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Homework Help: Dimensional analysis

  1. Jul 21, 2010 #1
    1. The problem statement, all variables and given/known data

    it is likely that the cross-sectional design of a new bridge will give arise to a vortex shedding from its downwind side. Such behaviour can creat harmful periodic forses on the bridge structure, the bridge designers is keen to determine the frequency of shedding. He expects the shedding frequency [tex]\omega[/tex], to depend on depth of the deck D, the length of the bridge in the direction of the flow L , the wind velocity V and the fluid density [tex]\rho[/tex] and the viscosity[tex]\mu[/tex]. Express this relationship in dimensionless form isolating[tex]\omega[/tex] ,L and [tex]\mu[/tex] in separate groups

    2. Relevant equations

    buckingham pi theorem; k-r=how many pi groups

    3. The attempt at a solution

    3 pi groups

    variable;
    [tex]\omega[/tex]
    D
    V
    [tex]\rho[/tex]
    [tex]\mu[/tex]
    L

    firstly are the units of frequency for this T^-1 and what units do i choose for L the direction of flow?
     
  2. jcsd
  3. Jul 21, 2010 #2
    just realised L is lenght!!! is depth L aswell then
     
  4. Jul 21, 2010 #3
    Take the following 3 groups of 4 physical quantities each:

    [tex]
    \omega, D, v, \rho
    [/tex]

    [tex]
    L, D, v, \rho
    [/tex]

    [tex]
    \mu, D, v, \rho
    [/tex]

    Each one has 4 physical mechanical quantities (with dimensions with respect to length time and mass only!). Therefore, you should be able to "construct" a dimensionless quantity from each of the groups.

    As a first step, you will need the dimensions of each of the physical quantities involved. I'll start. Dynamical viscosity enters in Newton's law for the viscous force:

    [tex]
    F = \mu \, A \, \frac{\partial v}{\partial l}
    [/tex]

    and the dimension of force is obtained from Second Newton's Law ([itex]F = m \, a[/itex]):

    [tex]
    [F] = [m] \, [a] = \mathrm{M} \, \mathrm{L} \, \mathrm{T}^{-2}
    [/tex]

    Then, we have:

    [tex]
    [F] = [\mu] \, [A] \, \frac{[v]}{[l]}
    [/tex]

    [tex]
    \mathrm{M} \, \mathrm{L} \, \mathrm{T}^{-2} = [\mu] \, \mathrm{L}^{2} \, \frac{\mathrm{L} \, \mathrm{T}^{-1}}{\mathrm{L}}
    [/tex]

    [tex]
    [\mu] = \mathrm{M} \, \mathrm{L}^{-1} \, \mathrm{T}^{-1}
    [/tex]
     
  5. Jul 21, 2010 #4
    i kinda know most of the dimensions of the groups of by heart.
    How did u make those equtions so fast !!!
    i have for the first pi group

    [tex]\omega[/tex]DV[tex]\rho[/tex]

    =T-1(ML-3)a(L)b(LT-1)c

    i am unsure wether that is the correct dimision for [tex]\omega[/tex] though.
     
  6. Jul 21, 2010 #5
    In the first group, only [itex]\rho[/itex] has a dimension of mass, so it cannot cancel with anything else. Therefore, it cannot enter in the dimensionless combination ([itex]a = 0[/itex]). Then, consider which of the remaining quantities have dimension of time. In what combination do you have to take them so that time cancels? This combination has only remaining dimension of length. How should you combine it with the remaining quantity from the group so that you cancel the length as well?

    I claim that the second group has a trivial dimensionless combination (there are two quantites with the same dimension).

    For the third group, you need to do a similar procedure as for the first one, but you need to cancel mass this time as well.

    EDIT:

    You were right about the dimension of [itex]\omega[/itex] and I use LaTeX for the equation editing.
     
  7. Jul 22, 2010 #6
    thanks. but i still can get my head around it. do i need to use the specific density to get rid of the mass otherwise i can't solve it.
     
  8. Jul 22, 2010 #7
    In the third group, yes, definitely.
     
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