# Calculating CFM from PSI and C.S.A.

by tk11
Tags: None
 P: 4 I'm currently designing an air filtration system for the filling compartment of a beverage filler. The goal is to introduce HEPA filtered air at a rate of 2 CFM through a circular pipe opening with diameter of x inches. The air is provided by a compressed air system at a pressure of 120PSI (I plan on reducing this pressure through an air regulator) My problem is that I do not know how to calculate the air flow rate in CFM using the information I have (PSI and diameter/CSA). Is there a formula or method of determining the CFM from the above information and possibly other information I could easily find? Thanks for your help. T.K.
 P: 1,600 120 PSI would be your starting or static pressure. Once the air begins to flow this will change, it will drop. The flow rate through your diameter will depend on how much pressure you want to allow the system to drop and the restrictions in your piping and fixtures.
 P: 927 Assuming no change in height, the pressure would be equal to 1/2 the density * (velocity of air) squared, or velocity = sqrt (2*Pressure/density) and flow rate equals velocity times cross sectional area, so cfm = area of pipe * sqrt (2*Pressure/density)
P: 4

## Calculating CFM from PSI and C.S.A.

There is a gauge a negligible distance away from where the air outlet will be that reads 110 PSI.

I'm basically just trying to find a relationship between pressure (in PSI), pipe diameter (where the air will exit into the chamber) and CFM.

I know that Q (CFM) = Area (ft^2) * Velocity (ft/min)

I just need to figure out a way to convert my air pressure to air velocity knowing the pressure, density and temperature. I'm hoping it is that easy as I do not have a gauge to measure the velocity of the air exiting the pipe.
P: 4
 Quote by daveb Assuming no change in height, the pressure would be equal to 1/2 the density * (velocity of air) squared, or velocity = sqrt (2*Pressure/density) and flow rate equals velocity times cross sectional area, so cfm = area of pipe * sqrt (2*Pressure/density)
From where and how did you derive that Velocity = sqrt (2*Pressure/density) ?
 P: 1,600 Compression ratio will have an effect on your system. The compression ratio indicates the effect of the high pressure (120 PSI) on the flow, due to turbulance created in the pipe. You will need to divide your flow rate by the compression ratio to determine the actual flow. (Or in your case multiply the compression ratio by 2 CFM to determine what the actual CFM needed at high pressure will be, since you know the flow rate you want to achieve.) Compression Ratio, CR, is calulated as the line pressure (P)+ local atmospheric (Patm)/ local atmospheric (Patm). CR = (P + Patm)/Patm Lets say your area has standard conditions of 14.7PSI add this to 120PSI = 134.7PSI / 14.7PSI = 9.16 CR Which means you will need to move about 18.32 CFM to obtain 2 CFM to your point of use using 120 PSI air. 9.16 * 2 = 18.32 CFM So keep in mind you are not trying to move 2 CFM through your pipe, you need to move 18.32 CFM at 120 PSI.
 P: 230 Why do you want to calculate the flowrate when you know that it is 2 cfm? Perhaps you want to know the pipe size for 2 cfm. Ideally, you should know the available pressure drop with respect to the pipeline length and accordingly adjust your regulator. Crane Technical Paper 410 is a good reference. However, 2 cfm is very low flowrate and you can get this by a 6 mm tubing. It is better to use flowmeters than precisely sizing the pipe. Artman, 2 scfm (i.e at 14.696 psi) is 0.218 acfm (i.e at 120 psig). It is 2/9.16
P: 4
 Quote by quark Why do you want to calculate the flowrate when you know that it is 2 cfm? Perhaps you want to know the pipe size for 2 cfm. Ideally, you should know the available pressure drop with respect to the pipeline length and accordingly adjust your regulator. Crane Technical Paper 410 is a good reference. However, 2 cfm is very low flowrate and you can get this by a 6 mm tubing. It is better to use flowmeters than precisely sizing the pipe. Artman, 2 scfm (i.e at 14.696 psi) is 0.218 acfm (i.e at 120 psig). It is 2/9.16
I was looking at the problem backwards in hopes that somebody knew a formula for flow rate off hand.

My CFM value is fixed at 2 CFM and my pipe size at the moment is 1". Now I just want to find the pressure I need to drop the 120PSI to using a regulator in order to get the 2 CFM through an orifice of 1".

The fill chamber is 2 cubic feet in volume so my 2 CFM would create an air change every minute. This could be increased but I'm trying to limit the usage of compressed air for this application. This could be increased but shouldnt effect what I'm looking for at the moment.

http://www.efunda.com/formulae/fluid..._flowmeter.cfm

The above link in the closest thing I've found to what I'm looking for but I am not dealing with an orifice plate as my upstream pipe diameter is the same diameter as my orifice.
P: 1,600
 Quote by quark Artman, 2 scfm (i.e at 14.696 psi) is 0.218 acfm (i.e at 120 psig). It is 2/9.16
I thought he was trying to achieve 2 CFM at 120 PSI. For which he would require 18.32 scfm.
 PF Patron HW Helper Sci Advisor P: 2,796 Hi tk11. One of the most common methods I've seen to control flow rate is to set up a system very much like what you've proposed. You have a source of air, you regulate it to a given pressure, but then you pass it either through an orifice which is sized to give you a specific flow rate or a rotometer. For the orifice method, I'd suggest setting your regulator at 30 psi and installing an orifice downstream of that with a 0.062" diameter hole drilled through a flat plate (ie: a flat plate orifice). This should give you very close to a 2 CFM flow rate. Alternatively, you could put a rotometer downstream of your regulator with an adjustable valve on it. These are very inexpensive, \$30 or so, give or take. You can get one at McMaster Carr. All you'd need to do is adjust the rotometer valve to give you 2 CFM. Hope that helps.
 P: 1 I am trying to figure time it takes to fill X volume (gallons) to X pressure (PSI) at X rate of flow (CFM. NOW, what I really want to know is how long it takes to fill 500 gallons to 150 PSI at 20 CFM. if there is a such formula, I would like to know it
 P: 5 I know this thread is very old. But as I was reading, I couldn't help but ask myself "What are the time units when referring to the velocity in the equation: velocity = sqrt (2*Pressure/density)?"
 Mentor P: 21,674 There are no units implied by the equation - it depends entirely on what system you want to use. The most common systems, however, use seconds (m/s or ft/s).
 P: 5 With this formula, are we supposed to assume that velocity is in seconds? Obviously, I do not know what the background and proofs behind this formula are. And in my ignorance, I can't feel comfortable assuming what the units of a variable are expressed in.
 P: 2 Some of the suggestions are correct to some degree but you cannot convert CFH to psi that is mixing apples and oranges! However, there is a method often used to achieve a predetermined flow that is called “choked flow.” It relies on the physical fact that gas cannot not exceed the speed of sound velocity in a small orifice! That is why you see lightening before you hear the thunder! Therefor regardless of the downstream pressure the flow (which can be measured in CFH) is only dependent on the gas density above the orifice which is dependent on pressure (assume constant temperature.) As someone mentioned, use the proper very small orifice and you can define the upstream pressure that will flow 2 CFH regardless of the downstream pressure restrictions. I made literally thousands of tests of pressure and flow when inventing our patented welding gas saver system. For example it takes about 3 to 8 psi in the gas hose going into a MIG welder to flow 30 to 35 CFH. However, if "choked flow" is not used it will vary while welding as the small gas hose in the MIG gun cable is bent, as spatter builds in the gun nozzle and for other reasons! That is why all proper regulator/flowmeters for regulator/flowgaugues (those with 2 gauges, one calibrated in CFH) operate above 25 psi. They both use a choked flow approach. The needle valve or very small orifice in a regulator/flowgauge operates above 25 psi to achieve what is called “automatic flow compensation.” The folks developing MIG and TIG welding in the 1950’s knew about flow variations and used this approach.

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