# Homemade spray dryer - how to generate powerful hot air?

I am currently trying to create homemade spray dryer,

Which is basically pumping hot air in a chamber where targeted liquid is sprayed, so that the water content could instantly evaporate, leaving the powder behind.

However, I couldn't find a way to generate hot air powerful enough to evaporate the droplets. What would you suggest?

I am using 0.5mm nozzle

• So, first how much water has to be evaporated? one litre? Two? How much energy is needed to do that? Then how much air is needed at the temperature you want, or need, to work at? Oh and consider you will need more air as some won’t get near the water droplets… Aug 14 '21 at 18:17
• It's easy enough to get as much hot gas as you like. A propane or natural gas burner has a flame temperature of about 1900 C. Aug 15 '21 at 14:47
• Hot air can be made from a hair blow dryer, a furnace, a gas turbine or any number of things in between. Give us a volume/mass and temperature and you'll get better help. Aug 16 '21 at 13:35
• I imagine you'll want to have really small droplets. Big ones will hit the side of the container before they fully evaporate and then you'll just get the stuff caked onto the side of the container instead of as a powder. Aug 18 '21 at 13:14

## 2 Answers

Start with specifications. What temperature, what flow rate, from those two and the specific heat capacity of air you can work out the power required.

$$P = \Delta T \times m \times SHC$$

where $$P$$ is the power in watts, $$m$$ is the mass per second and $$SHC$$ is the specific heat capacity of air. You will also need the density of air if you want to convert to volumetric flow rather than mass flow.

I would start from the amount of liquid being sprayed, because your goal is to evaporate the liquid and leave the solute(powder). Once you know that you can calculate the heat required to add to the air, and calculate the approximate temperature of the hot air (so that you can monitor with a control system).

Assuming you have a liquid of 1 kg that you need to evaporate what you need to do is

calculate the energy to change 1 kilogram of liquid at temperature T to 1 kilogram of liquid vapor at the same temperature (T).

In order to estimate the heat energy required for 1 kg, the substance needs to go through the following processes.:

1. Warm the liquid from T to the boiling temperature. Use the standard specific heat for liquid; you will be adding heat energy $$Q_{boil}$$

2. Convert the liquid at boiling point to liquid vapor at boiling point. You need to use the standard latent heat of vaporization for the liquid; you will be adding heat energy $$Q_{latent}$$

3. Cool the the vapor from boiling point to T. Use the standard specific heat of vapor; you will be removing heat energy $$Q_{vapor,T}$$.

If you know that energy (per kg), and you know the rate the liquid is spayed then you can do a calculation on the energy requirements per second (and therefore obtain a ball park figure) for the power requirements.

Then you need to also factor in the mass of the air that you will provide (however usually the energy requirements for the air are much less than the liquid).