Question

A tank, shaped like a cone has height 6 meters and base radius 5 meters long. It is placed so that the circular part is upward. It is full of water, and we have to pump it all out by a pipe that is always leveled at the surface of the water. Assume that a cubic meter of water weighs 10 000 N , i.e. the density of water is 10000 N m 3 . How much work does it require to pump all water out of the tank

Answer 1

The work required to pump out the liquid is 2.35 MJ.

What is work?

Work can be defined as the scalar product of the force and the displacement.

The formula to calculate the work done W is,

$W=F.d$

where F is the force and d is the displacement.

Note: It is assumed that the weight of one cubic unit of the liquid is $10000 \text{ N/m}^3$.

Break down the vertical tank into parallel disc elements whose thickness is dh and are located at a height h from the ground. The radius of the element at height h is r.

It is given that the height of the tank is 6 m and the base radius is 5 meters. Using these, the ratio of the radius r at height h to the height h by using the triangle similarity criterion is,

$\frac{h}{r}=\frac{6}{5} \\ r=\frac{5h}{6}$

Using the above value of r, the volume of the element is given by,

$V= \pi r^2dh\\V=\pi (\frac{5h}{6})^2dh\\V= \frac{25\pi h^2}{36} dh$

The weight of the volume element is the force required.

The weight of one cubic unit of the liquid is $10000 \text{ N/m}^3$.

So the weight F of the volume element at height h is given by,

$F=10000 (\frac{25 \pi h^2}{ 36})dh\\F= 6944.44\pi h^2dh$

At the height h, the difference in the height of the top of the tank and the volume element is (6-h). So the work required to pump out the volume of liquid at height h is given by the integrating the equation,

$W=6944.44\pi\int\limits^6_0 {(6h^2-h^3)} \, dx \\\\W=69.44\pi \left[ 2h^3-\frac{h^4}{4}\right]^6_0\\W=2,.35\times10^6\text{ J}$

The value of the W obtained is 2.35*10^6 J or 2.35 MJ.

Learn more about work here:

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