# Differential pressures in a chamber, revisited and clarified

1. Jun 3, 2013

### LowEParticle

This is a clearer, better drawn, and shorter version of a question I asked a few days ago. I apologize for not initially reducing it to this state - sometimes when reducing a question to its essence I stop too soon!

The diagram below shows two cross-sections of an axis-symmetric cylindrical chamber that has a valve armature passing through its center. The valve armature descends to close the chamber (left-hand drawing) and rises up to open the chamber (right-hand drawing). Where the valve armature's shaft passes through the bottom of the chamber it is sealed off by a frictionless, air-tight seal. The air pressure outside the chamber is always Phigh. If the chamber is closed, then the air pressure inside of it is Plow.

My problem is the total force Ftotal on the valve armature due to differential pressure when the chamber is closed. Ftotal is the sum of three terms:

F1 is the pressure inside of the chamber pressing upwards on the bottom of the valve armature plate (14mm radius):
F1 = Plow * $\pi$ * (14)2 (upwards)

F2 is the pressure outside of the chamber pressing upwards on the very bottom of the valve armature shaft (4mm radius):
F2 = Phigh * $\pi$ * (4)2 (upwards)

F3 is the pressure outside of the chamber pressing downwards on the top of the valve armature plate (18mm radius):
F3 = Phigh * $\pi$ * (18)2 (downwards)

So Ftotal = -F1 - F2 + F3

I showed all the significant radii on the right-hand drawing for easy identification, however, the solution above does not use the 20mm, 12mm, or 8mm radii. I have two questions:
1. Have I picked the correct radii to correctly calculate the relevant forces?
2. Are the 3 forces I've identified the only ones produced by differential pressures that are relevant to the caclulation of Ftotal?

Thank you very much for reading this and considering my problem.
Dave

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2. Jun 3, 2013

### 256bits

well hello there,
You should re-think F1.
Does the low pressure act on the whole of the 14mm radius, or just part of it?
If you examine F2, that may give you a clue.

3. Jun 4, 2013

### LowEParticle

Thank you very much for finding this error! I've written the corrected description of F1 below:

F1 is the pressure inside of the chamber pressing upwards on the bottom of the valve armature plate (14mm radius) excepting the 4mm radius of the armature shaft:
F1 = Plow * $\pi$ * (142 - 42) (upwards)

I appreciate your help very much!
Dave