🔒 Closed Calculus

[Kuroidonian]

A conical reservoir has an altitude of 3.6 m and its upper base radius is 1.2 m. If it is filled with a liquid of unit weight 9.4 kN/m^3 to a depth of 2.7 m, find the work done in pumping the liquid to 1.0 above the top of the tank.​

 

[🤖]

We can start by finding the volume of liquid in the reservoir. Since the reservoir is conical, we can use the formula for the volume of a cone:

V = (1/3)πr^2h

where r is the radius of the base, h is the altitude (height), and π is approximately 3.14.

Plugging in the given values, we get:

V = (1/3)π(1.2)^2(2.7) ≈ 3.44 m^3

Now, we can find the weight of the liquid:

W = (unit weight)(volume)

W = (9.4 kN/m^3)(3.44 m^3) ≈ 32.46 kN

To find the work done in pumping the liquid to 1.0 m above the top of the tank, we need to find the height the liquid needs to be lifted. Since the new height is 1.0 m above the top of the tank, the total height the liquid needs to be lifted is:

h_total = 3.6 m + 2.7 m + 1.0 m = 7.3 m

The work done is equal to the weight of the liquid times the height it is lifted:

work = W(h_total) = (32.46 kN)(7.3 m) ≈ 237.4 kNm

Therefore, the work done in pumping the liquid to 1.0 m above the top of the tank is approximately 237.4 kNm.