dear JayBee,

I do not know how big your problem is. You should know best.

Water condensation can easily be calculated. In some of the threads in this forum a preadsheet calculation is available for free.

Your blower is delivering approx 250 ltrs/sec.

Atlas Copco sells a refrigerant dryer for 285 lts/sec (FD285)

Visit their website www.atlascopco.com.

A dew point of approx. 3 degrC is normal.

success

teus ■

Dear JB,

I have seen a water cyclone separator after a cooler in one pneumatic conveying installation and it works well.

The water drain was set a little bit open. That costs of course some conveying air, but that was more than available.

In your case an automatic drain valve is sufficient.

When the aftercooler manufacturer states that there will be no condenstation because the air is cooled down to ambient temperature, he forgets that in the mean time the pressure is increased.

The rule is that:

higher temperature makes the air dryer

higher pressure makes the air wetter.

The strongest effect of the two wins.

If you cannot find the humidity/condensation spreadsheet calculation in one of the earlier replies in this forum, let me know and I will paste it again.

have a nice day

teus ■

Hi Jaybee,

We have had excellent luck with API Basco Separators

http://www.apiheattransfer.com/us/Pr...Separators.htm

If you do not intend to install a dryer after the separator you can still expect to drop out enough liquid that may cause a 1-3% reduction in flow. If a dryer is installed it sometimes causes a loss up to 10%, which cause the pneumatic conveying system not to function properly any more.

And you are absolutely correct about the liquid drop out; depending on your dew point after the after cooler you will drop out moisture. Are you intending to use a shell and tube air to water cooler, or an air to air cooler? Customarily the air to water coolers are more effiecient in dropping out moisture especially when you have chilled cooling water handy. Again, it really depends on the dew point. We have had good success with air to air coolers from Xchanger http://www.xchanger.com/applications.htm

or API Basco for liquid cooled applications. Both companies have shown expertise to help caculate the dew points and water knock out. Good luck! ■

Dear TEUS

HI HOU R U

WELL I GONE THRU THIS MAIL WHIE IWAS FOR A MOMENT ONLINE. ACC TO MR JB HE WOULD MAY HAVE HAVE HIS PROBLEM EXPELLED BY JUST INSTALLING MOISTURE SEPERATOR. BUT PNEUMATIC HANDLING SYSTEM IS ONE OF THE CRITICAL FIELD TO CONSIDER. JUST INSTALLING THE AFTERCOOLER HE MAY DROPOUT THE MOISTURE BUT SUBJECT OF THE MATTER IS UNTIL THE AIR DOESNT BEING BROUGHT TO ITS DEW POINT(AIR TEMP CORRESPONDING TO SATURATION PRESSURE OF WATER VAPOUR) IT WONT WORK REALLY , BCOZ AS THE WATER TRAVEL THROUGHOUT THE LINE ITS TEMP GETS DECREASED ND MOISTURE COMES DOWN , WHICH AGAIN LET THE PROBLEM TO REMAIN IN SYSTEM.IT HOEEVER UPTO SOME EXTENT AFTERCOOLER MAY WORK BUT THAT IS PATHETIC AS FOR PNEUMATIC CONVEYING

ANY WAY MR TEUS CAN U LET ME HAVE THE SHEET TO PRDICT THE MOISTURE AT PARTICULAR TEMP

CHEERS

RAHUL

MACAWBER BEEKAY

■

Dear Rahul,

You are right that the air coming from the air after cooler is at 100 % humidity at the local pressure.

From that point on, the temperature will not change much (stays close to material temperature)

But the pressure is decreasing along the pipe line and the the air gradually gets dryer.

Attached you will find the ZIP file “condensation.xls”

have a nice day

teus

Attachments

condensation_3 (ZIP)

■

Dear Sir Teus,

When the conveying air comes in contact of hot Material ( to be conveyed) its temperature will change then..

Regards ■

Dear sachin,

When air and material come together, each having a different temperature, heat is exchanged until the temperatures of air and material are equal.

Expansion of the air cools the air and heat energy is lost (or gained) through the pipewall.

This is influencing the mixture temperature along the conveying pipe and thereby the velocities.

The changing velocity affects the pressure drop and thereby the capacity.

success

teus ■

Awesome, please let us know how it worked once you are up and running.

Ralf ■

Originally Posted by Teus Tuinenburg

Dear Rahul,

You are right that the air coming from the air after cooler is at 100 % humidity at the local pressure.

From that point on, the temperature will not change much (stays close to material temperature)

But the pressure is decreasing along the pipe line and the the air gradually gets dryer.

Attached you will find the ZIP file “condensation.xls”

have a nice day

teus

Dear sir

I find this file quite useful and is able to eradicate few misconception of various clients. But one thing , iam not able to understand

a) The vapor pressure of comptrssed air in condition 2 is decreasing with incerase in temperature , but is also gets changes due to change in temperature at condition first. For instance , when i increase temperature in condition 1 teh vapour pressure at condition 2 gets increase instead of decreasing.Please make me understand about inter relation of temperture in condition 1 with vapur pressure at condition 2 ■

Dear kj,

Condition 1 is the starting condition of the considered air.

Condition 2 is the same air + water vapor at a different pressure and temperature.

That relation is always maintained and if you change the original situation (condition 1), the related condition (condition 2) changes also.

Calculated condition 2 is based on the entered condition 1.

If you change the original condition 1, then condition 2 changes

If you change condition 2, then the original condition 1 does not change.

If you increase the temperature in condition 1, then the amount of water in condition 1 increases.

The increased amount of water vapor is accounted for in the air of condition 2 by an increased water vapor pressure or RH. (Obviously, the RH was not 100 %)

I attached the latest version of the file.

Best wishes

Teus

Attachments

condensation_4 (ZIP)

■

Originally Posted by Teus Tuinenburg

Dear kj,

Condition 1 is the starting condition of the considered air.

Condition 2 is the same air + water vapor at a different pressure and temperature.

That relation is always maintained and if you change the original situation (condition 1), the related condition (condition 2) changes also.

Calculated condition 2 is based on the entered condition 1.

If you change the original condition 1, then condition 2 changes

If you change condition 2, then the original condition 1 does not change.

If you increase the temperature in condition 1, then the amount of water in condition 1 increases.

The increased amount of water vapor is accounted for in the air of condition 2 by an increased water vapor pressure or RH. (Obviously, the RH was not 100 %)

I attached the latest version of the file.

Best wishes

Teus

Sir

1) If vapur pressure in condition 1 incerases with incerase in temperature , then why not the same behaviour s followed in condition 2. I mean , if the vapour pressure incerased in cond 1 with incerase in temp the same (vapour pressure ) in condition 2 shall also incerase, instead it is decreasing

2) What does the dew point signifies in this spread sheet . I guess , if at that local condition , if the air temperaures equal to resulted dew point the vapour will start condense. ■

Dear kj,

1)

If you increase the temperature in condition 1, maintaining the RH, then the amount of water vapor increases also.

The temperature and pressure of air condition 2 does not change.

However, as the water content of condition 1 and condition 2 are considered the same, the RH of condition 2 must change or even condensation can occur.

2)

You are correct.

see:

http://en.wikipedia.org/wiki/Relativehumidity

Greetings

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

1)

If you increase the temperature in condition 1, maintaining the RH, then the amount of water vapor increases also.

The temperature and pressure of air condition 2 does not change.

However, as the water content of condition 1 and condition 2 are considered the same, the RH of condition 2 must change or even condensation can occur.

2)

You are correct.

see:

http://en.wikipedia.org/wiki/Relativehumidity

Greetings

Teus

Sir

Didnt understand your first point . Can you explain it onceagain

As i understrand if , incerase in temp incerases the vapour pressure of condition 2, than then if the temp of condition 2 is incerased , tthe vapour pressure shall incerase too ■

Dear kj,

If you manipulate a mass of air by changing the condition 1 temperature, pressure or RH, then you will get another result for condition 2.

Imagine a compressor.

Intake is condition 1:

The intake temperature = 30 degrC

The intake pressure is 1 bar absolute

The relative humidity = 100 %

Compressed condition 2:

The compressed air is 50 degrC

The compressed pressure is 8.5 bar absolute

Then the resulting RH = 100% and 0.018134 kg water/kg dry air has condensed.

Now the compressor faces a different intake condition 1 and the compressed condition 2 stays the same.

Intake is condition 1:

The intake temperature = 10 degrC

The intake pressure is 1 bar absolute

The relative humidity = 50 %

Compressed condition 2:

The compressed air is 50 degrC

The compressed pressure is 8.5 bar absolute

Then the resulting RH = 42.18% and 0.00000 kg water/kg dry air has condensed.

That is all.

The mass sum of dry air and water in condition 1 is the same as the sum of dry air and water in condition 2

Do not mix up the actual water vapor pressure with the saturated pressure at the specific temperature and pressure.

Have you studied the link I gave you?

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

If you manipulate a mass of air by changing the condition 1 temperature, pressure or RH, then you will get another result for condition 2.

Imagine a compressor.

Intake is condition 1:

The intake temperature = 30 degrC

The intake pressure is 1 bar absolute

The relative humidity = 100 %

Compressed condition 2:

The compressed air is 50 degrC

The compressed pressure is 8.5 bar absolute

Then the resulting RH = 100% and 0.018134 kg water/kg dry air has condensed.

Now the compressor faces a different intake condition 1 and the compressed condition 2 stays the same.

Intake is condition 1:

The intake temperature = 10 degrC

The intake pressure is 1 bar absolute

The relative humidity = 50 %

Compressed condition 2:

The compressed air is 50 degrC

The compressed pressure is 8.5 bar absolute

Then the resulting RH = 42.18% and 0.00000 kg water/kg dry air has condensed.

That is all.

The mass sum of dry air and water in condition 1 is the same as the sum of dry air and water in condition 2

Do not mix up the actual water vapor pressure with the saturated pressure at the specific temperature and pressure.

Have you studied the link I gave you?

Have a nice day

Teus

Sir

I agree with you with above stated example. But as you said , the vapour pressure in condition 2 is the sum of vapour pressure in condition 1 and vpressure at condition 2, my concern is , if we incerase the temperature in condition 1 keeping the condition 2 parameter fix, the vaour pressure in condition 1 & condition 2 increases. But if we incerase the temperaure in condition 2 , which is the displaced air (with elevated temperaure) why the vapour pressure in condition 2 decreases? ■

Originally Posted by guddu

Sir

I agree with you with above stated example. But as you said , the vapour pressure in condition 2 is the sum of vapour pressure in condition 1 and vpressure at condition 2,

Dear kj,

I did NOT say that

What I said is:

The effect you describe is true as long as the RH is 100% and there is still condensed air.

As soon as you increase the temperature where there is no condensed water anymore to vaporize, then the vapor pressure starts decreasing again.

Try in the example the following temperatures for condition 2:

Temperature-------- vapor pressure--------saturated vapor pressure

50.0 degrC-------------- 0.1217 bar --------------0.1217 bar

73.0 degrC-------------- 0.3499 bar --------------0.3499 bar

73.1 degrC-------------- 0.3515 bar --------------0.3515 bar

73.2 degrC-------------- 0.3529 bar --------------0.3529 bar

73.3 degrC-------------- 0.3544 bar --------------0.3544 bar

73.4 degrC-------------- 0.3557 bar -------------- 0.3558 bar

73.5 degrC-------------- 0.3557 bar -------------- 0.3573 bar

73.6 degrC-------------- 0.3556 bar -------------- 0.3588 bar

73.7 degrC-------------- 0.3555 bar -------------- 0.3603 bar

73.8 degrC-------------- 0.3555 bar -------------- 0.3617 bar

73.9 degrC-------------- 0.3554 bar -------------- 0.3632 bar

74.0 degrC-------------- 0.3554 bar -------------- 0.3647 bar

75.0 degrC-------------- 0.3547 bar -------------- 0.3806 bar

80.0 degrC-------------- 0.3509 bar -------------- 0.4673 bar

The effects, you describe, depend on the occurrence of condensation.

Further problems are easier to understand when you refer to an example , shown in an attachment file.

In case you disagree with the results from the calculation, can you provide the accurate result from a different calculation.

This would be very important, as we do not want to make mistakes.

have a nice day

Teus ■