Dehumidifier

[Josh2008]

If the outside maximum temperature is 104F / 84.5% rh and I want to reduce the relative humidity to 50% via a dehumidifier, what indoor temperature would I use?

This is for an electrical room that produces about 100 kW (hrly avg) heat load.

Is there a requirement for ACH? Once I know that I can calculate the CFM to use in my latent heat calculation.

I'm told the electrical equipment is liquid cooled which transfers heat to fan fins, which transfers by convection via the air flow. Does this dictate the CFM, or is it primarily the ACH?

Do I take sensible heat into account for the dehumidifier? I believe I do not since sensible heat accounts for heating and cooling loads to change temperatures, which I'm not concerned about.

Please provide as much information as possible. I've been searching through Chapters 28 and 29 of ASHRAE 2001.

Thanks!

 

[AbbyNormal]

104F and 84% RH would mean a dew point of 98F, this would be a world record

The way we build has a far greater impact on our comfort, energy consumption and IAQ, than any HVAC system we install

 

[walz]

Your are asking too many questions...

here i list my replies as below.

1. A dehumidifier could either be a cooling coil, a heater or a hygroscopic material.

2. To dehumdify the outside air, cooling coil is used as a dehumidifier. But as abbynormal says, your outside condition is also beyond my psychrometric chart.

3. The room CFM depends on the room sensible load ( including the sensible load from the electric equipment, lighting or walls etc.), the supply air temperature and the room condition Dry Bulb temperature.

4. No ACH is not the driving factor for CFM

5. No there is no ACH requirement for electrical room air conditioning.

5. Normaly electric rooms should be kept not more than 75 F.

6. As far as taking into account the sensible load for dehumidifer is concerned, you need to tell waht kind of system are you using. Is it one system mixing the fresh air with the room air and cooling from the cooling coil / dehumidifier, or you are going to use a dedicated treated fresh air unit. You need to explain the whole system.

One thing is for sure is that you need to run the load calcuations first. Identify how much room load and fresh air load you have and then identify the kind of system that you'll be going to use. Any kind of load conditions can be beated but you need to know the load and the system

 

[ChasBean1]

Heat incoming air to about 119°F to get to 50% RH.

 

[cjw81]

Heat to 119F and then what? Replace the equipment?
Your sensible load from the equipment:
1kw = 3412 btu x 100kw = 341,200 btu. sensible load.

 

[Josh2008]

Let me start over. Hopefully this is more realistic.

Outdoor condition 89F DBT and 83%rh
Required indoor condition 80F and 50%rh

Can a dehumidifier both cool from 89F to 80F and reduce the rh from 83% to 50%??

If so then here is my procedure...
I have calculated with the temperature difference that via conduction (walls, roof, floor, doors, etc) q=11373 btu/hr and electric equip 93 kW (317,600 btu/hr).

Total Sens Load = 11,373+317,600= 328,973 Btu/h

Using qs = 1.08 x cfm x deltaT I can adjust the dehumidifier temperature to calculate cfm

So... using 65F incoming air

cfm = qs/1.08xdelta T = 328,973 / 1.08*(80-65)=20,306 cfm

Now if this is correct so far where do I stand?

I've found a two methods of which I dont know if either are correct.

http://www.pdhonline.org/courses/m199/m199content.pdf

Page 7 (but i think this is for air conditioning)

http://www.pdhonline.org/courses/m227/m227content.pdf

Page 16

Any thoughts are appreciated.

 

[walz]

Ok this is very simple thing... as you gave everything and hence solved the problem.

Now first thing is that you dont have to dehumidify anything as major of the room air can be recirculated. As Ashrae standard 62.1-2007 specifies a minimum ventialtion of 0.06 CFM/ft2 for electric equipment room. Since i dont know the area so i base my analysis on 200 CFM as outside air.

Now your room condition is 80F @ 50% RH. the best off-coil temperature for you should be straight horizontal line starting from mixed air condition towards a RH of neighbhourhood of 92-94%. which is 61/60 (DB/WB)

Now adding an avarage of heat of fan to which will raise the temperature of air by 2F to make it 63. Now 63 should be the supply temperature rather than 65. (PLease note that you dont have room latent load). which means that your supply air condition (63 DB) and room condition (80F) should lie on almost the straight horizontal line.

Now based on delta T 80-63=17. the supply calculates to be somewhere near 18,000 CFM.

Now this can take care of your room load.

As far as outside air dehumidification is concerned, you dont have to worry at all because very little outside air 200 CFM is mixed with this room air, which creates no humidity problem. Infact you dont need enough of outside air. 200 CFM which i have considered is enough for 200/0.06 = 3333 ft2 area of your electric room. If your room area is lesser than this, even than your Fresh air sufficient enough to meet the standard requirement.

Please find attached the file which contains the psychrometric chart as well as the process calculations report

 

[Josh2008]

Thanks Walz

So I can't say I understand everything here but I have picked up a few things.

Originally I was told the design would be about 20% OA / 80% RA. I totally understand the concept that the more return air you send back into the room, the less work the dehumidifier has to do conditioning it to room requirements. From this I would assume that using the applicable regulations (in this case ASHRAE 62.1-2007, electrical room 0.06 cfm/ft2), I would spec out the minimum ventilation necessary per the regs for a number of reasons, including power savings. So in my case the room is 20mx55m = 11842 ft2 x 0.06 cfm/ft2 = 711 cfm.

Then from my ~18,000cfm range, 711cfm would put me approximately 711/(18711)=4% OA / 96%RA

Is off-coil temperature just the condenser coil temp required to condition the air per the requirements?

I thought that the coil temperature would be a horizontal line from the point of air temperature leaving the dehumidifier all the way to the 100% rh?

My last question here would be what about taking into account latent heat due to infiltration & permeation and intermitten openings (doors, garage door)? The latter of the two links I posted before touches on this...
How do I incorporate the calculated lb/hr moisture?

"If your room area is lesser than this, even than your Fresh air sufficient enough to meet the standard requirement." Please elaborate.

Much Appreciated!! Its nice to gain knowledge from others, thank you.

 

[walz]

Okies... Now here you got the New Ventilation CFM

Let me tell you my figures...

Supply Air (SA) = 17915 CFM
Outside Air (OA) = 715 CFM
Return Air (RA) = 17915-715 = 17200 CFM

The off-coil temperature is the temperature of air leaving the coil. Its is normally is the range of 90-95% RH. (not 100%)

Now because of OA = 715 CFM. This helps you to have a positively pressurized room. Instead of being your room infiltrated, it would be rather exfiltrated because of excess outside air.

What i meant by fresh air meeting the requirement is that if you size your fresh air considering a bigger room whereas the room is smaller, you'll get more fresh air than required. No need to worry, its meeting the standard requirement of minimum.

Coil Process wont be a straight line, because it has to remove a little latent heat (due to mixing of room air with little of fresh air). Since your room has primarily the sensible load only, its the room process line which would rather be horizontal.

please find the new chart as well as the process report