Find the work W required to pump the water out of the spout.

Question

A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s² for g. Use 1000 kg/m³ as the weight density of water. Assume that a = 4 m, b = 4 m, c = 12 m, and d = 3 m.)

Answer 1

If the tank has a 4mx4m =16m^2 base, is 12m tall and the spout is at 3m above the ground then all the water above the spout would be flowing out by itself. What would require work would be the water below the spout which ch has the volume of 4mx4mx3m=48m^3, with the masd of 1000kg/m^3 x 48m^3=48,000kg and weight 48,000lg x 9.8m/s^2=470,400N

A simplified way to answer the question:

Work on pumping out that an amount of water is the same as work needed to lift up the waters' center of gravity up to the level of the spout. In our case (water body in the form of a parallelepiped) the center of gravity is midway between the bottom and the top surfaces or at 1.5m

Therefore, the work would be equal to weight x 1.5 m or

470400N x 1.5 m==705,600J. Retaining 2 sig.figs it should be 710,000J=7.1 x 10^5J

Same result we get if breaking up the body of water in thin layers (parallel to the base of the tank) of thickness dx (x - distance of the layer from the spout level and integrate work W=Int (9.8m/s^2 16m^2 xdx) from x=0 to x=d