Negative pressure for EQ cr 5

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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

deepa chandrashekaran's picture
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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

James V. Dirkes II  P.E.'s picture
Joined: 2011-10-02
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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

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Joined: 2011-09-30
Reputation: 200

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

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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

deepa chandrashekaran's picture
Joined: 2011-10-02
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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

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Thanks Cheney !

deepa chandrashekaran's picture
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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

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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

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