Just an idea...
if the problem is simplified as force balance (downward by the mass of droplet and upward direction by airstrem force), below equation can be written.
we can assume the water droplets have spherical shapes (dia,R) ,lifted by air stream and hanged by zero velocity, airstream has velocity V and dynamic pressure V²/2g ,
simply balancing , F(air stream)=F(mass of droplet) ;
mass of droplet ,having force downward,F=ma
F=(4/3)x(3,14)xR³x density x 9,81 m/s²...(1)
pressure is force by unit surface (cross-section of sphere surface faced with airstream ,3,14*R²)
P=F/A
F=3,14 x R² x P ...................(2)
P= V²/(2x3,14) .....................(3)
3,14 x R² x V²/2x(3,14) = 4/3)x(3,14)xR³x density x 9,81 m/s²
V²=(8/3)x (3,14)² x density(1000 kg/m³)x R
V= 162,1 x R^(1/2)
assuming the droplets of water have 3 mm diameter;
R=0,003 m ; V =8,88 m/s ; V = 1574,8 ft/min
is the min velocity of air that can prevent water droplet to fall down