Positive Pressure

I may have missed the point here, but, why use two air handling units: one for each room?

Why not just use one fan in a duct between the two rooms? Use the fan manufacturers data for zero air flow against power to set the speed of the fan.

If you still need air flow in and out of each room individually you can still use a fan in a duct between the two rooms, and, control its power by using a differential pressure gauge between the two rooms.

Sorry if this is just being stupid.

Hello,

I want to ask something: can you explain me this "Use the fan manufacturers data for zero air flow against power to set the speed of the fan' ?

Thank you

See answer to #20

Just put the AHU in room # 2. and let it blow into room #1.

Do your own homework.

CR4 is not a homework cheat site; however, if you have questions about understanding concepts or how a portion of a problem is derived, these types of questions will be accepted.

I don't want that somebody else to make my homework, but I don't know which is the answer to this 2 questions:

-which is the relation between flow and this diferential pressure (20-30 Pa) ?

- how can i calculate how much should be the flows for Room1 (positive pressure room) and Room 2 (negative pressure) ?

Ask the instructor for help with the formulas.

Where does the air have to go?

If you put the inlet of one airhandling unit into room one and the suction of the second unit into room two and do NOT connect the rooms, room one will be slightly over pressured than room two.

Beware forces on your plastic sheet as you might blow it dependend on size and pressure differential.

What is this all about?

Remove the plastic...................

The text book doesn't allow it.

There is just one AHU for both rooms.

And I still have the same questions:

- which is the relation between flow and this diferential pressure (20-30 Pa) ?

- how can i calculate how much should be the flows for Room1 (positive pressure room) and Room 2 (negative pressure) ?

Thank you

The one with the outlet will have the higher pressure! But there is something missing in your question.

I think somebody left the door open .

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wait!

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Ups sorry forgot to say what I wanted . . .

3.1468 x KVA -RPM of the fan X 25 degrees C, multiplied by the square meterage of room No 1 (with positive pressure) all divided by room No 2 (volume) at 22 degrees C with no sunlight.

3.1468 x 0.67 - 2500 x 25 x R1m / R2v-3C = -10Pa @ 25C But you must allow 9.7% deviation on the final answer for the plastic sheeting if it is PVC.

Hope this helps.

Thank you for the answer. But can you explin it ? And '...R2v-3C' = room 2 volume - 3x 22 degree ?

Flow is a function of differential pressure and the sizes and discharge coefficients of the apertures between the two rooms. The situation as described is too complex to lend itself to calculation.

Thank You for the answer.

But how are calculated the flows and the pressures for rooms which had to be in positive (or negative) pressure ? Which is the rule ? What are doing the HVAC engineers ?

That should do it!

for exemple at 0.6 cfm the pressure is 1.1- 1.2 (on the red line) :

What should I do to obtain the results (positive pressure in room 1) ?

at 0.2 is 2; at 0.4 is 1.5- that means that the flow in room 2 is 0.4cfm with a pressure of 1.5 ? and in room 1 is a lower flow but the pressure is higher=2 ???

Can You explain me please ?

If the fan is in a duct directly between the two rooms, then, once you reach stability: there will be no air flow; there will only be one differential pressure, and, the red line fan above running at full power will give you just over 2 mm water.

If both rooms are leaky or there is flow through the plastic sheet then there will be air flow (through the fan) and you will need to measure the differential pressure in order to control a more powerful fan.

Well, in the end, the air flow of room 1 will be the same as the air flow of room 2, except that in the opposite direction; this is if you use only one fan to control both pressures.

You'll need a speed controller fed back by a barometer with linear output.

Now you do the numbers of that. Me too numb.

Hi AP#1,

You have not provided enough information to give sufficient guidance. Maintaining a pressure differential between two adjacent rooms depends on what happens in each room. For example, does all the air from the AHU only pass between the rooms, through the plastic sheet, or is some of the air returned from the room via other return air ducting? Is the plastic sheet fixed at the edges to form a screen, or does it simply hang loose to form a "curtain"? What is the expected pressure in room 2 relative to the rooms around it or relative to the outside pressure? What is the average air changes per hour for each room? In any case, the required flow across (or through) the plastic screen necessary to create the specified pressure differential is highly dependent on how the screen is fixed. Large openings or gaps require greater flows, smaller gaps require less flow. So we need more info in order to offer further assistance. 20 to 30 Pa is fairly typical for pressure differentials between adjacent rooms, but you have to be certain that this is the actual specification.

Hello,

- there will be a duct supply for room 1 and another duct supply for room 2 (both ducts commected to the same AHU)

-the plastic sheet is fixed at the edges to form a screen

- the only request is this 'pressure in room 1'>pressure in room 2 (there isn't any specific request)

Still not clear. Are you suggesting that both rooms get supply from the ducts originating from one AHU? Do you want to direct the airflow as such that room one has that little bit more pressure?

A drawing could help to understand what you try to achieve.

Also I suggest you do some basic research.

Here is one link to try: http://en.wikipedia.org/wiki/Cleanroom#Air_flow_principles

yes, both rooms get supply from the ducts originating from one AHU (is just one AHU for both rooms)

i need to make positive pressure in room1 (pressure in room 1>pressure in room 2), and i ask how can I do that ? usually in your work how you handle this ?

I can't really see how this is going to work. The pressure behind the AHU will be the same for both rooms regardless of volume directed into each room.

If it was my project I would make a duct to draw air in room 2 put the suction of the AHU into same room and push it into room 1.

But not knowing what this all is about (clean room, production, air quality requirements) and given the wall is not air tight there is nothing to go on about.

Good luck!

Hi AP.

IdeaSmith is correct - a drawing will help. The air entering the rooms will have to go somewhere. The air entering room 1 might all flow into room 2, but where does the air entering room 2 go to? Does it return to the AHU, is it exhausted via some other means, does it simply leak out through doors or other openings? We need to have the full picture. To create the pressure differential one can add a damper in the supply duct feeding room 2 to increase the pressure drop, or design the flow into room 1 to be higher than that into room 2. Your original question was about determining the flow to create the pressure drop but that is difficult to answer without understanding the big picture.