Calculate Work to Pump Water out of Tank | Joules Required

[fubag]

Calculate the work (in joules) required to pump all of the water out of tank. Assume that the tank is full, distances are in meters, and the density of water is 1,000 kg/m^3.


The conical tank (pointing downwards) has water exiting a spout feet above the tank. Find the work down in removing the water. Radius of tank = 5 meters, height = 10 meters, and then spout is 2 meters above this.

Well, I am not sure how to start this.

How can I find the volume of a slice? I also noticed that we have enough information to just solve for the Volume of the cone = 1/3pir^2h= 250pi/3.

If Work = Force * distance.

Force = g*density*volume
so can I just plug it in here?

F = (9.8)(1000)(250pi/3)

And then distance it needs to travel is 12 feet? 10 to go to the top of the conical tank, and plus 2 to exit the spout?


I am not sure, any help will be deeply appreciated.

 

[Orodruin]

And then distance it needs to travel is 12 feet? 10 to go to the top of the conical tank, and plus 2 to exit the spout?

All of the water does not have to be moved from the bottom to 2 meters above the tank, this is only true for the water at the bottom of the pool. Instead, should write down the work needed to lift a thin slice of thickness $dh$ at a height $h$ over the bottom to the exit spout. The volume of this slice is given by
$$
dV = \pi r(h)^2 dh,
$$
where $r(h) = h/2$ is the radius of the tank at height $h$. The work required to lift this to 2 meters above the pool is given by
$$
dW = (12-h) \rho g dV
$$
by the exact same physical reasoning you hinted at. Here, the $12-h$ is the height that the slice needs to be lifted. Integrating over all slices, you would obtain
$$
W = \int dW = \int_0^{10} (12-h) \rho g \pi \frac{h^2}{4} dh.
$$
Solving this integral and inserting the appropriate values for $\rho$ and $g$ will give you the correct answer.