Pressure in a Cylindrical Tank

[SeannyBoi71]

Homework Statement

On the afternoon of January 15, 1919, an unusually warm day in Boston, a 26.0 m high, 27.4 m diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded the streets in a 9 meter deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 kg/m3. If the tank was full before the accident, what was the total outward force the molasses exerted on its sides?

Hint: Consider the outward force on a circular ring of the tank wall of height dy and depth y below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside.

Homework Equations

F=\Deltap \times A
A = 2\pir^2 + \2 pirh
where A is area of the cylinder, h is height, r is radius

The Attempt at a Solution

From the hint my professor gave and from an example he did in class I tried Integrating the force with respect to y, the depth below the the surface, and got (from 0 to h)\intpi gy(2pir^2 + 2 pi r y) dy . I'm pretty sure I did the integration right, but I got an answer 1.42x10^10 and he says the answer should be in Mega-Newtons :S any help??

P.s. sorry if the integral is sloppy, I find it very hard to use the symbols on this text editor :P

 

[ehild]

You should not include the base of the cylinder. The force is ∫(ρgy dA)
and dA=(2rπ) dy. You wrote y twice.
ehild

 

[SeannyBoi71]

OK, I got a much more reasonable answer of 457 MN, but why do I not include the base of the cylinder? There would still be a pressure acting there, would there not?

 

[ehild]

I understood the problem that the force of the fluid exerted on the side wall was asked. Maybe, I am wrong, then you have to add the term r² π ρ g h (h is the height). Ask your professor what he meant.

There are two equal forces acting on the base, that from the fluid and the normal force from the ground, and they are equal. These forces compress the base but will not cause rupture. Against the pressure of the fluid on the wall, there is only the atmospheric pressure. The net outward force will stretch the wall out till it ruptures.

ehild

 

[SeannyBoi71]

I think you were right the first time. Thank you very much!