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anonymous

  • 4 years ago

Anyone around to help?

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    I have an equation but i dont know if it is correct

  3. anonymous
    • 4 years ago
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    ...

  4. anonymous
    • 4 years ago
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    |dw:1328518737850:dw|

  5. anonymous
    • 4 years ago
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    umm can you read it?

  6. anonymous
    • 4 years ago
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    |dw:1328518861214:dw|

  7. anonymous
    • 4 years ago
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    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?

  8. dumbcow
    • 4 years ago
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    forgotten physics little bit, i believe Work = Force *distance

  9. anonymous
    • 4 years ago
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    umm yes

  10. anonymous
    • 4 years ago
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    I suck at this

  11. anonymous
    • 4 years ago
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    in other words i suck at everything

  12. anonymous
    • 4 years ago
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    That looks correct to me, but don't quote me on it.... like dumbcow, it's been a while :P

  13. anonymous
    • 4 years ago
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    but like for pressure i used 62.4 i dont know if that is correct

  14. anonymous
    • 4 years ago
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    How did you arrive at that number?

  15. anonymous
    • 4 years ago
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    ohhh i forgot to include that the oil weighs

  16. anonymous
    • 4 years ago
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    Hold on a second, is that the formula for pressure volume work?

  17. anonymous
    • 4 years ago
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    umm ya

  18. anonymous
    • 4 years ago
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    I totally guessed this

  19. anonymous
    • 4 years ago
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    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

  20. anonymous
    • 4 years ago
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    ummm no when you are removing the oil the oil wieghs 50 lbs/ft^3

  21. anonymous
    • 4 years ago
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    i think you must take that into account

  22. anonymous
    • 4 years ago
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    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)

  23. anonymous
    • 4 years ago
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    ya but what happens to the integral?

  24. anonymous
    • 4 years ago
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    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

  25. anonymous
    • 4 years ago
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    Disclaimer: I'm conjecturing here, as I'm really not sure.

  26. anonymous
    • 4 years ago
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    lol ok

  27. dumbcow
    • 4 years ago
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    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 :)

  28. anonymous
    • 4 years ago
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    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.

  29. anonymous
    • 4 years ago
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    umm no i think you gotta pump it upwards

  30. anonymous
    • 4 years ago
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    Yeah, that's what the Wl arrow is supposed to indicate. The P is the pressure from the oil's weight.

  31. anonymous
    • 4 years ago
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    lol physics is too confusing for me

  32. anonymous
    • 4 years ago
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    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.

  33. anonymous
    • 4 years ago
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    r u sure?

  34. anonymous
    • 4 years ago
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    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 ;)

  35. anonymous
    • 4 years ago
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    I do second what dumbcow said though - double check with someone in the physics group to be safe.

  36. anonymous
    • 4 years ago
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    lol i willl. thanks :D

  37. JamesJ
    • 4 years ago
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    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?

  38. anonymous
    • 4 years ago
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    yes

  39. JamesJ
    • 4 years ago
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    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 ...

  40. JamesJ
    • 4 years ago
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    \[ \Delta V = (Area \ of \ circle)\Delta y = \pi r^2 . \Delta y = 25 \pi \Delta y \ \ [ft^3] \] 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?

  41. anonymous
    • 4 years ago
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    yes

  42. JamesJ
    • 4 years ago
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    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 \[ \Delta W = \Delta m.gy \] \[ = (1250\pi g) y . \Delta y\]

  43. anonymous
    • 4 years ago
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    ok

  44. JamesJ
    • 4 years ago
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    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., \[ W = \int_4^{10} 1250\pi g y \ dy \] \[ = 1250\pi g \int_4^{10} y \ dy \] \[ = \frac{1250 \pi g}{2} [ y^2 ]_4^{10} \] \[ = 625 \pi g (100 - 16) \] \[ = 52,500 \pi g \]

  45. anonymous
    • 4 years ago
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    ohh i used a diff equation

  46. anonymous
    • 4 years ago
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    and got a diff answer

  47. JamesJ
    • 4 years ago
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    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.

  48. JamesJ
    • 4 years ago
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    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.

  49. anonymous
    • 4 years ago
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    |dw:1328552206561:dw|

  50. JamesJ
    • 4 years ago
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    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.

  51. anonymous
    • 4 years ago
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    ya i was confused with the set up. but lik e y are my bounds diff?

  52. JamesJ
    • 4 years ago
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    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?

  53. anonymous
    • 4 years ago
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    umm let me see what i understood

  54. JamesJ
    • 4 years ago
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    Decide which it is and use the appropriate bounds.

  55. anonymous
    • 4 years ago
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    |dw:1328552517884:dw|

  56. anonymous
    • 4 years ago
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    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.

  57. JamesJ
    • 4 years ago
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    Yes, I think you're right. So integrate from 19 to 25

  58. anonymous
    • 4 years ago
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    ohh ok thanks guys :D

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