Evaporative cooling: principle, efficiency and the h-x diagram

Evaporative cooling evaporates water into an air stream: evaporation removes heat and the air cools down, and it does so with no compressor, no refrigerant and far lower energy use than conventional air conditioning. Thermodynamically it is an adiabatic process: the total enthalpy does not change, only the split between sensible heat (temperature falls) and latent heat (humidity rises). On the h-x diagram it therefore runs along a line of constant enthalpy, and its natural limit is the wet-bulb temperature of the inlet air.

The physical principle: evaporation and latent heat

When water evaporates into air, it consumes the latent heat of vaporization — about 2,501 kJ/kg at 0 °C, or 2,442 kJ/kg at 25 °C. This energy comes from nowhere other than the heat content of the air itself. The sensible heat the air releases as its temperature falls is exactly the heat that evaporation consumes, so the total enthalpy stays (approximately) the same.

Where the process runs on the h-x diagram

The initial state is point A (for example 35 °C / φ = 20%). The air passes through a humidifier and advances approximately along an isenthalp (a line of constant enthalpy, more precisely along a line of constant wet-bulb temperature) toward a lower temperature. The process ends either at the required point or, at the latest, on the saturation curve ($\varphi = 100\%$) — this is the theoretical maximum, the so-called adiabatic saturation temperature, which corresponds to the wet-bulb temperature of the inlet air. Real equipment reaches 70–95% of this maximum, depending on the adiabatic effectiveness $\eta_{a}$ (saturation efficiency, $\varepsilon$ in ASHRAE).

Calculating the saturation efficiency

The adiabatic effectiveness $\eta_{a}$ (saturation efficiency) is the basic parameter of every evaporative cooler:

$$\eta_{a} = \frac{t_{\text{in}} - t_{\text{out}}}{t_{\text{in}} - t_{wb}}$$

where $t_{wb}$ is the wet-bulb temperature of the inlet air.

Example: inlet air 35 °C / φ = 15%, wet-bulb temperature 17.5 °C, a cooler with $\eta_{a} = 0.85$:

$$t_{\text{out}} = 35 - 0.85 \cdot (35 - 17.5) = 35 - 14.9 = 20.1\ ^{\circ}\text{C}$$

Cooling by almost 15 °C without a single watt from a compressor.

Direct vs. indirect evaporative cooling

Direct evaporative cooling

The air comes into direct contact with water or a water mist. Simple, cheap, very effective in a dry climate. The drawback: the air entering the room is more humid. In a Central European summer, where the inlet φ is usually 50–60%, there is less room for cooling and the resulting indoor humidity can be uncomfortable.

Indirect evaporative cooling

Evaporation cools a secondary air stream, whose heat is passed to the primary stream through an exchanger without the two mixing. The primary air cools without picking up moisture — on the diagram it moves horizontally to the left (sensible cooling). The most advanced variant is the Maisotsenko cycle (M-cycle), which can bring the primary-air outlet temperature below the wet-bulb temperature. From a hygiene standpoint, indirect cooling is a common way to bring partial air cooling into a project at low capital cost.

The limits of evaporative cooling

The fundamental limit is clear: cooling works better the drier the inlet air is. Once the inlet φ exceeds 60–70%, the room for cooling drops dramatically. In Czech summers, with a midday relative humidity of 45–55%, direct evaporative cooling is less effective than in a dry continental climate; the indirect variants remain competitive even here. Another limit is water quality: hard water fouls the systems with limescale, soft water can cause corrosion. A further drawback is the hygiene risk from Legionella, which is kept in check by adequate water treatment, including blowdown and, where appropriate, UV disinfection. On direct systems this contamination control is mandatory. Direct evaporative cooling is not suitable for supplying spaces with elevated cleanliness requirements. Water consumption during operation is a further constraint.

Comparison with vapor-compression cooling

An evaporative cooler typically uses 10–25% of the energy of a vapor-compression system of the same capacity. The ratio of cooling output to the fan and pump power input (a simplified “COP” in this sense, not the thermodynamic COP of a refrigeration cycle) is on the order of 10–40 for evaporative systems, whereas the actual COP of a split air conditioner is usually 2.5 to 5. The trade-off is the dependence on the humidity of the inlet air and the inability to guarantee precise conditions without a supplementary system. For data centers, industrial halls and outdoor cooling in a dry climate, evaporative cooling is the first choice; for precise control in Central European offices, more of a supplement.

Frequently asked questions

Why can’t an evaporative cooler cool the air below the wet-bulb temperature? Because the process runs along a line of constant enthalpy, and the wet-bulb temperature is the point where this line crosses the saturation curve. Beyond it, the air cannot be cooled further, because it is saturated. The exception is advanced indirect cycles (the M-cycle).

Is evaporative cooling worthwhile in the Czech Republic? Direct cooling has limited potential in a humid Central European summer, but the indirect variants remain energy-efficient. It works best where the inlet air is relatively dry.

How does the saturation efficiency relate to design? It determines how close to the wet-bulb temperature the outlet gets. From $\eta_{a}$ and the inlet conditions you compute the outlet temperature directly, which can be verified on the h-x diagram.