100% Outside Air
100% Outside Air
(OP)
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
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





RE: 100% Outside Air
RE: 100% Outside Air
RE: 100% Outside Air
RE: 100% Outside Air
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
RE: 100% Outside Air
You better follow CB's comments.
RE: 100% Outside Air
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.
RE: 100% Outside Air
RE: 100% Outside Air
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
RE: 100% Outside Air
RE: 100% Outside Air
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.
RE: 100% Outside Air
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
www.biberthermal.com
RE: 100% Outside Air
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?)
RE: 100% Outside Air
RE: 100% Outside Air
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
RE: 100% Outside Air
RE: 100% Outside Air
RE: 100% Outside Air
RE: 100% Outside Air
RE: 100% Outside Air
DH
RE: 100% Outside Air
First assume a particular outside air quantity and do the load calculations, find out what the dehumidifed air quantity is. If it matches the assumed outside air quantity, then the initial assumption is right. Otherwise, do another load calculation absed on the dehumidified air quantity obtained and keep doing this (iteration !!!) till you get the same figure.
Alternatively, if you have a software like E20-II of Carrier, you can just input all the relevant data and put 100% outside air and the program gives an output which will be 100% outside air.
RE: 100% Outside Air
Here is web address for mentioned article and complete article name. If you still can not get it, give my your email address. Then send it to u by email. My email: naymo@greenaire.com.sg
"http://www.trane.com/commercial/library/archived_newsletters.asp"
Design tips for effective, efficient dedicated ventilation systems by Dennis Stanke, 2001, volume 30-3. PDF or HTML