HVAC airflow and load strategy: sensible versus latent dynamics

When we design HVAC systems, we quickly learn that supply airflow is not a free choice. It is fixed by the sensible cooling load in the space. The main control loop in a Fan Coil Unit or Air Handling Unit simply holds the dry-bulb temperature at the setpoint. Let us walk through exactly how this works and why the latent load sometimes creates problems.

Determining airflow from the sensible load

We begin with the sensible cooling load because that is what sets the airflow. The thermodynamic relationship for sensible heat removal is straightforward. The air we supply is cooler than the room air, so it absorbs heat as it warms up. The equation that describes this process is:

$Q_{s} = \dot{V} \cdot \rho \cdot C_{p} \cdot (T_{r} - T_{s})$

Here is what each term means in plain language:

$Q_{s}$ is the sensible cooling load in kilowatts.
$\dot{V}$ is the volumetric airflow rate in cubic meters per second.
$\rho$ is the density of air, roughly 1.20 kilograms per cubic meter at standard conditions.
$C_{p}$ is the specific heat capacity of moist air, about 1.02 kilojoules per kilogram per kelvin.
$T_{r}$ is the return or room air temperature in degrees Celsius.
$T_{s}$ is the supply air temperature in degrees Celsius.

At sea level we can combine the constants $\rho \cdot C_{p}$ into a single value of about 1.21. This gives us the practical equation we actually use:

$\dot{V} = \frac{Q_{s}}{1.21 \cdot (T_{r} - T_{s})}$

You simply choose a supply-to-room temperature difference, usually between 8 K and 10 K. That range keeps the air from dumping straight down out of the diffusers. Once you pick the temperature difference, the required airflow is locked in. You cannot change it without changing the sensible load or the supply temperature.

Fig. 1 — Required airflow against temperature difference

Sensible load Q_s30 kWTemperature difference Tr − Ts9 K

Required airflow

2.75

m³/s

Same airflow

2755

L/s

Diffuser behaviour

Comfortable spread

within the 8 K to 10 K range

Set the sensible load and the temperature difference. The required airflow falls as the difference widens. The shaded band is the usual 8 K to 10 K design range: below it the air volume grows quickly, above it the cold supply jet detaches and dumps out of the diffuser.

Satisfying the latent load

Once we have fixed the airflow based on the sensible load, we still need to remove moisture from the space. Latent load is not something you can measure with a dry-bulb thermometer. It is the energy stored inside water vapor that has evaporated into the room from people sweating and breathing, from cooking, or from outside air leaking in.

To remove this latent load, we must physically take the water vapor mass out of the space. We do this by supplying air that is drier than the room air. The supply air has a lower humidity ratio, $w_s$ , than the room air, $w_r$ . This drier air acts like a sponge. It absorbs the moisture generated in the space until its humidity ratio rises to match the room condition. Then the return air carries that moisture back to the cooling coil, where the water condenses and drains away.

The amount of latent cooling we achieve is given by:

$Q_{l} = \dot{V} \cdot \rho \cdot h_{fg} \cdot (w_{r} - w_{s})$

This equation comes from a basic mass balance that we convert into an energy balance. Let us walk through the derivation so you can see exactly how it is built.

Step 1: Calculate the mass flow rate of the carrier (dry air). We first convert the volumetric airflow into a mass flow rate of dry air by multiplying it by air density:

$\dot{m}_{da} = \dot{V} \cdot \rho \quad \text{[kg of dry air / s]}$

Step 2: Determine the moisture absorption capacity per unit mass. The humidity ratio tells us how much water rides along with each kilogram of dry air. The difference between the room humidity ratio and the supply humidity ratio shows how much moisture one kilogram of dry supply air can absorb:

$\Delta w = w_r - w_s \quad \text{[kg of water vapor / kg of dry air]}$

Step 3: Calculate the total mass of moisture removed. We multiply the dry air mass flow by this difference to find the total mass of water vapor swept out of the room per second:

$\dot{m}_{w} = \dot{m}_{da} \cdot \Delta w = \dot{V} \cdot \rho \cdot (w_{r} - w_{s}) \quad \text{[kg of water vapor / s]}$

Step 4: Convert moisture mass removal to energy removal (latent load). Loads in HVAC are expressed in power (kilowatts), not mass flow rates. To turn the mass of water removed into its energy equivalent, we multiply by the latent heat of vaporization. This constant is the energy that was originally needed to evaporate one kilogram of liquid water into vapor at room temperature:

$Q_{l} = \dot{m}_{w} \cdot h_{fg}$

When we substitute the expression from Step 3, we arrive at the fundamental equation we use in practice:

$Q_{l} = \dot{V} \cdot \rho \cdot (w_{r} - w_{s}) \cdot h_{fg}$

You can also think of humidity ratio in a very practical way. In HVAC we treat air as two separate things traveling together: dry air (the nitrogen and oxygen) and water vapor. The mass of dry air stays constant all the way around the loop from the AHU through the ducts, into the room, and back. It never condenses or gets created by people. So dry air works like a perfect conveyor belt that carries both heat and moisture.

The humidity ratio simply answers one question: for every one kilogram of dry air on that conveyor belt, how many kilograms of water molecules are riding along with it?

To see why the difference $\Delta w = w_r - w_s$ matters, picture one kilogram of dry air as a single bucket moving through the system.

When the bucket leaves the cooling coil and enters the room through the supply grille, it holds a certain amount of water: $w_s$ . For example, it might carry 0.008 kilograms of water per kilogram of dry air.

As the bucket travels through the room, people are sweating and breathing. They add new water molecules into the air. The bucket picks up these extra molecules.

When the bucket is pulled into the return grille, it must leave the room at the exact humidity ratio that matches the room design condition: $w_r$ . For example, 0.010 kilograms of water per kilogram of dry air.

The exact mass of water the bucket picked up inside the room is simply the difference: $w_r$ minus $w_s$ . In the example, that is 0.002 kilograms of water per kilogram of dry air. This difference is the moisture absorption capacity of the supply air. It tells us precisely how much of the room latent load each kilogram of supply air carries away and brings back to the coil.

Fig. 2 — The bucket on the conveyor belt

Supply humidity ratio w_s8.0 g/kgRoom humidity ratio w_r10.0 g/kgAirflow V̇2.0 m³/s

Absorption Δw

2.0

g water / kg dry air

Moisture removed

4.8

g/s

Latent cooling

11.8

Each bucket is one kilogram of dry air. It leaves the coil holding the supply humidity ratio (8 g/kg in this example), fills up as the occupants add moisture, and returns to the coil at the room humidity ratio (10 g/kg), where the difference condenses and drains away.

The water vapor already in the room is superheated. Cooling that vapor down to the dew point takes only a tiny amount of energy, about one percent of the total, and we already account for it when we use the specific heat of moist air in the sensible load calculation. That is why the latent equation uses only the latent heat of vaporization for the phase change itself.

This way of looking at the latent load helps you understand why we sometimes need special strategies when the moisture load is high compared with the sensible load.

The latent bottleneck and overcooling

We run into trouble when a space has a low Sensible Heat Ratio. This situation appears in densely occupied rooms or commercial kitchens where people and processes release a lot of moisture relative to the heat load.

If the sensible load is small, the airflow we calculated above is also small. To remove a large latent load with only a little air, the supply air must be extremely dry. That requirement forces the Apparatus Dew Point on the coil to drop to a very low value, sometimes below freezing. The coil ices up and airflow collapses.

The opposite choice also fails. If we size the airflow to meet the latent load instead, we deliver a large volume of very cold air. Moisture removal improves, but the space now receives far more sensible cooling than it needs. Room temperature drops below the setpoint. We call this condition overcooling, and occupants feel cold and clammy.

Fig. 3 — The two failure modes at a low sensible heat ratio

Sensible load Q_s15 kWLatent load Q_l25 kW

Sensible heat ratio

0.38

Qs / (Qs + Ql)

Airflow

1.38

m³/s

Apparatus dew point

-0.8

°C on the coil

Example space: 15 kW sensible, 25 kW latent (SHR 0.38). Sizing the airflow for the sensible load (left) gives only 1.38 m³/s, and carrying 25 kW of moisture in that little air pushes the apparatus dew point below freezing. Sizing for the latent load (right) needs 1.77 m³/s of deeply cooled air, which delivers about 19 kW of sensible cooling against a 15 kW load and pulls the room down to roughly 22 °C.

Solutions for high latent loads

When the standard sensible-driven airflow cannot provide enough dehumidification, we need different strategies.

Subcooling and reheat

We break the link between sensible airflow and latent capacity. We select the airflow volume and the coil temperature, typically 10 °C to 12 °C, specifically to condense the required amount of moisture. The air leaving the coil is now colder than the room requires for sensible cooling. We therefore install a reheat coil downstream. This reheat coil, either electric or served by low-pressure hot water, raises the supply air temperature back to a neutral value before the air enters the space. The space stays at the correct temperature while the moisture is removed.

Desiccant dehumidification

We can also remove moisture without relying on a cold coil at all. Air passes through a desiccant wheel, commonly filled with silica gel. The desiccant adsorbs water vapor chemically, independent of the air temperature. The process releases heat, however, so the air stream gains sensible heat. We then pass the air over a conventional cooling coil to remove that added heat before delivery. This approach works well when we need very low humidity levels that a standard cooling coil cannot reach.

Fig. 4 — Two ways to break the link

Coil temperature

°C, chosen for moisture

Air off the coil

°C, dry but cold

After reheat

°C, neutral supply

Follow each air stream from left to right. In the subcooling path the coil condenses the moisture and a reheat coil restores a neutral supply temperature. In the desiccant path the wheel adsorbs the moisture, the adsorption heat warms the air, and a cooling coil removes that heat before delivery.

System architecture: DOAS versus mixed air

A practical example shows why these choices matter. Consider a restaurant with a sensible load of 38 kW, a latent load of 27 kW, and a fresh air requirement of 1128 liters per second.

A Dedicated Outdoor Air System that supplies only the 1128 liters per second of ventilation air does not have enough total mass flow. When we deliver this volume at 10 °C, we obtain only about 19 kW of sensible cooling. Meeting the 27 kW latent load with such a small airflow would require a sub-zero dew point, which freezes the coil. The DOAS-only approach therefore fails.

Strategy 1: Central mixed-air Air Handling Unit

We mix the 1128 liters per second of fresh air with recirculated room air to reach a higher total flow, for example 3500 liters per second. We cool the entire mixed stream deeply to an Apparatus Dew Point of 8 °C. This single cooling process strips out the latent load. Because the high volume of 8 °C air also removes extra sensible heat, we add terminal reheat at the diffusers or in the ductwork to restore the correct room temperature.

Strategy 2: DOAS plus wet-coil terminal units

We let the DOAS handle the outdoor air. It deep-cools the 1128 liters per second, removing the outdoor latent load plus a portion of the internal latent load. The remaining sensible load and the balance of the internal latent load are handled by recirculating Fan Coil Units located in the space. Because these FCUs must remove moisture, they operate with standard chilled water temperatures of 6 °C supply and 12 °C return to keep the coils wet. They also require proper condensate drainage so water does not spill into the space.

In both strategies we separate the ventilation and dehumidification tasks from the main space temperature control. This separation prevents the airflow conflict that appears when we try to satisfy both loads with a single air stream sized only for sensible cooling.

Fig. 5 — Two working architectures for the restaurant

Ventilation air

1128

L/s at 10 °C

Sensible delivered

kW of the 38 needed

Latent requirement

sub-zero

dew point freezes coil

Restaurant loads: 38 kW sensible, 27 kW latent, 1128 L/s fresh air. Strategy 1 mixes fresh and recirculated air in one central AHU cooled to an 8 °C apparatus dew point, with terminal reheat. Strategy 2 lets a DOAS deep-cool the ventilation air while wet-coil FCUs in the space handle the rest.

Understanding these relationships lets you size equipment correctly from the start and avoid the common problems of coil icing or overcooling. The key is always to decide early whether the latent load or the sensible load will govern the airflow, then choose the architecture that matches that decision.