Most HVAC engineers I see understand that the buildup of pressure in a space is because of the outside air. You supply more air than you bring back, therefore you build up pressure. If a fan supplied air to a box and the same amount of air came back (no outside air), would the box pressure be 0? Assuming it is a large box and the ductwork loss for supply and return is equal. I would think the space would be still positive pressure. If you look at air balance or pump reports they measure the supply from a fan is high positive while the return is a low negative. Is it that the fan pushes air all the way through the system back to the fan, or does the fan blow air into the system and draw it back from the negative pressure on the return side?
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Theoretical Question
- Thread starter Dymalica
- Start date
DonPhillips
Structural
Don't forget exfiltration, doors, windows, toilet room exhaust, etc, all help reduce that differential pressure.
Don Phillips
Don Phillips
When HVAC guys talk about static pressure we are really talking about differential pressure.
Typically we are lazy, and do not fully state what the two pressures we are measuring between are.
Normally, if speaking about building pressure we mean interior pressure with respect to outside.
If speaking about duct pressure we are mean wrt the conditioned space.
In your theoretical closed system, if the question is "what is the differential pressure between the box and outside the box?" the answer is "it depends."
What does it depend on? Where and how large are the pressure drops in the closed circulation system.
With the fan off, all parts of the system are at the same pressure, and it is possible for that pressure to be the same as the pressure outside the system
Let's imagine that the fan can produce a pressure of 1 unit. Since it is closed system, the sum of all the pressure drops in the system must also equal 1.
For simplicity, let's let our imaginary system consist of four segments:
A. Supply duct.
B. Diffuser.
Box
C. Return grill.
D. Return duct.
Let's imagine that A=0.1, B=0.5, C=0.3 and D =0.1. In this case, the pressure in the box wrt outside would be 1-.1-.5=.4.
We can assign any arbitrary value of pressure drop to A, B, C and D, provided that the sum to 1.
With the exception of the limiting and impossible case of A=1, we can see that the pressure in the box must always be positive wrt outside the box.
Typically we are lazy, and do not fully state what the two pressures we are measuring between are.
Normally, if speaking about building pressure we mean interior pressure with respect to outside.
If speaking about duct pressure we are mean wrt the conditioned space.
In your theoretical closed system, if the question is "what is the differential pressure between the box and outside the box?" the answer is "it depends."
What does it depend on? Where and how large are the pressure drops in the closed circulation system.
With the fan off, all parts of the system are at the same pressure, and it is possible for that pressure to be the same as the pressure outside the system
Let's imagine that the fan can produce a pressure of 1 unit. Since it is closed system, the sum of all the pressure drops in the system must also equal 1.
For simplicity, let's let our imaginary system consist of four segments:
A. Supply duct.
B. Diffuser.
Box
C. Return grill.
D. Return duct.
Let's imagine that A=0.1, B=0.5, C=0.3 and D =0.1. In this case, the pressure in the box wrt outside would be 1-.1-.5=.4.
We can assign any arbitrary value of pressure drop to A, B, C and D, provided that the sum to 1.
With the exception of the limiting and impossible case of A=1, we can see that the pressure in the box must always be positive wrt outside the box.
I think that I will have to disagree with MintJulep and agree with imok2 if I understand him correctly.
I am going to use psia instead of " of water in my explanation. I will also assume that the system consists of a box, the fan, the supply duct, and the return duct.
Assume first that the system is at 14.7 psia with everything off. Also assume that the supply duct and the return duct have equal pressure drops of 0.5 psia and the fan will increase the pressure 1.0 psia from suction to discharge. When the fan starts the pressure drop between the box and the fan suction is 0.5 psia so the fan suction is at 14.2 psia. Add 1.0 psia and the fan discharge is 15.2 psia. Then the flow through the discharge duct drops the pressure .5 psia back to 14.7 psia. The box is still the same pressure as the outside.
Another way to look at this is to assume that the box, the duct, and the fan are rigid and sealed. Therefore the volume will not change or V=constant. Assume that the pressure inside is equal to the pressure outside before the fans starts. If you also assume that the fan does not heat the air (or that the heat transfer to the outside equals the amount of heat produced by the fan) then T=constant. Since the system is sealed the mass (or n, number of moles) cannot change. Therefore by the perfect gas law (PV=nRT OR P=nRT/V), P must remain constant since n,R (the perfect gas constant), T, and V are constant.
The pressure drops in the supply and return ducts ALWAYS equals the pressure added by the fan, therefore the pressure in the box does NOT change (dPsuction + dPdischarge = dPfan OR dPfan-dPsuction-dPdischarge=0). If the pressure in the box does not change and it was originally the same pressure as the outside pressure then it remains the same pressure as the outside pressure after the fan starts.
I am going to use psia instead of " of water in my explanation. I will also assume that the system consists of a box, the fan, the supply duct, and the return duct.
Assume first that the system is at 14.7 psia with everything off. Also assume that the supply duct and the return duct have equal pressure drops of 0.5 psia and the fan will increase the pressure 1.0 psia from suction to discharge. When the fan starts the pressure drop between the box and the fan suction is 0.5 psia so the fan suction is at 14.2 psia. Add 1.0 psia and the fan discharge is 15.2 psia. Then the flow through the discharge duct drops the pressure .5 psia back to 14.7 psia. The box is still the same pressure as the outside.
Another way to look at this is to assume that the box, the duct, and the fan are rigid and sealed. Therefore the volume will not change or V=constant. Assume that the pressure inside is equal to the pressure outside before the fans starts. If you also assume that the fan does not heat the air (or that the heat transfer to the outside equals the amount of heat produced by the fan) then T=constant. Since the system is sealed the mass (or n, number of moles) cannot change. Therefore by the perfect gas law (PV=nRT OR P=nRT/V), P must remain constant since n,R (the perfect gas constant), T, and V are constant.
The pressure drops in the supply and return ducts ALWAYS equals the pressure added by the fan, therefore the pressure in the box does NOT change (dPsuction + dPdischarge = dPfan OR dPfan-dPsuction-dPdischarge=0). If the pressure in the box does not change and it was originally the same pressure as the outside pressure then it remains the same pressure as the outside pressure after the fan starts.
Doesn't Bernoulli's Law say that there has to be a pressure differential ALL the way from fan outlet back to the fan inlet, just by virtue of having air flow?
The Central Value Theorem ought to apply here, so that somewhere along that path is a point where the absolute pressure is equal to atm pressure in the system. That would then put the fan outlet at above atm pressure, and the fan inlet at below atm pressure. The point of atm pressure would most likely be in the box, since that has the largest volume and slowest air velocity.
TTFN
FAQ731-376
The Central Value Theorem ought to apply here, so that somewhere along that path is a point where the absolute pressure is equal to atm pressure in the system. That would then put the fan outlet at above atm pressure, and the fan inlet at below atm pressure. The point of atm pressure would most likely be in the box, since that has the largest volume and slowest air velocity.
TTFN
FAQ731-376
AbbyNormal
Mechanical
Yes ge, I would go with n, V,T being constant then so would P
Take the "V" out of HVAC and you are left with a HAC(k) job.
Take the "V" out of HVAC and you are left with a HAC(k) job.
PV=nRT defines the total enthalpy of the system.
Ignoring fan heat, the total enthalpy of a closed system will remain constant.
That does not mean that the enthalpy at each point in the system remains constant. There must be pressure differences from point to point within the system for there to be flow.
However, after thinking further about this on the train, I think I am ready to change my position and say that the static pressure at any point of the closed system wrt outside the system is indeterminate.
Consider:
If we add two holes between the system and the outside, the net flow into or out of the system must be zero. Making no other changes to the system, the holes may be infinitesimally small so that the volume of air passing through the holes will also be infinitesimally small and therefore will not significantly alter the operation of the system
If (assuming constant pressure in the box) both holes are in the box, there will be no flow through either hole. If there is no flow between the box and the outside, then the pressure in the box must be equal to the pressure outside the box.
The pressure at the return must be lower than the pressure in the box. So, if one hole is in the box, and one hole is in the return, air must leak out from the hole in the box, and an equal amount of air must leak in to the hole in the return. If air can leak out of the box, the box must be positive wrt outside.
But, the pressure in the supply must be higher than the pressure in the box. If one hole is in the box, and the other hole is in the supply then air will leak out of the hole in the supply, and an equal amount of air will leak in at the hole in the box. If air can leak into the box, the box must be negative wrt outside.
So here we have a contradiction. The holes, assumed to make no change in the operation of the system, lead us to three mutually exclusive conclusions. The system is indeterminate.
I think it has to do with the cat that is inside the box.
Ignoring fan heat, the total enthalpy of a closed system will remain constant.
That does not mean that the enthalpy at each point in the system remains constant. There must be pressure differences from point to point within the system for there to be flow.
However, after thinking further about this on the train, I think I am ready to change my position and say that the static pressure at any point of the closed system wrt outside the system is indeterminate.
Consider:
If we add two holes between the system and the outside, the net flow into or out of the system must be zero. Making no other changes to the system, the holes may be infinitesimally small so that the volume of air passing through the holes will also be infinitesimally small and therefore will not significantly alter the operation of the system
If (assuming constant pressure in the box) both holes are in the box, there will be no flow through either hole. If there is no flow between the box and the outside, then the pressure in the box must be equal to the pressure outside the box.
The pressure at the return must be lower than the pressure in the box. So, if one hole is in the box, and one hole is in the return, air must leak out from the hole in the box, and an equal amount of air must leak in to the hole in the return. If air can leak out of the box, the box must be positive wrt outside.
But, the pressure in the supply must be higher than the pressure in the box. If one hole is in the box, and the other hole is in the supply then air will leak out of the hole in the supply, and an equal amount of air will leak in at the hole in the box. If air can leak into the box, the box must be negative wrt outside.
So here we have a contradiction. The holes, assumed to make no change in the operation of the system, lead us to three mutually exclusive conclusions. The system is indeterminate.
I think it has to do with the cat that is inside the box.
Those examples are not valid for describing the closed system, since the system is now open to the outside. In the second example, additional mass is added to system. Conservation of mass says that eventually, the added mass MUST cause a sufficient pressure gradient to result in mass being expelled back to the outside. At that NEW equilibrium point, net mass flow must equal zero. In the 2rd example, removal of mass at the supply reduces the total mass of the system, hence the mean pressure of the system is lower than atm pressure, so mass can be reintroduced elsewhere in the system. Again, at the NEW point net mass flow is zero.
The holes, by virtue of introducing additional mass into the system no longer the same as the original system. You can't change the inputs and then claim that the system is unchanged.
In any system, mass flow is driven by a concentration gradient, essentially Fick's Law. So in a closed system, since mass is conserved, a concentration surplus MUST be balanced by a concentration deficit elsewhere in the system. Therefore, if the system, unpumped is at atm pressure, pumping must raise the pressure above atm pressure in one place, and decrease the pressure somewhere else in the system. Since the pump is the motive engine, the greatest pressure gradient exists across that pump, which is in the opposite direction of the pressure gradient in the rest of the system.
TTFN
FAQ731-376
The holes, by virtue of introducing additional mass into the system no longer the same as the original system. You can't change the inputs and then claim that the system is unchanged.
In any system, mass flow is driven by a concentration gradient, essentially Fick's Law. So in a closed system, since mass is conserved, a concentration surplus MUST be balanced by a concentration deficit elsewhere in the system. Therefore, if the system, unpumped is at atm pressure, pumping must raise the pressure above atm pressure in one place, and decrease the pressure somewhere else in the system. Since the pump is the motive engine, the greatest pressure gradient exists across that pump, which is in the opposite direction of the pressure gradient in the rest of the system.
TTFN
FAQ731-376
IRstuff:
Yes, I considered that my little thought experiment causes the system to no longer be a closed system.
As you correctly note, and as I stated, the net mass flow through the holes must equal zero at the new equilibrium steady-state condition. I *believe* that with infinitesimally small holes that the mass flow through each will be so small wrt the total mass in the system as to have no perceivable effect on the system. However I am not *sure* that this is true, because yes, it does change the system, even if that change is small.
I agree totally that there must be a pressure gradient from point-to-point within the system. Furthermore, it is trivial to measure the pressure difference between any two points internal to the system.
The problem arises when you try to measure the pressure at any point inside the system wrt outside. Since the system is isolated, there is no common reference point. In electrical terms, it is a floating ground.
Any attempt to measure the pressure inside wrt outside will introduce a change to the system. Consider a simple "U" tube manometer. The pressure differential (if there is one at the point of measurement - more on this in a few lines) will either draw liquid into the system, or force liquid away from the system. In either case, the volume of the system has changed.
Now, I think we are all in agreement that there must be a pressure gradient within the system. And that sum of the pressures at each point will remain unchanged between at rest and with flow (total enthalpy remains constant).
So there must be some points within the system that are negative wrt outside, some points that are positive, and one point where the pressure is equal to outside. The location of those points is determined by the magnitude of the pressure drops within the system.
It seems obvious that the point of highest relative pressure in the system is at the fan discharge, and that the point of lowest relative pressure is at the fan inlet.
Let's simplify the system a bit. Let's imagine that there is only a single pressure drop in the system with a value equal and opposite to the pressure provided by the fan.
If that pressure drop is coincident with the fan inlet, then the entire system beyond the fan discharge must be positive wrt outside.
If the single pressure drop is coincident with the fan discharge, then then the entire system prior to the fan inlet must be negative wrt outside.
If we place the pressure drop at any other location then the pressure in the system between the fan discharge and the drop will be positive with a value of half of the fan's capacity. The portion of the system between the drop and the fan inlet will be negative with the same value.
The pressure exactly AT the drop will be equal to outside.
Now imagine that we have two drops, each half of the fan's capacity. The region of the system will be at equal pressure to the outside.
However, there is no reason to assume that the two pressure drops must be equal. Let's say that the fan provides 10 pressure units, drop 1 is 3 and drop 2 is 7.
In relative terms, between fan discharge and drop 1 the pressure wrt fan inlet will be 10. The pressure between drop 1 and drop 2 will be 10 - 3 = 7. The pressure between drop 2 and fan inlet will be 7 - 7 = 0. We can play this game with any set of arbitrary numbers.
Notice however that I had to resort to specifying the reference point for the measurement. There is no reason that the fan inlet has to be the arbitrary zero. The math works out with any arbitrary reference point. Let's say that fan discharge is "zero", and work backwards. The fan provides 10, so the fan inlet would be at -10. Between drop2 and drop1: -10 + 7 = -3. And that is dissipated by drop1 so we are back to a "zero" at the fan discharge.
Now notice that by explicitly discretezing the drops I was able to make a system where NO point was necessarily numerically equal to the outside. This is contrary to intuition, which tells us that there must be a point where the pressure inside is equal to the pressure outside.
We get around this apparent contradiction by noting that real pressure drops are not discrete, and that with real flow there is always an associated pressure drop over any length.
So, to get to the point. There has to be a point in the system that is equal to outside. The location of that point may be anywhere, and the location is determined by the magnitude and distribution of the pressure drops throughout the system.
Yes, I considered that my little thought experiment causes the system to no longer be a closed system.
As you correctly note, and as I stated, the net mass flow through the holes must equal zero at the new equilibrium steady-state condition. I *believe* that with infinitesimally small holes that the mass flow through each will be so small wrt the total mass in the system as to have no perceivable effect on the system. However I am not *sure* that this is true, because yes, it does change the system, even if that change is small.
I agree totally that there must be a pressure gradient from point-to-point within the system. Furthermore, it is trivial to measure the pressure difference between any two points internal to the system.
The problem arises when you try to measure the pressure at any point inside the system wrt outside. Since the system is isolated, there is no common reference point. In electrical terms, it is a floating ground.
Any attempt to measure the pressure inside wrt outside will introduce a change to the system. Consider a simple "U" tube manometer. The pressure differential (if there is one at the point of measurement - more on this in a few lines) will either draw liquid into the system, or force liquid away from the system. In either case, the volume of the system has changed.
Now, I think we are all in agreement that there must be a pressure gradient within the system. And that sum of the pressures at each point will remain unchanged between at rest and with flow (total enthalpy remains constant).
So there must be some points within the system that are negative wrt outside, some points that are positive, and one point where the pressure is equal to outside. The location of those points is determined by the magnitude of the pressure drops within the system.
It seems obvious that the point of highest relative pressure in the system is at the fan discharge, and that the point of lowest relative pressure is at the fan inlet.
Let's simplify the system a bit. Let's imagine that there is only a single pressure drop in the system with a value equal and opposite to the pressure provided by the fan.
If that pressure drop is coincident with the fan inlet, then the entire system beyond the fan discharge must be positive wrt outside.
If the single pressure drop is coincident with the fan discharge, then then the entire system prior to the fan inlet must be negative wrt outside.
If we place the pressure drop at any other location then the pressure in the system between the fan discharge and the drop will be positive with a value of half of the fan's capacity. The portion of the system between the drop and the fan inlet will be negative with the same value.
The pressure exactly AT the drop will be equal to outside.
Now imagine that we have two drops, each half of the fan's capacity. The region of the system will be at equal pressure to the outside.
However, there is no reason to assume that the two pressure drops must be equal. Let's say that the fan provides 10 pressure units, drop 1 is 3 and drop 2 is 7.
In relative terms, between fan discharge and drop 1 the pressure wrt fan inlet will be 10. The pressure between drop 1 and drop 2 will be 10 - 3 = 7. The pressure between drop 2 and fan inlet will be 7 - 7 = 0. We can play this game with any set of arbitrary numbers.
Notice however that I had to resort to specifying the reference point for the measurement. There is no reason that the fan inlet has to be the arbitrary zero. The math works out with any arbitrary reference point. Let's say that fan discharge is "zero", and work backwards. The fan provides 10, so the fan inlet would be at -10. Between drop2 and drop1: -10 + 7 = -3. And that is dissipated by drop1 so we are back to a "zero" at the fan discharge.
Now notice that by explicitly discretezing the drops I was able to make a system where NO point was necessarily numerically equal to the outside. This is contrary to intuition, which tells us that there must be a point where the pressure inside is equal to the pressure outside.
We get around this apparent contradiction by noting that real pressure drops are not discrete, and that with real flow there is always an associated pressure drop over any length.
So, to get to the point. There has to be a point in the system that is equal to outside. The location of that point may be anywhere, and the location is determined by the magnitude and distribution of the pressure drops throughout the system.
The fan, as far as a mathematical exercise, can be treated as a discontinuity, so there's not necessarily a "point" therein which is equal to the outside.
However, the central limit theorem constrains the remainder of the system to have such a point. Since the outlet is above and the inlet is below atm pressure, somewhere along the path is a point that is equal to atm pressure, most likely in the box, but not necessarily. Bernoulli's equation incorporates the continuity equation, and there is no reason for the remainder of the system to have discontinuities, so there must be a point that traverses between above and below atm pressure, and the continuity equations guarantees that the pressure is continuous across that traversal, so the point exists in the system.
From a physics perspective, assuming that we're not cooking the system, there's no need to have an arbitrary reference. P = NRT/V and Bernoulli's equation allows one to absolutely determine the pressure in the system at each point, except at the discontinuity that is the fan.
You cannot have pressure everywhere in the system be higher or lower than atm pressure, for that would imply either a temperature or mass change to the system as a whole. If the temperature is unchanged, then positive pressure everywhere must mean higher air density everywhere, which means more mass entered the system. Since no mass enters the system, the resultant NRT/V must equal atm pressure.
TTFN
FAQ731-376
However, the central limit theorem constrains the remainder of the system to have such a point. Since the outlet is above and the inlet is below atm pressure, somewhere along the path is a point that is equal to atm pressure, most likely in the box, but not necessarily. Bernoulli's equation incorporates the continuity equation, and there is no reason for the remainder of the system to have discontinuities, so there must be a point that traverses between above and below atm pressure, and the continuity equations guarantees that the pressure is continuous across that traversal, so the point exists in the system.
From a physics perspective, assuming that we're not cooking the system, there's no need to have an arbitrary reference. P = NRT/V and Bernoulli's equation allows one to absolutely determine the pressure in the system at each point, except at the discontinuity that is the fan.
You cannot have pressure everywhere in the system be higher or lower than atm pressure, for that would imply either a temperature or mass change to the system as a whole. If the temperature is unchanged, then positive pressure everywhere must mean higher air density everywhere, which means more mass entered the system. Since no mass enters the system, the resultant NRT/V must equal atm pressure.
TTFN
FAQ731-376
Ok, I'll go with that.
So for a real system:
We know that P will vary from point to point.
Closed system, so n is constant.
R is a constant.
Therefore T must change from point to point.
Absolute pressure may be determined by measurement of temperature.
So for a real system:
We know that P will vary from point to point.
Closed system, so n is constant.
R is a constant.
Therefore T must change from point to point.
Absolute pressure may be determined by measurement of temperature.
Well looking at this from a mechanics point of view and going by the fact that no outside influences are involved, and the distriubtion system is equally balanced then I felt that if I was putting say 500 CFM into the box and I was removing 500 CFM from the box then the pressure in the box would be "0". The temperature from any source would have to be negligable.
After still more consideration (train ride home):
The absolute pressure at any point in the system may be determined by measuring the temperature of a known number of moles occupying a known sub-volume around the point of interest.
I'm not sure that instrumentation to do that exists.
The absolute pressure at any point in the system may be determined by measuring the temperature of a known number of moles occupying a known sub-volume around the point of interest.
I'm not sure that instrumentation to do that exists.
Well, n is constant for the system, but not necessarily for each point in the system. I would think there's a bit of both T and n increasing and decreasing together along the path. The fan provides mechanical compression, so there ought to be a change in n as well as T from the compression.
The box would have to have the lowest pressure gradient, since it's bigger than the ducts attached to it. However, since the flow in must travel across the box to the outlet duct, there has to be a pressure gradient, albeit small, and possibly too low to measure accurately. But, it HAS to be there to get flow through the box. Since the box has the least amount of compression, it's certainly most likely to close to atm pressure, but again, the inlet of the box will be a bit higher, and the outlet of the box will be a bit lower.
TTFN
FAQ731-376
The box would have to have the lowest pressure gradient, since it's bigger than the ducts attached to it. However, since the flow in must travel across the box to the outlet duct, there has to be a pressure gradient, albeit small, and possibly too low to measure accurately. But, it HAS to be there to get flow through the box. Since the box has the least amount of compression, it's certainly most likely to close to atm pressure, but again, the inlet of the box will be a bit higher, and the outlet of the box will be a bit lower.
TTFN
FAQ731-376
The box need not be at atmospheric pressure though the inflow and outflow rates are same, in a dynamic condition. If we fully open the supply damper to the box and totally close the return damper then air flows into the room and the room pressure equals the fan static less the pressure drop in the duct work. At equilibrium condition, the inflow and outflow both will be zero, yet you have pressure in the box.
Now, if we open the return air damper to allow 100 cfm, there will be an inflow of 100 cfm and the box pressure will be the static pressure of the fan less the drop in supply and return duct works, corresponding to 100 cfm flowrate. Likewise, we can have any pressure set up in the box.
Perfect atmospheric pressure can be maintained only when the fan is off but not when the supply and return dampers are fully open. This is because, there should be some pressure drop for the air to flow from one point of the room to the other point, however less it may be.
All the above conditions are true for a totally leak proof system. In the presence of leakage outside the system, the same quantity (as that of input) of air can't be returned to the fan. That is why fresh air is being added to the system.
For atmospheric and poistive pressure rooms, room exit can be considered as a starting point and then deducting duct losses in the return system, one can get at what point the pressure becomes atmospheric and the negative.
For a negative pressure room, the exercise should start right from the fan discharge.
Now, if we open the return air damper to allow 100 cfm, there will be an inflow of 100 cfm and the box pressure will be the static pressure of the fan less the drop in supply and return duct works, corresponding to 100 cfm flowrate. Likewise, we can have any pressure set up in the box.
Perfect atmospheric pressure can be maintained only when the fan is off but not when the supply and return dampers are fully open. This is because, there should be some pressure drop for the air to flow from one point of the room to the other point, however less it may be.
All the above conditions are true for a totally leak proof system. In the presence of leakage outside the system, the same quantity (as that of input) of air can't be returned to the fan. That is why fresh air is being added to the system.
For atmospheric and poistive pressure rooms, room exit can be considered as a starting point and then deducting duct losses in the return system, one can get at what point the pressure becomes atmospheric and the negative.
For a negative pressure room, the exercise should start right from the fan discharge.
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