How Much Weight Is Needed to Increase PSI in a Water-Filled Cylinder?

[Kalagan]

Some numbers seem a little weird to me, I just want to get a second or third opinion on it.

Problem: How much Weight does it take to increase the PSI in a Cylinder filled with water by 10 PSI, 100 PSI, etc?

Assuming that there is a 1 ft tall Cylinder with 6 inches of water in it, with the AREA of the Cylinder floor being exactly 1 square inch; if you were to drop a Weighted Piston of exactly 10 lbs. on top of the Water, will that increase the PSI in the Cylinder by 10 PSI?

What I am really trying to figure out is, if there is a 110 ft tall by 20 ft wide Cylinder with 100 ft of water in it. How heavy will a Weighted Piston dropped on top of the water have to be to increase the PSI in the Cylinder 1 PSI, 10 PSI, 100 PSI? (Assuming that the Piston makes a perfect seal with the inner wall of the Cylinder and that Friction is nonexistent)

Are my calculations correct?

1) The area of the top of the water in Square Inches would be A = Pi x r squared, which would be A = 3.14 x (120 inches x 120 inches), so the Area of the top of the Water = 45,216 Square Inches
2) So to increase the PSI in the Cylinder by 1 PSI, you would have to drop a 45,216 lb Weighted Piston on top of the water?
3) To increase the PSI in the Cylinder by 10 PSI, you would have to drop a 452,160 lb Weighted Piston on top of the Water?

I think I am missing something, because those numbers seem TOO high for such a small increase in PSI. Can someone please help me here?

Thanks
Kalagan

 

[Dick]

You are thinking about the problem correctly. The weight is large because the area of the piston is large.