# Homework Help: How do i figure out the hydtostatic pressure in the tank

1. Dec 16, 2009

### Dell

an unpressurised cylyndrical storage tank has a thin wall of 5mm thickness and is made of steel with a maximum strength of 400Mpa in tension
the tanks height is 20m
(the tank is standing on its base like a tower)
Determine the maximum height it may be filled with water if the density of water is 1Mg/m^3

i first thought i could use hydrostatic pressure here, saying that the pressure in the tank is uniform ? but now im not so sure, thats the only kind of problem i have had so far, one where the pressure is uniform. if not how do i find the pressure? i think that logically the pressure at the bottom of the tank must be the highest, but how do i know exactly?

Last edited: Dec 16, 2009
2. Dec 16, 2009

### Dell

this is definitely not hydrostatic pressure like i thought before, the lower down i g the higher the pressure will be,

since the tank is unpressured, the force on the tank at the surface of the water will be 0, and the force will increase linearly

F(y)=9800*(pi)*(R^2)*y
(where R is the inner radius and y=0 is at the surface of the water and y=h at the base of the tank)

now that i have my Force i i need to find the pressure.
as far as i know, the pressure on the base will always be much higher than the pressure on the sides of the tank,

the force on the bottom of the tank = 9800*(pi)*(R^2)*h
the area of the bottom of the tank = pi*R^2

the pressure on the bottom of the tank= F/A=9800*h

i was given the maximum strength, can i compare this to the pressure i found at the bottom of the tank?

but i get a massive number as a result since my maximum strength is Mpa and here i have 9800*h (Pa)

also where do the radius radius and thickness come in, my result was independent of both

i know that for hydrostatic pressure the stress at the base of such atank is equal to

P*r/(2t)
where P is the hydrostatic pressure, can i use this here too, according to the way i proove the expression i would think so
i disconnect the base and use equilibrium equations on it.

if so

9800*h*r/(2t)=400e6

h=400e6*2*t/(9800*r)

but i still get a large answer
h= 108.8435m which is higher than the tank

this is possible, but means that the tank is built to withstand the pressure of a full tank

but am i doing this correctly??

3. Dec 17, 2009

### pongo38

Assume side can slide over the base without water leakage! Consider lowest "ring" say 10mm high. The tension in that ring comes from pressure d g h where d is density of water, h is height of water. Equilibrium of half a ring in plan will reveal the tension in it. Then to stress etc. When I did it I made a lot of mistakes with units; so be careful.

4. Dec 17, 2009

### Dell

i still get an answer larger than the height of the tank, using the same principle .