Infiltration Cooling Load?

[Rankin755]

Hi all, I have a school project and I need to calculate the overall cooling load of a given office building. I already calculated the exterior and interior cooling loads, however when I calculate the infiltration load I find q to be well over 100000 Btu/hr. and that cant be right.

So here is what I have. 0.5ACH from 8am to 6pm, and 0.2 ACH the rest of the time. The site is Denver Colorado in July. I have been using the formula


q=4840(cfm)(deltaW). From the Psychrometric charts I found the indoor W=0.0122, and the outdoor W=0.029. so the difference or delta W = 0.0168. And I get a value for unoccupied + occupied = 6830.21Btu/hr.

my total cooling load was calculated to be 36015.09Btu/hr. This factors in people, appliances, radiation, and heat transfer through the walls and roof.

I understand there is alot more you need to know before you can give me the exact answer, however a ball park estimate would work.

First of all does this sound relatively close? and is there some other factor I am overlooking??


any help would be appreciated

Rankin755

UW-Milwaukee
Mechanical Engineering

[Abby Normal]

For starters Denver is high altitude and in hot weather the ambient dew point is low. Under a worst case design condition infiltration or deliberate ventilation will dry your building out.

A worst case dehimidification design at the 0.4% level for Denver still has a dewpoint of only 60.2F and that just barely exceeds the definition of neutral air.

Usually with commercial buildings the ventilation will pressurize or depressurize a building to the point that it over powers infiltration.

So you would handle the fresh air as follows. Determine your fresh air ventilation rate and your required supply air rate.

The return air is the difference of the two

Work out the properties of the mixture of the return air rate and the fresh air rate, this is the air entering your cooling coil.

Determine the required cooling capcity of coil.

[Abby Normal]

because denver is high altitude, you cannot use standard air equations

you will need a pyschrometric chart for the altitude

[Rankin755]

To find the humidity ratio I used psychrometric chart #4, used for high altitudes. I found the average temperature @ .4% to be 93 deg. Is this what you are referring to?

Thanks for the help

[Abby Normal]

.4% cooling design would be somewhere around the 93 db and perhaps a 60 something wet bulb

The amount of mositure in that air is going to be a lot lower than would be found inside the room. You would only be worried about the sensible heat, any infiltrating air or ventialtion air will have a LOWER humidity ratio than what you want to maintain indoors, so it is going to dry your building out. Delta W is going to be NEGATIVE, the delta W formula is a latent heat- moitsure removed formula, you may have it corrected for the altitude, at sea level the constant is 4760


0.4% dehumidification level would have that air at a 60.2 dew point and a coincidental dry bulb of about 68F. As on a worst case day the delta W here would be only slightly positive