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Unit reduction is not making sense, what's going on?

  1. Jul 28, 2011 #1
    I am using an equation that states that critical sonic velocity is:

    [itex]a_{cr}=\sqrt{\frac{2\gamma}{\gamma+1}gR_{gas}T'}[/itex]

    The units should result in velocity ([itex]\frac{m}{s}[/itex])

    Yet, when I do the unit reduction I do not get that answer.

    Here is what I get:

    Lets ignore the square root for now.
    Also, [itex]\frac{2\gamma}{\gamma+1}[/itex] has no units, so that makes our job easier.

    We are left with [itex]gR_{gas}T'[/itex]

    With this we have: [itex]\left( \frac{m}{s^{2}}\right)\left( \frac{kJ}{kg\bullet K}\right)\left(K\right)[/itex]

    Right away we cancel the Kelvin unit.

    We now have: [itex]\left( \frac{m}{s^{2}}\right)\left( \frac{kJ}{kg}\right)[/itex]
    Next we conver kJ to N∙m: [itex]\left( \frac{m}{s^{2}}\right)\left( \frac{1000 N \bullet m}{kg}\right)[/itex]

    And N to [itex]\frac{kg∙m}{s^{2}}[/itex]: [itex]\left( \frac{m}{s^{2}}\right)\left( \frac{1000 \frac{kg∙m}{s^{2}} \bullet m}{kg}\right)[/itex] = [itex]\left( \frac{m}{s^{2}}\right)\left( \frac{1000 m^{2}}{s^{2}}\right)[/itex] = [itex]\left( \frac{1000m^{3}}{s^{4}}\right)[/itex]

    Now if we throw the square root back on we see that:

    [itex]\sqrt{\left( \frac{1000m^{3}}{s^{4}}\right)}[/itex]≠[itex]\frac{m}{s}[/itex]

    I'm sure I am making a silly mistake somewhere, but I cannot find where. The text I got this from used English units and it checked out. I though that SI units would be even easier...

    Thanks!
     
  2. jcsd
  3. Jul 28, 2011 #2
    it's almost as if you did not need g to be in there...are you sure that g is supposed to be there? and is it really gravity? Are you sure is not just supposed to mean gamma? ...maybe you have doubled up in your constants...Wikipedia shows speed to be sqrt(gamma X Rgas X T)
     
  4. Jul 28, 2011 #3
    What is R?

    If it's the gas constant your units for it are wrong.
     
  5. Jul 28, 2011 #4
    Critical sonic velocity does not have anything to do with the planet's gravity. Therefore, g (the way you used it) should not enter the formula.
     
  6. Jul 28, 2011 #5
    The unit for the universal gas constant is:

    [tex]
    [R] = \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}
    [/tex]
     
  7. Jul 28, 2011 #6
    unreal89: The g in your equation isn't gravity; it's a unitless constant. I have a hunch it's a conversion factor that makes the conversion from lbm to lbf easier when you're dealing with English units. You can ignore it when dealing in metric units, since gc=1 in that case. I would like to see the equation from the original text to make sure I'm telling you right.

    About the R, unreal89 has the right units, but it's for a specific gas. The universal gas constant has units of J/mol*K, but moles is a very inconvenient measurement for problems like this, so we divide the universal constant by the molar mass of the gas to get a constant specific to whatever gas we're analyzing. The accepted units for this constant are kJ/kg*K. This equation insists on a constant for a specific gas, because as you can see, it wouldn't yield the right units otherwise.
     
  8. Jul 28, 2011 #7
    You are correct, Tim. The g in this equation is used for converting lbm to lbf. Once you go into SI units, it is unnecessary.

    Thanks for everyone's help.
     
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