Volume Flow Rate CFM question 3

[Twoballcane]

Hi guys and gals,

I have a quick question on volume flow rates, if air is flowing thru a louver, will the CFM be the same at the inlet and outlet of the louver? And, the only thing that will change is the velocity (ft/min) of the air from the inlet to the outlet of the louver? Am I thinking right?

Thanks in advance for your time.


Tobalcane
"If you avoid failure, you also avoid success."

 

[rb1957]

well for sure there is conservation of mass (density*Volume = density*area*velocity). if there is no compressibility effects (likely in a duct, sounds like an HVAC question) means that the density doesn't change, then CFM is constant. if the areas are the same, then the velocity will be the same.

but i don't get "the only thing that will change is the velocity (ft/min) of the air from the inlet to the outlet of the louver?"

 

[MintJulep]

If air can only go in at the goesinta, and can only come out at the goesoutta, how could the cfm change?

The total pressure will change.

 

[Twoballcane]

Thanks for responding...
rb1948:
Well would the louver restrict air flow? As in slowing it down? Does louvers have a % that is open? Say for an example that the louver is 90% open, does that mean the air velocity will be reduced by 10%?

MintJulep:
I see what you are saying, but if a duct ends at a louver, the geometry chages so it either slows the air down because so the obstruction (louver) or what comes in goes out. Back pressure is another concern...but Ill figure that out later...

Thanks

Tobalcane
"If you avoid failure, you also avoid success."

 

[katmar]

Because air is compressible, if there is a pressure drop through the louvre then the density of the air on the downstream (low pressure) side will be less than the density on the upstream side. As the density decreases the volume increases and the velocity will increase.

The conservation of mass means that the same mass rate (i.e. lb/hour) of air flows on both sides, but the actual CFM will be higher downstream. The flow in SCFM will also be the same both sides (because SCFM is actually a unit of mass flow and not volumetric flow).

For normal HVAC applications I would expect the change in CFM to be very small.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Twoballcane]

Katmar,
Thaks for responding. I should post my real intent, Im trying to model a duct and louver in a room in Flotherm, but it is difficult to model the louver. So does having the louver in the model make sense? Is the differance in CFM and ft/min befor and after the louver neglectable?

Tobalcane
"If you avoid failure, you also avoid success."

 

[JStephen]

For typical AC duct applications, air could be treated as incompressible.

The terms are being confused above. The louver "slows air down" in the sense that the velocity along the entire duct may be reduced from what it would otherwise be. That doesn't necessarily mean that air slows down between the inlet and outlet of the louver itself, though. Depending on the open area, the velocity could go up or down- but CFM should be about the same.

 

[MintJulep]

Just model it as a pressure drop.

 

[rb1957]

i had pictured the lourve as a deflector, rather than as a throttle ...

i think you're asking if i get X cfm with the louvre open, how much would i get if the lourve was 1/2 closed. personally (not being an HVAC type) i'd expect something close to X/2 ...

 

[ko99]

Twoballcane,

Since the vent is not wide open, your model should include a resistance that's equivalent to the louvers. Some modeling methods avoid the complexity of a detailed louver model. The best approach depends on the importance of the vent and the data you have available.

1. Estimate a rough percentage open and model it as an equivalent planar resistance.

2. Measure the flow-pressure curve across the louvers using a wind tunnel, then use flotherm to create an advanced planar resistance to match.

3. If you don't have the hardware to #2, do it in software. That is, create a detailed model of the louver by itself in a numerical wind tunnel, then use the simplified 2D resistance in your room model.

4. Find published data for louvers. Possible sources:
-- The manufacturer
-- I.E. Idelchik "Flow Resistance: A Design Guide for Engineers"
-- HVAC forum on eng-tips


ko (

http://www.ecooling.biz)

 

[WoodyPE]

Less than 10% the speed of sound is incompressible (5,000 fpm or less). The velocity of air coming out of a duct should be less than that or else you'll be blowing paper all over the room as well as generating a lot of noise. I give you another clue. You are only talking about a small pressure drop anyway.

So to answer your question, the cfm into the louvre should equal to the cfm out.

http://www.angelfire.com/planet/gadgetforum/index.html

 

[WoodyPE]

Oh just one more thing to complicate matters. The air velocity in a duct is 0 fpm at the wall and the velocity profile is somewhat of a square polynonomial on the flow cross-section. This changes to a more parabolic shape as you go down the duct. Don't worry about that though! It doesn't change what you are doing.

a freebie on my PE exam!!!

http://www.angelfire.com/planet/gadgetforum/index.html

 

[Twoballcane]

Thanks guys for responding!

Ko99:
Thanks for the suggestions. That was what I wanted to do, but I have louvers that will direct air at certain directions and wanted to see how the reduction of air speed will affect the air flow within the room. If you are familure with flotherm, I am using the recirculator. With this model I can direct the air, but I (in its dialog box) can not implement resistance. I can use another feature that will restric the air, but when I add that right at the supply of the recirculator, it does not like it and over rides the resistance feature.

Thanks guys for the help, I have a better understanding of the situation, but I think I will break down and model in the louver. Pretty pics convice certain people anyways.





Tobalcane
"If you avoid failure, you also avoid success."

 

[ko99]

The resistance doesn't have to be coincident with the recirculator, you should be able to just put in series and get the same results.

ko (

http://www.ecooling.biz)

 

[Twoballcane]

Ko99

Thanks Ko99, I tried that yesterday, instead of collasping it towards the recirculator, I collasped it to the middle of the resistance volume. And guess what! The velocity is faster. Which makes sense.

If we looked at conservation of mass, yes the densities are the same but not the cross area (resistance).

M1=M2
v1p1a1 =v2p2a2 <densities are the same>
v1a1=v2a2 <at this point the CFMs equal>
v2=(v1a1)/a2 <when area becomes smaller the velocity goes faster and CFM does not chage>




Tobalcane
"If you avoid failure, you also avoid success."

 

[Compositepro]

I believe that you are all missing a significant point about air entering a room from a duct and it has to do with conservation of momentum. The air from the duct is slowing down because it is entraining ambient air and accelerating it. So, yes, as velocity goes down cfm goes up.

 

[Twoballcane]

Compositepro,
Thanks for responding.

For my application, I have two really big fan cooling units for a realy small room. So once in steady state, the fan cooling unit is supplying and returning air so the flow profile looks well...fluid. In flotherm once I added the resistance, the velocity increased. I think this is because of a small room with high CFM fan cooling units.



Tobalcane
"If you avoid failure, you also avoid success."