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?

1 Answer
Jan 6, 2018

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)