The Air-to-Solids Ratio in DAF Design, From First Principles

What exactly is the air-to-solids ratio?

Dissolved air flotation works by precipitating micro-bubbles out of a pressurised, air-saturated recycle stream as it is throttled to atmospheric pressure. Those bubbles (typically 30–100 µm) attach to and become enmeshed in the conditioned flocs, lowering their bulk density below that of water so they rise. The single dimensionless parameter that decides whether there is enough bubble mass to do this is the air-to-solids ratio:

Because bubble-floc attachment is a surface phenomenon, A/S is really a proxy for the total bubble surface area (and buoyant volume) presented per unit mass of solids. Below the required A/S, a fraction of flocs never acquire enough attached air to reach positive buoyancy and are lost to the underflow. Above it, the float blanket is robust but you are paying for air — and recycle pumping — you do not need. The whole of DAF air design is finding that target and delivering it economically. This article goes deep on the equation the brief A/S mention in our DAF sizing guide only touched on.

What is the full A/S design equation?

The released air mass is set by how much air the pressurised recycle holds in excess of what it can retain at atmospheric pressure. Expanding ṁ_air,released in terms of air solubility and pressure gives the classic flotation design equation (Metcalf & Eddy / Eckenfelder form):

The term (f·P − 1) is the physics: Henry's law makes dissolved air proportional to absolute pressure, so f·P is the air held under pressure and the −1 subtracts what stays dissolved at 1 atm after release. Only the difference comes out as bubbles. Note the units cancel to a pure mass ratio: S_a (mL/L) × 1.3 (mg/mL) gives mg/L, times R (m³/h = kL/h) gives g/h of air, over S_a,inf (g/m³) × Q (m³/h) = g/h of solids.

When R is expressed as a recycle ratio r = R/Q, the flow terms simplify and A/S becomes independent of absolute scale — it depends only on r, pressure, saturation fraction, solubility and the influent concentration.

Worked example 1: compute A/S from operating conditions

Take an industrial DAF treating an influent of S_a,inf = 600 mg/L TSS at Q = 40 m³/h. The saturator runs at 5 bar gauge (so P = 6 atm absolute, since 1 bar ≈ 0.987 atm, taken as ~6 atm), packed to f = 0.85, at 20°C where S_a = 18.7 mL/L. Recycle is R = 12 m³/h (r = 30%).

• Air released per litre of recycle: 1.3 · S_a · (f·P − 1) = 1.3 × 18.7 × (0.85×6 − 1) = 1.3 × 18.7 × 4.10 = 99.7 mg/L.
• Air released mass rate: 99.7 mg/L × 12,000 L/h = 1.196 × 10⁶ mg/h ≈ 1.20 kg air/h.
• Solids mass rate: S_a,inf · Q = 600 g/m³ × 40 m³/h = 24,000 g/h = 24 kg solids/h.
• A/S: 1.20 / 24 = 0.050 kg air/kg solids.

An A/S of 0.05 sits at the strong end of the normal band — appropriate for a light, well-conditioned chemical floc but generous for a dense oily sludge. If float testing showed the target was only 0.03, you could cut the recycle, drop the saturator pressure, or accept the margin as robustness against peak solids.

How do you rearrange it to find the recycle ratio?

In design you usually know the target A/S (from bench float tests) and must find the recycle that delivers it. Solve the governing equation for R:

Worked example 2. Same effluent and saturator as above (S_a,inf = 600 mg/L, S_a = 18.7 mL/L, f·P − 1 = 4.10), but float testing fixes the target at A/S = 0.035. Then:

• Denominator: 1.3 × 18.7 × 4.10 = 99.7 mg air per L recycle.
• Numerator: 0.035 × 600 = 21.0 mg air needed per L feed.
• r = 21.0 / 99.7 = 0.211, i.e. a ~21% recycle. At Q = 40 m³/h that is R ≈ 8.4 m³/h.

If the delivered air per litre is fixed by the saturator, r scales linearly with both the A/S target and the influent concentration. That linearity is what makes the next effect so important. To push more air without more recycle, raise (f·P − 1) — see our note on DAF saturator design for the pressure and packing trade-offs.

How does influent solids concentration change A/S?

Look again at the denominator S_a,inf·Q. For a fixed air supply (fixed R, P, f, T), the achieved A/S is inversely proportional to the influent solids concentration. This is the effect operators feel most sharply, because feed strength swings far more than temperature or pressure.

Worked example 3. The DAF from example 1 delivers 1.20 kg air/h and ran at A/S = 0.050 on 600 mg/L. A process upset lifts the feed to 1,200 mg/L at the same 40 m³/h:

• New solids load: 1,200 g/m³ × 40 m³/h = 48 kg/h (double).
• Air unchanged at 1.20 kg/h, so A/S = 1.20 / 48 = 0.025 — halved.
• To restore A/S = 0.050 you need 2.40 kg air/h. With everything else fixed, released air is linear in R, so recycle must double from 12 to 24 m³/h (r from 30% to 60%).
• Alternatively hold R and raise pressure: released air ∝ (f·P − 1). Going from 4.10 to 8.20 means f·P = 9.2, i.e. P ≈ 10.8 atm (~9.8 bar gauge) at f = 0.85 — often impractical, which is why recycle is the usual lever.

The lesson: size air on the peak solids load, not the average, and give the recycle pump and saturator headroom. A unit tuned only to mean strength will collapse its float blanket the first time the feed doubles.

What A/S ratio should you target?

The right target comes from bench flotation tests on the real effluent, but the floc character sets the ballpark. Denser, hydrophobic solids (oil, grease) attach air readily and need little; light, hydrophilic or highly hydrated chemical flocs need much more bubble surface per unit mass.

Use the table to bracket the design, then confirm with jar-and-float tests before sizing the recycle. Selecting the dissolved air flotation package against a tested A/S — rather than a generic catalogue figure — is what separates a unit that meets consent from one that carries over solids on the first strong feed.

Where does temperature enter the A/S equation?

Air solubility S_a falls with temperature, so the same saturator pressure delivers less released air in summer. This enters the equation directly through S_a:

Between 10°C and 30°C the deliverable air per litre of recycle drops by about a third at constant pressure. Because A/S is linear in S_a, a saturator sized only for cool winter operation can undershoot its A/S target by 15–20% on a hot day. Always evaluate the governing equation at the warmest expected operating temperature, and add recycle or pressure margin accordingly.

Setting the air-to-solids ratio in five steps

1. Fix the A/S target. Run bench float tests on the real effluent, or bracket from the application table (0.005–0.06), to fix the design A/S.
2. Establish the solids load. Compute m_solids = S_a,inf x Q at the PEAK influent concentration and flow, not the average.
3. Set saturator physics. Choose pressure P (typically 4–6 bar gauge), saturation fraction f (0.5–0.9), and evaluate S_a at the warmest operating temperature.
4. Solve for recycle. r = (A/S x S_a,inf) / [1.3 x S_a x (f.P - 1)]. This gives the recycle ratio that delivers the target air mass.
5. Add margin and verify. Size recycle pump and saturator with headroom for peak solids and summer temperature, then confirm A/S on the commissioned unit.