How do I calculate the dew point of undried, compressed air? 1

[c2sco]

I'm compressing ambient, undried air so that I can sparge it into a tank to aerate the contents. Hence the air could have any RH at inlet conditions, even 100%. I'm also installing a cooler as the compressed air is too hot for my process. At what temperature will my air start to form dew as it is cooled? Psychrometric tables don't go much above atmospheric pressure, whereas this process will use air at about 1 barg.
Thanks,
Stuart

 

[skoutso]

Use the following equations:

Pw=xw * Pv
Pw=yw * Pt

Pv = vapor pressure of water at system temp.
xw = mole fraction of water in liquid
Pt = total gas pressure
yw = mole fraction of water in gas
Pw = water partial pressure in gas

Since xw=1.0 (only water in liquid phase), combining and solving vs. Pv you get:

Pv = yw * Pt

The gas pressure (Pt) and water mole fraction (yw) is known. Compute Pv.

Look at the saturated steam tables for the temperature that corresponds to Pv. This is the dew point temperature you are looking for.

 

[Guidoo]

As an addition to Skoutso:

At compressor inlet you know pressure Pt (atmospheric?) and temperature. You can look-up saturation pressure Pv in steam table or estimate using:

Pv = 0.0179 * EXP(0.0404*temperature)

Pv in bara, temperature in °C

Since you assume RH=100%, you can calculate yw using:

yw=Pv/Pt

At compressor outlet, you know Pt, and yw remains unchanged, so now you can calculate Pv using:

Pv=yw*Pt

You can now look-up condensation temperuture in steam tables or use:

temperature=24.8ln(55.7*Pv)

temperature in °C, Pv in bara

Note that the equations are fitted in temperature range 10-100 °C, so be careful if compressor inlet or condensation temperature are outside this range.

 

[c2sco]

Many thanks to both of you. It's amazing what you forget - fancy me forgetting steam tables. Oh well, it's been a long time since I used them...
Using skoutso's method, for 20°C, 100% RH air, humidity = 0.0146 kg/kg from a psychrometric chart I have. Converting this to mole fraction I get 0.02295 water, hence 0.02295 bara partial pressure. This compares OK with interpolating Perry's steam table which gives me 0.02337 bara. If I compress to 2 bara total pressure, if hot enough my water vapour pressure becomes 0.04674 bara. Looking down the steam table again I get 31.8°C dew point. Sounds good.

Assuming these are correct, trying Guidoo's correlations, I get:

1)
Pv = 0.0179 * EXP(0.0404*temperature)
Pv in bara, temperature in °C
So
Pv = 0.0179 * (EXP(0.0404*20) = 0.04016 which is a bit high

2)
temperature=24.8ln(55.7*Pv)
temperature in °C, Pv in bara
So
temperature=24.8ln(55.7*0.04674) = 23.7 which is a bit low

or, trying to work from the first correlation result,
temperature=24.8ln(55.7*2*0.04016) = 37.16 which is a bit high

So I'm not confident in these correlation unless you can spot my mistake? If we can correct this, they would be useful.

Either way, thanks for helpful answers.

Regards,
Stuart

 

[Guidoo]

Stuart,

After I submitted my previous post, I also realized that my simple correlations may be a bit too simple for your calculations. This is basically because the relative error is quite large at the lower temperature range (so around 20°C). Since your compressor inlet temperature will be in this range this results in an error that is too large...

In case you want to use a correlation i.s.o. using the steam tables, you could use the Antoine equation:

Pv=1/750*EXP(18.3036-3816.44/(temperature+227.02))

Pv in bara, temperature in °C