Work required to pump tank full of kerosene

[m0gh]

Homework Statement

Right cylindrical tank measures 30ft high and 20ft in diameter. How much work is required to fill the tank with kerosene which has a weight of 51.2 pounds per cubic feet

Homework Equations

I know that W=Fx_i
and F=mg

The Attempt at a Solution

My professor did a similar problem involving draining a cone but I honestly have no clue where to start on this one. I think if I can get it set up I can finish it

 

[Mikex007]

Depends on how far you 'lift' the kerosene to get it into the tank?

 

[m0gh]

I'm assuming the problem is asking from the bottom of the tank. The problem reads: "It is full of kerosene weighing 51.2 lb/ft^3. How much work does it take to pump the kerosene to the level of the top of the tank"

If you ask me the problem states that the tank is already full and 0 work is required but I know that would be too easy.

 

[Tanya Sharma]

Consider a cross section of the cylinder at height y having thickness dy .What is the volume of water contained in this cross section ?Please work with variables only.You may fill in the values later.

The question doesn't mention but I am assuming that water is pumped in from ground level.

 

[m0gh]

If you mean a cross section that creates a circular disc then the Volume would be

V = ∏(r²)dy Substitute 10 for r which gives V=∏ 100 dy?

 

[Tanya Sharma]

Right...

What is the weight of this volume of water ? Assume density to be ρ i.e ρ = 51.2

 

[m0gh]

If the volume of one disc is 100∏dy then I'm guessing (30-y)*100p∏dy would equal the weight?

 

[Tanya Sharma]

No...it is simply density times volume .

So weight of water W = ρV .

Now how much work is done in order to fill this slice of the tank i.e work required to raise this amount (weight) of
water from ground level to level y ?

 

[m0gh]

I'm sorry the wording of these problem types always gets me confused.

So if the weight W = 100∏p*r^2*dy

Then the work is determined by the integral of the weight from 0 to y?

 

[Tanya Sharma]

No...

Work is force times displacement .Weight is force so how can you integrate weight and get work done .

Read post#8 and think again .

 

[m0gh]

Oh I think I see. If it is displaced y units then it should be weight*y ?

 

[Tanya Sharma]

Right...this is what you integrate under proper limits and get total work done.

 

[m0gh]

Thank you so much. I took the integral from [0,30] of ∫51.2*100*pi*y*dy

which gave: 5120∏[(y^2)/2] from [0,30]

5120∏[450-0] = 2,304,000∏

I hope this is correct I feel a lot better about these problems if so.

EDIT: This answer is not correct. I just searched for this problem online and found the solution here: http://calclab.math.tamu.edu/~belmonte/m152/E/E_2009c/x1a_sols.pdf It is the last problem on the page (#7)

The only difference is my attempt uses just (y) while theirs uses (30-y) can you explain this?

 

[m0gh]

Even using the (30-y) yields the same result after integrating. I think I'm more confused now than I was before.

 

[Tanya Sharma]

Yes... the two integrals give same value .But no need to get confused.

We have calculated the work done in filling the tank by pumping kerosene in from the ground level.Each slice of kerosene was raised from ground level i.e y=0 to level y i.e raised through a a height y.Hence multiplication by y.

In the link you have given ,the question asks to calculate work required to pump the kerosene to the level of top of the tank .Each slice of kerosene was raised from level y to level y=30 ,i.e raised through a a height 30-y.Hence multiplication by 30-y.

 

[lurflurf]

Wow I am confused. We are pumping the kerosine (not water) out (not in).
We will need to know

work=weight*average distance
weight=weight density*volume
volume cylinder=(pi/4)*diameter^2*height

find the average distance either with an integral or similar.

 

[BiGyElLoWhAt]

Ahhh, the 30 - y comes into play because as you pump, the work required to pump it in changes (linearly)
Your trying to figure out the work required to pump from 0 to dy, then from dy to 2dy, then from 2dy to 3dy...

This makes sense because as you pump more kerosene into the tank it fills up and the contained kerosene feels a weight force and exerts that force as W on the kerosene you're pumping in, which requires you to do more work to get it in there.

It appears as though they are saying there is no work required to pump it in (without the weight of the kerosene in there. So basically you're integrating to find the work done on the "pump" by the kerosene in the container.

$MATH$
What's the weight force from an arbitrary amount of kerosene?

$\rho V = \rho Ay$
but, for any given amount of kerosene what is the work required to "lift" that kerosene to the top?
keeping in mind $W = \int F (dot) dr$
we have $W =\int \rho A dy (dot)$ (distance from the height of the kerosene [y] to the top)
This happens to be a displacement parallel to the force, so we can drop the dot product and what we have is
$W= \rho A \int_{a}^b (30 - y)dy$
more specifically
$W = \rho A \int_{0}^{30}(30-y)dy = 5120\pi (30y-\frac{y^2}{2})|_0^{30}$
$=5120(450) \pi \approx 7238229.474 \approx 7.24x10^6$

 

[m0gh]

I'm really sorry. I just forgot to multiply the pi and that's why I thought it was wrong.

Thank you, Tanya. You have been very helpful and patient. I really enjoy when I ask for help and someone actually helps me to understand instead of just spitting out the answer.

 

[BiGyElLoWhAt]

volume cylinder=(pi/4)*diameter^2*height

No.
a cylinder has a circle for a cross section and a height.
V=Ah
and for a cylinder this is pi r^2 h

[edit*] Retracted. Didn't realize that said diameter. Lol

 

[BiGyElLoWhAt]

I'm really sorry. I just forgot to multiply the pi and that's why I thought it was wrong.

Thank you, Tanya. You have been very helpful and patient. I really enjoy when I ask for help and someone actually helps me to understand instead of just spitting out the answer.

hahaha yea I did that too, then I went back a step and saw the pi and was like oh ****.