DAF Bubble Dynamics and Bubble-Particle Attachment

Why are DAF bubbles 10-100 micrometres?

In a dissolved air flotation (DAF) unit a side-stream of clarified effluent is pressurised to 4-6 bar in a saturator, where air dissolves to near the Henry’s-law limit. When this stream is injected into the flotation tank through a nozzle or needle valve, the pressure collapses to atmospheric in milliseconds. The water is now grossly supersaturated: the dissolved-gas concentration far exceeds what atmospheric pressure can hold, and the excess must leave solution as gas.

The degree of supersaturation is the ratio of saturator to release pressure. For a saturator at 5 bar absolute releasing to ~1 bar, the supersaturation ratio S ≈ 5, so roughly four-fifths of the dissolved air (that above the 1 bar equilibrium) is precipitated. Because nucleation is heterogeneous — occurring on the nozzle surface and on suspended particles rather than in the bulk — a very large number of small bubbles form almost simultaneously. High local nucleation density plus rapid pressure release starves each nucleus of gas, so bubbles stay small: typically 40-70 µm, with a distribution spanning 10-100 µm. This is the defining feature of DAF versus coarse-bubble flotation, where millimetre bubbles rise too fast to capture fragile flocs.

The internal (Laplace) overpressure of a bubble, ΔP = 2σ/r, is modest at these sizes — for a 60 µm bubble (r = 30 µm) with surface tension σ ≈ 0.072 N/m, ΔP ≈ 4.8 kPa (0.05 bar) — so the bubbles are mechanically stable once formed and do not readily coalesce in the quiescent contact zone.

How fast does a micro-bubble rise? (Stokes law)

A single micro-bubble rising in still water reaches terminal velocity almost instantly. Because the Reynolds number is well below 1 (creeping flow), the rise velocity follows Stokes’ law, driven by the density difference between water and air:

Worked example 1 — a 60 µm bubble. With d_b = 60×10⁻⁶ m and Δρ = 998 − 1.2 ≈ 997 kg/m³:

• Numerator: 9.81 × (60×10⁻⁶)² × 997 = 9.81 × 3.6×10⁻⁹ × 997 = 3.52×10⁻⁵.
• Denominator: 18 × 1.002×10⁻³ = 1.804×10⁻².
• v_b = 1.95×10⁻³ m/s ≈ 1.95 mm/s ≈ 7.0 m/h.

Reynolds check: Re = ρ_w v_b d_b / µ = (998 × 1.95×10⁻³ × 60×10⁻⁶) / 1.002×10⁻³ = 0.12. Since Re « 1, the Stokes (laminar, no-slip-equivalent) assumption holds and no drag correction is needed. Stokes remains valid up to roughly 120 µm (Re ≈ 1); above that a rigid-sphere or Hadamard-Rybczynski correction applies.

The strong d² dependence is the central design tension in DAF: halving bubble diameter quarters the rise velocity, as the table below shows.

How much bubble surface is available per litre of air?

Attachment is a surface phenomenon, so the useful quantity is not the mass of air but the interfacial area it provides. For a monodisperse cloud, the specific surface area per unit gas volume is a = 6/d_b, and the bubble number concentration follows from the gas volume fraction:

Worked example 2 — one litre of released air at 60 µm. Take V_air = 1 L = 1×10⁻³ m³.

• Volume of one bubble: V₁ = (π/6) d_b³ = 0.5236 × (60×10⁻⁶)³ = 0.5236 × 2.16×10⁻¹³ = 1.131×10⁻¹³ m³.
• Number of bubbles: N = 1×10⁻³ / 1.131×10⁻¹³ = 8.8×10⁹ bubbles per litre of air.
• Area of one bubble: A₁ = π d_b² = π × 3.6×10⁻⁹ = 1.131×10⁻⁸ m².
• Total interfacial area: N × A₁ = 8.8×10⁹ × 1.131×10⁻⁸ ≈ 100 m² — equivalently a_b = 6/d_b = 6 / 60×10⁻⁶ = 1×10⁵ m²/m³, and 10⁵ × 10⁻³ = 100 m².

Per unit air mass: at 20°C and ~1 bar, air density is ~1.20 kg/m³, so 1 L weighs 1.2 g and this litre delivers roughly 83 m² of interface per gram of air (~83,000 m²/kg). That is why DAF needs only a few grams of air per m³ of feed: the interfacial area, not the air mass, does the work. It also explains the penalty of coarse bubbles — at 600 µm the same litre would offer just 10 m². Fixing the bubble-size distribution at the nozzle is therefore a hydrodynamic design problem best interrogated with CFD analysis of the nozzle and contact zone.

How do bubbles attach to particles?

Bubble-particle capture in the contact zone is a heterocoagulation process, mathematically analogous to deep-bed filtration: the rising bubble is the “collector” and the flocs are the particles it collects. Edzwald’s white-water collector model treats the bubble blanket as a suspension of collectors and gives a first-order removal of particles as they pass through it:

Two probabilities are multiplied. First the collision efficiency η_T — the fraction of particles in the bubble’s projected path that actually contact it — is the sum of three transport mechanisms, exactly as in filtration theory:

• Interception (η_I): a particle following a streamline passes within one particle radius of the bubble surface; scales with (d_p/d_b)².
• Sedimentation / gravity (η_G): a denser particle settles across streamlines onto the rising bubble; scales with the particle settling velocity relative to v_b.
• Brownian diffusion (η_D): sub-micron particles diffuse to the surface; scales with the inverse Peclet number and matters only below ~1 µm.

Second, the attachment efficiency α_pb is the probability that a collision leads to a stable bubble-particle bond. It is governed by DLVO-type surface chemistry: a bubble carries a slightly negative charge, so raw negatively charged flocs are repelled (α → 0). This is precisely why coagulation to near charge-neutrality is a prerequisite for DAF — it lifts α toward 1. See our companion guide to coagulation and flocculation and the overview of what a DAF system is.

Does smaller bubble size help or hurt?

Bubble size acts in opposite directions, which is why 40-70 µm is the sweet spot. From the equations above, decreasing d_b:

• Helps collision: η_T rises (interception scales with 1/d_b²) and the collector term Φ_b/d_b in the rate law grows, so more particles are captured per unit air.
• Hurts transport: the Stokes rise velocity v_b falls as d_b², so both the aggregate’s rise and the contact-zone throughput drop. Below ~20 µm bubbles barely rise and can be swept out with the effluent.

The net collection rate scales roughly as v_b/d_b ∝ d_b, so per-bubble capture actually improves with size — but the total interfacial area per unit air (6/d_b) and the number of collectors fall. Optimising a real unit balances these against buoyancy of the finished aggregate, the subject of the next section.

What is the rise velocity of a bubble-floc aggregate?

Capture is only useful if the resulting aggregate is buoyant enough to reach the surface within the flotation zone. The aggregate rises by Stokes’ law again, but now with an effective density found from a volume-weighted mass balance of floc plus attached bubbles:

ρ_agg = (m_p + n·m_b) / (V_p + n·V_b), with the equivalent diameter d_agg = [6(V_p + n·V_b)/π]^1/3.

Worked example 3 — one 60 µm bubble on a 50 µm floc. Take a coagulated floc d_p = 50 µm at ρ_p = 1050 kg/m³ (a light alum/hydroxide floc, only slightly denser than water):

• Floc volume: V_p = 0.5236 × (50×10⁻⁶)³ = 6.55×10⁻¹⁴ m³; mass m_p = 1050 × 6.55×10⁻¹⁴ = 6.87×10⁻¹¹ kg.
• Bubble volume: V_b = 1.131×10⁻¹³ m³; mass m_b = 1.2 × V_b = 1.36×10⁻¹³ kg (negligible).
• Aggregate volume: V_agg = 6.55×10⁻¹⁴ + 1.131×10⁻¹³ = 1.786×10⁻¹³ m³.
• Effective density: ρ_agg = 6.89×10⁻¹¹ / 1.786×10⁻¹³ = 386 kg/m³ — far below water, so the aggregate is strongly buoyant (Δρ = 998 − 386 = 612 kg/m³).
• Equivalent diameter: d_agg = [6 × 1.786×10⁻¹³ / π]^1/3 = (3.41×10⁻¹³)^1/3 = 70 µm.
• Rise velocity: v_agg = 9.81 × (70×10⁻⁶)² × 612 / (1.804×10⁻²) = 1.63×10⁻³ m/s = 1.6 mm/s ≈ 5.9 m/h.

Comparison to design rise rate. A DAF is designed for a surface (hydraulic) loading of 5-15 m³/m²·h. At 5.9 m/h a single-bubble aggregate only just clears the low end — it would wash out at a typical 8 m/h. Attaching a second and third bubble drops ρ_agg to ~171 kg/m³ and raises d_agg to ~92 µm, lifting v_agg to about 13.6 m/h, comfortably above design. This is the quantitative reason DAF aims for several bubbles per floc — and why undersized air or poor coagulation (few attachments) produces a slow, unstable float. The sizing consequences are developed in our guide to how to size a DAF system.