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force exerted by air

force exerted by air

force exerted by air

(OP)
I am trying to size an actuator for an air damper. My entrance velocity is 30 MPH. I would like to know how to calculate the force the air will exert on the dampers when they are closed.

A similar problem to this would be to try to calculate the force on a butterfly valve when only the velocity is known.

Thanks
 

RE: force exerted by air

In general, one can use the following equation for force:
force = velocity * mass flow rate [N]

where

velocity [m/s]
mass flow rate [kg/s]  

http://www.engineering-4e.com

RE: force exerted by air

F = mass * velocity is the momentum of a mass.

Fluid drag force is proportional to velocity [bold]squared[/b], more specifically,
http://hyperphysics.phy-astr.gsu.edu/Hbase/airfri2.html
 

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air

That is correct and, in my opinion, a force created by the air if it is coming at the right angle at an object ...

I do not see anything wrong with what I have put out there ...

Please let me know if something else needs to be considered or done in this case ...

Thanks!

http://www.engineering-4e.com

RE: force exerted by air

I think you need to use the total pressure capability of the fan.   

RE: force exerted by air

Total pressure capability of the fan?  How do you figure that?   The fan isn't touching the dampner.  Only the air at the velocity of 30 mph is touching the dampner!

FOURe  Do you see the velocity SQUARED term.  You only put V^1.  Little bit of a difference, don't you agree.  Why not have a look at the link I posted to reassure yourself.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air

Maybe I misunderstood the application - I was thinking the damper was on the discharge of a duct.  In that case does it not have to hold against that pressure?

 

RE: force exerted by air

BigInch:

If I may add, velocity is squared in my input.  It is contained in the mass flow rate.

mass flow rate = velocity * density * cross sectional area

force =  velocity * mass flow rate

Therefore, you and I are in agreement -- it is a function of ~ v^2 ...

I am OK with where we are ...

Thanks!

http://www.engineering-4e.com

RE: force exerted by air

You don't have a fan curve or anything else that would tell you the static pressure in the duct?

Your velocity pressure is .5*(density of air)*(velocity of air)^2

so if you just want the force that the velocity exerts on the dampers, just multiply that velocity pressure by the cross sectional area of the dampers.

P = F/A... F = P*A

If you want the actual force on the dampers, then you need the (static pressure + the velocity pressure)at full open multiplied by cross sectional area.

-Mike

RE: force exerted by air

How about:

force = velocity * mass flow rate + (air total pressure - pressure after the dampers) * cross sectional area

http://www.engineering-4e.com

RE: force exerted by air

I have to take it back somehow:

How about:

force = (air total pressure - pressure after the dampers) * cross sectional area   

http://www.engineering-4e.com

RE: force exerted by air

Is this an existing system? Can you stick a pressure tap in it?  If you can, just do a static pressure reading and add it to the calculated velocity pressure reading, multiply by area.

-Mike

RE: force exerted by air

Ya where pressure = Cd density * V^2

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air

Its not missing.  Its on the link.  And.. its in "my" Cd factor too.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air

Wow, so much back and forth over this...

I'd vote for mjpetrag a few posts above. It's just a stagnation pressure reduced from Bernoulli's:

P = (rho * V^2)/2

There's 0.016 psi above atmospheric acting on the dampers at a 30 mph wind. Just multiply by the damper area to get total force. If it's a 40 ft2 damper bank it would be around 90 lb force.
 

RE: force exerted by air

Then you'd get it wrong, because of your missing Cd.  Cd is 1.4 for a flat plate perpendicular to flow and almost 0 if its a knife edge profile against the flow.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air

Is this "flow" analagous to that with a branch of a piping system where say a gate valve is shut and flow continues to other branches?

RE: force exerted by air

NO  Its EXACTLY like a swing disk check valve opening.

If the damper were closed (static pressure only on one side (fan discharge pressure x area), then the total force on the damper would reduce to only be pressure x area of the damper.  

If the damper is at least partially open, the static component of pressure on each side of the damper is assumed equal (since there is no elevation change across the damper).  Any differential pressure therefore must be caused by flow around the damper, which makes drag the only mechanism of interest, hence the Cd.

Come on guys.  Differential pressure across a valve isn't figured only with P1, its P1-P2.  
For a valve,
Q = Cv dP^0.5
(Q/Cv)^2 = dP    
   since  Q/A = V
(V/Cv)^2 = dP
 V^2 / Cv^2 = dP
dP = V^2 / Cv^2
P1-P2 = 1/ Cv^2 * V^2
so, looks like Cd is pretty much physically equivalent to the inverse of 2 IRstuff x Valve_Coefficient^2

Do you see how Cv for a valve is analogous to Cd/2 for some shape being hit by a flow.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air

If the damper is like a symmetric butterfly valve (OP) wouldn't your pressure on both sides of the pivot axis cancel? A1 = A2.  Then all you're dealing with is friction torque, plus a fudge factor. Except the OP has an air velocity with the damper closed, somehow...

RE: force exerted by air

There's enough information to solve it any way he wants.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air

Bribyk:

That was my way of thinking ...

BigInch:

I agree with you.  There is plenty of information and input provided.  Number crunching needs to be done and the results need to be shown and compared with different ways o thinking  and looking at the problem that is on our hands ...

To All:

Thanks!

http://www.engineering-4e.com

RE: force exerted by air

Use of Cd.  My understanding of its use is that--
 experimental Cd is based on the stagnation pressure far up and downstream as being equal.

 

RE: force exerted by air

Hmm...I just re-read the OP.  If the damper is fully closed, then there shouldn't be any airflow across that interface.  

If that's the case, then the drag equation isn't applicable, regardless of value of the drag coefficient, because the airstream velocity would be zero.  

This is where I think you need the fan curve, which gives you the static pressure at zero flow.  

You would then still need to calculate the converse case, which is what the actuator needs to move the damper when there is air flowing, which would involve the drag equation, but in a condition where the drag coefficient is unknown.  

For a standard butterfly valve, I would think that you'd also need to split the area across the axis of rotation, since the air pressure on one side is helping you turn the valve, while the pressure on the other side is working against you.  The drag coefficient will be different on the two sides of the valve, since the separation point of the airflow will be different.

TTFN

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RE: force exerted by air

That's true with the two drag coefficients but this is getting pretty technical for an HVAC flapper.

RE: force exerted by air

Because the OP said it's closed is why I said use the fan total pressure, which is static pressure at zero flow (assuming fan inlet and duct discharge are at the same atmospheric pressure).  

Really though he needs to check w/ the damper manufacturer, or google "sizing damper actuators" - there appears to be plenty of guidelines out there.

RE: force exerted by air

Ya, it was one of those trick questions.  So, tell me how do you have a velocity of 30 mph when the damper is closed?  I keyed off the "30 mph", djv on the "closed" half.  So then Julian what is it, 30 mph or closed?

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air



I repeat my previous question?

Is this "flow" analagous to that with a branch of a piping system where say a gate valve is shut and flow continues to other branches?  
 

RE: force exerted by air

Hey guys...this is just a wind load problem...

30 mph "air speed"....

Unfactored pressure exerted by the "air speed" = 2.30 psf

0.00256xV^2

 

RE: force exerted by air

It must be one heck of a leaky damper with that type of entering velocity.  Somehow, I think that that is the velocity at design flow and the OP is asking about shut off pressure for his/her damper actuator.  Of course, a little static pressure operating and closed would make this much easier.

I notice that the OP has the thread marked so he/she is watching all this engaging conversation go by as we gnash our gums over his problem.

rmw

RE: force exerted by air

sailo, As far as I'm concerned, its all your's.
I'm outa' here.
 

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air

How did a stagnation pressure become such a matter of opinion? There's no drag because there's negligible flow-though. Cd = N/A. There is about 0.016 psi at 30 mph. If you're worried about multiplication factors, size the actuator for 0.024 psi. Multiply by damper area for total force. Guys (not all of you), 30 replies and no clear answers makes this an AOL chat room vs. a place where people can trust in good engineering advice.
 

RE: force exerted by air

Well looking back, that's a fair comment.  
BTW, where were you on 28 April?
 

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air

Look back. I chimed in a 29 April. That's not good enough? Jeez...  smile

RE: force exerted by air

Ah.. yes.  I also see that was when I still thought it was not a stagnation pressure problem. blush I admit I had a hard time picturing how you could have 30 mph and a closed damper at the same time.  Actually, I still don't see how, but apparently that IS the case.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: force exerted by air

stagnation pressure is in the equations that is no surprise, but the greatest force on the damper is not when it is closed, but partly opened. if the actuator is not sized properly to get you past that you literally have to shut the system down to reposition it, assuming that you've not already broken the shaft!

 

RE: force exerted by air

Now that makes sense.  Now back to Cd ....

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

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