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Pumping gasoline

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Gasoline is stored in a cylindrical tank buried on its side, with the highest part of the tank 5 ft below the surface. The tank is 6 ft in diameter and 10 ft long. Density of gasoline is 45 lb/cubic ft. Assume that the filler cap of each automobile gas tank is 2 ft above the ground.
    (a) How much work is done in emptying all the gasoline from the tank, which is initially full?

    (b) Recall that 1 hp is equivalent to 33,000 ft-lbs/min. For electrical conversions 1 kW (1000W) is the same as 1.341 hp. The charge for use of electricity generated by a power company is about 7.2 cents per kWh. Assume that the electrical motor in the gas pump is 30% efficient. How much does it cost to pump all the gasoine from this tank?


    2. Relevant equations

    see above

    3. The attempt at a solution
    (a)volume of a generic "slice" = 10 * 2x[tex]\Delta[/tex]y = 20[tex]\sqrt{9-y^2}[/tex][tex]\Delta[/tex]y
    Force acting on that slice = 45 * 20[tex]\sqrt{9-y^2}[/tex][tex]\Delta[/tex]y
    W to pump that slice up =
    900[tex]\sqrt{9-y^2}[/tex][tex]\Delta[/tex]y(10-y)
    Total work = [tex]\int900\sqrt{9-y^2}(10-y)[/tex] from -3 to 3 [tex]\approx[/tex] 127,234.5 ft-lbs

    (b)I know this isn't right... And it's probably very confusing as well.
    127,234.5 ft-lbs * (1 hp/ 33,000 fl-lbs/min) = 3.8559 hp * min * (1 kW/ 1.341 hp) = 2.875 * 1000 J/s * min = 2.875 * 1000 J/s * 60 s = 175,209.66 J * (1 kWh/ 3,600,000 J) = .0479 kWh * 7.2 cents = .345 cents
    for efficiency, I divided the final answer by .3 which gives me 1.15 cents, which is obviously not right.

    The first part may be right, but I really don't know what to do on the second part. Any help would be greatly appreciated.
     
  2. jcsd
  3. Dec 7, 2009 #2

    Dick

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    Ok, so 1 cent worth of power will buy you 1.341*33000*60*(0.3)/7.2 ft-lbs of energy counting the efficiency factor. If I divide 127234 by that I get 1.15 cents. I don't think that's obviously not right. Though I've got to confess the units conversions are making me dizzy.
     
    Last edited: Dec 8, 2009
  4. Dec 8, 2009 #3
    Thanks for your help.
    So it appears that 1.15 cents is correct? I just thought it would cost so much more for all of that work! If you're up to it, would you mind checking if I calculated the work correctly?
    Man, calculating work is a ton of work!:eek:
     
  5. Dec 8, 2009 #4

    Dick

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    It actually easy to do in this case. The work is just the work required to raise the center of mass of the underground tank to the level of the automobile, 10 ft.
     
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