The work done by emptying a well full of water upto a certain height [closed]

Question

Lets say we have a well full of water (For now lets say it is cylindrical) .Lets say its height is $h$_$well$. The water is filled upto a height $h$_$water$. The radius of the well (cylindrical) is $r$ .Here , $h$_$water$ $\lt$ $h$_$well$ . A Pump has to bring all of that water from the well upto the height $h$_$target$. The question is , after the pump is done emptying the entire well , How much work $W$ the pump has done in total ?

With that , there is also some more general question I can ask along the way.Lets say , I have a function $f(x)$ .Now I will rotate it $360$^$o$ .To create a solid of revolution , and hollow it out. Lets say its height is $h$_$well$. The water is filled upto a height $h$_$water$. Here , $h$_$water$ $\lt$ $h$_$well$ .A Pump has to bring all of that water from the well upto the height $h$_$target$.With this general setup ,How much work $W$ the pump has done in total after the pump is done emptying the entire well ?

Maybe , If I don't want to pull all the water , and let there have some leftover water with height $h$_$leftover$ in the general well , then how much work $W$ I will do?

Answer 1

A simple approach will be to treat the water you're moving as a point mass - to do this, find the height of the center of mass of the water you're going to move. It's very straightforward to calculate the work done to raise a point mass, as it's simply $mgh$. Since water is of uniform density, you just need to find the mean value of the function that defines the well width to find the mean height - for a cylindrical well, the center of mass will be halfway up the column of water. To empty a well of height $h$ that's filled to the brim, each drop of water will have to be moved by $h/2$, on average.

In the example of a cylindrical well with water height below the top of the well, you'll find the center of mass of the water at $h_{water}/2$, which needs to be moved a distance of $h_{well} - h_{water}/2$. The total work that needs to be done is $mg(h_{well} - h_{water}/2)$. Since the mass of the water being moved is also related to the water column height, this may be rewritten as $(\pi r^2h_{well}\rho) g(h_{well} - h_{water}/2)$, where $\rho$ is the density of water.

If you don't drain all the water, you'll need to run these calculations for only the water that's actually being moved, and not the full column. The center of mass of the water being moved will be raised by the height of the water that remains.