Water Loss

Is there a psychrometric chart to hand? From the chart, one can work out the total water in the air. The answer is no more than that value.

1 drop....

Can you convert this to Moles? Tks

Look at this link to start:

http://en.wikipedia.org/wiki/Dew_point#Calculating_the_dew_point

It might help to realise the complexity of the requested info.

You need to convert to absolute humidity then work out the 2nd state absolute water content (from tables/charts or calculation formulas(??), and get the difference to obtain the water loss.

None, increasing the pressure of the mixture of air and water increases the air's capacity to carry water. Going the oposite way would drop water out as shown by the vapour trails you see when an aircraft wing or formula 1 car wing is operating in high level of humidity.

Now you make me wonder why I have water in the tank of my air compressor. Perhaps you can make this clear also?

Removing air from the compressor's tank causes the pressure to drop and the air to cool, and the water precipitates out of the air. If the pressure and the temperature don't change -- and the tank walls are the same temperature as the air -- the inside of the tank will stay dry.

Your compressor takes in atmospheric air and water vapour.
The actual amount of water vapour depends on the temperature and RH on the day. Compressing the air causes the temperature to rise, and and hot air holds a lot of water vapour
The act of compression also causes the vapour to condense, but not until the hot air cools in the tank, or just as likely in the pipework and tools.

By compressing 8 times do you mean reducing the volume to 1/8th?If so, as an example, your unit of air say, when saturated could contain 100g water vapour per m3, but in fact is 60%RH and thus contains only 60g per m3.When the air volume is reduced to 1/8th the water content now becomes 60g per 1/8th m3, or the same as 480g per m3.At the original temperature, the air cannot hold more than 100g of vapour per m3, so the excess 380g condenses to become liquid water.You need to know the original temperature to put exact figures to answer your question.

In reply to 8 and 10, it should be made clear that the condensed surplus water is 380g per m3 when compressed. The dimensions of the actual 'Unit' volume (or the pressure) of the starting amount of air has not been specified, nor the temperature.

I have used water vapour density to explain the point based on change of volume.
If you start with 8m3 of air, and 100g/m3 saturated vapour, there wil be 800g of water, but at 60%RH will reduce to 480g. Now when the air is compressed to 1/8th volume, you will have 1 m3 of air, but still containing the original 480g of water.

Back at the original temperature (that can only hold 100g/m3) there will be 380g of surplus vapour that will condense.

Go back and re-do your math, the poster before you mentions per cubic meter compressed, means the intake was 8 cubic meters.

That definitely depends on how you read the question.

the original question was: "How much water can be separated from 1 unit air @ 60 % R.H. by compressing it 8 times? At the same temperature (unit= m3..."

I read that to mean one cubic meter of original humid air.

You say "per cubic meter compressed". I interpret that to mean "per cubic meter [of humid air, which is then] compressed" You apparently interpret it to mean "per cubic meter [of already] compressed [humid air].

This is a good start for thinking. Only the math doesn't match. The amount of water hasn't been there from the start. In your model you need to end with 1 m3 compressed from 8, I think.

The temperature at which the air is at 60% RH is also needed as the absolute water content depends on the temperature at each humidity.