Indoor pool dehumidification load: take out the moisture with a coil or with outdoor air
The water surface of an indoor pool keeps adding moisture, and the hall air tends toward saturation. We convert that moisture into a supply humidity ratio on the chart and compare the heat needed when a coil removes it in summer and when dry outdoor air does in winter.
Situation
Say an indoor pool hall with 312.5 m² of water surface (25 m × 12.5 m) is to be held at 30 °C and 55 % RH. The water at 28 °C has a saturated surface, so its vapour pressure is higher than the hall air's, and the difference keeps evaporating water. If humidity is not controlled, glazing and roof structure get condensation and the finishes deteriorate. Three things to find: what supply humidity ratio carries the evaporated moisture away, what does the work in summer and winter (the coil or outdoor air), and how much the load grows if hall humidity is raised by 5 points. The conditions are illustrative; in design the evaporation rate comes from a pool-specific equation or manufacturer data.
Given conditions
| Item | Value |
|---|---|
| Hall air R | 30 °C, 55 % RH (2 K above the water temperature of 28 °C, example value) |
| Water surface | 312.5 m², water at 28 °C. The design evaporation rate of 0.15 kg/(m²·h) is an example input (total 46.9 kg/h) |
| Supply | 20,000 CMH (at hall state), dry-bulb 32 °C |
| Coil leaving | 95 % RH when dehumidifying |
| Three outdoor cases | Summer 32 °C / 60 % with 20 % outdoor air; winter -5 °C / 50 % with 20 % and with 50 % outdoor air |
| Pressure | 101.325 kPa (sea level) |
Evaporation is taken as proportional to the difference between the vapour pressure at the water surface (saturated at 28 °C, 3,782.207 kPa) and the partial pressure of vapour in the hall air (2.335 kPa at 55 %). This article takes the design evaporation as the value at 55 % hall RH and adjusts for other humidities by that pressure difference ratio.
Reading it on the chart
- Plot the hall air R (30 °C, 55 %) and the water-surface saturation point W (28 °C, 100 % RH). The line from R to W is the direction moisture moves from the water into the air.
- Supply S lies near R's dry-bulb but 2.06 g/kg drier in humidity ratio. The supply must be that much drier so that, after taking up the evaporated moisture, it becomes R.
- If the mixed point M of outdoor air and R has a higher humidity ratio than S, the coil cools M to the 95 % RH point C, lowering the humidity ratio, and reheats from C to S.
- If M is drier than S, no dehumidification is needed; the air only needs heating to S's temperature.
Step-by-step calculation
1) Hall air and evaporation. 30 °C / 55 %: WR = 14.67 g/kg, specific volume 0.8790 m³/kg, dew point 20.0 °C. Evaporated moisture mw = 0.15 × 312.5 = 46.9 kg/h.
2) Supply humidity ratio. Dry air mass flow ṁ = 20,000 / 3600 / 0.8790 = 6.320 kg/s. The supply must become R after taking up the moisture, so WS = WR − mw / ṁ = 14.67 − 46.9 / 3600 / 6.320 × 1000 = 12.61 g/kg.
3) Summer (32 °C / 60 %, 20 % outdoor air). M is 30.40 °C, 15.34 g/kg, which is 2.73 g/kg above WS. The coil leaving state C is 95 % RH at W = 12.61 g/kg, which is 18.42 °C. The coil removes 62.2 kg/h and its load is Q = 6.320 × (69.83 − 50.51) = 122.1 kW. Reheat from C to 32 °C is 88.3 kW.
4) Winter with 20 % outdoor air (-5 °C / 50 %). The outdoor air is very dry at 1.24 g/kg, which brings the mixed humidity ratio down to 11.99 g/kg. It is below WS, so no dehumidification is needed and only 67.8 kW of heating to 32 °C is required.
5) Winter with 50 % outdoor air. M is 12.71 °C, 7.95 g/kg. It is drier than WS, so no coil is needed and only 199.8 kW of heating is required.
| Case | Mixed M (°C / g/kg) | Treatment | Moisture removed by coil (kg/h) | Coil (kW) | Reheat or heating (kW) |
|---|---|---|---|---|---|
| Summer (32 °C · 60 %) | 30.40 °C · 15.34 | dehumidify | 62.2 | 122.1 | 88.3 |
| Winter 20 % OA (-5 °C · 50 %) | 23.14 °C · 11.99 | heating only | 0.0 | 0.0 | 67.8 |
| Winter 50 % OA (-5 °C · 50 %) | 12.71 °C · 7.95 | heating only | 0.0 | 0.0 | 199.8 |
Interpreting the result
In summer the outdoor air is humid, so on top of the 46.9 kg/h from the pool the outdoor air brings moisture, and the coil removes 62.2 kg/h. In winter the outdoor air is dry: with only 20 % outdoor air the mixed humidity ratio (11.99 g/kg) is already below the supply humidity ratio needed (12.61 g/kg), so no coil is needed and only 67.8 kW of heating to 32 °C remains. The dehumidification is already done, so raising outdoor air to 50 % just raises heating to 199.8 kW. Dry outdoor air dehumidifies for free, but the price is the heat to warm −5 °C air. That is why pool air handling often uses an exhaust heat recovery exchanger or a dehumidifying heat pump (which uses the heat taken at the coil for reheat and for pool water).
Raising the hall humidity changes things as follows (20 % outdoor air, summer outdoor 32 °C / 60 %).
| Hall RH | Hall W (g/kg) | Hall dew point | Evaporation ratio (55 % = 1) | Evaporation (kg/h) | Summer coil (kW) |
|---|---|---|---|---|---|
| 50 % | 13.31 | 18.4 °C | 1.15 | 53.8 | 145.6 |
| 55 % | 14.67 | 20.0 °C | 1.00 | 46.9 | 122.1 |
| 60 % | 16.04 | 21.4 °C | 0.85 | 40.0 | 100.1 |
| 65 % | 17.42 | 22.7 °C | 0.71 | 33.1 | 79.4 |
As hall RH rises from 50 % to 65 %, evaporation falls from 53.8 kg/h to 33.1 kg/h, because the vapour pressure difference shrinks. But the dew point rises from 18.4 °C to 22.7 °C, so condensation starts on glazing and exterior structure at a higher surface temperature. Raising humidity to cut evaporation trades against condensation risk. Check structure surface temperatures with the method in Winter exterior wall condensation check.
What the calculation leaves out
This is a heat and moisture balance with the evaporation rate as an input. The real evaporation depends strongly on the number of bathers, activity such as waves and fountains, and air movement over the water, so take it from a design equation and manufacturer data. Also left out are the hall's sensible loads (glazing and wall losses, lighting), the heat to warm the pool water, and the ventilation needed for water chemistry and air quality. Building codes set a minimum outdoor air rate, so the outdoor-air fraction cannot go below it.
Humid air rising from the water readily condenses on ceilings and glazing above. Blowing supply air across glazing and exterior walls to warm the surfaces prevents condensation at the same hall humidity. The supply point from the chart has to be read together with this airflow plan.
Common mistakes
- Calculating the dehumidification load from an RH difference. The load is humidity ratio difference times dry air mass flow. The summer coil load of 122.1 kW comes from humidity ratio.
- Taking more outdoor air than needed in winter. At 20 % outdoor air no dehumidification is needed already; going to 50 % raises heating from 67.8 kW to 199.8 kW.
- Assuming higher humidity is a free way to cut evaporation. Evaporation drops, but the dew point climbs to 22.7 °C and condensation risk grows.
- Using one design evaporation figure for every condition. This article adjusts only by the pressure difference ratio from the 55 % value. Look at peak-occupancy values separately.
- Matching the supply dry-bulb but ignoring glass surface temperature. A good average hall humidity still condenses on cold glass.
Try this calculation on the site
Open the summer dehumidifying process on the chart
The chart link plots the summer outdoor air, hall air, mixed point, coil leaving point and supply point, with cooling-dehumidifying and reheat process cards that give the coil kW and reheat kW. For the opposite direction see Winter office humidification load, for outdoor-air decisions see Economizer changeover, for coil leaving state and reheat see Summer cooling coil selection and reheat, and for plotting see How to read a psychrometric chart.
References
- ASHRAE Handbook Fundamentals 2017 (SI), chapter 1 Psychrometrics: humidity ratio, enthalpy, specific volume and saturation vapour pressure (Hyland-Wexler). The psychrometric numbers here were computed with this site's engine, which implements those equations at 101.325 kPa.
- ASHRAE Handbook HVAC Applications, the natatoriums section: the design principle that evaporation from a pool surface is proportional to the difference between the vapour pressure at the water surface and the partial pressure of vapour in the air and varies with bather activity, and how indoor pool air handling treats dehumidification and heat recovery. This article does not use that equation and takes the design evaporation as an input.
- Water surface 312.5 m², water 28 °C, hall 30 °C / 55 %, design evaporation 0.15 kg/(m²·h), airflow and outdoor-air fractions are example values, as is the margin of the hall temperature above the water.