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Calculate Differential pressure
- Thread starter juan2025
- Start date
dlm1919
Mechanical
- Sep 11, 2002
- 1
My experience says no, there is no way to accurately pre-determine the differential pressure of a space based on supply and exhaust flows. The reason for this is that all rooms are "leaky", meaning some air leaks out of door jambs, window sills, wall joints, ceiling tile, etc. Also, the reference point for checking the diffential pressure probably fluctuates.
But there are still some basic rules-of-thumb to follow:
1. First, choose a stable reference point to measure your static pressure to. If the reference point is outside, try adding a dampner to it and try to have it equalize for wind loads by tying-in multiple spots around the building. If the reference spot is indoors, keep it away from doors or other spots where the pressure can change easily.
2. Second, if you want a space slightly positive to another space, make sure the supply air flow is greater than the air removed from the space (this includes return and exhaust air). My company usually designs at 0.5 cfm/sq ft of extra supply air to give about 0.04" wg static pressure. Though this is variable based on the "tightness" of the room shell.
3. If you want a space that is slightly negative, make sure the supply air is slightly less than air removed from the space.
4. Ultimately a field balance will set the extra outside air flow or exhaust air flow to obtain the differential pressure your trying to obtain. In order to maintain the pressure you will need an automatic control to measure the pressure and adjust the supply air or exhaust.
5. Finally, items to consider with static pressure are: Will the exit doors still open when the room is positive or negative to adjacent space? Will the room pressure cause moist air to enter wall cavities and cause ice or mold problems?
Good luck.
But there are still some basic rules-of-thumb to follow:
1. First, choose a stable reference point to measure your static pressure to. If the reference point is outside, try adding a dampner to it and try to have it equalize for wind loads by tying-in multiple spots around the building. If the reference spot is indoors, keep it away from doors or other spots where the pressure can change easily.
2. Second, if you want a space slightly positive to another space, make sure the supply air flow is greater than the air removed from the space (this includes return and exhaust air). My company usually designs at 0.5 cfm/sq ft of extra supply air to give about 0.04" wg static pressure. Though this is variable based on the "tightness" of the room shell.
3. If you want a space that is slightly negative, make sure the supply air is slightly less than air removed from the space.
4. Ultimately a field balance will set the extra outside air flow or exhaust air flow to obtain the differential pressure your trying to obtain. In order to maintain the pressure you will need an automatic control to measure the pressure and adjust the supply air or exhaust.
5. Finally, items to consider with static pressure are: Will the exit doors still open when the room is positive or negative to adjacent space? Will the room pressure cause moist air to enter wall cavities and cause ice or mold problems?
Good luck.
Use the ASHRAE orifice equation to estimate:
Q = 2610 * A * dP^.5
Q is differential air flow (supply/exhaust, units of cfm)
A is net open area in ft2
dP is pressure difference across boundary, in inches w.c.
Often the net open area is unknown. Establish a common-sense differential supply/exhaust flow, adjust the net open area as needed by caulking, sealing, gaskets, and door sweeps.
Quark, your formula does not account for net open room area which is a principle variable in this application. Also, flow changes with the square root of pressure change - the relationship is not linear...
Best regards, -CB
Q = 2610 * A * dP^.5
Q is differential air flow (supply/exhaust, units of cfm)
A is net open area in ft2
dP is pressure difference across boundary, in inches w.c.
Often the net open area is unknown. Establish a common-sense differential supply/exhaust flow, adjust the net open area as needed by caulking, sealing, gaskets, and door sweeps.
Quark, your formula does not account for net open room area which is a principle variable in this application. Also, flow changes with the square root of pressure change - the relationship is not linear...
Best regards, -CB
CB!
That formula gives the minimum amount of supply air required more than exhaust air. You have to add air leakage through openings to it as per the formula you mentioned above. (and as for duct work if the flow is constant so the differential pressure)
The formula is given by Sauters who are specialised in contamination control systems for vaccine manufacturing. (upon thinking I came to know that it is the same simple formula that I used to use for air receiver fill method)
This formula quantifies to some extent at design level and higher redundancy generally practised in the HVAC design will help you further![[wink]](/data/assets/smilies/wink.gif "[wink] [wink]")
.
dlm1919!
Does your formula 0.5cfm/sq.ft for 0.04"static apply for a room of height 8', 10' and 12' also? I strongly feel no.
Regards,
That formula gives the minimum amount of supply air required more than exhaust air. You have to add air leakage through openings to it as per the formula you mentioned above. (and as for duct work if the flow is constant so the differential pressure)
The formula is given by Sauters who are specialised in contamination control systems for vaccine manufacturing. (upon thinking I came to know that it is the same simple formula that I used to use for air receiver fill method)
This formula quantifies to some extent at design level and higher redundancy generally practised in the HVAC design will help you further
.dlm1919!
Does your formula 0.5cfm/sq.ft for 0.04"static apply for a room of height 8', 10' and 12' also? I strongly feel no.
Regards,
Room height is not applicable. Neither is volume. Whether you supply a net 100 cfm greater amount than the air exhausted from an office or the Boeing factory out in Everett, WA, your space pressure is a function of the net open crack area.
Assuming (to simplify, for airflow inward only) the office were completely leak free, the space would reach equilibrium with the fan discharge static pressure. The same would occur in a hypothetically leak-free Boeing plant, although it would take longer to reach that point as such a large volume of air would need to be compressed to the equilibrium fan outlet pressure.
Sauters' equation as given above can't be true in any application I've seen unless the room volume is not in cubic feet. I think there might be an assumption built into it that elimintes it from applying to this application. By this equation, a typical 20 ft x 20 ft room (10 foot ceiling) with 600 cfm in and 300 cfm out would calculate to 7.4 x 10^-7 Pa (or 7,600 Pa if the bracket was in the wrong place). Neither of these are anything close to typical or can really be arrived at based on a room volume without known net open crack area.
Assuming (to simplify, for airflow inward only) the office were completely leak free, the space would reach equilibrium with the fan discharge static pressure. The same would occur in a hypothetically leak-free Boeing plant, although it would take longer to reach that point as such a large volume of air would need to be compressed to the equilibrium fan outlet pressure.
Sauters' equation as given above can't be true in any application I've seen unless the room volume is not in cubic feet. I think there might be an assumption built into it that elimintes it from applying to this application. By this equation, a typical 20 ft x 20 ft room (10 foot ceiling) with 600 cfm in and 300 cfm out would calculate to 7.4 x 10^-7 Pa (or 7,600 Pa if the bracket was in the wrong place). Neither of these are anything close to typical or can really be arrived at based on a room volume without known net open crack area.
CB!
Here are some concerns from my side. For a leak proof room (including no exhaust or return) the room pressure equals the fan discharge static. I totally agree with you. But your comment "although it would take longer to reach that" indicates a parameters which is not constant during that period.For a smaller room it takes less time and for a bigger room it takes more time to get filled up to the maximum pressure. (i.e. fan discharge static ideally)
That means you have to pump up more air into the room if it is bigger.
So if you keep constant volume input then you have to pump for longer time for bigger rooms. Or if you increase flow rate accordingly then you can match both the times. That is why it is room volume dependant.
Once you maintain the pressure in the room then you have to take care of leakage only.
Note: Your further discussion will open up my thoughts if I am wrong.
Regards,
Here are some concerns from my side. For a leak proof room (including no exhaust or return) the room pressure equals the fan discharge static. I totally agree with you. But your comment "although it would take longer to reach that" indicates a parameters which is not constant during that period.For a smaller room it takes less time and for a bigger room it takes more time to get filled up to the maximum pressure. (i.e. fan discharge static ideally)
That means you have to pump up more air into the room if it is bigger.
So if you keep constant volume input then you have to pump for longer time for bigger rooms. Or if you increase flow rate accordingly then you can match both the times. That is why it is room volume dependant.
Once you maintain the pressure in the room then you have to take care of leakage only.
Note: Your further discussion will open up my thoughts if I am wrong.
Regards,
Quark,
If we are considering the time it takes to adjust to changes in room pressurization, then yes, the room volume is a relavent variable. But the original problem only looked at the steady-state condition of maintaining a certain room pressure by supply/exhaust differential. This problem is not volume dependent.
---KenRad
If we are considering the time it takes to adjust to changes in room pressurization, then yes, the room volume is a relavent variable. But the original problem only looked at the steady-state condition of maintaining a certain room pressure by supply/exhaust differential. This problem is not volume dependent.
---KenRad
CB and Kenrad!
Here I put my point in simpler way. Suppose there is xcfm leakage from various openings (cracks, doors etc.) and you are supplying only x cfm into the room, you can never be able to maintain the pressure. You have to supply more air than x cfm. (ie return cfm+x+some extra cfm)
Once the pressure is built up and fan gets adjusted to the room pressure then you need not supply the extra cfm and the system will automatically adjust to it.
This is what I am thinking so far. Anything wrong in it? Let me know.
Regards,
Here I put my point in simpler way. Suppose there is xcfm leakage from various openings (cracks, doors etc.) and you are supplying only x cfm into the room, you can never be able to maintain the pressure. You have to supply more air than x cfm. (ie return cfm+x+some extra cfm)
Once the pressure is built up and fan gets adjusted to the room pressure then you need not supply the extra cfm and the system will automatically adjust to it.
This is what I am thinking so far. Anything wrong in it? Let me know.
Regards,
Quark,
Let’s look at the simple case of a room with a single fan bringing outside air into the room to maintain a positive room pressure (ie, no recirculation of air). When the system is first started up, the flow of this fan would be relatively high, because the room starts out in a non-pressurized condition -- there is less resistance to flow into the room, so the fan is operating low on its curve. And since the d/p across the walls starts out low, there is relatively little flowing out of the boundary cracks. So at the start, the air flow into the room is higher than the flow out of the room, and the room pressure begins to climb. The rising room pressure then causes the fan flow to drop and the crack flow to rise, until the room reaches a steady state condition where the two flows are equal, and the room pressure is constant. This pattern would hold true regardless of the room volume, although the time to reach steady state would be volume dependant.
The formula that CB is using does not take into consideration the time it takes to reach steady state, and for most purposes, it doesn't matter much. I believe that the formula you are using doesn’t either, and that the volume parameter is just a way of estimating the “net open area” or crack area that CB is talking about.
---KenRad
Let’s look at the simple case of a room with a single fan bringing outside air into the room to maintain a positive room pressure (ie, no recirculation of air). When the system is first started up, the flow of this fan would be relatively high, because the room starts out in a non-pressurized condition -- there is less resistance to flow into the room, so the fan is operating low on its curve. And since the d/p across the walls starts out low, there is relatively little flowing out of the boundary cracks. So at the start, the air flow into the room is higher than the flow out of the room, and the room pressure begins to climb. The rising room pressure then causes the fan flow to drop and the crack flow to rise, until the room reaches a steady state condition where the two flows are equal, and the room pressure is constant. This pattern would hold true regardless of the room volume, although the time to reach steady state would be volume dependant.
The formula that CB is using does not take into consideration the time it takes to reach steady state, and for most purposes, it doesn't matter much. I believe that the formula you are using doesn’t either, and that the volume parameter is just a way of estimating the “net open area” or crack area that CB is talking about.
---KenRad
CB!
It is a great reminder (f in = f out). But one parameter that is blocking my thought is for a room of 1 Cu.M volume(mathematical), if filled with air of 1 cu.m it's pressure is atmospheric. If you want to rise the pressure we have to increase the volume of air inside. The higher the pressure more is the volume of air required.
Note: Your second statement, somehow, enhanced my confusion. Curiosity did kill the cat. :-(
It is a great reminder (f in = f out). But one parameter that is blocking my thought is for a room of 1 Cu.M volume(mathematical), if filled with air of 1 cu.m it's pressure is atmospheric. If you want to rise the pressure we have to increase the volume of air inside. The higher the pressure more is the volume of air required.
Note: Your second statement, somehow, enhanced my confusion. Curiosity did kill the cat. :-(
Absolutely. But your comments refer to tank technology and miss the relation of room air pressure with respect to steady-state flow in and pressure building from friction at cracks due to leakage.
"If you want to raise the pressure we have to increase the volume of air inside" is not true. The room volume (and air volume within the room) remains constant. Think back to Bernoulli's - All terms cancel except for pressure 1 is pressure 2 plus energy from friction losses. Physical volume remains constant, flow volume in (cfm) = flow volume out (cfm), but room pressure builds from friction of air passing through the open crack area.
I totally see your point about room volume, but it holds true with a static environment and not the one in this problem.
"If you want to raise the pressure we have to increase the volume of air inside" is not true. The room volume (and air volume within the room) remains constant. Think back to Bernoulli's - All terms cancel except for pressure 1 is pressure 2 plus energy from friction losses. Physical volume remains constant, flow volume in (cfm) = flow volume out (cfm), but room pressure builds from friction of air passing through the open crack area.
I totally see your point about room volume, but it holds true with a static environment and not the one in this problem.
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