Sizing Swimming Pool Heaters

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The necessary heat to increase and maintain the temperature of an outdoor pool can be expressed as

h_total = h_surface + h_heat-up         (1)

where

h_total = total heat load (btu/hr)

h_surface = heat loss through surface - mainly evaporation of water from the surface (btu/hr)

h_heat-up = heat load necessary to move the pool from the initial temperature (btu/hr)

Heat-Up Load

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

h_heat-up = V 8.34 (lbs/gal) dT_w 1.0 (Btu/lb ^oF) / dt         (3)

where

dT_w = difference between initial temperature and the final temperature of the water (^oF)

dt = heat pick-up time (hr)

Surface Heat Loss due to Temperature Difference

The heat load necessary to cover up the surface loss due to temperature difference between pool surface and 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 (^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.

Example - Pool Heating

A pool with dimensions length 30 ft, width 20 ft and dept 6 ft is heated from 50^oF to 75^oF. 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

 h_heat-up = (27000 gal) (8.34 lbs/gal) ((75 ^oF) - (50 ^oF)) (1.0 Btu/lb ^oF) / (24 hr)

    = 234562 Btu/hr

• 1 Btu/hr = 2.931x10^-4 kW = 0.252 kcal/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 65 ^oF 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 ^oF) - (65 ^oF)) (30 ft) (20 ft)

    = 30000 (Btu/hr)

Heat-up Load in SI-units

The heat-up load in SI-units can be calculated as 

h_heat-up =  ρ c_p l w d dT / dt          (5)

where

h_heat-up = heat flow rate required (kW, kJ/s)

ρ = 1000 - density of water (kg/m³)

c_p = 4.2 - specific heat capacity of water (kJ/kg^oC)

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)

Example - Heat-up Load in SI-units

The heat-up load for a swiming pool 12 m x 6 m x 1.5 m, heated from 10^oC to 20^oC in 10 hours, can be calculated as 

h_heat-up =  1000 (kg/m³) 4.2 (kJ/kg^oC) 12 (m) 6 (m) 1.5 (m) (20 (^oC) - 10 (^oC)) / (10 (hours) 3600 (sec/hours))

            = 126 (kW)

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• Evaporation from Water Surfaces - The amount of evaporated water from a water surfaces - like swimming pools or an open tanks - depends on the temperature in the water and in the air, and the humidity and velocity of the air above the surface - online calculator