Cooling Tower Heat Rejection

Q=500RΔTQ = 500 \, R \, \Delta T

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

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This is the water-side sensible-heat equation wearing trade clothing. Q = ṁ·c·ΔT with mass flow in pounds per hour and c = 1 BTU/(lb·°F) becomes Q = 500 × gpm × ΔT°F, because a gallon of water weighs 8.34 lb and there are 60 minutes in an hour: 8.34 × 60 = 500.4, rounded to 500 forever. A 1000 gpm tower on a 10 °F range is therefore rejecting 500 × 1000 × 10 = 5,000,000 BTU/h — about 1465 kW, or 417 tons of refrigeration.

Two habits keep this honest. First, the constant assumes plain water near ambient temperature; a 30% propylene glycol loop has a specific heat near 0.9 BTU/(lb·°F) and a different density, so the real constant is closer to 460 and using 500 overstates the duty by roughly 8%. Second, on a chiller the tower rejects the evaporator load plus the compressor work, which is why condenser water flow is sized at 3 gpm/ton against 2.4 gpm/ton on the chilled-water side. The solver applies the 500 constant per gallon per minute per Fahrenheit degree regardless of the units you type, converting first.

Cooling Tower Heat Rejection
Q=500RΔTQ = 500 \, R \, \Delta T
Where
  • QQ= Heat rejection rate
  • RR= Recirculation rate
  • ΔT\Delta T= Cooling range
Missing one of these? Work it out first, then come back