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Engineering ToolBox > Sizing Swimming Pool Heaters
The necessary heat to increase and maintain the temperature of an outdoor pool can be expressed as
htotal = hsurface + hheat-up (1)
where
htotal = total heat load (btu/hr)
hsurface = heat loss through surface - mainly evaporation of water from the surface (btu/hr)
hheat-up = heat load necessary to move the pool from the initial temperature (btu/hr)
The heat-up load depends on the volume of the pool which can be expressed as
V = l w d 7.5 (gal/ft3) (2)
where
V = volume (Gal)
l = length (ft)
w = width (ft)
d = depth (ft)
The heat-up load can then be expressed as
hheat-up = V 8.34 (lbs/gal) dTw 1.0 (Btu/lb oF) / dt (3)
where
dTw = difference between initial temperature and the final temperature of the water (oF)
dt = heat pick-up time (hr)
The heat load necessary to cover up the surface loss due to temperature difference between pool surface and ambient air can be expressed as
hsurface = ks dTaw A (4)
where
ks = surface heat loss factor - for sheltered positions with average wind velocity 2 to 5 (mph), the surface heat loss factor is in the range 4 to 7 (Btu/hr ft2 oF)
dTaw = temperature difference between the air and surface water in the pool (oF)
A = surface area of the pool (ft2)
Note that a major part of the heat loss from the swimming pool surface is due to evaporation of water from the surface.
A pool with dimensions length 30 ft, width 20 ft and dept 6 ft is heated from 50oF to 75oF. The volume in the pool (in gallons) can be calculated as
V = 30 (ft) 20 (ft) 6 (ft) 7.5 (gal/ft3)
= 27000 (gal)
The heat-up heat required to heat the pool up in 24 hours can be calculated as
hheat-up = 27000 (gal) 8.34 (lbs/gal) (75 - 50) (oF) 1.0 (Btu/lb oF) / 24 (hr)
= 234562 Btu/hr
Note that this value is without the increased heat loss through the surface due to increased temeperature difference and increased evaporation when the temperature increase.
If the ambient temperature is 65oF and the wind conditions is modereate, the heat loss through the surface due to temperature difference (when the pool is heated up) can be calculated as
hsurface = 5 (Btu/hr ft2 oF) (75 - 65) (oF) 30 20 (ft2)
= 30000 (Btu/hr)
The heat-up load in SI-units can be calculated as
hheat-up = ρ cp l b d dT / dt (5)
where
hheat-up = heat flow rate required (kW, kJ/s)
ρ = 1000 - density of water (kg/m3)
cp = 4.2 - specific heat capacity of water (kJ/kgoC)
l = length of swimming pool (m)
w = width of swimming pool (m)
d = dept of swimming pool (m)
dT = difference between initial and final temperature (oC)
dt = heat up time (sec)