How would you calculate the pressure drop of makeup air passing through a door undercut and through the gap around a door? I have tried modeling it using the ASHRAE DUCT FITTING DATA BASE, but have found for the small airflows and dimensions of the openings, I get an out of range error. Also, what would you suggest as the maximum allowable pressure drop? Is there any code outside of a door fire rating that would limit the undercut size or pressure drop?
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Door Undercut Pressure Drop 1
- Thread starter belz
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Look at thread 403-98883. I asked a similar question on airflows through doors etc and got a lot of useful assistance.
As regards max PD across the door. I don't think there is a recommended value. The PD would be limited by the ability to open the door. You wouldn't want to impede the escapees of a fire. 15 Pa seems a value that can be opened without too much stress.
(Assuming a door for instance 0.8x2m The force on it would be:
F=PxA pressure x area
15N/m2(or Pascals) x 2 x 0.8= 24 Newtons
Convert this to kg Force by F=MA force = mass x acceleration due to gravity (9.81):
Mass = 24N/9.81 = 2.5kG which equates to 5 lB's
I belive that fellow engineers in the Forum have used PD's of up to 25-30Pa without a problem in opening the doors, although I wouldn't like to be doing it on a regular basis.
15Pa is considered adequate in clean rooms for containmant.
In the UK we undercut doors but only when permitted by the nice Building Control Officer. Usually non vision door grilles are fitted with built in fire dampers. Depending on the situation, the dampers can be either intumescent or mechanical. (i.e. if the door is to a plant room, the damper is usually required as intumescent..this is because if the grille is used for fresh air make up, then if a mechanical one closed, there would be no combustion air and a CO problem might occur).
We undercut doors by about 20mm 3/4" so that it maintains fire integrity but gives a reasonable vol flow. I would expect 100l/sec of air through such a gap without any problem.
Again look at the above thread and there are some very useful equations from our colleagues.
Friar Tuck of Sherwood
As regards max PD across the door. I don't think there is a recommended value. The PD would be limited by the ability to open the door. You wouldn't want to impede the escapees of a fire. 15 Pa seems a value that can be opened without too much stress.
(Assuming a door for instance 0.8x2m The force on it would be:
F=PxA pressure x area
15N/m2(or Pascals) x 2 x 0.8= 24 Newtons
Convert this to kg Force by F=MA force = mass x acceleration due to gravity (9.81):
Mass = 24N/9.81 = 2.5kG which equates to 5 lB's
I belive that fellow engineers in the Forum have used PD's of up to 25-30Pa without a problem in opening the doors, although I wouldn't like to be doing it on a regular basis.
15Pa is considered adequate in clean rooms for containmant.
In the UK we undercut doors but only when permitted by the nice Building Control Officer. Usually non vision door grilles are fitted with built in fire dampers. Depending on the situation, the dampers can be either intumescent or mechanical. (i.e. if the door is to a plant room, the damper is usually required as intumescent..this is because if the grille is used for fresh air make up, then if a mechanical one closed, there would be no combustion air and a CO problem might occur).
We undercut doors by about 20mm 3/4" so that it maintains fire integrity but gives a reasonable vol flow. I would expect 100l/sec of air through such a gap without any problem.
Again look at the above thread and there are some very useful equations from our colleagues.
Friar Tuck of Sherwood
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I dug this up from the CIBSE
Q=A Cd (2 DP/rho) sqrt
A=area of opening m2
Cd = 0.61 typically (depends on the sharpness of the edges of the opening. If its more aerodynamic, it will be higher.
DP is the pressure differential (Typically 15Pa)
Rho = 1.305 kg/m3 at atmospheric pressure/NTP
sqrt = square root of all in brackets
Q= m3/sec
so for a gap of 20mm under a 800mm wide door, the flow will be 49l/sec (gives a velocity of 3.04m/sec)
Friar Tuck of Sherwood
Q=A Cd (2 DP/rho) sqrt
A=area of opening m2
Cd = 0.61 typically (depends on the sharpness of the edges of the opening. If its more aerodynamic, it will be higher.
DP is the pressure differential (Typically 15Pa)
Rho = 1.305 kg/m3 at atmospheric pressure/NTP
sqrt = square root of all in brackets
Q= m3/sec
so for a gap of 20mm under a 800mm wide door, the flow will be 49l/sec (gives a velocity of 3.04m/sec)
Friar Tuck of Sherwood
belz, looks like Quark and I were on vacation. There's a thread referenced in that thread where we go over this item pretty painstakenly. Direct link to it is:
If the hyperlink doesn't work it's thread 403-43928.
The principles are in ASHRAE Applications Ch. 51:
Q = 2610 A dP^.5
Although this may be a dead issue(?)
If the hyperlink doesn't work it's thread 403-43928.
The principles are in ASHRAE Applications Ch. 51:
Q = 2610 A dP^.5
Although this may be a dead issue(?)
I have a nomograph that I have been using. It was published in HP&AC in June 1970. I have no idea where I picked it up. If I ever run across a critical application I will check it using some of the info above. On the other hand if it were important I would not be using or relying on the door crack.
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