force exerted by air

[Yulian24]

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

 

[FOURe]

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

 

[BigInch]

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/

 

[FOURe]

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

 

[ARenko]

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

 

[BigInch]

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/

 

[ARenko]

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?

 

[FOURe]

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

 

[mjpetrag]

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

 

[FOURe]

How about:

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

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

 

[FOURe]

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

 

[mjpetrag]

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

 

[BigInch]

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/

 

[IRstuff]

The big question is the missing factor of 2.

The typical drag equation looks like:
1/2 * Cd * rho * V²


TTFN

FAQ731-376

 

[BigInch]

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/

 

[ChasBean1]

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.

 

[BigInch]

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/

 

[sailoday28]

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?

 

[BigInch]

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/

 

[Bribyk]

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...