Date: Wed, 10 Dec 2008 16:10:18 -0500
Reply-To: List Moderator <ecgrants**At_Symbol_Here**UVM.EDU>
Sender: DCHAS-L Discussion List <DCHAS-L**At_Symbol_Here**LIST.UVM.EDU>
From: List Moderator <ecgrants**At_Symbol_Here**UVM.EDU>
Subject: 5 RE: [DCHAS-L] laboratory ventilation question

From: "Norwood, Brad" 
Date: Wed, 10 Dec 2008 15:28:18 -0500
Subject: RE: [DCHAS-L] laboratory ventilation question

Russ,

My back-of-the-envelope calculation gives me 9000 cubic feet in the  
room (assuming 9' ceiling)

For a 6' hood with 3' opening **At_Symbol_Here** 100 ft/min linear velocity, I get  
approximately 900 cu. Ft./ min. (not atypical for a hood).

That would give you six air changes in an hour (so my calculation  
isn't far off from theirs).

That being said, the air brought in to the room to make up for what  
the hoo= d takes out typically comes in AT the hood.  This will do NO  
good for the students at the other end of the room.  In fact, IMO this  
does no good for any students not working IN the hood.

So, again in my opinion, this is NOT adequate for the protection of  
the students or the faculty.

Brad

Bradley K. Norwood, PhD
Laboratory Director
Arista Laboratories
1941 Reymet Road
Richmond, VA  23237
(804) 271-5572 ext. 307
(804) 641-4641 (cell)
brad.norwood**At_Symbol_Here**aristalabs.com

==

From: "Bell,Martin" 

Russ,

To determine the number of air changes you divide the total room  
exhaust volume (cfm) by the room volume. This will give you the air  
changes per minute. Convert that to air changes per hour by  
multiplying by 60. For example let's use you numbers and some  
assumption to determine the number of air changes per hour in this lab:

Room volume - 50'x20'x10' = 10,000 cf - assuming the ceiling height is  
10 feet.  Room Exhaust Volume - I assuming the hood is the only  
exhaust in the room so if you have a hood that is 6' with an 1.5' (18  
inches) opening and a face velocity of 100 feet per minute -  
6'x1.5'x100 fpm = 900 cfm

Air change per minute = room exhaust volume/room volume = 900/10000 =  
= 0.09 air changes/minute x 60 = 5.4 air changes per hour.

Martin W. Bell, CHMM

Drexel University
Department of Environmental Health and Safety 
Environmental Health and Safety Manager

Telephone   : 215-895-5892
Cell Number: 215-778-4278
Fax Number: 215-895-5926

==

Subject: RE: [DCHAS-L] laboratory ventilation question
Date: Wed, 10 Dec 2008 15:38:07 -0500
From: "Wawzyniecki Jr, Stefan" 

Russ-  Here's  a quick calculation:

Assuming 18" sash opening (1.5')

6' X  1.5' X 100 FPM  =   900 CFM

Assuming 10' ceiling

50' X 20' X 10' = 10,000 ft^3

Dividing 10,000 ft^3  by 900 cfm = 11.1 min,  the time it would take for
one air change.

....or, roughly 5 ACH

So, theoretically, that's how they calculated it.  BUT-  The best mgmt
practice would be to provide local exhaust.

Someone else can poke holes in my math skills or in my reasoning.

-Stefan Wawzyniecki

==

From: "Debbie M. Decker" 
Date: December 10, 2008 3:43:49 PM EST (CA)
To: chas list 
Subject: RE: [DCHAS-L] laboratory ventilation question

The University of California Laboratory Design Guide (http://ucih.ucdavis.edu/docs/labdesignguideSept2007.pdf 
) recommends at least 6 air changes per hour or 1 cmf per square foot  
of (gross) lab space.  This recommendation is based on ANSI Z9.5 and  
long years of arguing with design engineers over acceptable  
ventilation in a lab space .

For 1000 sq ft lab, the fume hood would need to exhaust at least 1000  
CFM.  With a 6 foot fume hood at 100 fpm with a wide open sash, you  
might be close.  But it's a lazy engineering solution, IMHO.  What if  
the noise from the fume hood makes it difficult for the instructor to  
teach?  You said that the hood is in the corner of the room.  There  
will not be good mixing of make up air nor will there be good sweep of  
contaminated air from the entire space.  Contaminated air will pool  
and eddy, conditioned supply air won't be sufficiently mixed and hot/ 
cold spots will develop - occupant comfort will be compromised and  
occupant safety may be compromised as well.

Relying on a fume hood for all of the ventilation into and out of a  
space is risky and a really poor engineering solution.  At the very  
least, there should be a minimum of 4-6 ach all the time out of the  
lab, through a general laboratory exhaust.  Operating a fume hood at a  
full open sash height and at a (relatively) high face velocity wastes  
energy and doesn't necessarily increase safety.



Debbie M. Decker, Campus Chemical Safety Officer
Environmental Health and Safety
University of California, Davis
1 Shields Ave.
Davis, CA  95616
(530)754-7964/(530)752-4527 (FAX)
dmdecker**At_Symbol_Here**ucdavis.edu
Co-Conspirator to Make the World A
Better Place -- Visit www.HeroicStories.com and join the conspiracy

==
From: "Lazarski, Peter M." 
Date: December 10, 2008 3:52:22 PM EST (CA)
Subject: RE: [DCHAS-L] laboratory ventilation question

Looking at ANSI/ASHRAE 62-1989, Ventilation for Acceptable Indoor Air  
Quality. Table 2 shows for a laboratory with an estimated maximum  
occupancy of 30/1000 sq. ft., 20 cfm/person as a requirement for  
outdoor air ventilation. If these criteria are met, indoor air quality  
can be considered acceptable.

I would think local exhaust ventilation in the form of 'snorkels'  
would have been more appropriate for this area. Out of curiosity, was  
this space designed specifically to be a dissection laboratory or was  
the use decided after the building plans were finalized and well under  
construction?

Two suggestion: 1st - calculate the room height to see whether the 5  
air changes/hr meets the total occupants x 20 cf/m requirement. 2nd -  
set off a smoke candle in the hood to determine whether its working at  
all.

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