Atmospheric water generator

I have a simple project of a water-from-air generator.

The vertical cylinder tank is embedded in the ground to a depth of up to 4 meters. The electric fan sends ambient air into the cylinder. The difference between air temperature and temperature of the cylinder wall will lead to condensation.

Could someone explain how can I estimate the condensed amount of vapor from the air?

• What is the tank embedded in? – Bob D Aug 25 at 16:16
• The tank is embedded in the ground. – solaryasha Aug 25 at 16:37
• So you are doing the project in the summer when the ground is cooler than the air? – Bob D Aug 25 at 16:49
• Exactly. But I don't know how to design condensation. I guess it should be flimwise, but I have no clue how to write an energy balance. – solaryasha Aug 25 at 17:46
• It has to do with dew point. Let me see what I can put together for you. Stand by. – Bob D Aug 25 at 18:42

Water will begin to condense from air when the temperature of the air drops to the saturation (boiling/condensation) temperature of the water vapor in the air. This temperature is called the "dew point" because it's typically the night time or early morning temperature following a hot humid afternoon when dew forms on grass. So the first thing you need to confirm is that the temperature on the walls of the tank (which should be the same as the air originally in the tank) is equal to or less than the dew point for the air you are pumping in.

You can determine the dew point in a couple of ways. The most convenient is to use the ASHRAE Psychrometric chart. It comes in english and SI units. The chart contains a number of useful thermodynamic properties, but the ones you would use are temperature (dry bulb), relative humidity, saturation temperature (dew point), and humidity ratio.

On the chart find the intersection of the lines for the dry bulb temperature and relative humidity. From that point moving horizontally to the left locate the saturation temperature, which is the dew point. As an example, if the dry bulb temperature is 26.7 C (80 F) and the relative humidity is 70% (conditions common here on Long Island in summer), the chart will give you a saturation temperature of about 20.7 C (69 F).

Once you know the saturation temperature you can compare that with the temperature of the walls of your buried tank. Here on Long Island the average all round temperature 6 feet (about 2 meters) below the ground surface is about 20.8 C (55 F), so you would be well below the dew point.

Determining the actual amount of condensation (liquid water) is another matter and can get complicated. For one thing you are pumping in air from the atmosphere into the tank, driving some of the existing air out and mixing with the remainder. But we can determine what the maximum amount of water you can theoretically obtain from a given volume of air at a given temperature and humidity, knowing that the dew point has been reached.

Let's say you have 1 $$m^3$$ of my Long Island summer air. We already determined that the dew point of 20.8 C has been reached for 6 meters. So it is a matter of how much water we can get by the ground extracting heat from the air such that all the water vapor is condensed to liquid form. If we can consider the earth around the tank as a constant temperature heat sink, then all of the water vapor should theoretically condense.

To do this we return to the Psychrometric chart. Taking the intersection point of the dry bulb temperature and relative humidity but now moving horizontally to the right we can find the humidity ratio of the air in kg of moisture per kg of dry air. For our example we have a humidity ratio of 0.155 kg of water per kg of dry air. Since most of the volume of the tank is occupied by air of the saturated air (you can confirm it by looking at specific volumes from the steam table), we can approximate the mass of 1 $$m^3$$ of dry air in the tank as 1.293 kg. Thus we get (0.155)((1.293)= 0.2 kg (200 g) of water.

Hope this helps.