Compressed Air CFM 2

[k4thomas]

I am currently designing a HEPA filtered air system for a beverage filling machine. The goal is to complete an air change every 2 minutes within a 6.5 cubic foot compartment (not sealed and at atmospheric pressure). The SCFM would thus be 3.25

This is to be accomplished using a 1" inner diameter compressed air line (120PSI) passing through a 0-150 PSI regulator and then exhausting into the top of the fill compartment with open vents on the sides and bottom.

My question is how do I determine what pressure setting I restrict my regulator to in order to achieve my 3.25 SCFM?

(I can measure the length of the piping from the regulator to the fill compartment if necessary as well as the temperature.)

 

[BigInch]

You need a regulator you can set at a discharge flow of 3.25 scfm with an outlet pressure of 14.696 psia.

You know the 120 psig inlet pressure, the outlet pressure and the flow. Flow across the regulator is Q = Cv (P1^2-P2^2), solve for Cv and set the regulator so that the Cv is the same. Find the Cv vs Position curve for the regulator and see what %Open gives you the Cv you need.

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[sailoday28]

BigInch-I'm just curious and not familiar with control valve dp flow characteristics- is your statement of
Q = Cv (P1^2-P2^2), a control valve characteristic?

Regards

 

[BigInch]

Thank You.
Please add a ^0.5 to the end of that.

I purposely left out the conversion factor, valve correction factor and compressibility effects, and don't mention you must check for non-critical flow.

Q = Cv * N1 * f_dP * ((P1^2-Pd^2)/Z/G/T)^0.5

Q = Standard Flow
N1 = 120/ºR/PSI
f_dP = Valve correction coefficient
P1 = Upstream pressure absolute
P2 = Downstream pressure absoute
Z = compressibility factor at upstream temp and avg press
G = specific gravity of gas relative to air
T = upstream temperature absolute

When not in critical flow.

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[katmar]

Setting the pressure to control the flow is the wrong way of going about it. A simple rotameter is all you need. This will give you control and indication of your flow rate. You will still need the pressure regulator to get the rotameter into its correct range, but don't use the pressure regulator to control the flow.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[k4thomas]

I will look into your suggestion katmar.

BigInch, I believe I will have critical flow as I expect the drop in pressure to be greater than 50%.

I have found several different variations of the Cv computation formula and was wondering which is best. I found the follow equation from

http://www.engineeringtoolbox.com/flow-coefficients-d_277.html

for critical flow.

Cv = q*[SG*(T + 460)]^(1/2) / [1360*(dp*po)^(1/2)]

q = free gas per hour, standard cubic feet per hour (Cu.ft/h)
SG = specific gravity of flowing gas gas relative to air at 14.7 psia and 60oF
T = flowing air or gas temperature (oF)
pi = inlet gas absolute pressure (psia)
dp = (pi - po)
po = outlet gas absolute pressure (psia)

 

[BigInch]

You must avoid critical flow. At critical flow you cannot change the flow through the valve, ie. you have NO control. Save your money buying a valve and just punch a hole in a end cap. Critical flow is basically telling you that the outlet pressure is too low or your inlet pressure is too high. I'd suggest you reduce the inlet pressure, perhaps by installing a hand adjustable plug valve.



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[k4thomas]

Does the outlet pressure have to be 14.696 psi as you recommended though? Is there a way to have an outlet pressure greater than 60 PSI in order to have non critical flow while ensuring an air change every 2 minutes within the 6.5 cubic foot compartment at standard conditions?

I'm just trying to find the simplest way of using that 120 psi compressed air line to achieve my required air changes.

 

[BigInch]

The outlet has to be 14.696 psia (more or less the difference between that and actual barometric pressure) if you're near the Earth's surface at sea level and you're letting N2 out of the pipe to some room at atmospheric pressure. You said your compartment had open vents at the bottom, I just assumed it was an atmospheric pressure room of some kind.

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[k4thomas]

The regulator or valve would not be feeding directly into the compartment though, there will be at least 7-10 feet of 1" pipe between the exhaust hole into the compartment and that valve/regulator. Dont know if that changes anything.

And is critical flow the same as choked flow? Because from what I understand about choked flow the volumetric/mass flow rate can be controlled independently from the downstream pressure and temperature.

 

[zdas04]

You have a lot of choices. The one I prefer is use a regulator to drop the pressure to about 65 psig. Then use PSV calculations to size a choke nipple to give you your desired flow rate.

This does result in critical flow through the choke nipple, but so what? In spite of what has been said above, critical flow is a very precisely understood thing. With a known orifice and a known upstream pressure and temperature, critical flow has exactly one mass flow rate regardless of downstream pressure (until downstream pressure increases enough to stop the critical flow).

With a properly sized hole (around 0.056 inches) you will get exactly the flow rate you want without excessive noise or a lot of complexity. Just set a cheep Big Joe regulator to 65 psig and drill a hole in a pipe-sized billet and you'll have it. There is nothing magic about the 65 psig. I would play with the upstream pressure until I got to a hole size that matched a standard drill size (e.g. at 13 psig you would get your flow rate with a 1/8 hole). Big Joe regulators are regularly used to cut pressure from over 100 psig to under 20 psig for burners without problems.


David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

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[sailoday28]

BigInch Regarding your formulation
Q = Cv * N1 * f_dP * ((P1^2-Pd^2)/Z/G/T)^0.5
Should't for non choked flow- adiabatic(and low mach no), the pressure drop be proportional to the square of flow?
Regards

 

[insult2injury]

Q = Cv * N1 * f_dP * ((P1^2-Pd^2)/Z/G/T)^0.5 <-- it is proportional to the square of flow...notice the ^0.5

I2I

 

[insult2injury]

scratch that...didn't see the squares on the individual pressure terms

I2I

 

[KenRad]

The combineed Engineering manhours that have been spent responding to this post have already exceeded what it would cost to follow katmar's suggestion -- install a variable area flowmeter. No guesswork involved; adjust the regulator until the flowmeter reading is right, and you're done. McMaster-Carr has flowmeters in this range for less than $100.

---KenRad

 

[BigInch]

That's why I told himm to punch a hole in the end cap.

Going the Big Inch!

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