For example, a half-full cylindrical tank that holds a liquid that is 12 pounds per cubic foot(4ft radius by 8 ft tall). How do we find the work done in pumping the fluid out of the tank from the top outlet?
I tried using Work = Force * distance:
- determined the force is the volume and weight of the liquid, $\pi*(4)^2 * 8/2 * 12 = 768\pi$
- Because work is the integral of force: Work = $\int_{4}^{8}768 \pi dx = 3072\pi$ , which is incorrect (Ans: 4608ft-lb)
Where have I gone wrong here, did I get the value of the force wrong, where is my knowledge gap?
The work done to pump out liquid from the top outlet is different for the liquid at different depths.
Volume of infinitely thin layer of liquid ($dx$) at depth $x$ from top is,
$dV = \pi \cdot 4^2 \cdot dx = 16 \pi ~ dx ~ $ and this liquid needs to be taken $x$ distance against gravity.
So work done in pumping out the liquid in unit ft-lb is,
$ \displaystyle \int_4^8 \rho~ x ~ dV $
where $\rho = 12 $ lb / ft$^3$