1. ## work by pumping

A circular swimming pool has a diameter of 16 m, the sides are 4 m high, and the depth of the water is 3.5 m. How much work (in Joules) is required to pump all of the water over the side? (The acceleration due to gravity is 9.8 and the density of water is 1000 .)

so i tried
volume=64pi dx

weight= vol *density

64000pidx

integral .5 to 4 64000pidx

2. Originally Posted by dat1611
A circular swimming pool has a diameter of 16 m, the sides are 4 m high, and the depth of the water is 3.5 m. How much work (in Joules) is required to pump all of the water over the side? (The acceleration due to gravity is 9.8 and the density of water is 1000 .)

so i tried
volume=64pi dx

weight= vol *density

64000pidx

integral .5 to 4 64000pidx
did you learn anything from this post?

http://www.mathhelpforum.com/math-he...k-pumping.html

3. ya but its different i followed the exact way my teacher did it but its not working

4. Won't the work required to be done just be the gain in in gravitational PE of the entire water?

The total mass of the water= Volume of the water(density)
= 1000(pi)(64)(3.5)=m(say.)

Now when the water is pumped over the side, the gain in gravitational PE is mg(4-3.5) = mg(0.5)
=3448220.16 J ??????

5. If interested in general problems which involve work to empty a tank

See Work on my web site

6. Originally Posted by bandedkrait
Won't the work required to be done just be the gain in in gravitational PE of the entire water?

The total mass of the water= Volume of the water(density)
= 1000(pi)(64)(3.5)=m(say.)

Now when the water is pumped over the side, the gain in gravitational PE is mg(4-3.5) = mg(0.5)
=3448220.16 J ??????
that is not right though

7. Originally Posted by dat1611
that is not right though
Oh god that was a monumental blunder!!!!

Ignore my first post, please. Or preserve it as an example of easily avoidable human stupidity!!!!!!

8. can anyone else try to solve it

9. Since the density is constant and the cross sections are constant

we don't need an integral.

So we think of the mass as being concentrated at the center of gravity

here y= 1.75 and the distance 4 - 1.75 = 2.25 m

weight is pi * 64*1000*9.8*3.5

Calculate weight * dist = 64*1000*9.8*3.5*pi*2.25

If you do integrate the limits are 0 to 3.5

you have 64*1000*9.8*pi integral[(4-y)dy]