Evaporation heat loss

Energy2green Wind And Solar Power System

Get Instant Access

The following analysis is for a still pool as per ISO/TR 12596:1995 (1995a). The evaporative heat loss from a still outdoor pool is a function of the wind speed and of the vapor pressure difference between the pool water and the atmosphere, given by qe = (5.64 + 5.96vo3)CPw - Pa) (5.18)

where qe = heat loss by evaporation (MJ/m2-d).

Pw = saturation water vapor pressure at water temperature, tw (kPa).

Pa = partial water vapor pressure in the air (kPa).

v0.3 = wind speed velocity at a height of 0.3 m above the pool (m/s).

If the wind velocity above the pool cannot be measured, it can be obtained from climatic data by the application of a reduction factor for the degree of wind shelter at the pool. Usually, the wind speed is measured at 10 m from the ground (v10); therefore,

For normal suburban sites, v = 0.30v10 For well-sheltered sites, v = 0.15v10

For indoor pools, the low air velocity results in a lower evaporation rate than usually occurs in outdoor pools, and the evaporative heat loss is given by qe = (5.64 + 5.96vs)(PW - Penc) (5.19)

where

Penc = the partial water vapor pressure in the pool enclosure (kPa). vs = air speed at the pool water surface, typically 0.02-0.05 (m/s).

Partial water vapor pressure (Pa) can be calculated from the relative humidity (RH),

where Ps = saturation water vapor pressure at air temperature, ta (kPa).

Saturation water vapor pressure can be obtained from

Ps = 100(0.004516 + 0.0007178tw - 2.649 X 10-6 t2w + 6.944 X 10-7 t3w)

The presence of swimmers in a pool significantly increases the evaporation rate. With five swimmers per 100 m2, the evaporation rate has to increase by 25-50%. With 20-25 swimmers per 100 m2, the evaporation rate has to increase by 70-100% more than the value for a still pool.