# Thread: how much work is required to pump the water

1. ## how much work is required to pump the water

a cone shaped reservoir is 20 ft in diameter across the top and 15ft deep. if the reservoir is filled to a depth of 10 ft, how much work is required to pump all of the water to the top of the reservoir? can you tell me if my setup is correct? thanks in advance..IMG_0003.pdf

2. Originally Posted by slapmaxwell1
a cone shaped reservoir is 20 ft in diameter across the top and 15ft deep. if the reservoir is filled to a depth of 10 ft, how much work is required to pump all of the water to the top of the reservoir? can you tell me if my setup is correct? thanks in advance..IMG_0003.pdf
The mass of the shaded slice is
dm = ρ*π*r^2*dy
Work done to pump out this slice is
dW = ρ*g*π*r^2**(15-y)dy......(1)
r and h are related to R and H by
r/h = R/H. So r = (R/H)*h
Substitute this value in the eq(1) and find the integration.

3. ok can you explain what is g? and your relationship r/h = (R/H)*h? yeah im not quite following all the way...thanks

The mass of the shaded slice is
dm = ρ*π*r^2*dy
Work done to pump out this slice is
dW = ρ*g*π*r^2**(15-y)dy......(1)
r and h are related to R and H by
r/h = R/H. So r = (R/H)*h
Substitute this value in the eq(1) and find the integration.
Yesterday 10:26 PM

4. Work done in lifting a body through a height h is m*g*h where g is the acceleration due to gravity.
If r is the radius of the cone and H is the height, the R, H and the side of the cone forms a right angled triangle. As the level of the water decreases the height also decreases. At any instant r, h and vertex , and R, H and vertex form similar triangles. So the sides are proportional. Hence
R/H = r/h.