Question

A conical tank is buried 1 foot underground with the apex pointing up, base down. The height of the cone is 5 ft and the radius is 6 ft. It is filled with liquid with density k pounds per cubic foot. Work involved in pumping the fluid to ground level?

Answer 1

Let the radius of the conical tank buried 1 foot underground with the apex pointing up be `r =6ft` and its height be `h=5ft`.

Given that the tank is filled with liquid with density k pounds per cubic foot. So the weight liquid per unit volume will be `m=k` lbswt.

Now for the sake of our calculation let us consider the apex of the the tank `(O)` as origin and height `h` along Y-axis.

It is obvious that the rate of increase of radius of the conical tank with depth will be given by `r/h` .

Hence at an arbitrary depth `y` its radius will be `(ry)/h`

So the volume of an imaginary circular disk of infinitesimal thickness `dy` of the liquid layer at this depth `y` will be given by

`dv=pir^2(y/h)^2dy`

So weight of this thin disk of liquid will be

`=m*dv=m*pir^2(y/h)^2dy`

`=pikr^2(y/h)^2dy`

So work done against gravitational pull for lifting this imaginary thin disk to a height of `y` will be given by

`dw="weight"(mdv)xx"height"(y)=pikr^2*(y/h)^2dy*y`

The total work done `W` to lift the liquid at the apex will be obtained by integrating this `dw` as follows where `y` varies from `0toh`

`W=pikr^2int_0^h(y/h)^2ydy`

`=>W=(pikr^2)/h^2[y^4/4]_0^h`

`=>W=(pikr^2)/h^2*[h^4/4]`

`=>W=(pikr^2*h^2)/4`

Again weight of tankful liquid is `1/3pir^2hk` lbs wt. Adding the work done to lift liquid to the ground level ,1ft higher from apex we get total work done

`W_"total"=(pikr^2*h^2)/4+ 1/3pir^2hk*1` ftlbs

`=(pik*6^2)/4*5^2+ 1/3pi*6^2*5k*1` ftlbs

`=285pik` ftlbs

Answer 2

`W = 285 pi \ k " lb ft"`

Explanation:

A simpler approach as this has been moved from Calculus to physics.

The centre of mass of a cone is 3/4 along the way from vertex to base. So an equivalent mass is being lifted and gaining potential energy as follows:

`W = Delta U = m g Delta h`

`= rho V g Delta h`

`= k 1/3 pi r^2 h \ g \ ( 1 + 3/4 h)`

`= k 1/3 pi \ 6^2 * 5 \ g \( 1 + 3/4 * 5)`

`= 285 pi \ g \ k`

I think in these old units you then drop the g, and conclude that:

`W = 285 pi \ k " lb ft"`