Room Pressurisation Continued! 1

[JohnnyHS]

thread403-43928

Hi Guys

I had just found this thread looking for room pressurisation calcs - v useful. Thanks! In the petrochem industry we not only need a pressurised "room" (for safety aspects so nasty/explosive gases do not enter) but also air changes as well (to stop them staying there if they do or if generated in the house).

I had an application for a house 4m x 2.5m x 3m with a 50 PA pressure requirement and 12 air changes per hour. We need to use effective "leakages" to ensure this air change happens. this is where vents come into it.

From the calcs i have worked out that with a 2 x 200mm x 200mm vents i get 12.74(!)air changes whilst maintaining a 50PA overpressure if the HVAC unit supplies 106 litres per second.

I have generated an excel sheet to help me calc this in the future. if any of y'all would like this then let me know

Cheers!

 

[Zesti]

Reading your post I do get interested about your calculation-sheet.

Can you post it somewhere for download?

On the other hand, do you really need a spreadsheet to do the multiplication 4m x 2.5m x 3m x 12 = 360 m3/h = 100 litres per second?

Supply 360 m3/h to the room and regulate the return/exhaust airflow so that room pressure is at 50Pa.

So therefore I'm interested to see your sheet.

 

[Dymalica]

So does anyone know the formula for room pressure in in w.c. as related to CFM in and out of a room? I read that entire thread and it did not make sense. The formula q=2610 gives some huge number. Thanks

 

[JohnnyHS]

Hi guys,

please find the xl doc attached. yes Zesti you are right it is a pretty simple equation for that calc. I like the sheet because it allows me to play with the vent sizes and HVAC supplier air flow requirements so i can possibly reduce the vent sizes to reduce the duty (therefore cost of the HVAC)

Yellow is the user to enter the data and red is the formula results..

in the eg attached i have worked out that i need 106 l/s to have the required pressure (50PA) for the vents i have. as my HVAC (i have already) delivers 135 l/s i will have a safety margin (13PA!) for any other leaks to keep it above 50PA (for info i do have some doors in the house but have a 3 minute delay on the low pressure alarm)

Please do let me know your comments and/or amendments you think may be useful.

Dymalica i am not sure what you are asking. have a look at the sheet that might help. Just put the PA into a converter to get the Inches H20 any q's let me know.



Johnny H-S BSc(Hons) MInstMC

Engineers like to solve problems. If there are no problems handily available, they will create their own problems.

 

[Zesti]

There is an error on line 17:

You multiply by 0.5 but this should be "to the power 0.5" or "square root".

I did not quite get the point of the vent size that you have in your sheet but I do now.

However, in order to get a 50Pa pressure difference over 2 holes each sized 200mm x 200mm will require quite a bit more airflow than 360 m3/h... as you will see when you change the equation in cell N17.

Besides, you don't specify any resistance factor for the vents.

dP = zeta x 1/2 x rho x v^2

rho = density of the air (kg/m3)
v = airspeed (m/s)
zeta = resistance factor for the opening (dimensionless)
dP = pressureloss over the opening (Pa)

The square root function that should be in cell N17 comes from the "^2" term in the equation above.

As for the value of zeta...
(somewhere around 1 usually)

 

[Zesti]

Dymalica wrote:

"So does anyone know the formula for room pressure in in w.c. as related to CFM in and out of a room?"

Well, there is no formula to describe the relation you mention.

Lets say you supply 1000 m3/h to a room and that the extraction is 600 m3/h.

The difference of 400 m3/h would leave the room through all leaks and openings in the room. The driving force for this flow would be the pressure in the room being higher than the surroundings.

If the pressure is higher, the flow through the leaks will be higher and the extraction flow will be lower.

If the pressure is lower, the flow through the leaks will be lower and the extraction flow will have to be higher.

If cracks/leaks are narrow they have more resistance, if there is a large opening in the wall somewhere then resistance is low and the flow at a given pressure difference will be high.

There is no way to know what the leak rate through all the cracks, openings and leaks will be in de the design stage.
You will only know when the construction is complete.

People who think that they can calculate their way through the problem should think again about the actual proces/physics involved and what the practical implications are.

The only test that can be done is the blowerdoor test that has been mentioned in other threads and like I said before, this again can only be done when construction is complete.

Reading these threads I get the feeling that there is a lot of misunderstanding out there about what is involved in creating a pressure difference in a room...

And on a more general note: it is my experience that many people in HVAC tend to want to make EXACT calculations on subjects where each and every input for the calculation is a guess, a flawed measurement, a value inherent to some shoddy construction made by the building contractor etc etc etc.

Believe me, you can not make EXACT calculations on many things that are involved in HVAC. Then all you can do is make calculations that satisfy you, as an engineer, that you can be comfortable with your design and that the end result will be satisfactory to the customer.

And to digress even further from the original topic: maybe there should be more room in education for "making educated guesses, based on rough calculations, figuring out which issues are important and which are not" instead of every school-problem ending in a neat and tidy answer accurate to the third decimal place...

(Dymalica : of course all of this is not meant to you personally)

Sorry if this ended up in a bit of a rant

 

[ChrisConley]

I was involved with a project, years ago, where a client wanted to keep their building pressurized to prevent infiltration of corrosive salts from a nearby mine.

Much calculation (to two decimal places) was done and hundreds of thousands of CFM of air were used to pressurize the building.

At the end of the day the largest problem they had wasn't the approximate crack length, or wind pressure, but the fact that the operators leave the 30' overhead doors open all summer to cool the building. With a nortwest wind at 25 mi/hr and a 5 x 30' door, I didn't even start the calculation on what would be required to prevent infiltration.

 

[ChasBean1]

JohnnyHS, back to your original post and the spreadsheet (good catch by Zesti on the .5 being multiplied vs. to the .5 power) - it's important to note that the leakage you now calculate (really, estimate) is 1007 cfm (475 lps). This value is the difference between supply and exhaust. If you have only supply, it's fine to base your air exchange rate on that value, but in most circumstances you'd have both mechanical supply and exhaust, so you'd want to base your air exchange rate on the higher of those values, not on the calculated leakage rate.

Just thought I would add that note... CB

 

[Zesti]

By consequence the same error also appears in cell F28.

Here "/0.5" should be "^2" ("squared").

 

[Zesti]

JohnnyHS, can you explain what you mean by "vent"?

Is this just a hole in the wall, in this case sized 200mm x 200mm ?

If you vent is just a square hole in the wall then I am quite sure you will not get 50Pa inside the room, not even at 475 l/s as ChasBean correctly calculated using the corrected equation.

Half of 475 l/s through a hole 200x200 gives a speed of about 5.9 m/s

Using dP = zeta x 1/2 x rho x v^2 we get

dP = zeta x 1/2 x 1.2 x 5.9^2 = zeta x 21 Pa

If dP is supposed to be 50 Pa then zeta must be around 2.5 which is certainly not the case for just an opening.

So, my question is: what is contained in the number "2610" in the equation?

Is this just a conversion factor for all those weird imperial units or is there more to/in it?

 

[ChasBean1]

Interesting point. You're right - they don't cross check with each other. The equation is from ASHRAE Applications 52.5 (in 2007 ver.). They start with Q = 776 C A (2 dP / rho)^.5 and assume flow. coeff. C is .6-.7 (.65 avg) and density is .075 lbm/ft3 and draw that out.

I have to read more into the 776 (they don't appear to derive it in this chapter).

Without too much additional homework now, I'd guess that there are assumed relative roughnesses of crack openings with net higher friction loss, versus 2 big square openings the poster mentions. That's my guess now, but I'd like to look at it some more...

CB

 

[JohnnyHS]

Hi Guys,

Zesti, Vent in this case means a weighted louvre so you can weight the vents to create more of a back pressure in the room.

http://www.waterloo.co.uk/products_list.php?ID=6

Thanks for the points regarding the formula errors! i'll change that and come back to y'all. The HV unit is installed soon so i'll have actual results too rather than this theoretical stuff! as we've heard before "it should work in theory!!"

To be honest with ref to the 2610 Zesti i got that from the previous forum

http://www.eng-tips.com/viewthread.cfm?qid=43928

Johnny H-S BSc(Hons) MInstMC

Engineers like to solve problems. If there are no problems handily available, they will create their own problems.

 

[Zesti]

I can not find any "weighted louvres" among the Waterloo products.

Why would you want such a louvre anyway? I would think a louvre with damper should be alright: supply a fixed flow to the room and adjust the damper on the exhaust vent so that room pressure is at the desired level.

Can you explain to me why you chose a "weighted louvre"?

 

[JohnnyHS]

its down to the dreaded terminology Zesti. you are quite correct the louvres i have in the house can be weighted simply by adding or removing washers on the flaps to dampen if required

http://www.waterloo.co.uk/dpdfs/YG.pdf

the calcs all came about with needing the pressurisation AND the air changes in the house. In the figure on the link in this post (page 2) the air velocity is on the left which is what the HVAC unit is supplying. drawing the line through the 50PA (what i need) gives me the vent sizes required.

The calc sheet was designed to allow me to play quickly with differing vent sizes, pressure requirements and flow inlet rates...


Johnny H-S BSc(Hons) MInstMC

Engineers like to solve problems. If there are no problems handily available, they will create their own problems.

 

[AbbyNormal]

perhaps call it a barometric damper, or a back draft damper with counterweights on it

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[Michael09]

The 'weighted louvres' you are looking for are called in UK air pressure stabilisers.

If you type on the net you will see some USA based manufacturers.

I personally used them for pressurised staircased or operating theatres in UK.

 

[Waramanga]

Air pressure stabilisers? Are they calibrated to give a certain constant pressure drop? Anyone have a link to info?

 

[Zesti]

A quick search found these:

http://www.rdvent.com

http://www.aercon-pul.com

http://www.apreco.co.uk

All three are in the UK and the main markets for these pressure stabilisers seem to be hospitals and clean rooms.

I have never seen these kind of pressure stabilisers over here in The Netherlands.

 

[mauricestoker]

The application is fairly common for economizers, VAV, and pressure relief dampers in general. If you look up the Greenheck or Ruskin sites, it may be what you are looking for. Depending on the application, the actuation will differ, hence how it is titled. I use them mainly on differential static and rarely on temperature.
The application, fixed volume and minimum required ACH while maintaining differential pressure, is typical for an ABSL-3+ laboratory by the Dept of Ag standards. I've used a 120V actuator and static setpoint set at balance (trial and error) to go with VFD ramp speed to provide a failsafe. With low draw on exhaust ampereage, or trial and error setpoint on dP, the pressure relief damper diverts supply air to exterior of secondary containment with exhuast failure, providing some relative dP while exhaust (interior neutral, exterior high positive).

 

[SimonWatkins]

Morning All

On the original post, once corrected the spreadsheet is a useful one but I can't help but feel it is a little over-complicated as on first glance, it converts your metric values to imperial, does the calc, and the converts back.

We seem all to be working in SI units here so the formula to use for pressurisation is as follows (normalised for average atmospheric conditions):

Q = 0.83 x A x P^(1/N)

Where:

Q = airflow rate <m3/s>
A = leakage area (not room area) <m>
P = pressure difference <Pa>
N = 2 for large openings, approx 1.6 for cracks, etc. When N=2 is used, it is essential square root P.

This is the standard formula used as a basis in BS 5588 - stairway pressurisation and is supported by ASHRAE Applications (1995 SI version) chapter 48.

Weighted louvres, etc are more commonly known as barometric dampers and are generally a very reliable means of achieving the internal pressurise levels specified. If you want 50Pa, they are configured to open above that pressure to release excessive air supply to atmosphere.
This is particularly important in stairway pressurisation applications when there is not only a requirement to pressurise a room but also to achieve a given velocity of air across a doorway when it is opened. If you needed say 2 m/s across an open door, this would require an airflow of around 4.5 - 5 m3/s which is far in excess of the airflow required to pressurise the room. The airflow difference will be exhausted through the barometric damper until the door is opened where the room pressure will drop and the damper will close.

Hope the above is of some use

Regards