anonymous
  • anonymous
Anyone around to help?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I have an equation but i dont know if it is correct
anonymous
  • anonymous
...

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anonymous
  • anonymous
|dw:1328518737850:dw|
anonymous
  • anonymous
umm can you read it?
anonymous
  • anonymous
|dw:1328518861214:dw|
anonymous
  • anonymous
yeah, more or less :P if i'm reading this correctly, it's assuming that we're pumping the oil from the top and not from the bottom? that's why the integral?
dumbcow
  • dumbcow
forgotten physics little bit, i believe Work = Force *distance
anonymous
  • anonymous
umm yes
anonymous
  • anonymous
I suck at this
anonymous
  • anonymous
in other words i suck at everything
anonymous
  • anonymous
That looks correct to me, but don't quote me on it.... like dumbcow, it's been a while :P
anonymous
  • anonymous
but like for pressure i used 62.4 i dont know if that is correct
anonymous
  • anonymous
How did you arrive at that number?
anonymous
  • anonymous
ohhh i forgot to include that the oil weighs
anonymous
  • anonymous
Hold on a second, is that the formula for pressure volume work?
anonymous
  • anonymous
umm ya
anonymous
  • anonymous
I totally guessed this
anonymous
  • anonymous
I think that's for calculating the work done by an expanding body... like when you start heating the oil it exerts pressure and hence does work
anonymous
  • anonymous
ummm no when you are removing the oil the oil wieghs 50 lbs/ft^3
anonymous
  • anonymous
i think you must take that into account
anonymous
  • anonymous
So there's a pressure from the volume of oil, i get what you mean... i'm still not sure, I've only ever seen the formula used for expanding bodies. I was thinking something simpler like this: The bottom of the cylinder is 25 feet below the surface. So you're trying to lift pi*r^2*h, so 471 cubic feet of oil by 25 feet. That's 471*50 = 23550lb you need to lift by 25 feet to get it from the bottom of the cylinder to the surface - assuming your hose is attached to the bottom of the cylinder (which makes sense to me)
anonymous
  • anonymous
ya but what happens to the integral?
anonymous
  • anonymous
So, that's an amount of work to lift the oil, Wl. If you'd want to take the pressure from the oil's weight into account, that would amount to work Wp, but that pressure would make lifting the oil to the surface *easier*, wouldn't it? It's pushing through the hose by its own weight, reducing the amount of pumping you'd have to do. So, taking the weight of the oil into account, it would be Wl - Wp as the final work - how to exactly figure out the Wp, I'm not sure - it may well be that the pressure volume work formula will work for that :P
anonymous
  • anonymous
Disclaimer: I'm conjecturing here, as I'm really not sure.
anonymous
  • anonymous
lol ok
dumbcow
  • dumbcow
that makes sense to me, then you could convert lbs to N for Force needed to lift that weight but yeah try physics forum to get help setting up the right equations...then we can help with the math :)
anonymous
  • anonymous
I hope what I said makes some sort of sense though - it's like this: |dw:1328520553141:dw| That's barely discernible, but may illustrate. The pressure P is actually resulting in a force in the same direction as the pumping, so it would subtract from the work needed to lift the oil.
anonymous
  • anonymous
umm no i think you gotta pump it upwards
anonymous
  • anonymous
Yeah, that's what the Wl arrow is supposed to indicate. The P is the pressure from the oil's weight.
anonymous
  • anonymous
lol physics is too confusing for me
anonymous
  • anonymous
Oh... I just saw something: you're actually starting to pump from 25-6 feet depth, and only end up at 25 feet when the tank is empty. So to be precise, you would have to integrate over the distance from 19 to 25 feet.
anonymous
  • anonymous
r u sure?
anonymous
  • anonymous
At the beginning, you need to lift the oil only 19 feet. When the tank is empty, the last drop, you'll have to lift 25 feet. I'm pretty sure that's what needs integrating here ;)
anonymous
  • anonymous
I do second what dumbcow said though - double check with someone in the physics group to be safe.
anonymous
  • anonymous
lol i willl. thanks :D
JamesJ
  • JamesJ
Ok, so I as I read the set up the oil is at a depth of 4 to 10 feet below the surface of the ground. Do you agree with that?
anonymous
  • anonymous
yes
JamesJ
  • JamesJ
Suppose now you want to move a thin layer of oil at the top of the tank to the surface. Suppose that thin layer has a thickness of \( \Delta y \) ft. Then the volume of that layer is ...
JamesJ
  • JamesJ
\[ \Delta V = (Area \ of \ circle)\Delta y = \pi r^2 . \Delta y = 25 \pi \Delta y \ \ [ft^3] \] and its mass is \[ \Delta m = \Delta V . (Mass \ per \ ft^3) \] \[ = 25 \pi \Delta y . (50) \ \ [lb] \] \[ = 1250 \pi \Delta y \ [lb] \] So far, so good?
anonymous
  • anonymous
yes
JamesJ
  • JamesJ
Now, the work required to move that mass to the surface will depend on its depth, let's call that depth \( y \). Then the work required to move that mass to the surface is \[ \Delta W = \Delta m.gy \] \[ = (1250\pi g) y . \Delta y\]
anonymous
  • anonymous
ok
JamesJ
  • JamesJ
Hence the total work is the sum of moving all of those layers of oil from y = 4 ft to y = 10 ft. I.e., \[ W = \int_4^{10} 1250\pi g y \ dy \] \[ = 1250\pi g \int_4^{10} y \ dy \] \[ = \frac{1250 \pi g}{2} [ y^2 ]_4^{10} \] \[ = 625 \pi g (100 - 16) \] \[ = 52,500 \pi g \]
anonymous
  • anonymous
ohh i used a diff equation
anonymous
  • anonymous
and got a diff answer
JamesJ
  • JamesJ
Now, there's a couple of interpretation questions here. In the way we've done the problem here, we've assumed the only thing we need to worry about is the net work moving the oil from the tank to the level of the ground. In pragmatic terms however, you might have to do work moving all the oil to the top of the tank and then let gravity move it from there to the ground, that second step being work you couldn't recover.
JamesJ
  • JamesJ
On the other hand, I can imagine a set up where the pump is at the level of the ground, in which case, the calculation as is stands, because you do recover the energy of the oil moving from the top of the tank--which is 5 ft above the surface of the ground--back to the level of ground.
anonymous
  • anonymous
|dw:1328552206561:dw|
JamesJ
  • JamesJ
Barring any other information, I'd go with the solution as I've laid it out. But if you get a chance to ask the question before submitting the answer, I'd ask.
anonymous
  • anonymous
ya i was confused with the set up. but lik e y are my bounds diff?
JamesJ
  • JamesJ
Yes they are. is the oil at a depth of 4 to 10 feet underground? Or at a depth of 14 to 20 feet underground?
anonymous
  • anonymous
umm let me see what i understood
JamesJ
  • JamesJ
Decide which it is and use the appropriate bounds.
anonymous
  • anonymous
|dw:1328552517884:dw|
anonymous
  • anonymous
Didn't the problem state that the *top* of the tank is 10 feet below ground level? That would put the oil between 19 and 25 feet below ground level.
JamesJ
  • JamesJ
Yes, I think you're right. So integrate from 19 to 25
anonymous
  • anonymous
ohh ok thanks guys :D

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