How to calculate potential energy of water in a tank?

AI Thread Summary
To calculate the potential energy (PE) of water in a cylindrical tank, the formula U = 1/2 mgh can be used, where m is the total mass of water, g is the gravitational acceleration, and h is the height of the water. The potential energy is relative to the base of the tank, meaning it can be considered zero at that point. The discussion emphasizes that potential energy is not an absolute quantity and can change when water flows to another location. The derivation involves integrating the differential potential energy dU = grAydy, where A is the area of the water surface and r is the water density. Understanding these calculations is essential for assessing the stored potential energy in the tank.
fuzzylogic
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Suppose you have a cylindrical tank of water that is filled to height h. Total mass of water is M. How do you calculate the potential energy of water relative to the base of the tank?
 
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fuzzylogic said:
Suppose you have a cylindrical tank of water that is filled to height h. Total mass of water is M. How do you calculate the potential energy of water relative to the base of the tank?

Welcome to the PF.

What is the context of the question? What do you intend to use the PE for?
 
I just want to know what the stored p.e. is
 
fuzzylogic said:
I just want to know what the stored p.e. is

How come? And how did the water get into the tank? And where is it going to go as it flows out of the tank?

PE is relative to something, it is not an absolute quantity. It is valid to say that the stored PE of water in a tank is zero. And then if the water flows out to some other tank, there may be a change in the stored PE...
 
does this make sense:
g=gravitational acc
y=water level from base
A=area of water surface
r=water density
m=total mass of water
U=PE

then dU=grAydy
U=1/2grAh^2=1/2mgh (m=rAh)
 
fuzzylogic said:
does this make sense:
g=gravitational acc
y=water level from base
A=area of water surface
r=water density
m=total mass of water
U=PE

then dU=grAydy
U=1/2grAh^2=1/2mgh (m=rAh)
Yes, that's fine.
 

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