# Thread: Work Problem: Help Needed.

1. ## Work Problem: Help Needed.

I need some guidance for this problem:

--x--

Find the work needed to empty the horizontal cylindrical tank if it is full of water. The pump is 2 meters abovce the tank, the tank is 8 meters long (lying on its side), and ther radius of the cylinder is 2 meters.

--x--
Here's what I have so far:
Since work = force * distance, I've gotten the distance as (7-y). The cross section is a rectangle witrh a lenth of 8 meters. To get the width of the rectangle, I used x^2+y^2=25. I solved for y and got: squareroot(25-y^2).
So, the area of the cross-section is 8*squareroot(25-y^2).

It's given by my book that water has a density of 9800 Newtons.

To get the equation of the force, I did: 9800*(7-y)*[8*squareroot(25-y^2)].

Am I supposed to integrate now with boundaries of -5 to 5?
I got stuck in the middle (after factoring out the constant) while trying to integrate it:

78400*[integral from -5 to 5](7-y)*[squareroot(25-y^2)].

Can someone check my work and help me? Thanks.
The answer is 43.1 million Joules, but I'm not sure how to get it.

2. Originally Posted by shiroubi
I need some guidance for this problem:

--x--

Find the work needed to empty the horizontal cylindrical tank if it is full of water. The pump is 2 meters abovce the tank, the tank is 8 meters long (lying on its side), and ther radius of the cylinder is 2 meters.

--x--
It helps if you write the problem correctly. The radius of the tank is 5m.

The minium work needed to empty the tank is equal to the change in
potential energy when the water is lifted by 7m.

The density of water is $\approx 1000\ \mbox{kg/m^3}$, so the mass
of the water in the tank is $m=\pi \times 5^2\times 8 \times 1000 \approx 628000\ \mbox{kg}$

The change in potential energy is:

$
\Delta E=628000 \times 9.91 \times 7 \approx 43.1 \times 10^6\ \mbox{J}
$
,

RonL

3. ## Change in potential energy

Originally Posted by CaptainBlack
It helps if you write the problem correctly. The radius of the tank is 5m.

The minium work needed to empty the tank is equal to the change in
potential energy when the water is lifted by 7m.

The density of water is $\approx 1000\ \mbox{kg/m^3}$, so the mass
of the water in the tank is $m=\pi \times 5^2\times 8 \times 1000 \approx 628000\ \mbox{kg}$

The change in potential energy is:

$
\Delta E=628000 \times 9.91 \times 7 \approx 43.1 \times 10^6\ \mbox{J}
$
,

RonL
I have a doubt about the formula you used.
You have used:
Change in potential energy = m*g*(change in height(h))
The doubt is that is 'h' is same for every element of water?(It is possible only if water is pumped in an identical tank in the same position)

4. Originally Posted by malaygoel
I have a doubt about the formula you used.
You have used:
Change in potential energy = m*g*(change in height(h))
The doubt is that is 'h' is same for every element of water?(It is possible only if water is pumped in an identical tank in the same position)
It is true that the potential energy required to pump a volume/mass element
out of the tank is proportional to the height that the element has to be
raised to reach the height of the pump.

However some elements need to be raised by greater and some by lesser
heights. If you write the resulting integral out you will find you are effectively
defining the height that the centre of mass has to be raised to be on the
same level as the pump.

Alternatively you can observe that the symmetry of the tank ensures that
every volume element x metres above the axis of symmetry can be matched
by an equivalent element the same distance below the axis. The sum of
the potential energies required to raise the elements to the level of the pump
is twice the potential energy required to raise one element of the same mass
from the axis of symmetry to the level of the pump.

The nice thing about the symmetry argument is that it still works when the
tank is tilted!

RonL

5. Originally Posted by shiroubi
I need some guidance for this problem:

--x--

Find the work needed to empty the horizontal cylindrical tank if it is full of water. The pump is 2 meters abovce the tank, the tank is 8 meters long (lying on its side), and ther radius of the cylinder is 2 meters.

--x--
Here's what I have so far:
Since work = force * distance, I've gotten the distance as (7-y). The cross section is a rectangle witrh a lenth of 8 meters. To get the width of the rectangle, I used x^2+y^2=25. I solved for y and got: squareroot(25-y^2).
So, the area of the cross-section is 8*squareroot(25-y^2).

It's given by my book that water has a density of 9800 Newtons.

To get the equation of the force, I did: 9800*(7-y)*[8*squareroot(25-y^2)].

Am I supposed to integrate now with boundaries of -5 to 5?
I got stuck in the middle (after factoring out the constant) while trying to integrate it:

78400*[integral from -5 to 5](7-y)*[squareroot(25-y^2)].

Can someone check my work and help me? Thanks.
The answer is 43.1 million Joules, but I'm not sure how to get it.
Here is one way.

If the x^2 +y^2 = 25 is correct, then the radius of the cylinder is 5m, not 2m as posted.

Then the idea of your solution is correct.

Work = force*distance
The force here is the weight of the dV.
The distance here is 2m plus the distance of the dV from the top of the cylinder. The water will not go higher than the level of the pump---the water will be pumped off at the pump level.

You placed the (0,0) at the center of the circle of the cylinder. So the boundaries of the dV is from -5m to 5m.
And so, also, the distance traveled by dV, for the measure of work done, is 2 +(5-y), or (7-y).

The dV is a horizontal plate that is (2x) wide, (8m) long, and (dy) thick.
dV = (2x)(8)(dy) = 16x dy
Since the integration is for dy, you expressed the x in terms of y,
x^2 +y^2 = 25
x^2 = 25 -y^2
x = sqrt(25 -y^2).
Hence, the dV is 16sqrt(25-y^2)dy
So, the force due to that is
dWeight = (volume)*(density) = (16sqrt(25-y^2)dy)*(9800) = 156000sqrt(25-y^2)dy.

Then, dWork = (7-y)*(156000sqrt(25-y^2)dy)
Integrate both sides,
Work = (156000)INT.(-5 -->5)[(7-y)sqrt(25-y^2)]dy
Separating that,
Work = (156000)INT.(-5 -->5)[7sqrt(25-y^2)]dy - (156000)INT.(-5 -->5)[y *sqrt(25 -y^2)]dy
Work = (1,097,600)INT.(-5 -->5)[sqrt(25-y^2)]dy - (156000)INT.(-5 -->5)[y *sqrt(25 -y^2)]dy

Using the Table of Integrals, (for easy solution. If you need to derive the integral without using the Table, then the solution is just longer),
INT.[sqrt(a^2 -x^2)]dx = (x/2)sqrt(a^2 -x^2) +(a^2 /2)[arcsin(x/a)] ---**

The INT.[y*sqrt(25-y^2)]dy will come to
= INT.[(25-y^2)^(1/2)](-2y / -2)(dy)
= -(1/2)INT.[(25-y^2)^(1/2)](-2y dy)
= -(1/2)(2/3)(25-y^2)^(3/2) +C
= -(1/3){sqrt[(25-y^2)]}^3 +C -----***

So,
Work = (1,097,600){[(5/2)sqrt(25 -y^2) +(25/2)arcsin(y/5)](-5 -->5)} +(156000)(1/3){[sqrt(25-y^2)]^3}(-5 -->5)}
Work = (1,097,600){0 +(25/2)arcsin(1) -0 -(25/2)arcsin(-1)} +(156000/3){0 -0}
Since arcsin(1) = pi/2, and arcsin(-1) = -pi/2, then,
Work = (1,097,600)(25/2){pi/2 -(-pi/2)}
Work = (1,097,600)(12.5){pi}