Dedicated outdoor air system (DOAS): dry the outdoor air below the room dew point so it carries the latent load
A dedicated outdoor air system (DOAS) treats only the ventilation air. If it dries that air below the room dew point, the outdoor air also soaks up the moisture released in the room, and the fan coils or radiant panels handle only sensible heat and can run dry. We work out how dry the air has to be, and why it gets hard when the outdoor airflow is small.
Situation
Say an office floor is designed with a DOAS plus fan coil units (FCUs) in the rooms. The ventilation requirement gives 3,000 CMH of outdoor air, and the room loads are 30 kW sensible and 6 kW latent, mostly from occupants. To avoid running condensate drains to every fan coil, the designer wants the fan coils not to dehumidify, so the dry outdoor air from the DOAS has to absorb all of the room latent load. Three questions: what humidity ratio (dew point) must leave the DOAS, what are the DOAS coil load and the sensible load left for the fan coils, and how does this compare with drying the air only to the room humidity ratio and reheating it. The conditions are illustrative; the real outdoor airflow and design weather come from the codes and data for the site.
Given conditions
| Item | Value |
|---|---|
| Outdoor O | Dry-bulb 32 °C, RH 65 % (humidity ratio 19.58 g/kg, dew point 24.6 °C, enthalpy 82.3 kJ/kg) |
| Room R | Dry-bulb 24 °C, RH 50 % (humidity ratio 9.30 g/kg, dew point 12.9 °C) |
| Room load | Sensible 30 kW, latent 6 kW (the outdoor air load is handled separately at the DOAS coil) |
| Outdoor airflow | 3,000 CMH at intake conditions; also compared at 2,000 · 3,000 · 4,000 · 5,000 CMH |
| DOAS coil leaving | Assumed at 95 % RH and supplied as is, with no reheat |
| Pressure | 101.325 kPa (sea level) |
Reading it on the chart
- Plot outdoor O (32 °C, 65 %) and room R (24 °C, 50 %). A horizontal line left from R meets the saturation curve at the room dew point, 12.9 °C.
- Work out the supply humidity ratio WS that absorbs the latent load and draw a horizontal line at that height. It lies below R.
- Where that line meets the 95 % RH curve is the coil leaving state C. The line from O to C is the DOAS coil cooling and dehumidifying.
- The line from C to R is the supply air picking up sensible and latent heat in the room until it becomes R. Its upward slope to the right shows it takes both.
- The alternative path cools from O to N, the 95 % point at the same height as R, then reheats horizontally to R.
Step-by-step calculation
1) Outdoor air mass flow. At the intake specific volume of 0.8917 m³/kg, ṁ = 3,000 / 3600 / 0.8917 = 0.9346 kg/s.
2) Required supply humidity ratio. For the supply to carry the 6 kW room latent load, ṁ × (WR − WS) × (2501 + 1.86 × 24) = 6. So WS = 9.30 − 6 / (0.9346 × 2,545.6) × 1000 = 6.78 g/kg. Its dew point is 8.26 °C, 4.7 K below the room dew point of 12.9 °C.
3) Coil leaving state. The point at 6.78 g/kg and 95 % RH is at 9.02 °C dry-bulb with an enthalpy of 26.13 kJ/kg.
4) DOAS coil load. ṁ × (hO − hC) = 0.9346 × (82.34 − 26.13) = 52.5 kW. The condensate is ṁ × (WO − WS) = 43.1 kg per hour. The enthalpy carried away by the condensate (under 1 %) is ignored.
5) Sensible load taken off the room. The 9.0 °C supply warming to 24 °C absorbs ṁ × cp × (24 − 9.02) = 0.9346 × 1.0186 × 14.98 = 14.3 kW. The fan coils are left with 30 − 14.3 = 15.7 kW sensible and no latent load.
6) Changing the outdoor airflow. With less outdoor air carrying the same 6 kW of latent load, the humidity ratio difference must be larger, so the leaving dew point drops.
| Outdoor airflow | Supply W (g/kg) | Dew point (°C) | Coil leaving (°C) | DOAS coil (kW) | Room sensible taken (kW) | Fan coil sensible (kW) |
|---|---|---|---|---|---|---|
| 2,000 CMH | 5.52 | 5.3 | 6.0 | 38.9 | 11.4 | 18.6 |
| 3,000 CMH | 6.78 | 8.3 | 9.0 | 52.5 | 14.3 | 15.7 |
| 4,000 CMH | 7.41 | 9.6 | 10.3 | 66.4 | 17.4 | 12.6 |
| 5,000 CMH | 7.79 | 10.3 | 11.1 | 80.3 | 20.6 | 9.4 |
Interpreting the result
At 3,000 CMH the DOAS handles all of the room latent load if it delivers 9.0 °C air with a dew point of 8.3 °C. Cooling air to 9.0 °C needs chilled water (or refrigerant) colder than that, so coil selection has to confirm that the plant's chilled-water temperature can do it. At 2,000 CMH the coil would have to reach 6.0 °C, which is hard for an ordinary chilled-water coil; a direct-expansion coil or desiccant dehumidification is worth looking at. More outdoor air makes the dew point target easier but raises the DOAS coil load.
The fan coils see only room air (dew point 12.9 °C), so if their coldest coil surface (the chilled-water inlet side) stays at or above that dew point they run dry. That reduces the need for condensate piping and the dirty drain pan problem at the fan coils, and allows warmer chilled water.
The alternative (room-neutral supply) dries the outdoor air only to the room humidity ratio of 9.30 g/kg. The coil leaving point N is 13.7 °C, the DOAS coil takes 42.1 kW, and reheating to 24 °C takes 9.8 kW. The fan coils then carry both the 30 kW sensible and 6 kW latent load, 36.0 kW in all, and every fan coil condenses.
| Approach | DOAS coil (kW) | Reheat (kW) | Fan coils (kW) | Total cooling (kW) | Fan coil condition |
|---|---|---|---|---|---|
| Latent handled by DOAS (cold supply) | 52.5 | 0 | 15.7 | 68.3 | can run dry |
| Room-neutral supply (reheat) | 42.1 | 9.8 | 36.0 | 78.1 | condensing |
The 9.8 kW difference in total cooling equals the 9.8 kW of reheat (exactly, with these definitions): the air is cooled, warmed back up, and the room then has to cool it again (and the reheat needs its own heat source). Same outdoor air, same room load; the difference comes purely from where the supply temperature is set.
What the calculation leaves out
Supply air at 9.0 °C is colder than the room dew point of 12.9 °C. Duct and diffuser surfaces exposed to room air can sweat, so insulate the DOAS supply ducts and use high-induction diffusers or deliver the air into the fan coil inlet. Duct insulation thickness can be worked out the same way as in Duct insulation condensation.
This looks only at one design outdoor condition. On hot, dry days little dehumidification is needed; on cool, humid spring and autumn days the sensible load is small but the latent load is not, and the room can be overcooled. Reheat, an energy recovery wheel or airflow reset may then be needed, and year-round operation should be checked with hourly weather data. When occupancy falls, both the latent load and the required outdoor air change together.
Common mistakes
- Setting the DOAS leaving condition to the room humidity ratio. The room latent load then goes to the fan coils and they condense. To carry the latent load, the leaving dew point here has to be 4.7 K below the room dew point.
- Assuming the same coil still works with less outdoor air. At 2,000 CMH the coil must reach 6.0 °C.
- Drying the outdoor air only to the room humidity ratio and reheating it to room temperature. The 9.8 kW of reheat comes straight back in the total cooling. With the DOAS carrying the latent load, the 14.3 kW of sensible cooling that the cold supply provides should be subtracted from the fan coil capacity.
- Converting outdoor airflow to mass with the room specific volume. If the airflow is stated at intake conditions, use the outdoor specific volume of 0.8917 m³/kg.
- Forgetting duct and diffuser condensation. 9.0 °C supply is below the room dew point.
Try this calculation on the site
Open the DOAS process on the chart
The chart link plots the outdoor, DOAS leaving and room points with a process card from outdoor to leaving, so you can check the coil load and condensate. For coil bypass factor see the cooling coil calculator, for coil selection in an ordinary mixed-air unit see Cooling coil selection and reheat, for dry coils and high sensible heat ratio rooms see Data center high SHR load, 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.
- Expressing latent heat as ṁ × ΔW × (2501 + 1.86 t), and the moist air specific heat 1.006 + 1.86 W, follow from the enthalpy equation in the same chapter.
- Splitting a system into a dedicated outdoor air unit and separate room sensible equipment is a common arrangement covered in ASHRAE Handbook HVAC Systems and Equipment. This article does not quote chapter numbers or specific design values from it.
- Outdoor 32 °C / 65 %, room 24 °C / 50 %, the loads, 3,000 CMH of outdoor air and the 95 % leaving RH are example values. The real outdoor airflow comes from the ventilation code that applies to the building, and the design weather from local data.