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Thread: i need big help with this "work" problem (tank pumping)

  1. #1
    Sep 2009

    i need big help with this "work" problem (tank pumping)

    i have an exam tomorrow, and am now luckily comfortable with everything but these stupid tank questions. the professor went over an example using a cone, which im still kind of shaky with, but apparently there are other shapes too..
    ok this is pretty hard to explain, because its given by a picture. it says "a tank is filled with water. find the work required to pump the water out of the outlet."

    ill try my best to explain the diagram. Its an upside-down triangular prism. the height of the triangular face is 3 m and its width is 3 m... the length of the rectangular middle region is 8 m... there is a 2m pump extending from the top of a triangle...

    to express visually Google Image Result for

    imagine that picture was inverted, and h and b are both equal to 3m, and l is 8m. there is a small 2m extending from the point where b is....

    EDIT :::::: ok...i just got my work done (which turned out to be incorrect but everything seemed logical, so hopefully im on the right working on typing it up- it should be up within 5 mins or so)

    RE-EDIT:::::::: since mass= density * volume, i did 1000, which is the density of water, times volume which i found to be 8*r(5-h).... then, i multiplied these two by 9.8 for gravity to get weight...

    so we now have (9.8)(1000)[8r(5-h)] ....the 5-h might be wrong, i had 3-h+2, which simplified to 5-h. the +2 i put in because the little pump is an additional 2m.
    so now i want to get an integral in terms of h, so using similar triangles, i found 1.5/3 = r/h or r=1.5h/3
    i plugged that into (9.8)(1000)[8r(5-h)] to get (9.8)(1000)[8(1.5h/3)(5-h)]

    i then put that into an integral from 0 to 3 in terms of dh. i cleaned it up and got 39200 times the integral of h(5-h) from 0 to 3... that was pretty easy to evaluate and i got 529,200 J. this is wrong as the book says the answer is ~1,060,000 J. everything i did made sense to me, so please tell me where i went wrong. i appear to have gotten right around exactly half of the actual answer, so it appears im just missing a 2 multiplier somewhere, but i have no idea where
    Last edited by twostep08; Jul 22nd 2010 at 05:33 AM. Reason: added my work
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  2. #2
    Sep 2009
    anybody have any idea where i went wrong? i tried cleaning it up, but as i said, my answer is half of the actual answer, so im wondering where the extra multiplier is coming from
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  3. #3
    MHF Contributor
    skeeter's Avatar
    Jun 2008
    North Texas
    note the end-view sketch of the trough ...

    right side is $\displaystyle y = 2x$ ...

    horizontal cross-section of liquid has width = 2x = y , length = 8, thickness = dy

    volume of a horizontal cross-section is $\displaystyle 8y \, dy$

    weight-density of water is $\displaystyle 9800 \, N/m^3$

    $\displaystyle \displaystyle W = \int_0^3 9800(5-y)(8y) \, dy = 1,058,400 \, J$
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