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Blanking on simple math.

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Just a quick question. This is fairly terrible that I'm blanking on something as fundamental as this. It's to do with units and dimensions. So, Area, for example, = L * L = L^2. If I have a quantity, R, for example and let's say it's units are L. If I use R in some arbitrary equation and square it does it then become the |R|^2 L or |R|^2 L^2 (where |R| is the scalar component of R)???


    2. Relevant equations
    The following is what got me confused about this. u(the average velocity) = 14/R2 * umax(The max velocity) * (R(1/4) - R(1/15).



    3. The attempt at a solution
    So, the units don't check out in the above equation for velocity, do they?
     
  2. jcsd
  3. Oct 20, 2013 #2

    Pythagorean

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    Yes, units suffer the same operation as their values.
     
  4. Oct 20, 2013 #3

    arildno

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    Well, if "R" in your problem is already NON-dimensional (say a ratio between two radii), it is wholly unproblematic.
     
  5. Oct 20, 2013 #4

    HallsofIvy

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    Certainly [tex]R^{1/4}- R^{1/15}[/tex] would make no sense if R is a "distance". Perhaps, as arildno suggests, R is really a ratio of distances. Could you give us more information about that formula?
     
  6. Oct 20, 2013 #5
    My apologies, R is a radius, therefore it has units of metres.
     
  7. Oct 20, 2013 #6
    Thanks for clearing that up.
     
  8. Oct 20, 2013 #7

    Borek

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    Could be it is an approximate formula that doesn't follow strict derivation, in such case units can be off.
     
  9. Oct 20, 2013 #8
    It's only for calculating the average fluid flow in a pipe, nothing overly complicated. We weren't told that it was an approximate solution so I couldn't say if it is or isn't.
     
  10. Oct 20, 2013 #9

    pasmith

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    No. Most likely the formula is valid only if length is measured in metres (and only for a particular fluid), and that the necessary constants of proportionality would have to be put back in if other length units were to be used.
     
  11. Oct 20, 2013 #10
    Could you explain that? The length would be measured in metres and there are no constants of proportionality evident in the equation, so the units still wouldn't check out?
     
  12. Oct 20, 2013 #11

    pasmith

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    For the dimensions to be consistent, you would need
    [tex]
    u_{avg} = u_{max}\frac{k}{R^2} (R^{1/4} - \alpha R^{1/15}) = u_{max} \frac{k}{R^{7/4}}(1 - \alpha R^{-11/60})
    [/tex]
    where [itex][k] = L^{7/4}[/itex] and [itex][\alpha] = L^{11/60}[/itex].

    It may be that for the fluid in question [itex]k = 14\,\mathrm{m}^{7/4}[/itex] and [itex]\alpha = 1\,\mathrm{m}^{11/60}[/itex].
     
  13. Oct 20, 2013 #12
    Gotcha. Thanks for that.
     
  14. Oct 20, 2013 #13

    Ray Vickson

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    This can get confusing; it all depends on where the units are "attached". If you say "radius = R meters", the quantity R is dimensionless and you can make immediate sense of things like sin(πR) or log(R), etc. However, if you say that "R = 2 meters", for example, then R has dimensions and you need to be a lot more careful.
     
  15. Oct 20, 2013 #14
    That's a great explanation. Thanks. I find that the more you think about it, the more muddled up you can get. So, basically, in your example, R is essentially some arbitrary scalar?
     
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