Hello all,
One of the requirements for EQ Credit 5 is to prove that the areas such as
the janitor's closet need to be negatively pressurized to a minimum
of 5Pa when doors are closed and 1Pa when doors are open. Generally, in
mechanical drawings, the cfm values are mentioned for exhaust fans.
I wanted some insight as to how this pressure difference information can be
obtained for both cases. Is there a straightforward calculation?
Thanks,
Deepa
Personal opinion:
. 5 Pa is difficult to (accurately) measure
. 1 Pa is impossible outside of a lab
I don't know what they were thinking when making these requirements, since
they're not too clear, but I suspect that if you demonstrate .5cfm / sq.ft.
and no recirculation you'll be OK.
James V. Dirkes II, P.E., LEED AP
Obtaining a negative pressure difference also requires sealing the space.
Door seals, and sealing walls. If you exhaust 0.5 cfm/sf from a room with
chicken wire walls, you won't achieve any kind of measurable pressure
difference.
I usually steer clients away from this credit for this very reason
(difficulty coupled with measurability)
--
Karen
I've documented many EQc5s and have never been asked to prove anything.
For what it's worth, back when I used Contamw for stair pressurization
calcs, even a modest airflow of about 200 CFM per 3 foot door created
pressure differences of 0.1-0.15 in. That's on the order of 25-30 PA.
So as long as you're in the range of 50 CFM per door, you'll be OK
(assuming the door undercut and crack are the only openings into the
room).
This is a major generalization, and there are a lot of parameters that
would need to be included to accurately calculate, but it helps me sleep
at night knowing 50 CFM / door usually meets this credit.
And Karen, I don't follow why you would steer clients away from this
credit. Putting full height walls around janitor's closets and print
shops and negatively exhausting them is a good thing, right?
James Hansen, PE, LEED AP
We have an exhaust rate of 1.31cfm/sf (or 75cfm per door) which exceeds the
minimum requirement greatly. According to what you have
mentioned, this should be good as far as creating a negative pressure.
Just to know, are there any equations to calculate this or is it best
determined by physical measurement?
Deepa
Hi Deepa,
My two cents.
For your question, you need to know:
1. the air volume in room and its pressure
2. To achieve 5Pa, how much more air should be introduced -
Higher pressure x air volume in room/atmospheric pressure
Now you get the number for the static condition... ;)
to maintain the pressure in a leaky room you have to pump in more air,
right?
3. from the ASHRAE Handbook of Fundamentals:
Q = 2610 x A x (DP)1/2
? Q is the room leakage flow rate in cfm,
? 2610 is a conversion factor,
? A is the net open crack area of the room (generally not measurable), and
? DP is the differential pressure.
4. If you'd like to maintain certain pressure in the room, you need to
supply 2+3 first. When the pressure is established in dynamic condition,
you need to supply only 3 to maintain the pressure you want.
This is only the theoretical calculation for your reference.
Cheers,
Cheney
Thanks Cheney !
Cheney, can you post the chapter in the ASHRAE Handbook of Fundamentals where
I can find the equation
Johannes
Von: bldg-sim-bounces at lists.onebuilding.org
[mailto:bldg-sim-bounces at lists.onebuilding.org] Im Auftrag von Cheney
Gesendet: Mittwoch, 6. Oktober 2010 22:43
An: creations.deepa at gmail.com
Cc: Bldg-Sim
Betreff: Re: [Bldg-sim] Negative pressure for EQ cr 5
Hi Deepa,
My two cents.
For your question, you need to know:
1. the air volume in room and its pressure
2. To achieve 5Pa, how much more air should be introduced -
Higher pressure x air volume in room/atmospheric pressure
Now you get the number for the static condition... ;)
to maintain the pressure in a leaky room you have to pump in more air, right?
3. from the ASHRAE Handbook of Fundamentals:
Q = 2610 x A x (DP)1/2
* Q is the room leakage flow rate in cfm,
* 2610 is a conversion factor,
* A is the net open crack area of the room (generally not measurable), and
* DP is the differential pressure.
4. If you'd like to maintain certain pressure in the room, you need to supply
2+3 first. When the pressure is established in dynamic condition, you need to
supply only 3 to maintain the pressure you want.
This is only the theoretical calculation for your reference.
Cheers,
Cheney
Hi Johannes,
I am sorry for my typo error. The equation can be found in ASHRAE Handbook
- HVAC Applications. My version is 2003, page 52.5. It is actually under
Fire an Smoke Management.
Cheers,
Cheney