4. Heating System
Pool Heaters can increase the comfort level of your pool even if the environment naturally dictates a pool a little cooler than you would like.
The total required heat to increase and maintain the temperature in an outdoor pool can be calculated as
h_{total} = h_{surface} + h_{heatup} (1)
where
h_{total} = total heat load (btu/hr, W)
h_{surface} = heat loss from pool through the surface – mainly evaporation of water (btu/hr, W)
h_{heatup} = heat load required to increase the pool temperature (btu/hr, W)
HeatUp Load
The heatup load depends on the volume of the pool. The volume can be calculated as:
V = 7.5 l w d (2)
where
V = volume (Gal)
l = length (ft)
w = width (ft)
d = depth (ft)
The heatup load can be calculated as
h_{heatup} = 8.34 V dT_{w} / dt (3)
where
dT_{w} = difference between initial temperature and the final temperature of the water in the pool (^{o}F)
dt = heat pickup time (hr)
Swimming Pool HeatUp Load or HeatUp Time Calculator – Imperial Units
Add the required data for your swimming pool below. Note that you must use either “HeatUp Load” or “Heat Pickup Time” in the calculator.
Note that heat loss during heatup is not included in the calculation.
Surface Heat Loss due to Temperature Difference
The heat load required to replace the surface heat loss due to the temperature difference between the pool surface and the ambient air can be expressed as
h_{surface} = k_{s} dT_{aw} A (4)
where
k_{s} = 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 ft^{2} ^{o}F)
dT_{aw} = temperature difference between the air and surface water in the pool (^{o}F)
A = surface area of the pool (ft^{2})
Note that the major part of the heat loss from a swimming pool surface is a result of evaporation of water from the surface.
Example – Pool Heating
A pool with dimensions length 30 ft, width 20 ft and dept 6 ft is heated from 50^{o}F to 75^{o}F. The volume in the pool (in gallons) can be calculated as
V = (30 ft) (20 ft) (6 ft) (7.5 gal/ft^{3})
= 27000 (gal)
The heatup heat required to heat the pool up in 24 hours can be calculated as
h_{heatup} = (27000 gal) (8.34 lbs/gal) ((75 ^{o}F) – (50^{ o}F)) (1.0 Btu/lb ^{o}F) / (24 hr)
= 234562 Btu/hr
 1 Btu/hr = 2.931×10^{4} kW = 0.252 kcal/hr
Note that this value is without the increased heat loss through the surface due to the increased temperature difference and increased evaporation when the temperature increases.
If the ambient temperature is 65^{ o}F 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
h_{surface} = (5 Btu/hr ft^{2} ^{o}F) ((75 ^{o}F) – (65 ^{o}F)) (30 ft) (20 ft)
= 30000 (Btu/hr)
Heatup Load in SIunits
The heatup load in SIunits can be calculated as
h_{heatup} = ρ c_{p} l w d dT / dt (5)
where
h_{heatup} = heat flow rate required (kW, kJ/s)
ρ = 1000 – density of water (kg/m^{3})
c_{p} = specific heat water (4.2 kJ/kg^{o}C)
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 (^{o}C)
dt = heat up time (sec)
Example – Heatup Load in SIunits
The heatup load for a swimming pool with dimensions 12 m x 6 m x 1.5 m, heated from 10^{o}C to 20^{o}C in 10 hours, can be calculated as
h_{heatup} = (1000 kg/m^{3}) (4.2 kJ/kg^{o}C) (12 m) (6 m) (1.5 m) ((20 ^{o}C) – (10 ^{o}C)) / ((10 hours) (3600 sec/hours))
= 126 (kW)
Swimming Pool HeatUp Load or HeatUp Time Calculator – SI Units
Add the required data for your swimming pool below. Note that you must use either “HeatUp Load” or “Heat Pickup Time” in the calculator.
Note that heat loss during heatup is not included in the calculation.

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