Work required to pump tank full of kerosene

In summary, the conversation discusses a problem involving filling a cylindrical tank with kerosene and calculating the work required to do so. The conversation covers various equations and methods, including using weight, volume, and displacement to determine the work done. Ultimately, the solution involves multiplying the weight of the kerosene by the average distance it is raised, which is done through an integral. The correct answer is determined to be approximately 7.24x10^6.
  • #1
m0gh
27
0

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=Fxi
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
 
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  • #2
Depends on how far you 'lift' the kerosene to get it into the tank?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
Right...

What is the weight of this volume of water ? Assume density to be ρ i.e ρ = 51.2
 
  • #7
If the volume of one disc is 100∏dy then I'm guessing (30-y)*100p∏dy would equal the weight?
 
  • #8
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 ?
 
  • #9
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?
 
  • #10
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 .
 
  • #11
Oh I think I see. If it is displaced y units then it should be weight*y ?
 
  • #12
Right...this is what you integrate under proper limits and get total work done.
 
  • #13
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?
 
  • #14
Even using the (30-y) yields the same result after integrating. I think I'm more confused now than I was before.
 
  • #15
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.
 
  • #16
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.
 
  • #17
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##
 
  • #18
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.
 
  • #19
lurflurf said:
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
 
Last edited:
  • #20
m0gh said:
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 ****.
 
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FAQ: Work required to pump tank full of kerosene

What is meant by "work required" when pumping a tank full of kerosene?

Work required refers to the amount of energy needed to transfer the kerosene from an external source into the tank. This includes the force required to lift the kerosene, overcome friction and resistance, and any other factors that may affect the pumping process.

What factors affect the amount of work required to pump a tank full of kerosene?

The amount of work required to pump a tank full of kerosene is affected by the height and diameter of the tank, the viscosity of the kerosene, the pump efficiency, and the distance between the source and the tank. Other external factors such as temperature and pressure can also impact the amount of work required.

How is the work required to pump a tank full of kerosene calculated?

The work required to pump a tank full of kerosene can be calculated by multiplying the force required to lift the kerosene by the distance it needs to be pumped. This is known as the work-energy principle and is expressed as W = F x d, where W is work, F is force, and d is distance.

Is the amount of work required to pump a tank full of kerosene the same for all types of pumps?

No, the amount of work required can vary depending on the type of pump used. Different pumps have different mechanisms and efficiencies, which can affect the amount of work needed to pump the same amount of kerosene. For example, a centrifugal pump may require less work than a reciprocating pump due to its design and operation.

How can the work required to pump a tank full of kerosene be reduced?

The work required to pump a tank full of kerosene can be reduced by using a more efficient pump, minimizing the distance between the source and the tank, and reducing any obstructions or resistance in the pumping process. Additionally, maintaining proper maintenance and regularly cleaning the pump can also help improve its efficiency and reduce the amount of work required.

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