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Change of units problem

  1. Mar 19, 2015 #1
    I think this problem is straightforward , but I'm not used to these units. (Normally I use SI units)

    From a physics book (U.S version ) I have this equation:

    ##h=10 + 4.43*Q^2##

    where the unit of ##h=[ft]##, and ##Q=[ft^3/s]##

    now the book rewrites the above equation, such that ##Q=[gal/min]## and get

    ##h=10 + 2.2*10^{-5}*Q^2##


    I just can't see what the book does to get from the first equation to the secound. Can anyone help me with this?

    Best regard
    J. Maack
     
  2. jcsd
  3. Mar 19, 2015 #2

    RUber

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    Homework Helper

    1 Gallon is equal to 0.133681 ft3.
    Does that math work out?
     
  4. Mar 19, 2015 #3

    DrClaude

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    Staff: Mentor

    Knowing that 1 gal = 0.1337 ft3, can you do it?

    [Edit: beaten by RUber...]
     
  5. Mar 19, 2015 #4
    I can't get the same result as the book with that 1 gal = 0.1337 ft3

    What i do is
    ##4.43*(1/60^2)*0.1337##

    Isn't that correct?
     
  6. Mar 19, 2015 #5

    DrClaude

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    Staff: Mentor

    Nope.

    Personally, I don't like those kind of equations which work only for certain choices of units. So do as I do, start by figuring out what the units of the factor 4.43 have to be for the original equation to hold.
     
  7. Mar 19, 2015 #6
    The unit for 4.43 would be ##[s^2/ft^5]## for the equation to hold.
     
  8. Mar 19, 2015 #7

    DrClaude

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    Staff: Mentor

    That's correct, but to simplify further calculations, keep the volume in there (and the units of h): [ft (s/ft3)2]. Now you should be able to do the conversion to [gal/s].
     
  9. Mar 19, 2015 #8
    Perfect! Tanks
     
  10. Mar 19, 2015 #9

    HallsofIvy

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    First, in order for this to makes sense the different units must measure the same thing! That is the case here- both "cubic feet" and "gallons" measure volume, both "seconds" and "minutes" measure time. "cubic feet per second" and "gallons per minute" measure how fast volume is changing- perhaps how fast a liquid is moving through a pipe.

    h, in the first equation is in feet and Q is in cubic feet per second. But Q is squared so [itex]Q^2[/itex] has units of "cubic feet squared" (or feet to the sixth power) over "seconds squared". That tells us that the "10" must have units of feet and the "4.43" must have units of "seconds squared over feet to the fifth power": [itex]\frac{s^2}{ft^5}\times\frac{ft^6}{s^2}= ft[/itex] which can then be added to the "10 ft" to get "h ft".

    In the second equation, h is still in feet but now Q is in gallons per minute so the "[itex]2.2*10^{-5}[/itex]" must have units of [itex]\frac{min^2 ft}{gal^2}[/tex] so that [itex]\frac{min^2 ft}{gal^2}\times\frac{gal^2}{min^2}= ft[/itex] again. There are about 7.48 cubic feet per gallon (I had to look that up) and 60 seconds per minute (that I did not).
    That gives [itex]\frac{1}{7.48} \frac{gal}{ft^3}[/itex] and [itex]\frac{1}{60} \frac{min}{sec}[/itex] so [itex]\frac{1}{7.48^2} \frac{gal^2}{ft^6}[/itex] and [itex]\frac{1}{3600}\frac{min^2}{sec^2}[/itex] So our conversion, from [itex]\frac{s^2}{ft^5}[/itex] to [itex]\frac{min^2 ft}{gal^2}[/itex] can be done by
    [tex]\left(4.43\frac{sec^2}{ft^5}\right)\times\left(\frac{1}{60}\frac{min}{sec}\right)^2\left(\frac{1}{7.48}\frac{gal}{ft^3}\right)^2= \left(4.43\right)\left(\frac{1}{3600}\right)s\left(\frac{1}{55.95}\right)= 2.18\times 10^{-5}[/tex]

    That's not exactly the same as the given "[itex]2.2\times 10^{-5}[/itex]", perhaps because of round-off error, but that's the idea.
     
    Last edited by a moderator: Mar 19, 2015
  11. Mar 19, 2015 #10
    That should be 7.48 gal/ft3, not 7.48 ft3/gal.

    Chet
     
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