Question

A vertical right circular cylindrical tank has height h=8 feet high and diameter d=6 feet. It is full of kerosene weighing 50 pounds per cubic feet. How much work (W) does it take to pump all of the kerosene from the tank to an outlet which is at the level of the top of the tank?

Answer 1

Answer:

The work done to  pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$  

Step-by-step explanation:

The work is defined by:

$W=\int dFdx$ (1)    

The force here will be the product between the volume and the kerosene weighing, so we have :

$dF=\pi R^{2}dy*50$

This force will be in-lbs.

Where R is the radius (3 feet)                    

Then using (1), we have:

$W=\int \pi R^{2}dy*50(8-y)$  

Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.

$W=\int_{0}^{8} \pi 3^{2}dy*50(8-y)$

$W=450\pi \int_{0}^{8}dy(8-y)$

$W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)$

$W=450\pi(8y|_{0}^{8} -\frac{y^{2}}{2}|_{0}^{8})$  

$W=450\pi(8*8 -\frac{8^{2}}{2})$

$W=450\pi(64 -\frac{64}{2})$

Therefore, the work done to  pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$  

I hope it helps you!