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#11




SCFM vs. CFM, also air flow/pressure across a regulator
On Tue, 23 Dec 2003 09:41:46 0800, Grant Erwin wrote:
I am an electrical engineer and so I view things from that perspective. I considered the issue of the regulator and essentially my analysis agrees with Gary's 100%. I believe 2 things to be true: 1. Mass is conserved (what goes into the regulator must come out) 2. To first order, energy is conserved through the regulator That's why I don't see why flow x pressure wouldn't be constant across a regulator in steady state. Postulate a big air tank pressurized to 180 psi, with a long (long enough so the air has time to cool to ambient) pipe to an ideal regulator which regulates the pressure down to 90 psi. The regulator's output is a pipe of the same size which is connected to a constant load. The cfm going into the regulator is measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator at 90 psi? 10 CFM of course. As you note, mass is conserved, or as Kirchhoff's laws tell us, current is everywhere the same in a series mesh. There is nowhere else for the air to go. If it has a certain mass flow into the valve, it has to have exactly the same mass flow out of it. The thing people seem to be having trouble grasping is that CFM is a measure of mass flow. This is pretty obvious for an incompressible liquid like water, but for a gas, CFM has to be stated in terms of a standard temperature and pressure. That's been defined by the standards bodies to be 68 F and 1 atmosphere. Gary 
#12




SCFM vs. CFM, also air flow/pressure across a regulator
In article , Grant Erwin says...
Postulate a big air tank pressurized to 180 psi, with a long (long enough so the air has time to cool to ambient) pipe to an ideal regulator which regulates the pressure down to 90 psi. The regulator's output is a pipe of the same size which is connected to a constant load. The cfm going into the regulator is measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator at 90 psi? Don't do the problem in cubic feet, simply convert into number of molecules (and for simplicity, perform it with nitrogen) that you calculate using teh universal gas law. So many molecules of nitrogen enter the regulator, the same number leave it. If the temperature at the inlet and outlet are the same, then the volume will scale inversely like the pressure. P(1) X V(1) = P(2) X V(2) Pressure goes down by a factor of two, the volume will increase by two. The pressures btw should be absolute, not gage, not a problem to do in psig if the pressures are high. Jim ================================================== please reply to: JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com ================================================== 
#13




SCFM vs. CFM, also air flow/pressure across a regulator
In article , Gary Coffman says...
10 CFM of course. As you note, mass is conserved, But there is twice the mass flow in the upstream (hp) side in teh example. In one minute, 10 cu feet of air at 180 psi flows in. If you count the number of atoms (mass) you will find that the same number come out the downstream side at 90 psi. They just take up more room at a lower pressure. I would say that if 10 cubic feet of atoms at 180 psi flow in, then 20 cubic feet of atoms will flow out, at 90 psi. PV = PV, universal gas law and all. PV = nKT n is not varying, same number of atoms. KT is constant if you allow the gas to come to thermal equibrium at the outlet of the regulator. So P1/P2 = V2/V1 and all. Jim Jim ================================================== please reply to: JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com ================================================== 
#14




SCFM vs. CFM, also air flow/pressure across a regulator
Gary Coffman wrote: Postulate a big air tank pressurized to 180 psi, with a long (long enough so the air has time to cool to ambient) pipe to an ideal regulator which regulates the pressure down to 90 psi. The regulator's output is a pipe of the same size which is connected to a constant load. The cfm going into the regulator is measured to be 10 cfm @ 180 psi. What cfm will come out of the regulator at 90 psi? 10 CFM of course. As you note, mass is conserved, or as Kirchhoff's laws tell us, current is everywhere the same in a series mesh. There is nowhere else for the air to go. If it has a certain mass flow into the valve, it has to have exactly the same mass flow out of it. This doesn't sound right to me. Let's think about the amount of air molecules that go into the regulator during one minute. It's the number of molecules in ten cubic feet at some temperature at 180 psi (which isn't an absolute pressure to be sure). Now that many molecules have to come out the other side in that minute, right? (Kirchoff and all that.) The gas law is PV = nRT. If we call the input side 1 and the output side 2, then we can write P1V1 = nRT. Since the number of molecules, n, is the same, and the temperature is the same, and since R is a constant, then P2V2 = nRT. So once again I do not see why P1V1 shouldn't equal P2V2. I can (finally!) see why you can't plug in gage pressure into this equation. The absolute pressure is (I believe) gage pressure plus 14.7 psi. Therefore I predict the answer V2 = (180+14.7)*10/(90+14.7) = 18.6 cfm as long as the temperature on both sides is equal. Boy, I wish I had 2 flowmeters. Grant Erwin 
#15




SCFM vs. CFM, also air flow/pressure across a regulator
In article , Grant Erwin says...
Boy, I wish I had 2 flowmeters. You don't need them! You understand the physics behind it instead, which is better. Jim ================================================== please reply to: JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com ================================================== 
#16




SCFM vs. CFM, also air flow/pressure across a regulator
"jim rozen" wrote in message
... The regulator is a feedback controlled valve. ... it does not get perceptibly hot. Indeed because expansion is occuring inside the regulator, under certain conditions, they may start to chill preceptably. So where does the heat go? I'm going to guess that compressive systems such as this act as heat pumps, thus all the heat goes back to the compressor, while cooling occurs at the tools and regulator. Eh? Tim  "That's for the courts to decide."  Homer Simpson Website @ http://webpages.charter.net/dawill/tmoranwms 
#17




SCFM vs. CFM, also air flow/pressure across a regulator
Gary Coffman writes:
Mathematically, this has the same appearance as a dissipative resistance, but there is *no dissipation*. Energy not used is simply retained in the tank. Impossible. We agree that mass flow (CFM) is conserved (equal) on either side of the regulator. We know: Power = CFM * pressure. The regulator imposes: pressure (out) pressure (in). Thus, since CFM is equal on both sides of the regulator, but pressure decreases, there is a loss of power in the output compared to the input. This ends up as heat as a direct consequence of the restriction that creates turbulence and lowers the pressure. This waste heat is mostly added to the output flow, even though the output may be at a cooler temperature due to expansion. Some of the power lost is hissing noise that radiates away and also eventually becomes heat. None of this is "visible" to the source, so it is not "simply retained in the tank". Look, if you want an air analog to an electrical transformer, you would have to have an air motor (not a diaphragm regulator) doing the pressure reduction, with the motor output used to regeneratively compress more free air. That way, power would be conserved, like a transformer, subject to some mechanical inefficiency. 
#18




SCFM vs. CFM, also air flow/pressure across a regulator
OK. CFM does not equal mass flow. CFM means simply cubic feet of air per
minute. The amount of mass in that air is proportional to both the pressure and the temperature, as described by the gas law PV = nRT. I agree there are secondorder losses in a regulator. It might make some noise, and it might warm up a little or cool down a little. This is also true of a transformer. You know, what's *really* amazing is that I know Gary Coffman and Richard Kinch and I are all intelligent, good writers, and well educated, and yet we get THREE different answers. I'd just love to get to the point of agreement. So, Richard, I DISAGREE that cfm is equal on either side of the regulator. It cannot be because mass is conserved! (See my previous posts) I think it *is* like a transformer. Grant Erwin Richard J Kinch wrote: Gary Coffman writes: Mathematically, this has the same appearance as a dissipative resistance, but there is *no dissipation*. Energy not used is simply retained in the tank. Impossible. We agree that mass flow (CFM) is conserved (equal) on either side of the regulator. We know: Power = CFM * pressure. The regulator imposes: pressure (out) pressure (in). Thus, since CFM is equal on both sides of the regulator, but pressure decreases, there is a loss of power in the output compared to the input. This ends up as heat as a direct consequence of the restriction that creates turbulence and lowers the pressure. This waste heat is mostly added to the output flow, even though the output may be at a cooler temperature due to expansion. Some of the power lost is hissing noise that radiates away and also eventually becomes heat. None of this is "visible" to the source, so it is not "simply retained in the tank". Look, if you want an air analog to an electrical transformer, you would have to have an air motor (not a diaphragm regulator) doing the pressure reduction, with the motor output used to regeneratively compress more free air. That way, power would be conserved, like a transformer, subject to some mechanical inefficiency. 
#19




SCFM vs. CFM, also air flow/pressure across a regulator
In article , Richard J Kinch
says... We agree that mass flow (CFM) Stop right there. Mass flow and CFM are not the same thing. Once you realize this you can begin to think about the issue in a different way. Cubic feet per minute is only a time rate of volume. So in a given time (say one minute) a give volume of material will pass through your system. Depending on the pressure inside the volume, you can have all different amounts of material. For example, a liter of gas at atmosperic pressure has only a few atoms. If you image the same volume at scuba tank pressure, it has a lot more atoms inside. Jim ================================================== please reply to: JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com ================================================== 
#20




SCFM vs. CFM, also air flow/pressure across a regulator
In article , Tim Williams says...
So where does the heat go? In the case where large amounts of gas are expanding, the expansion valve will cool, and the heat from the surrounding area will flow into the region. Also the gas will exit the expansion valve at a lower temperature than it came in at. This is how gas liquifers are designed, either with direct expansion valves or with expansion turbines, where the gas actually performs mechanical work as it expands, and thus cools. ================================================== please reply to: JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com ================================================== 
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