Question

How much work does it take to pump the oil to the rim of the tank if the conical tank from `y=2x` is filled to within 2 feet of the top of the olive oil weighing `57(lb)/(ft)^3`?

Answer 1

`W= 1/16pirhogy^4 = 5730 lb*ft`, if y =2.0ft .

The height data provided is confusing: Is the oil level 2 ft high or 2 ft from the rim? The answer assumes y= 2ft high.

Explanation:

Work is defined as: `W = intvecf*dvecs`

The minimal work done to lift a fluid particle with mass `deltam` from the bottom of the cone to height y is:

`deltaW = int_0^y(deltam) g dy = deltam g y`

Hence the work to bring a layer of fluid at level y in the cone should be:

`deltaW =deltam g y=(rhodv)gy=rhogy(pix^2)dy`
where
`rho=` density of the fluid; `dv=Ady = pix^2dy`
`x =y/2` cross-section radius of a fluid layer in the cone

`W= int_0^y 1/4pirhogy^3dy = 1/16pirhog(y^4)_0^y=1/16pirhogy^4`

Let y = 2ft fills when the fluid fills the cone up to the rim

`W= 1/16pirhog (2)^4 =pirhog`

`W = pi (57 (lb)/(ft^3))(32 (ft)/sec^2)(ft^4)=5730 lb*ft`

Basically the work is equal to potential energy of the cone filled with oil. Applying the potential energy concept (mgh) yields the same result.

Note: the data on the fluid level is confusing. Is the fluid 2 ft high or 2 ft from the rim? If the later is the case, the height of the cone is not given. The solution if assumes the general form is:

`W= int_(y_1)^(y_2) 1/4pirhogy^3dy = 1/16pirhog(y^4)_(y_1)^(y_2)=1/16pirhog(y_2^4-y_1^4)`