Need help in calculating holding value for new product

[garry2]

We have developed a new product for an industrial application in large air ducts. The product is called a "duct balloon" which is a portable inflatable device that is equipped with a 110V positive pressure blower that operates at 10" w.g. The blower must remain on at all times to keep the balloon fully inflated. When the duct balloon reaches full inflation pressure, it exerts force around the interior perimeter of these large duct work, and creates a tight seal. The balloon material is nylon and the duct work is always carbon steel. Typical duct balloon sizes are 15' wide x 20 ' high x 3' thick. The 3' dimension is the part that seals up against the interior perimeter of the duct work.

Engineers love the concept but ask how much differential pressure it takes before the balloon will move. Is there any way to calculate this value?

Product information can be found on our website at www.ductballoon.com

 

[Danger]

A 20 x 15 foot duct? Wow!
Anyhow, you'll need to know the friction of the balloon surface as well as the duct. There are lots of different types of nylon surface.

 

[FredGarvin]

Honestly,something like that you could take a swag at, but it really is reason to make up a test rig to test it in real world conditions. It wouldn't be terribly difficult or expensive to do.

 

[berkeman]

I'd suggest either putting the balloon in a slightly enlarged section of ductwork (so it ends up being supported by the downstream lip), or putting some coating inside the duct at the balloon position to give you a higher coefficient of friction for the nylon balloon. Either option should mitigate concerns of the balloon dislodging, and might save you some money on the blower (by virtue of a lower pressure being needed to keep the balloon in place).

 

[garry2]

I did some research and the following are the coefficents of friction:
Balloon materila is a 420 denier nylon = .43
Duct work is carbon steel = .60

 

[berkeman]

I thought it took two materials in contact for there to be a coefficient of friction (but it's been a long while since I studied the subject). Are those numbers that you show for the material rubbing against itself?

 

[garry2]

Yes, you are correct. When the duct balloon is fully inflated its surface will press up against the carbon steel surface. It would be great if someone could help me calculate what type of differential pressure it would take in order for the duct balloon to loose its grip and start to move.

 

[Q_Goest]

The force of the air (Fa) you are trying to seal must be less than the frictional load (Fu) created. So Fa < Fu

Fa = dP * A
where dP = the pressure difference across this seal
A = cross sectional area being sealed

Fu = u*Fn
where u = coefficent of friction
Fn = Normal Force

This is where it gets tricky. The normal force could be the difference in pressure between the inside of the bag and the pressure of the air being sealed, or the inside of the bag and the pressure on the low pressure side of the seal. Neither is exactly correct. For most seals, we look at a linear pressure distribution across the seal from one side to the other. In this case, you could do that but it could be far from accurate due to leakage of air from the high pressure end to the low pressure end. To be conservative, you would want to use the high pressure end to do the calculation as I think you'll find that given sufficient length, the seal will be very stable. If you use the high pressure end of the seal, normal force becomes:

Fn = A * (Pb - Pa)
where A = area of contact between the seal bag and duct = circumference times length
Pb = Pressure inside the bag
Pa = Air pressure you are trying to seal.

Note that this also says you can't seal any pressure that is higher than the pressure inside your sealing bag.

You could improve this situation by applying all your sealing load at or near the end you are trying to seal, and then 'create a leak path' so to speak, to the low pressure end of the seal. This would allow all the sealing to be done at the higher pressure end, and the area between the seal bag and duct could be at the lower pressure, allowing Pa to be replaced by the pressure at the low pressure end of the bag. <patentable?>

I'd also suggest putting a 2 or 3 to 1 safety factor on to this at a minimum. I suspect there's enough frictional force holding the bag in place that you will end up with a very high safety factor, so long as you don't try to seal gas that is close to the bag pressure.

Just a side note: Seals work on contact stress such that the contact stress must be higher than the pressure difference trying to be sealed. If the dP you are trying to seal exceeds the contact stress of your seal, it will leak.

 

[garry2]

Using your formula for a duct balloon that measures 20' high x 15' wide x 3' deep:

A (210 sq. ft) * (Pb is 10" w.g. - Pa 5"
w.g)

Then Fn = 1050?

I'm lost!

 

[FredGarvin]

You need to convert your perssure units to pounds per square foot if you have the area in square feet.