Psychrometric Processes

Sensible heating and cooling, cooling and dehumidifying coils, humidifying, mixing air streams and evaporative cooling, with worked examples.

On this page

The rules for the chart
Sensible heating and cooling
Humidifying
Mixing air streams
Cooling and dehumidifying
Evaporative cooling
All the processes on one chart
FAQ

The rules for the chart

Every air-conditioning process is a movement of the state point on the chart, and the energy for it comes from one equation: Q = ṁ (h₂ − h₁), with ṁ the mass flow of dry air and h the enthalpy from Moist Air and the Psychrometric Chart. The direction tells you the process:

Process	Direction on the chart	Energy
Sensible heating	horizontally to the right (W constant)	Q = ṁ (h₂ − h₁)
Sensible cooling (above the dew point)	horizontally to the left	same
Humidifying with steam	vertically up (T almost constant)	steam flow = ṁ ΔW
Evaporative cooling	up and to the left, along constant enthalpy	no net energy: h constant
Cooling and dehumidifying	down and to the left, towards the saturation curve	Q = ṁ (h₁ − h₂) (sensible plus latent)
Mixing	on a straight line between the two states	mass-weighted average of W and h

Sensible heating and cooling

Worked example: heating winter air

Air at 0 °C and 80 percent (W = 3.02 g/kg, h = 7.54 kJ/kg) is heated to 21 °C at 0.1 kg/s.

The humidity ratio does not change, so the final state is 21 °C and 20 percent, h = 28.79 kJ/kg.
Heating power Q = 0.1 × (28.79 − 7.54) = 2.12 kW. The shortcut 1.2 × flow (L/s) × ΔT agrees: 3.24 kW.
The air arrives dry because the relative humidity falls so far; see the next section.

Cooling without condensation is the mirror image. If the coil surface is below the air’s dew point, water condenses and the process becomes cooling and dehumidifying instead.

Humidifying

To raise the humidity add steam or a fine water spray (adiabatic humidifier, which also cools the air). With steam the state moves almost vertically; the energy is the latent heat of the steam.

Worked example: bringing the heated air to 40 percent

Target: 21 °C and 40 percent, W = 6.17 g/kg.
Steam needed = ṁ ΔW = 0.1 × (0.00617 − 0.00302) = 0.315 g/s = 1.13 kg/h.
Heat in the steam (h_g about 2 676 kJ/kg): 0.84 kW, about 40 percent as much as the sensible heating. Humidification is a big energy user, which is why many buildings accept 30 percent in winter.

Mixing air streams

When two air streams mix, the humidity ratio and enthalpy of the mixture are the mass-weighted averages: W_m = (ṁ₁W₁ + ṁ₂W₂) ÷ (ṁ₁ + ṁ₂), and the same for h. On the chart the mixed state lies on the straight line between the two, at a distance in proportion to the mass flows. Re-circulating return air with some fresh outdoor air is the standard way to keep ventilation and energy in balance.

Worked example: 12 percent outdoor air and 88 percent return air

Return: 24 °C, 50 percent (W = 9.30 g/kg, h = 47.81). Outdoor: 33 °C, 40 percent (W = 12.61, h = 65.51).
The outdoor air is the 55 L/s of ventilation, 12 percent of the supply by mass. W_m = 0.120 × 12.61 + 0.880 × 9.30 = 9.70 g/kg; h_m = 0.120 × 65.51 + 0.880 × 47.81 = 49.94 kJ/kg.
Temperature from h = 1.006 T + W (2 501 + 1.86 T): T_m = 25.09 °C, 49 percent. The mixed temperature is close to, but not exactly, the weighted average of the temperatures (25.08 °C).
Warning: if the mixed state ends up above saturation (cold, humid outdoor air mixing with warm return air) fog forms in the duct.

Cooling and dehumidifying

A cooling coil with a surface colder than the air’s dew point removes heat and water at once. The coil apparatus dew point (ADP) is the average surface condition; the air leaves on a line from the entering state towards it, and the closer the leaving state is to the ADP, the more effective the coil. The fraction that bypasses the coil unchanged is the bypass factor.

Worked example: the coil of the example office

The office of this guide needs 425 L/s of supply air at 13 °C (Heating and Cooling Load Calculations). The coil receives the mixed air above (25.1 °C, 9.70 g/kg) and cools it to 13 °C and 95 percent. The result is close to the 7.27 kW of the load calculation (the supply-air condition is a little different) (W = 8.86 g/kg, h = 35.45 kJ/kg).

Mass flow of dry air = 0.425 ÷ 0.822 = 0.517 kg/s.
Total coil load = ṁ (h_in − h_out) minus the small enthalpy of the condensate = 0.517 × (49.94 − 35.45) − condensate = 7.47 kW (2.12 tons).
Sensible part = ṁ c_p ΔT = 6.39 kW; latent part = 1.08 kW; coil SHR = 0.86.
Condensate = ṁ ΔW = 0.517 × (9.70 − 8.86) g/s = 1.6 kg/h: it must drain away through a trapped drain pan.
With a coil surface at about 9 °C the bypass factor is (13 − 9) ÷ (25.1 − 9) = 0.25: about 25 percent of the air passes without contact.

If the room needs a lower humidity than the coil delivers, the air is overcooled and reheated (wasteful) or the coil is run colder; dedicated dehumidification with a desiccant wheel or a heat-pipe coil is the efficient answer in humid climates.

Evaporative cooling

Evaporating water into air cools it without any refrigeration. The process follows a line of constant enthalpy (adiabatic saturation), and the lowest temperature possible is the wet-bulb. A direct evaporative cooler reaches an effectiveness ε = (T_in − T_out) ÷ (T_in − T_wb) of 70 to 90 percent.

Worked example: a desert climate

Outdoor air 35 °C and 25 percent: W = 8.76 g/kg, wet bulb 20.2 °C.
With an 85 percent effective pad the supply is T = 35 − 0.85 × (35 − 20.2) = 22.4 °C at 81 percent relative humidity (W = 13.81 g/kg).
Water evaporated for 0.5 kg/s of air = 0.5 ÷ 0.885 × 5.05 g/s ≈ 10 kg/h.
In a humid climate (35 °C and 60 percent) the wet bulb is 28.2 °C and there is little to gain: evaporative cooling works where the air is dry.

All the processes on one chart

Each process is an arrow on the chart; the arrows are drawn from the computed states.

Previous: Moist Air and the Psychrometric Chart. Next: Heating and Cooling Load Calculations.
Psychrometric Calculator.
Refrigeration and Heat Pumps for the cooling source.
HVAC and Psychrometrics: the whole guide.

FAQ

Why does the coil dehumidify?

If the coil surface is colder than the air’s dew point, water condenses on it. The air leaves cooler and drier.

Why do I reheat after cooling?

To hold the humidity down, the coil makes air colder than the room needs; reheating brings it back. Using recovered heat (condenser heat or a heat pipe) avoids the energy penalty.

What is the sensible heat ratio of a coil?

Sensible cooling ÷ total cooling. A dry coil is 1.0; a humid-climate coil 0.6 to 0.75.

Can mixing cause fog?

Yes. If the straight line between two states crosses the saturation curve, the mixture is supersaturated and condenses.

References

ASHRAE, Handbook – Fundamentals, Chapter 1 and Handbook – HVAC Systems and Equipment, Chapter on air-cooling and dehumidifying coils.
McQuiston, F. C., et al., Heating, Ventilating, and Air Conditioning, Wiley (psychrometric processes).
CIBSE Guide B, Heating, Ventilating, Air Conditioning and Refrigeration.

Last reviewed: October 2026. Coil and pad performance are typical.