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Air pressure in hurricane

  1. Jun 21, 2007 #1
    1. The problem statement, all variables and given/known data
    Estimate the air pressure at the center of a hurricane with wind speed of 300 km/h at the center. Answer in Pa


    2. Relevant equations
    P + 1/2 (density)(velocity)^2

    density of air 1.29 kg/m^3

    3. The attempt at a solution
    P + 0.5 (1.29 kg/m^3)(90000)
    Pressure = 58050 atm

    1 atm = 101.3kPa so 58050 atm x 101.3 kPa = 5880465 kPa or 5.88 x 10^9 Pa

    but the answer is 9.7 x 10^4

    What am I doing wrong?
     
  2. jcsd
  3. Jun 21, 2007 #2

    bel

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    don't use km/h, convert to m/s, if i may hazard a guess

    P= .5*(1.29 kg m^(-3)) (83.33333 ms^(-1))^2
    = .5 *(1.29 kg m^(-3)) (6944.444444 m^2 *s^(-2))
    = 4479.167 kg m^(-1) s^(-2)
    = 4479.167 Pa
    = 4.48 E3 Pa
    if the equation was right, the answer was probably wrong, or i am, probably =).
     
    Last edited: Jun 21, 2007
  4. Jun 21, 2007 #3

    D H

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    What you are doing wrong is being sloppy with units and not stopping when you get a ridiculous answer. 58,050 atm??? When you get an answer like that it means you have done something wrong.

    You also have a sign error. What is the relation between static pressure, dynamic pressure, and total pressure? Which of these pressures is the question asking you to give?

    Edited to add:

    Is this truly how the question is phrased? If so, it is wrong. Much better question: Estimate the pressure at the calm center of a hurricane with winds of 300 km/h just around the center.
     
    Last edited: Jun 21, 2007
  5. Jun 21, 2007 #4
    I copied and pasted the question directly onto the post, so this is what we have to work with.

    I know I did something wrong with the units, but even when they are changed I still get nowhere near the right answer.
     
  6. Jun 21, 2007 #5

    D H

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    Look at bel's post. The dynamic pressure is 4.48 kPa.
     
  7. Jun 21, 2007 #6
    To answer your question about total pressure and relation of static/dynamic, isn't P = P + 1/2 density(velocity)^2 the relationship?

    To label it I think it's P(total) = P(static) + (dynamic)1/2 density(velocity)^2


    So, I guess what I've been calculating is the dynamic. How do I calculate static?
     
  8. Jun 21, 2007 #7

    D H

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    You already stated it:
    [tex]P_{total} = P_{static} + P_{dynamic}[/tex]
    where
    [tex]P_{dynamic} = \frac 1 2 \rho u^2[/tex]

    Now assume the static pressure is constant over the eye of the hurricane and that the total pressure does not exceed one atmosphere. What is the pressure total pressure at the dead center of the hurricane?
     
  9. Jun 21, 2007 #8
    I'm not sure ... I'm confused

    If P static is constant, wouldn't that mean its 0 ?

    To me, that would then mean Ptotal = P dynamic, which isn't the case
     
  10. Jun 21, 2007 #9

    D H

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    Why would think that? Constant means unchanging, not zero. In other words, assume [itex]P_{static}[/tex] has the same value at the center of the hurricane (no winds) as it does at the eye wall (max winds).
     
  11. Jun 21, 2007 #10
    So, if its unchanging and there is no formula for it, how do we calculate for it if its not a given?
     
  12. Jun 21, 2007 #11

    D H

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    Do something else that helps solve the problem. A big part of physics is knowing how to estimate things. For example, assume the pressure at the total pressure at the eye wall is one atmosphere.
     
  13. Jun 21, 2007 #12
    Not sure if this makes sense...but, if I assume that the total pressure is 1 atm or 101.3 kPa, and Pstatic is 100kPa, and the calculated dynamic is 4.48 kPa, maybe I should do something like this: (although, I'm not accounting for total pressure ?!)

    P1 + 1/2dv^2 = P2 + 1/2dv^2
    100kPa +4.48kPa = 100kPa +P2dynamic
     
  14. Jun 22, 2007 #13

    D H

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    From your last post,
    [tex]P_{dynamic} = \frac 1 2 \rho u^2 = 4.48 \text{kPa}[/tex]
    From post #11,
    [tex]P_{total} = 1 \text{atm}[/tex]
    From post #7,
    [tex]P_{total} = P_{static} + P_{dynamic}[/tex]
    so what is the static pressure? Assuming that the static pressure inside the eye wall is constant, what is the total pressure at the center (where the wind speed is zero?)
     
  15. Jun 23, 2007 #14
    If 1 atm = 101.3 kPa

    Then, I'd think static pressure is:

    101.3 = Pstatic + 4.48 kPA
    Pstatic = 96.8 kPa in Pa 9.7x10^4

    :smile:
     
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