×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Paint Room Dehumidification
2

Paint Room Dehumidification

Paint Room Dehumidification

(OP)
I am trying to determine the load on a system where 100% outside air is required.  I also want to use as small of system as possible.  My thought is to use bypass air to enable the reduction of humidity, which is the important factor, while reducing dry bulb temperature a small amount.  I have Entering air of 95 F DB / 86.5 WB.  I need the mixed air to be at 85 F DB / 72 F WB. The Total Air Volume is 60,000 CFM.   I am trying to determine the following:
% of air to be by passed
Required leaving air DB / WB from the air going thru the coil to yield target mixed air conditions
BTUH load on the coil

It has been a long time since I have done any Psychometric work, so I need some help to solve this problem.

RE: Paint Room Dehumidification

You need to cool 40,000 CFM down to 80*F DB and 64*F WB to get a mixed air of 60,000 CFM and 84.923*F DB and 72.171*F WB

RE: Paint Room Dehumidification

(OP)
Thanks for the help - you have saved me a lot of time.

RE: Paint Room Dehumidification

Imok2,

Just for a clarification. Are you calculating the weighted average for wet bulb as well? This may give you a wrong result. Better method is to take weighted average of enthalpies, rather and check wet bult temperature at the DB and enthalpy.

24500 cfm @95DB and 86.5WB + 35500 cfm @78DB and 57.7WB (treated air) will yield 60000 cfm @84.0DB and 71.5WB, which seems to be close.

RE: Paint Room Dehumidification

seco32 quarks numbers are better then mine because you save about 18,000 btus of cooling and bypass 500 CFM less air
Thanks quirk!! your the best

RE: Paint Room Dehumidification

quark,Point of correction on above my post I recalcuated again and it seems that your calculations and mine are a push but yours is better because of less bypass but you seem to use a bit more refrigeration then mine

RE: Paint Room Dehumidification

imok2,

Mixed air wet bulb corresponding to your data is 72.7F instead of 72.17 (that is the reason I guessed that you might be using weighted average for WBT at mixed condition. Infact, you should do the enthalpy balance for mixed condition and read corresponding WBT at given DBT and enthalpy).

The exact solution is 35400@78, 58.6 and 24600@95 and 86.5. This will yield mixed air condition of 60000@85 and 72.

This difference between the conditions you stated and my final one is about 12 tons (mine being higher, but that corresponds to actual required conditions). This is about 3% of total tonnage and that is why it seems to be insignificant.

I appreciate your efforts to make our posts accurate.
 

RE: Paint Room Dehumidification

OK, I withdraw the comment about weighted average of WBT as that doesn't seem to be the case. However, 72.7F WBT at mixed condition, in your case, is correct (hoping that my psychro calculator is precisesad).

To match 72F WBT, the 40000cfm air is to be treated to
80F DBT and 63F WBT. The Enthalpy difference is (51.1-28.5) = 22.6 btu/lb. For 40000cfm, the heat load is (40000 cu.ft/min*22.6btu/lb*60min/hr)/(13.79cu.ft/lb*12000btu/hr/ton) = 328 tons.

As per my calculation, enthalpy difference is (51.1-25.5) = 25.6 btu/lb.

For 35400 cfm, heat load is (35400cu.ft/min*25.6btu/lb*60min/hr)/(13.68cu.ft/lb*12000btu/hr/ton) = 331.22 tons

The difference of 3 tons can be a calculation round up (of enthalpy values) as tonnage should be same for a set of given inlet and outlet conditions.

Yet, I would go with 40000 cfm treated air and corresponding DBT and WBT as I need not cool the air to lower temperatures if a solution exists at higher temperatures.

Thanks to you, Imok2
 

RE: Paint Room Dehumidification

Thank you Imok and quark for taking the time to help out and crunching numbers for all.

My view is different though - I question the need to use a by-pass altogether, just size you system for the calculated 100% OA at 330-ton or whatever is calculated and that's it. This way, when the system operates in les critical outside conditions you get even better humidity control.


I even question the need for humidification altogether
I have to say, that's a lot of tonnage for a paint room make-up air - can't you confine the paint room into a place that is "unoccupied" so that make-up air is not dehumidifed at all? i.e push/pull ventilation within the paint booth.

RE: Paint Room Dehumidification

cry22,

They are all valid questions and should be analysed before trying to come to a conclusion. Like myself, imok2 might also have presumed that OP has come to a conclusion after looking into various alternatives.

If it is the not the case, he must take cues from your post. He can only tell us why RH is important (paint quality, painting quality, drying time?)

Full Dehumidification -> advantage - higher DP
Disadvantage -> higher investment on dehumidifying equipment (coil and AHU)

Face Bypass System -> advantage - lower investment
Disadvantage -> lower DP and thus higher power consumption by chiller

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources