Mixing compressed air with different dew points

[David Gin]

Two compressed air(A, B) at 100 PSIG:
Air A - Compressed with -70C dew point.
Air B - Compressed without entering an air dryer.

If this two compressed air is distributed into a tank together, would the the air mix to form Air C (same pressure with dew point between air A and B)?

I have read a little bit into air mixture and some says the process is slow enough to classify as not mixing. If it does not mix is there a reason. If it does, how would the dew point of the mixture air be found theoretically (psychrometric chart)?

 

[mfb]

If this two compressed air is distributed into a tank together, would the the air mix to form Air C (same pressure with dew point between air A and B)?

The pressure will depend on the amount of gas flowing in and its volume.
I'm not sure if you can avoid mixing - you would have to fill in the air extremely carefully for sure, and I doubt this would work at all.

If it does, how would the dew point of the mixture air be found theoretically (psychrometric chart)?

You can convert the dew point (or relative humidity for B) to absolute humidities, make a weighted average there, and transform back to a dew point.

 

[jim hardy]

use partial pressures and figure out molar mix ?
mw air ~29, water vapor ~18

 

[David Gin]

The pressure will depend on the amount of gas flowing in and its volume.
I'm not sure if you can avoid mixing - you would have to fill in the air extremely carefully for sure, and I doubt this would work at all.

If two of the air have the same pressure shouldn't the resultant mixture have the same pressure as well? So air mixing is inevitable?

use partial pressures and figure out molar mix ?
mw air ~29, water vapor ~18

Could you expand on this? Thanks.

 

[mfb]

I fixed the quote tags and merged the two posts.

If two of the air have the same pressure shouldn't the resultant mixture have the same pressure as well?

It depends on how you mix the air. If you have valves that you open for a while and close again, pressure in the third volume doesn't have to reach the pressure in the tanks, for example.

If your two tanks have different pressure, it is obvious that your mixture cannot be equal to both tank pressures at the same time.

 

[jim hardy]

If this two compressed air is distributed into a tank together, would the the air mix to form Air C (same pressure with dew point between air A and B)?

Yes. But the resulting dewpoint will be probably higher than one expects.

Could you expand on this? Thanks.

What goes into the receiver tank is a mix of air and water vapor.
Pressure in the tank is the sum of two pressures, one from the air and one from the water vapor.
Water vapor is just steam so we could find its pressure from a steam lookup table

The pressure of the water vapor in your mix is saturation pressure for the dewpoint
The weather bureau folks at NOAA have given us a handy online calcuator, here
http://www.srh.noaa.gov/epz/?n=wxcalc_vaporpressure
Where it says "If you want the actual vapor pressure enter the dewpoint:"
i enter 32 degf and get back 0.09 psi (and some other units but i learned English system)

so if i have a tank full of air at one atmosphere(14.7psia) and dewpoint 32degf,
the water vapor pressure is 0.09 psia
and the air pressure is 14.7 - 0.09 = 14.61 psia.
Mix is 0.09/14.7 = 6.12% water vapor

If pressure is 100 psig(114.7 psia) and dewpoint still 32degf
water vapor pressure still 0.09 psia
so air pressure is 114.7 - .09 = 114.61 psia
and mix is 0.09/114.7 = 0.078% water vapor

As you see, the compressed air contains a lot less %moisture than uncompressed air at same dewpoint.

Now - your dried air in tank A with dewpoint -70degf has so little water vapor in it you can consider it almost absolutely dry
At that NOAA calculator when i type in -70F dewpoint i get 0psi, which is '0.000something' rounded off.
So if i mix your tank B of moist air with an equal volume of that dry dry air from A , i'll halve the moisture content of the mix.

So the question becomes , how much moisture is in tank B's air ?
If we mix B's 32 degf dewpoint air with A's dry air , and tanks A and B are same size,
we'll have twice the volume of air-water vapor mix,
but the water vapor is now spread out through that double volume
so its pressure will be half.

So our resulting mix will have vapor pressure half as much, 0.045 psi.

I'm having trouble finding a calculator to go from vapor pressure in psi to temperature at that low range
So i had to resort to trial and error in that NOAA calculator linked above
typing in 15 degf gives 0.04 psi
typing in 16 degf gives 0.05 psi
so i'd say vapor pressure of 0.045 psi means a dewpoint between 15 and 16 degf.
And that lies between A's -70degf and B's +32degf dewpoints, but not halfway between them.

It'd be more accurate to use pressures in pascals , but i am too old for SI

And, I was off base - we didn't have to do molar fractions at all ! Partilal pressures worked fine.

.....................

Now - let us think for a moment about your tank B
this exercise shows why you get water in the bottom of your compressor tank.
When your compressor takes in local air at say 80 degf and dewpoint maybe a pleasant 60 degf, 1 atmosphere of course,
that air/water-vapor mix with 60degf dewpoint has water vapor pressure of 0.26 psia (NOAA )
and air pressure of 14.7 - 0.26 = 14.44 psia

When the compressor raises the mix to 114.7 psia
it raises both air and water vapor by the same compression ratio, 114.7/14.7 = 7.80
and it heats them - feel the compressor's head !

so air pressure becomes 14.44 X 7.80 = 112.67 psia
and water vapor becomes 0.26 X 7.80 = 2.03 psia
and that mix of hot air and steam flows into the tank where it cools down to room temperature.
Well, water vapor at 2.03 psia will condense when it cools down to its dewpoint of about 127 degf (trial and error on NOAA calculator)
so as the mix cools down the water vapor condenses on the tank walls
(and that's why you have to drain the tank daily.)

Result of this is, the air in tank B will acquire dewpoint same as room temperature where it's located. That's because the water vapor inside the tank can only cool to room temperature and no lower.

What's room temperature? 80 degf like we said ? Could be higher outside in summer.
Let's take 80...
vapor pressure at 80 deg dewpoint is 0.51 psi courtesy NOAA calculator
half that is 0.25 psi
which i trial and error in the NOAA calculator back to a dewpoint of 59 degf.

Interesting, eh ?
Intuitively you'd expect the dewpoint of the mix midway between -70 and ambient, maybe around zero, but it's higher due to water's nonlinear saturation pressure-temperature curve.
In fact our hypothetical mix has dewpoint only 1degf below that of the room air.

I urge you to try that calc at 100F dewpoint. We had some compressors located in 110degf ambient with steam wafting right past their air intake - no wonder the poor things made water by the ton .

If it does, how would the dew point of the mixture air be found theoretically

that's what i tried to address.

Now you can add in the practical details of how you mix. Plenty of expertise here at PF !

old jim