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anonymous
 4 years ago
Anyone around to help?
anonymous
 4 years ago
Anyone around to help?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have an equation but i dont know if it is correct

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328518737850:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328518861214:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0forgotten physics little bit, i believe Work = Force *distance

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in other words i suck at everything

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That looks correct to me, but don't quote me on it.... like dumbcow, it's been a while :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but like for pressure i used 62.4 i dont know if that is correct

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How did you arrive at that number?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhh i forgot to include that the oil weighs

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hold on a second, is that the formula for pressure volume work?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I totally guessed this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0ummm no when you are removing the oil the oil wieghs 50 lbs/ft^3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think you must take that into account

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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
 4 years ago
Best ResponseYou've already chosen the best response.0ya but what happens to the integral?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Disclaimer: I'm conjecturing here, as I'm really not sure.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that 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
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0umm no i think you gotta pump it upwards

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, that's what the Wl arrow is supposed to indicate. The P is the pressure from the oil's weight.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol physics is too confusing for me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh... I just saw something: you're actually starting to pump from 256 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
 4 years ago
Best ResponseYou've already chosen the best response.0At 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
 4 years ago
Best ResponseYou've already chosen the best response.0I do second what dumbcow said though  double check with someone in the physics group to be safe.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol i willl. thanks :D

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, 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?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Suppose 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \Delta V = (Area \ of \ circle)\Delta y = \pi r^2 . \Delta y = 25 \pi \Delta y \ \ [ft^3] \] 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?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Now, 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 \[ \Delta W = \Delta m.gy \] \[ = (1250\pi g) y . \Delta y\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Hence 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
 4 years ago
Best ResponseYou've already chosen the best response.0ohh i used a diff equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and got a diff answer

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 4 years ago
Best ResponseYou've already chosen the best response.0On 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 tankwhich is 5 ft above the surface of the groundback to the level of ground.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328552206561:dw

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Barring 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
 4 years ago
Best ResponseYou've already chosen the best response.0ya i was confused with the set up. but lik e y are my bounds diff?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Yes 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
 4 years ago
Best ResponseYou've already chosen the best response.0umm let me see what i understood

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Decide which it is and use the appropriate bounds.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328552517884:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Didn'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
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, I think you're right. So integrate from 19 to 25

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh ok thanks guys :D
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