100% Outside Air

[ChipFuller]

I'm designing an air conditioning system for 100% outside air. The area is a small lab room approximately 950 ft3. My heating loads are 6600 Btuh sensible and 1500 Btuh latent. I trying to design the cfm required. I'm working in a dry climate so I've neglected the latent. Some people suggest that I should:

6600/(1.08*(97-55)) where 97 is the outside air requirements and 55 is the leaving temperature off the coil. This delivers about 152 cfm. The reason I've been given is because that temperature of the air I want to deliver and am making up for the 6600 heat gain. This gives me about 10 Air Changes/Hour

Other people suggest that I should use:

6600/(1.08*(75-55)) where 75 is the temperature inside the room. This delivers about 305 cfm. I will size the coil from this formula: 1.08*(97-55)*305. I have also been told if I use this method that I will freeze up the coil because the air flow is too great.

Which method is right? Can I buy this system as a packaged unit? Also where can I find a psychometric chart for 4500 feet elevation?

Thanks

 

[FCIBSE]

Have you consider Ventilation Displacement system yet?

 

[ChipFuller]

What's a ventilation desplacement system?

 

[chicopee]

your reasoning escapes me. Is the sensible heat of 6600 btuh generated by lab equipment, lighting and the 10 or less air changes and heat transfer thru walls, ceiling and floor?

 

[ChasBean1]

Chip -

Here's how I see this. I wouldn't go by either option you've shown. I would analyze this from load to source, and there is a latent gain at the load. I would therefore use:

Q=4.5*cfm*dH in lieu of Q=1.08*cfm*dT

Total space load is latent + sensible, which is 8,100 BTU/hr. To obtain cooling airflow needed, use the Q=4.5*cfm*dH equation, which solves for total load (sensible and latent) based on enthalpy. With a saturated 55°F supply air stream, and to maintain the 75°F space temperature at 50% relative humidity, the load (room) would require 364 cfm.

364 cfm is the volumetric flow rate that the central AHU would need to cool from 97°F to 55°F. Using Q=1.08*cfm*dT, as you note that this is a dry area, the AHU coil would need to be sized for this reduction in temperature at this flow. This equates to an AHU cooling capacity of 16,500 BTU/hr.

Note that I'm skeptical of the lack of need for the AHU to remove moisture because this is a "dry climate." Dry is a relative term. 97°F and 24.6% humidity would produce saturated 55°F air. If the humidity were higher than 24.6%, the coil would also need to remove moisture.

I'm open to rebuttal on this - haven't gone back to thermo/HVAC books, but this way seems to make sense to me.

As a side note, be careful of sizing DX units. The load conditions stated will be rare so you need a unit that will perform well at partial load.

Best of luck. -CB

 

[quark]

I vote for CB. However if you want to go your way you should consider 75 to 55 deg.F because at no time you want to increase the room temperature above 75 deg.F. If you consider a TD of 95 to 55 for a load of 6600 Btu/Hr then when this heat load adds up from the room the leaving temperature will be higher than 75 deg.F.

You better follow CB's comments.

 

[anandHVAC]

I am sorry, Perhaps not agree with all above explanation.
My answer to CHIPFULLER's question is as follows.
I will start with basics. If you refer to carrier handbook or any other engineering book on Air-conditioning. You will find that the grand total heat, in your case it is 8100btu/hr i.e 0.675 tons, is used to calculate the tonnage requirement.This means that your coil should be capable to give you an output of 0.675. Now the question is how much air is needed to achieve this.This can not be decided with
above two equations.You will have to do the iterations.First assume some fresh air i.e 1 airchange/hr and calculate the dehumidified air quantity requied.Repeat this procedure till you balances your fresh air with dehumidified air.This will give you the quantity of air required to maintain the desired inside conditions.
Answer to your second question regarding pyschometric chart for different elevation.U will note that all the standard psychometric charts are designed at atmospheric pressure of 760mm of Hg.For pressure other than 760mm of Hg, you will have to physically draw it. Procedure on how to draw psychometric chart is available in any air conditioning book. U can refer to that.
I hope all above explanation clarified all your queries. In case of any query pls. let me know.

 

[ChasBean1]

Anand, could you clarify the iteration? What do you get for an answer? Thanks, -CB

 

[briand2]

ChasBean1's response of 20th April 2003 is absolutely perfect. I've sized countless air conditioning systems over the years, and I've never had to carry out some "iteration" to get the cooling requirement.

AnandHVAC: I too would be really interested to see why an iterative process should be necessary on this occasion (I might be experienced, but I never miss the opportunity to learn!!).

Regards,

Brian

 

[coolingunit]

anandHVAC: Your explanation is excellent. I would like to know the iterative procedure of calculating the fresh air requirement.

 

[anandHVAC]

"Iteration" means trial and error method.
During heatload calculation you will have to assume some fresh air quantity to calculte dehudified CFM requirement.
You will have to reiterate this procedure by varying fresh air quantity till it balances with dehumidified CFM.
Unless this both quantity matches you will not satisfy the condition of 100% fresh air.
It is imposible for me to explain the method of calculation via mail. However, u can refer to carrier handbook for
more information.
In case of any queries/cmments do let me know. I will try my best to address it.

 

[cbiber]

I have found several psychrometric charts online for elevations other than sea level. For example Google found this one for me using the search terms "psychrometric chart altitude":

http://www.heatcraftheattransfer.com/pdfs/Normal5000.PDF

It's for 5000 feet, not 4500, but it's close enough to get you started.

Biber Thermal Design

http://www.biberthermal.com

 

[quark]

Dr. Zekyl says, it is always better if you have time to practically design a system. (supplying air quantity to remove heat load and moisture load and incrementing it to reach balance)

Mr. Hyde says, well if you want to increase complexity of a simple solution go ahead. None of carrier's people nor any other engineer objects you.The fundamental is to check the total enthalpy of fresh air and the conditions you wanted at the room. Basically theair exiting the room should have same properties of required condition otherwise room conditions cannot be maintained. Now by the simple calculation given by CB and used widely everywhere, calculate the cfm. This is enough cfm that will carry heat and moisture out of the room.

This is more simpler than recirculation system because you need not further remove the moisture from the exiting air stream. Just you are throwing it away. Still, if you love math iterations, higher order differential equations and probability will always help you. Best of Luck.

(Bah! am I being too dramatic?)

 

[ChasBean1]

This could be a good exercise because we might be talking apples and apples here. Anand, did the reiterative process produce a result?

 

[energy7]

Lets keep it simple, your application is not rocket science

1st, your CFM required is dictated by your total sensible load and the temp you want to maintain in the room at the supplly air temp you plan to provide.

it is simply found as

Rm CFM = Room BTHU QS / 1.08 ( Trm - Tsa )

2nd, beause you specify 100% OSA ( I will not ask why 100% ),
your coils will never see this load.

Coil Capacity will need to be based on the air flow design above

at ( OSA EAT db / wb - SAT - (LAT) db/wb )

good luck

 

[lilliput1]

The cooling coil will need to cool 100%OA to the condition such that when this coil leaving air is discharged into the room. it will pick up heat and moisture & air is then exhausted. The condition of the air exhausted typically matches the room inside conditions. If the cooling coil cools the air to 50 deg F & the room design condition is 72 def F then the air quantity required Rm Load in BTU per hr/(1.1 x (72-50)). Now you must also satisfy the latent loads. The Coil must dehumidify the same quantity of air to the humidity such that when discharged into the room it will absorb the latent load & it's RH will be within the design range. You may have to reheat to maintain the required space humidity.

 

[lilliput1]

Labs typically require 100% outdoor air. Make sure you have enough for exhaust hood makeup. Simplest control is constant volume reheat. You can also have (2) position - occupied/unoccupied or variable volume, all with reheat. Usually humidity is cirtical so humidification is also required. Chilled water system is best. If DX you will find out that when you size the it for the load w/ 100% OA, the unit CFM (typically 350 CFM/ton) would be more than you need to do the room sensible load.What I have done is recirculate back some of the air. This unit should only serve a lab not a group of labs to avoid cross contamination. Part of the fan discharge is ducted direct back to the return. You have to recalculate & do trial & error of the coil entering air condition (mix of OA & recirculated air at coil discharge condition plus fan heat. The AC unit should have hot gas controls. A humidistat in addition to a thermostat should be used to bring on cooling. Final space temperature should be controlled by the reheat coil.

 

[greenaire]

Could you see "the dedicated units" from the news letter of

http://www.trane.com?

 

[ChasBean1]

No I couldn't. Link is to trane.com main page. Checked under "News" and didn't see "dedicated units." Tried a search for "dedicated units" and no luck. Could you be more specific?

 

[tmprider]

Try using an ERV sized for the 100% fresh air load. Locate it in the ceiling space of the lab for ease of maintenance and cleaning. This will most likley decrease the size of the unit that you go with, and will allow you to take advantage of "free cooling". Overall, your energy consumption will be less due to the lower delta T at your coils.

DH