Airflow coil condensate estimate for water collection

Use measured airflow and inlet humidity to calculate the water available, then use coil performance to predict the outlet condition. Pipe diameter and length alone cannot establish either outlet temperature or collection rate. The missing link is heat transfer: how much heat the coil removes while its surface remains below the incoming air’s dew point.

For atmospheric water collection, distinguish three quantities: moisture removed from the air, liquid reaching the reservoir, and cooling capacity required. They are related, but they are not interchangeable. The method below separates them and shows where measurements must replace assumptions.

What the model needs

Define the air path first. Air flowing inside an externally cooled tube behaves differently from air crossing a finned coil. Both arrangements require the same inlet properties, but their heat-transfer area and pressure losses differ.

Record these inputs before estimating collection:

A fan’s free-air rating is not the flow through the completed apparatus. Tubes, bends, filters and wet surfaces increase resistance. The operating flow comes from the intersection of the fan curve and system resistance curve.

The same distinction applies to fan-cooled enclosure airflow: a fan label does not establish delivered airflow after installation. Measure flow through the complete collection air path, not through an unrestricted fan opening.

Calculate the water available

Relative humidity is not a direct measure of water mass. Convert inlet and outlet conditions to humidity ratio, expressed as kilograms of water vapor per kilogram of dry air.

At ordinary atmospheric conditions:

w = 0.62198 × p_v / (P - p_v)

Here, P is absolute air pressure, and p_v is water-vapor partial pressure, in matching units. Obtain vapor pressure from a psychrometric calculator or chart using temperature and relative humidity.

For a liquid-condensing coil:

Condensate rate = m_da × (w_in - w_out)

Use dry-air mass flow, m_da, in kilograms per hour to obtain condensate in kilograms per hour. Divide inlet volume flow by inlet specific volume, expressed in cubic meters per kilogram of dry air.

Consider this example at approximately sea-level pressure:

These inlet conditions give a humidity ratio of approximately 0.0160 kg/kg dry air. The dew point is approximately 70.5°F [21.4°C].

Saturated air at the assumed outlet temperature has a humidity ratio of approximately 0.0106 kg/kg dry air. Inlet specific volume is approximately 0.881 m³/kg dry air, giving about 193 kg/h of dry air.

The resulting estimate is:

193 × (0.0160 - 0.0106) ≈ 1.0 kg/h of condensate

That is approximately 1.0 L/h before collection losses. It is a conditional result, not a prediction that any particular tube will achieve that outlet state.

A saturated outlet is also not universal. Air bypassing cold surfaces can mix with colder, drier air and leave below 100% relative humidity. Use the actual outlet humidity ratio whenever it is available.

Predict outlet temperature with coil performance

The water calculation becomes useful only when the outlet condition is credible. A controlled coolant temperature does not automatically make the leaving air reach that temperature.

For a dry coil with an approximately constant cooling-side temperature, a useful preliminary model is:

T_out = T_c + (T_in - T_c) × exp[-UA / (m_da × c_p)]

In this expression:

• T_c is the approximately constant cooling-side temperature.
• U is the overall heat-transfer coefficient in W/(m²·K).
• A is the corresponding heat-transfer area in m².
• m_da is dry-air mass flow in kg/s.
• c_p is moist-air sensible heat capacity on a dry-air basis, in J/(kg dry air·K).

Use consistent area definitions for U and A. For a simple tube, internal surface area is π × internal diameter × length. Wall resistance, external cooling and any fins must be represented consistently in the overall coefficient.

The equation shows the main trend. Increasing area or heat-transfer coefficient brings the outlet closer to the cooling-side temperature. Increasing airflow generally moves the outlet farther away, although airflow also changes the heat-transfer coefficient.

This sensible-only equation is not a complete model once condensation starts. A wet coil removes latent heat as well as sensible heat. Continuing the dry calculation through condensation can substantially understate the required cooling capacity.

A common preliminary wet-coil model uses an effective saturated surface state and a bypass factor, BF:

h_out = h_s + BF × (h_in - h_s)

w_out = w_s + BF × (w_in - w_s)

The subscript s identifies the effective saturated surface condition, often described using an apparatus dew point. Enthalpy, h, represents the air’s combined sensible and latent energy.

Obtain the bypass factor and effective surface condition from coil performance data or calibrated measurements. They are not arbitrary settings that can be inferred from diameter alone.

With enthalpy in kJ/kg dry air and temperature in °C, calculate outlet temperature from:

T_out = (h_out - 2501 × w_out) / (1.006 + 1.86 × w_out)

This model remains an approximation. Uneven wall temperatures, partial wetting, air leakage and changing coolant conditions can require a segmented heat-and-mass-transfer calculation.

Check the cooling capacity

Cooling capacity cannot be omitted from an outlet-temperature prediction. This is a physical limit, even when electrical consumption is outside the calculation’s scope.

For ordinary moist air:

h = 1.006 × T + w × (2501 + 1.86 × T)

Here, T is in °C, and h is in kJ/kg dry air. A practical first estimate of air-side cooling duty is:

Q ≈ m_da × (h_in - h_out)

Use mass flow in kg/s to obtain cooling duty in kW. This approximation neglects the relatively small sensible enthalpy carried away by drained liquid water.

For the 100 cfm example, inlet enthalpy is approximately 71 kJ/kg dry air. The assumed saturated outlet at 59°F [15°C] is approximately 42 kJ/kg dry air.

The resulting air-side cooling requirement is approximately 1.6 kW. That is heat removal, not compressor electrical input. Fan heat entering the process air and heat leaking through the apparatus add to the refrigeration duty.

If the cooling system cannot remove that load at its actual operating conditions, the assumed outlet state will not hold. The surface warms, collection decreases, or both occur.

What diameter and expansion actually change

Diameter affects velocity, pressure loss, surface area and heat transfer simultaneously. A smaller tube does not simply mean better cooling because air passes closer to the wall.

At an actual flow of 100 cfm:

These values assume the stated airflow actually passes through each tube. The smaller tube imposes much greater resistance, potentially reducing delivered flow substantially.

Higher velocity can improve the air-side heat-transfer coefficient. It also increases pressure loss and can carry droplets downstream. Increasing length adds surface area, but also resistance. Parallel paths may provide useful area without forcing all the air through one narrow passage.

A larger downstream chamber reduces mean velocity and may help droplets separate. It does not, by itself, provide a useful refrigeration stage.

An ordinary fan-driven area expansion does not create a controlled cold outlet below the air’s dew point. Useful expansion cooling requires a suitable pressure source and thermodynamic process, not merely a change in duct diameter.

Treat a collection chamber as a drainage and separation component unless a validated cooling mechanism is included. Any cooled chamber surface must also be included in the heat-transfer and cooling-load calculations.

What to measure before trusting the estimate

A performance test must distinguish vapor removal from liquid collection. A cold surface can remove moisture while poor drainage prevents that water from reaching the reservoir.

Measure inlet and outlet temperature, humidity and pressure during stable operation. Place sensors where they represent mixed airflow, away from direct contact with cold metal or entrained droplets.

Also record actual airflow, coolant conditions and collected water over a measured interval. Compare the reservoir gain with m_da × (w_in - w_out).

A persistent difference can indicate:

• Water retained on coil surfaces or in low points.
• Droplets escaping with the outlet air.
• Re-evaporation before water reaches the reservoir.
• Leakage around the intended air path.
• Humidity or airflow measurement error.

The mechanisms behind coil condensate carryover also matter in collection equipment: water leaving as droplets is not necessarily water captured in the tank. Provide drainage and separation without assuming every condensed droplet falls directly into the reservoir.

Repeat measurements across the expected inlet temperature and humidity range. A result obtained in warm, humid air does not establish production during cooler or drier operation.

This test establishes thermal and collection performance under the measured conditions. It does not establish drinking-water safety, hygienic construction or suitability of wetted materials. Those require separate water-quality assessment and treatment design. If surfaces freeze, collected liquid rate also depends on frost accumulation and defrost operation.

What to specify for the controls enclosure

Keep the collection air path, reservoir and electrical controls physically separated. A controls enclosure is not automatically a sanitary process vessel or water-storage tank.

For custom enclosures, specify material, width, height, depth and every penetration. We build custom electrical enclosures from one piece, with those details defined by your drawing.

Use 304 stainless steel for suitable general environments, and consider 316L where salt air or chlorides increase corrosion exposure. If surrounding cleaning involves hose-directed water, specify the appropriate manufacturer-rated NEMA Type 4X design. Our ratings are manufacturer ratings for the specified design; the enclosures are not UL listed.

Locate cable entries away from collection drains and specify sealed fittings for the installation. Use the cutouts, glands and HMI windows guide for penetration fundamentals rather than treating a drilled opening as a complete sealing detail.

Provide the controls heat load, ambient temperature range and expected moisture exposure with the enclosure drawing. Intentional process condensation belongs outside the controls compartment. If moisture appears inside it, use the checks for water inside a sealed electrical enclosure instead of assuming the collection process makes that condition acceptable.

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