Inertial pressure versus curvature pressure
The Weber number compares the pressure a flow exerts on an interface with the pressure surface tension uses to keep it rounded. Flow at speed U carries a pressure of order ρU²; surface tension on a curved surface of size L produces one of order σ/L (2σ/R inside a spherical drop of radius R). Their ratio is the formula.
The MIT surface-tension slides write it with the drop radius a as inertia over curvature pressure, and the MIT similitude notes call it the group that indicates the importance of surface tension. Far below 1, surface tension holds the interface shape; far above 1, the flow can deform or tear it. Where the boundary lies depends on how ρ and L are defined.
Which density and which length?
The calculator returns ρv²L/σ from ρ (kg/m³), v (m/s), L (m) and σ (N/m). The definition is your choice, and for one drop it can change the result several hundredfold.
| Problem | ρ | v | L |
|---|---|---|---|
| Secondary (aerodynamic) breakup of a drop in gas | Surrounding gas density | Slip speed between drop and gas | Drop diameter (sometimes radius) |
| Falling raindrop scaling | Air density | Fall speed | Radius a (MIT slides) |
| Liquid jet or sheet leaving a nozzle | Liquid density | Jet speed | Jet diameter or sheet thickness |
The Ansys CFX theory guide defines the secondary-breakup We with surrounding-fluid density, slip velocity and particle diameter, noting that the radius is sometimes used. A radius-based value is half the diameter-based one, so check the definition behind any threshold you borrow.
Open this calculatorHow large is We for water and air at 20 °C?
Air density 1.166 kg/m³ is the IAEA Table 3.1 value at 98.1 kPa; 1.2043 kg/m³ is that value scaled to 101.325 kPa as an ideal gas (1.166 × 101.325 ÷ 98.1). Water surface tension is 0.0728 N/m from OpenStax; IAEA Table 3.3 gives 72.70 mN/m (0.0727 N/m) for saturated water at 20 °C, about 0.1 % lower.
| Property | Water | Air |
|---|---|---|
| Density ρ | 998.2 kg/m³ (IAEA Table 3.3) | 1.166 kg/m³ at 98.1 kPa (IAEA Table 3.1); 1.2043 kg/m³ scaled to 1 atm |
| Viscosity μ | 1.0000 × 10⁻³ Pa·s (IAEA) | 1.819 × 10⁻⁵ Pa·s (IAEA) |
| Water-air surface tension σ | 0.0728 N/m (OpenStax), 72.70 mN/m (IAEA) | |
For a feel of magnitudes, the next table gives σ/L and the speed at which We = 1, v = √(σ/(ρL)), using 1.2043 kg/m³ for air and 998.2 kg/m³ for water.
| L | σ/L | We = 1 speed, air | We = 1 speed, water |
|---|---|---|---|
| 10 μm | 7,280 Pa | 77.7 m/s | 2.70 m/s |
| 100 μm | 728 Pa | 24.6 m/s | 0.854 m/s |
| 1 mm | 72.8 Pa | 7.77 m/s | 0.270 m/s |
| 10 mm | 7.28 Pa | 2.46 m/s | 0.0854 m/s |
Smaller drops have larger curvature pressure, so air needs a much higher relative speed to deform them. At equal length and speed, the water-density We is 998.2 ÷ 1.2043 ≈ 829 times the air-density We; mixing the definitions inverts the conclusion.
Spray drop, raindrop and tap-water jet
Example 1. Will a spray-nozzle drop break up in air?
Given. A 0.5 mm water drop moves at 30 m/s relative to still air (assumed diameter and speed). Air at 20 °C, 1 atm: ρg = 1.2043 kg/m³; σ = 0.0728 N/m.
Unit conversion. d = 0.5 mm = 0.0005 m; 72.8 mN/m ÷ 1000 = 0.0728 N/m.
Calculation. ρgv² = 1.2043 × 30² ≈ 1,084 Pa and σ/d = 0.0728 ÷ 0.0005 = 145.6 Pa, so We = 1.2043 × 30² × 0.0005 ÷ 0.0728 ≈ 7.44.
Interpretation. The Ansys CFX guide uses this definition: deformation becomes significant near We = 1, below 12 is vibrational breakup and 12 to 50 is bag breakup. At 7.44 the drop oscillates but stays below bag breakup. It would need v = √(12σ/(ρgd)) ≈ 38.1 m/s for We = 12 and about 77.7 m/s for We = 50. The radius gives We ≈ 3.72, not comparable with those limits; the water density by mistake gives We ≈ 6,170 and a false prediction of shattering. With water viscosity 1.0000 × 10⁻³ Pa·s, Oh = μ/√(ρσd) ≈ 0.0052, so viscosity barely delays breakup.
Example 2. How large can a raindrop grow?
Given. The MIT slides balance air drag ρaU²a² against weight of order ρga³, giving U ∼ √(ρga/ρa); the drop survives only if ρaU² ∼ ρga is below σ/a. Inputs: water 998.2 kg/m³, air 1.2043 kg/m³, σ = 0.0728 N/m, g = 9.81 m/s², radii a = 1 mm and 3 mm.
Unit conversion. a = 0.001 m and 0.003 m; the radius goes into L, as in the slides.
Calculation. For a = 1 mm, U ∼ √(998.2 × 9.81 × 0.001 ÷ 1.2043) ≈ 2.85 m/s, and the air-density We = 1.2043 × 2.85² × 0.001 ÷ 0.0728 ≈ 0.135. For a = 3 mm, U ∼ 4.94 m/s and We ≈ 1.21. Here We equals the Bond number Bo = ρga²/σ: for a = 1 mm, Bo = 998.2 × 9.81 × 0.001² ÷ 0.0728 ≈ 0.135. The radius at which We = 1 is the capillary length ℓc = √(σ/(ρg)) = √(0.0728 ÷ (998.2 × 9.81)) ≈ 2.73 mm.
Interpretation. Per the slides, drops smaller than the capillary length are held together by surface tension; larger ones are torn apart. They quote about 2 mm (σ = 70 dyn/cm), the same order as 2.73 mm. The speed omits a coefficient, so it is not a terminal velocity. On a diameter basis the values double (0.269 and 2.42).
Example 3. Liquid Weber number of a tap-water jet
Given. Water leaves a 10 mm outlet at 4 L/min (assumed); 998.2 kg/m³, σ = 0.0728 N/m.
Unit conversion. Q = 4 L/min = 4 × 10⁻³ m³ ÷ 60 s ≈ 6.67 × 10⁻⁵ m³/s. A = π × 0.01² ÷ 4 ≈ 7.85 × 10⁻⁵ m², so v = Q/A ≈ 0.849 m/s.
Calculation. With liquid density and jet diameter, We = 998.2 × 0.849² × 0.01 ÷ 0.0728 ≈ 98.8. ρv² ≈ 719 Pa against a cylindrical-surface curvature pressure σ/R = 0.0728 ÷ 0.005 = 14.56 Pa. With air density, We ≈ 0.119.
Interpretation. Inertia far exceeds curvature pressure, so the water leaves as a continuous stream rather than dripping. An air-density We near 0.1 means the air hardly tears it; a slow stream breaks into drops further down mainly through capillary (Plateau–Rayleigh) instability. The stream accelerates and thins as it falls, so state that We refers to outlet conditions.
Reading the drop-breakup regime table
The Ansys CFX guide, citing experiments, lists these regimes for We based on gas density, slip velocity and drop diameter.
| We range | Regime |
|---|---|
| We < 12 | Vibrational breakup |
| 12 < We < 50 | Bag breakup |
| 50 < We < 100 | Bag-and-stamen breakup |
| 100 < We < 350 | Sheet stripping |
| 350 < We | Catastrophic breakup |
Double a radius-based We before comparing. The same guide uses a separate correlation for large Oh; viscosity slows deformation, so expect a higher critical We for viscous liquids and compute Oh too. The table gives the breakup mode, not its speed: a drop that crosses a fast stream briefly may slow down before breakup completes, which is why CFD breakup models also compute a breakup time.
Common ways to misuse the Weber number
- Using the absolute speed. Breakup depends on speed relative to the gas; a drop carried with the stream has a small We even if fast over the ground.
- Treating surface tension as a constant. OpenStax lists water at 0.0756 (0 °C), 0.0728 (20 °C) and 0.0589 N/m (100 °C), and soapy water at about 0.0370 N/m, roughly half, so surfactant can nearly double We.
- Mixing nozzle diameter and drop diameter. Nozzle diameter describes primary breakup of the jet; drop diameter describes secondary breakup. Do not compare them with one threshold.
- Deciding from We alone. Oh, exposure time and gas turbulence change the outcome. Use We for screening and base design on data or validated models with the same definition.
Weber number questions engineers ask
For drop breakup, does ρ mean the water density or the air density?
For secondary breakup the usual definition, also used by the Ansys CFX table, takes the gas density. For water and air at 20 °C the two differ by about 829 times, so match your reference first.
Does a drop always break up once We exceeds 12?
No. Twelve is the typical bag-breakup onset for the diameter definition. It rises for viscous liquids, and a drop exposed for less than the breakup time slows before breaking. On a radius basis the same onset is about 6.
Can the Weber number explain why raindrops stop growing beyond a certain size?
In the MIT scaling, a falling drop’s air-density We equals its Bond number, reaching 1 at the capillary length, about 2.73 mm for water at 20 °C; larger drops are torn apart. Read it as an order of magnitude, not an exact maximum.
How do I convert surface tension given in dyn/cm?
1 dyn/cm = 1 mN/m; divide by 1000 for N/m. 70 dyn/cm = 0.070 N/m; 72.70 mN/m = 0.0727 N/m.
Related dimensionless numbers
Interface problems usually also need viscosity, gravity and oscillation periods checked.
- Reynolds number (Re): With the same v and L, We/Re = μv/σ is the capillary number: does viscosity rival surface tension?
- Froude number (Fr): We/Fr² = ρgL²/σ is the Bond number, used in Example 2 to weigh gravity against surface tension.
- Strouhal number (St): (capillary time √(ρL³/σ) ÷ transit time L/v)² = We; pair with St for drop oscillation or spray pulsation frequencies.
References
- MIT (J. W. M. Bush): Surface Tension, Lecture 3 slides. We = ρU²a/σ, Bond number, capillary length about 2 mm, raindrop scaling
- Ansys: CFX Theory Guide, Secondary Breakup Models. We from fluid density, slip velocity, diameter; regimes 12/50/100/350
- MIT OCW (2.154): Similitude lecture notes. We indicates the importance of surface tension
- UCF Pressbooks (OpenStax College Physics): Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action. Water σ at 0/20/100 °C, soapy water
- IAEA: Thermophysical Properties of Materials for Nuclear Engineering (2008). Tables 3.1 (air) and 3.3 (water)
- NASA: Dimensional Analysis Handbook / Dimensionless Numbers. We as inertia over surface tension