Finding Work of a tank full of water Please include details?????????

A tank is full of water. Find the work *W required to pump the water out of the spout. (Use 9.8 for **g and 3.14 for **π*. If you enter your answer in scientific notation, round the decimal value to two decimal places. Use equivalent rounding if you do not enter your answer in scientific notation.)

*W = ????????????????????? *

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Re: Finding Work of a tank full of water Please include details?????????

You can model the tank by where the center of the tank is at (0, 0, 0). For a given z between -r and r, we have a thin "sheet" of water covering the disk which has radius so area . Taking the sheet to have "infinitesmal" thickness dz it has volume and so weight where is the density of water in kilograms per meter. To lift that "sheet" to the top of the sphere, you have to lift it r- z meters and then you have to lift it that additional h meters to the outlet. That is you have to lift weight [tex]g\gamma\pi(r^2- z^2)[tex] a distance meters and so have to do work to lift that sheet out of the tank. To find the total work to empty the tank integrate that from z= -r to r.

Re: Finding Work of a tank full of water Please include details?????????

Yankees2694

This problem is not so easy as it appears to be.

I will try to give you an idea how to solve it using ordinary mathematics but you can also use calculus.

First of all what I understand is that you have one tank the shape of a sphere with radius 3m and above the sphere there is a pipe like a cylinder of 1.5 m in height.

If this data is correct then my solution will be correct.

to solve the problem you have to do the following :

1. calculate first the volume of the sphere V. There is a formula and it is easy.. V=4/3πr^3

2. find the mass of the water inside the tank ,I presume that the tank is full only up to the top of the sphere and the pipe on the top is empty....

you can find the mass of the water using the well known formula of the density....Mass = d x V

3. I presume that the sphere is fixed on a floor or on any horizontal plane . consider another plane parallel to the floor that touches the sphere at a point on the top

where the pipe begins... This I will consider as ground zero....and all the water of the sphere is beneath that plane which has a distance from the botom = 6m.

4. Now consider another plane (midplane) that intersects the sphere passes through the center and divides the sphere into two semispheres the upper one and the

down one... This plane has a distance from the top of the sphere 3m.

5. The work needed to pump out all the water from the sphere and put it at the beginning of the pipe is : Work = Mass x g x r !!!!!

you may be surprized but I will prove it for you. Consider a small mass of water m that is located at a distance d from the mid plane in the upper semisphere and the

same amount of mass that is exactly at a distance d after the midplane and in the same vertical direction. Then the mass in the uper semisphere needs an amount of

work w1= m x gx(r-d) to be pumped out while the amount of work needed by the same mass located in the second semisphere beneath the first is w2 = m x g (r+d)

Adding these two amount of work we get total work W = m x g r + m x g r . this means that mathematically it is sufficient to understand that the work needed

for any amount of mass to be pumped out of the sphere is equivalent the work needed to pump this amount of mass from the CENTER OF THE SPHERE!!!

6, You know now that the total amount of work needed to pump out of the sphere all the water is : Work = Mass x g x r . Add the extra work needed to transfer this amount of water through the pipe out at a height 1.5 mettres or any height you want and you have the total amount of work you are looking for.

MINOAS

Re: Finding Work of a tank full of water Please include details?????????

Let w = weight of water in tank.

Assume the water is emptied onto a large surface level with top of spout.

PE1 = wr

PE2 = w(d+h)

Work = w(r+h)

Re: Finding Work of a tank full of water Please include d

[QUOTE=Hartlw;778217]Let w = weight of water in tank.

Assume the water is emptied onto a large surface level with top of spout.

PE1 = wr

PE2 = w(d+h)

you cant just say PE = Wr without proof ...after all different parts of the water need different amounts of P.E to get out of the sphere...unless you prove that all the water can be pumped out from the center of the sphere!!!!

on the other hand the problem can be solved using the mean value theorem of the integral calculus......

MINOAS

Re: Finding Work of a tank full of water Please include d

Its elementary physics:

The PE of a mass is its weight times distance of CG above a reference line.

Re: Finding Work of a tank full of water Please include details?????????

Hartw

you dont understand ...

it is not a matter of elementary physics or not...but your mistake is that you take all the amount of water from one point .....thats not correct you have to prove it.

it is not a point! An amount od water located at a different distance from the spout needs a different amount of energy to be pumped out .

You can't just said mxgxr without proof....

You must prove that the work is Mxgxr and that it is considered as all the water is pumped out from the centre !!! you have to prove this......

Minoas

Re: Finding Work of a tank full of water Please include details?????????

y_{cg} = ∫ydv/V

PE = ∫ydw = ∫yρdv

PE = ρVy_{cg} = Wy_{cg }Then post #4

Re: Finding Work of a tank full of water Please include details?????????

I am not quite sure if Yankey understands your proof but ...... it asserts my previous findings that the total work is equal to the work MXgxr i.e all the water is pumped from the centre of the sphere which is of course its C.G ......No further comments .

Re: Finding Work of a tank full of water Please include details?????????

Considered as a problem in calculus (this IS a Math Forum), post #2 is correct. If I may offer a slightly different style:

Take a coordinate system at the bottom of the tank. Then the work to raise a slice of water at a height y to a height d+h is

dW = (h+d-y)ρdv, and dv=A(y)dy

Then W is the integral from 0 to d.

For a cylindrical tank of height d and A(y) = A,

the simple integral gives the result:

W = (ρAd)h + (ρAd)d/2

W = wh +wd/2

(For the sphere, HallsofIvy puts the origin of the coordinate system at the center of the sphere to simplify the algebra.)