Cooling Tower Rule of Thumb (1% evaporation, 10°F cooling)

Shows that evaporating 1% of the recirculating water lowers the bulk water temperature by about 10°F, by setting the latent heat removed equal to the sensible heat lost by the remaining bulk.

$$\Delta T=\frac{m_{evap}\times h_{fg}}{(m_{recirc}-m_{evap})\times c_p}$$

Variables

Symbol	Meaning	Unit
$\Delta T$	Bulk water temperature drop	°F
$m_{evap}$	Mass evaporated per minute	lb/min
$m_{recirc}$	Total recirculating mass flow	lb/min
$h_{fg}$	Latent heat of vaporization (about 970)	BTU/lb
$c_p$	Specific heat of water (1.0)	BTU/lb·°F

Worked example

For 1,000 gpm: 81,000 BTU/min = (8,340 - 83.4) lb x 1 BTU/lb·°F x ΔT, so ΔT = 81,000 / 8,256.6 = 9.8°F, about 10°F.

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Part of The Blueprint of Water