Table of Contents
What Is Evaporation Rate?
Evaporation rate measures how quickly water transitions from liquid to vapor at a free surface. It depends on the vapor pressure difference between the water surface and the surrounding air, wind speed, and surface area. Understanding evaporation is critical for pool maintenance, agriculture, climate science, and industrial cooling systems.
The evaporation process requires energy (latent heat of vaporization), which is why evaporation causes cooling. This principle is used in swamp coolers, cooling towers, and even human perspiration. The latent heat of water is approximately 2.45 MJ/kg at 25°C.
Formula
Where E is evaporation rate (kg/h), v is wind speed (m/s), Ps is saturation vapor pressure at water temperature, Pa is actual vapor pressure of the air, and A is surface area (m²). This is the simplified Penman equation.
Factors Affecting Evaporation
| Factor | Effect on Evaporation |
|---|---|
| Higher temperature | Increases (higher vapor pressure) |
| Lower humidity | Increases (larger vapor deficit) |
| Higher wind speed | Increases (removes saturated air) |
| Larger surface area | Increases proportionally |
| Higher altitude | Slight increase (lower air pressure) |
Frequently Asked Questions
How much water does a pool lose to evaporation?
A typical swimming pool (50 m²) in a warm, dry climate can lose 5-10 liters per day. Using a pool cover reduces evaporation by 90-95%. Heated pools evaporate faster due to higher vapor pressure at the water surface.
Why does wind increase evaporation?
Wind removes the thin layer of saturated air just above the water surface, maintaining a strong vapor pressure gradient that drives continued evaporation. In still air, this boundary layer becomes nearly saturated, slowing evaporation dramatically.
What is the latent heat of vaporization?
It takes approximately 2.45 MJ to evaporate 1 kg of water at 25°C. This is why evaporation is such an effective cooling mechanism. A pool losing 10 L/day through evaporation loses about 24.5 MJ of heat energy, equivalent to running a 280W heater continuously.