity | Pa·s |
| L | Channel length | m |
| Q | Volumetric flow rate | m³/s |
| D | Channel diameter | m |
Assumptions & validity
This tool assumes:
- Circular cross-section — this is the key assumption; rectangular channels need the resistance approximation.
- Fully developed, steady, laminar flow.
- Newtonian, incompressible fluid.
Worked example
Water (μ = 1.0 mPa·s) flows at Q = 1 µL/min through a circular channel L = 10 mm long and D = 50 µm across.
ΔP = 128 × 10⁻³ × 0.01 × (1.667×10⁻¹¹) / (π × (50×10⁻⁶)⁴) ≈ 1.09 kPa