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Which surface do I measure for pressure? Answered

So this is a fairly simple question, but here it is:  If you have a 5" pipe leading out of a pressurized chamber with a 6" inch disk covering the hole is the amount of pressure pushing the disk down that of 5" or 6".  I am pretty sure it is 5" because it has the equivalent pressure pushing the other way outside of that diameter.  Now if the 5" pipe had a bore of 5" and an outside diameter of 6" would the result be the same?


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Jack A Lopez
Jack A Lopez

Best Answer 9 years ago

I think you are wondering about how pressure, or differences in pressure, produce a net force on an object.

The way this works is, at every little point on the surface of an object, there is a small force dF= P*dA.  The direction of dF (and the direction of dA) is always normal to the surface.  Integrating over all those little dFs, and that gives the net force on the object.

From the text of your question, it sounds to me like you are building a comically large champagne bottle.  And the cork for this large bottle looks like a cylinder, or actually like two cylinders.  The part stuck into the neck of the bottle is a cylinder with a diameter of 5 inches, but it abruptly transitions to a cylinder with a diameter of 6 inches outside the bottle.

I drew a picture of this cork, here,
and I divided its surface area into five smaller areas, A1, A2, A3, A4, A5, and then calculated the force due to each area, F1, F2, F3, F4, F5,  due to pressure inside the bottle, P1, and pressure outside the bottle P2.

I claim that the net force on the cork is just F = A1*(P1-P2), where A1 is the area of the smaller (5 inch diameter) circle that fits inside the neck of the bottle.  A2 is a larger (6 inch diameter) circle on the top part of the cork.

F1 = P1*A1 pushes the cork upward due to pressure P1 inside the bottle.


F2 = P2*A2 pushes the cork down due to pressure P2 outside the bottle

but there is also F3 due to that little ring-shaped flat part.

F3 pushes upward with a force P2*A3 = P2*(A2 -A1)

And F2 and F3 added together is just P2*A1 downward.

Areas A4 and A5 do not put any net force on the cork since all  the little dFs on these areas are all radially pointed inward at each other, and taken together they sum to zero. That is F4 =0 = F5 =0.

Adding all the Fs together gives

F1 + (F2 + F3) + F4 + F5

= P1*A1 -P2*A1 + 0 + 0 = (P1-P2)*A1


9 years ago

"Yes" and "yes".


9 years ago

it is the 5 inches...unless there is some fluid leaking out of the gasket towards where the 6 inch plate is bolted on...

Assuming the 'seal' is at the bore diameter of the 5 inches its 5 inches.

Pressure is pushing radially on the circumference of the pipe, and axially on the end plate.