Pool Heater Output for a Target Heat-Up Time
The heater input rating (the nameplate Btu/h) needed to warm a pool by a rise in a target time.
Example
You enter
- Pool volume (gallons) 20000
- Temperature rise (°F) 10
- Target heat-up time (h) 5.2125
- Efficiency as a fraction (0.80 gas; 1.0 for a heat pump entered at its heat output) 0.8
You get
- Required heater input rating 400,000 Btu/h
Checked against published worked examples from: The Pool Class (CPO certification training).
Details, formula, and sources
Scopeinput = (gallons x 8.34 x rise) / (target_hours x efficiency); it delivers input x efficiency. Warming a 20,000 gal pool +10 F in 5.2 h at 80% needs a 400,000 Btu/h (input) heater; ask for it in 3 h and it jumps to 695,000 Btu/h. Answers 'what size heater do I need' instead of the time from one heater. Ignores cover/standby losses, so add margin. The equipment ratings govern.
Formula or tableQ_btu = gallons x 8.34 x dT; required_output_btu (the heater INPUT rating) = Q_btu / (target_hours x eff); the inverse of hours = Q_btu / (input x eff).
Source and editionThe sensible water-heating relation (1 Btu raises 1 lb of water 1 F) solved for the heater input rating, by name.
Free accessThe water-heating energy relation and the 8.34 lb/gal constant are standard published values. The equipment ratings and site conditions govern.
What governsEstimate. AHJ and licensed professional govern.
API field namesgallons, dT_F, target_hours, eff, required_output_btu
- Sensible heat Btu = gallons x 8.34 x dT; 1 Btu raises 1 lb water 1 Fthermodynamics
- Input-rating inverse input = energy / (target_hours x efficiency), delivered output = input x efficiency; for a heat pump enter its heat output in Btu/h with efficiency 1.0, not its COPequipment ratings
- No standby loss ignores cover/evaporation/standby losses; add margin for real heat-upscope of this tile