qty of hot air

1 kg water needs 550 kcals to evaporate. 1 cu.m. air can deliver about 0.25 kcal on cooling through 1 deg.C. 35 cu.m. air will have to cool through about 60 deg.C to supply this. Add some more for contact efficiency. If heat damage is an issue the solution may be to use more volume of air. Bioramani

You got to specify the air and water temperature. In SI units, you would need a thermal heat input of 2420 KJ to evaporate 1 kg of water (Latent Heat). Considering air has sp heat of 1.015 kJ/kg and assuming you have 60 C temperature drop available for air, then

2420 kJ/kg = mass of air (kg) x 1.015 kJ/kg/C x 60 C, or

Mass of air required to evaporate 1 kg of water = 40 Kg

Air has a specific volume of 22.4/29 = 0.772 Nm3/Kg

So, the volume of air required = 31 Nm3.

Note- this assumes no effciency loss or specific heat input to raise the temperature of water, both of which increase the air quantity required.

It depends upon the moisture content of the air being used. Hot or not, if the air is being introduced without a wet-bulb depression, no drying can take place. The greater the wet-bulb depression, the more evaporation takes place for the same volume of air. So, until the air inlet conditions are known, the question is unanswerable.