Building an atmospheric chamber

[mtrip]

Building an "atmospheric chamber"

I have to build a chamber (ok, large box) to do some atmospheric pressure tests. The box will be 48 inches square. I have to simulate going from sea level (0 feet, 14.696 psia) to 7000 feet (11.34 psia). I'm thinking of building the "box" from 3/4" plywood reinforced with cross-braces (if need be). I've looked at "wet/dry shop vacs" and think one of these might be able to pull the "vacuum" needed. The box is 64 cubic feet and the shop vac can pull 170CFM.

The box will have a vent on it to make it "leaky" to control the pressure inside (I'm thinking the vacuum can suck while the vent can leak letting in air). This way I can control the pressure without resorting to flapper valves or the like. It should also prevent the shop vac getting into a "clogged inlet" and racing the motor.

Does this sound reasonable? I've been looking to find the way to calculate all this but haven't found a reference yet.

Thanks.

 

[pallidin]

Just my take(I'm no expert):

Plywood is not a good choice. It has porosity issues. Gases can leak through, especially during pressure differential.

The shop vac rated at 170 CFM draw assumes relatively non-vacuum conditions. That is, the CFM for a store-bought shop vac will substantially reduce VERY quickly in drawing air out of a sealed container.

Just some basic thoughts...

 

[AlephZero]

Forget trying to make this from wood.

Supppose your 64 cu ft box is a 4 ft cube. The area of each face is 16 sq ft = 2300 sq in.
A pressure difference of 14.7 - 11.3 = 3.4 psi will give you a total force on each face of 2300 x 3.4 = 7800 lb which is about 3.5 tons.

If you really want to build this, you need to do some proper design work on it, not something "reinforced by cross braces if need be".

 

[jack action]

I agree with the calculations of AlephZero.

Plus, concerning air pump requirement, the power needed is the airflow times the pressure differential.

With the appropriate unit conversion:

Power = 170 CFM X 3.4 psi X 3.254 = 1880 W = 2.5 hp

This is without accounting for inefficiencies (motor, pump, duct restrictions, etc.). That power is related to the concept of "http://en.wikipedia.org/wiki/Airwatt" ". You will need a motor a lot more powerful than that.

You will have to consider that power when selecting your controlled «leakage».