Math teachers,profs, tutors or smart math people please help..?

Question

A tank in the form of a hemispherical bowl of radius 2m is full of water. Find the work required to pump all of the water to the point 3m above the tank.


I need to correct my midterm exam and i cant figure this out . Please show a few steps so i can understand .
Thanks

Answer 1

Think of thin layers of water that have to be moved up to 3 m above the tank. The top surface has to be moved 3 m, the last little bit on the bottom has to be moved up 3+ radius m up.

The little dV elements are pi*z*dt, where

z is the radius of the surface area which is a function of h, the height measured from the bottom or the top, your choice.

Basically:

dW = dV*g*h is the work required to move that dV slice up to h.

You'll need a trig function to get the radius of the surface as a function of the height of water in the tank.

These dV are very squat cylinders., their height is dh.

Draw a few sketches to get the geometry down.

+ add
To check you work you can get bounds on your answer.
The correct answer should be between these cases:

1. assume all the water is at the surface of the bowl and is raised 3 m

2. assume all the water is at the bottom of the bowl and is raised 3 + 2 = 5 m

This is an example of how engineers learn to 'bound' their answers. If you were going to size a pump to do the job, you'd take the higher work value, right?

Answer 2

muna,
The work required to lift something is W=m*g*h. Define the density of water as D, (which you can look up online, make sure it is in the right units.) V is a volume, so W=V*D*g*h . The volume is cut up into differential slices, which we will integrate over the bowl from x=0 (the top) to R (the bottom, pardon my sign convention). V=dx*A(x) and h=(3+x) (ie, the bottom of the bowl is 5 meters away from the final point). The area at the top of the bowl is pi*R^2, and it gets smaller towards the bottom of the bowl, so A(x)=pi*(x^2-R^2).

Therefore,
W=D*g*pi*integrate( dx*(3+x)*(x^2-R^2) ,x from from 0 to R)
R=2 so
W=D*g*pi*integrate( dx*(3+x)*(x^2-4) ,x from from 0 to 2)
The integration is trivial, I'll leave it to you.

Answer 3

There is no way we should do this for you. Look at all the samples in your text. Work through simpler possibilities to get an idea of the pattern in solving the problem.

Answer 4

mass of water :m=water density * volume=1*(1/2(4/3 pi*r^3))=16/3 pi

w=mgh=16/3pi * 9.8* 3=16*9.8*pi

pi=3.14