If 3000 ft3
of air is crossing an evaporator coil and is
cooled from 75°F to 55°F, what would be the volume of
air, in ft3
, exiting the evaporator coil

Respuesta :

Charles law:
V1/T1=V2/T2
where T is in deg. K.
75°F=297.04°K
55°F=285.92°K
=>
V2=V1/T1*T2
=3000*(285.92/297.4)
=2887.8 c.f.