1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Wind Speed from isobars

  1. Dec 3, 2017 #1
    1. The problem statement, all variables and given/known data
    A weather map shows a shallow, stationary depression centred over a point on the Earth’s surface at latitude 50◦ N. The isobars corresponding to pressures of 998, 1000 and 1002 mbar are concentric circles of radius 50, 200 and 350 km respectively. Estimate the wind speed on the 1000 mbar isobar, neglecting the effects of friction between air and ground. You may assume that the wind direction follows the isobars (i.e. rotating anti-clockwise in circles around the depression as viewed from above). [1 mbar = 100 Pa; you can take the density of air to be 1.2 kg m−3 .]

    2. Relevant equations
    Pressure: ##p = F/A##
    Centripetal force: ##F =mv^2/r ##
    Coriolis force*: ##F = -2m (\vec{\omega} \times \vec{r}) ##

    * According to my lecture notes, this leads to the Coriolis force for motion of speed v on Earth's surface to be ##F= 2m\Omega v sin \lambda## to the right (northern hemisphere), where ##\lambda## is the latitude and ##\Omega## the angular velocity of the Earth. I understand that this is the case for motion along a single longitude (N-S), but cannot see why this holds for east-west motion too. I would be glad for some explanation.
    3. The attempt at a solution
    I thought about it the following way:
    There are two forces acting on the air: the Coriolis force ##F_1## acting outwards radially and a force ##F_2## due to the pressure gradient acting inwards radially. The resultant of these two forces must be the centripetal force. By the information given the pressure gradient may be assumed linear.

    If we take the depth of the depression to be z and consider a strip of width ##\delta r## centred on ##r_2## we have:

    ##F_1 = 2 \Omega v sin\lambda 2\pi z r_2 \delta r \rho ##

    ##F_2 = 2\pi z ((r_2+\delta r) (p_2+ \frac{dp}{dr} \delta r) - r_2 p_2) \approx 2\pi z ( p_2\delta r + r_2 \delta r \frac{dp}{dr}) ##

    Setting the difference between these two equal to the centripetal force gives after some algebra:

    ##v^2 - 2r_2 \Omega sin\lambda v - p_2/\rho - r_2\rho \frac{dp}{dr} = 0 ##

    with the solution ##v \approx 289 m/s##, however the answer is supposed to be ##28 km/h##. Where am I going wrong? What's wrong with my model? Thank you. :)
     
  2. jcsd
  3. Dec 3, 2017 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I am confused about what your "strip" is. Is it the entire circular region between ##r_2## and ##r_2 + \delta r##? You will be better off considering an arbitrary infinitesimal volume element and applying the divergence theorem to find the influence of the pressure gradient. You may also want to double check the dimensional analysis of your final expression.

    I suggest that you start with the Coriolis effect in a rotating plane (it really is no different in the 3D setting when rotating around an axis. Do you understand why both radial and polar motion in the rotating plane leads to a Coriolis effect?
     
  4. Dec 3, 2017 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    In your relevant equation for the force, it should be ##\vec v##, not ##\vec r##.
    If you resolve the Earth's rotation vector into components normal and parallel to the local surface, the normal component provides the same magnitude of radial force around the whole isobaric circle. The parallel component is parallel to the local NS direction, so has no effect in the NS sections of the gyration; in the EW sections it leads to a vertical force, but the consequences of that are constrained.

    I believe you can ignore centripetal force (something to do with the Rossby number). Just balance the Coriolis force with the pressure gradient. (But I got 36km/h. Haven't tried checking whether centripetal force accounts for the difference.)
     
    Last edited: Dec 3, 2017
  5. Dec 4, 2017 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The difference is actually significant. I too get 36 km/h when ignoring the centripetal force, but 27 km/h when not doing so. The Rossby number for this system is not very large (order unity).
     
  6. Dec 4, 2017 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Thanks for following up.
     
  7. Dec 4, 2017 #6
    Yes, my strip is an annulus from ##r_2## to ##r_2 + \delta r##, sorry about the inprecise formulation.
    This question comes from a course in rotational dynamics (which I have just begun) and we have not yet studied any fluid dynamics so I am unsure about how to apply the divergence theorem (as I have only met this in maths). Is there any other way to go about it? Does it really make a difference to deal with an infinitesimal element instead of an infinitesimally thin annulus considering that we have circular symmetry?

    Thank you for pointing out my typo in the final equation. It should of be:

    ##v^2 - 2r_2 \Omega sin\lambda v - \frac{p_2}{\rho} - \frac{r_2}{\rho} \frac{dp}{dr} = 0 ##
     
  8. Dec 4, 2017 #7
    Yes that is a typo, so sorry!

    Thank you! That finally explains it: so only the force in the plane is the same in magnitude, the overall force is not.
     
  9. Dec 4, 2017 #8

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Then your force equation is incorrect. As follows from symmetry, the total force on the annulus is zero. You need to study an infinitesimal volume element.

    Yes. The velocity of the annulus is not uniform, its overall momentum is zero, and the way you are attempting to do it you are ending up with an additional pressure term that should not be there.

    The argument for the pressure force is rather simple. You can follow these steps:
    1. For an arbitrary volume, write down the pressure force acting on it in terms of a surface integral over the volume’s surface.
    2. Apply an integral theorem (closely related to the divergence theorem).
    3. The above was completely general. Now consider a small volume dV.
    (Fun fact: Essentially the first two steps above are the basic steps behind deriving Archimedes’ principle.)
     
  10. Dec 4, 2017 #9
    Thank you, I have now attempted to do this by considering a cuboid at ##r_2##. In this case, I assumed the area of opposing faces to be constant, hence getting rid of the extra pressure term due to the curvature of the surfaces. Other than that, the situation was as previously and I ended up with:

    ##v^2 + 2r_2\Omega sin\lambda v - \frac{r_2}{\rho}\frac{dp}{dr} = 0 ##

    giving ##v \approx 7.47 m/s \approx 26.9 km/h ##. I believe this is correct.

    Thank you for all the help!
     
  11. Dec 4, 2017 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You can maybe get there a little faster by thinking in terms of the pressure gradient having to overcome centrifugal force and balance the Coriolis force. Pressure gradient is force/volume, giving your equation above.
    Note that in a high pressure system the pressure gradient is assisted by the centrifugal force. Looks like that leads to somewhat lower speeds for the same pressure gradient.
     
  12. Dec 4, 2017 #11

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    This is what I got. Would you mind posting your new argumentation for posterity?

    I think it is essentially what she did. Her original problem being how to write the pressure term as a force per volume using a pressure gradient. This is why I think it is still important to post the argumentation. Even if the end result is correct, we could then check it for inconsistencies and possible improvements.
     
  13. Dec 4, 2017 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, I see that... just saying that it is not necessary to get into the deltas.
     
  14. Dec 4, 2017 #13

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    But that requires you to know that the pressure gradient is a force per volume, which I think is what the OP had some issues with. I also do not think we can be completely sure what she did without seeing her argumentation.
     
  15. Dec 6, 2017 #14
    Well the only thing that really changed was my ##F_2## by neglecting the curvature on the cuboid:

    ##F_2 = 2\pi z (r_2 (p_2+ \frac{dp}{dr} \delta r) - r_2 p_2) \approx 2\pi z r_2 \delta r \frac{dp}{dr}) ##

    My previous solution also had a sign error I corrected.
     
  16. Dec 6, 2017 #15

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You actually do not need to specify the shape of the test volume if you go down the route described in #8. Do you feel confident enough with integral theorems in vector calculus to do the derivation for the general case or would you need more pointers?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Wind Speed from isobars
  1. Speed of Wind (Replies: 1)

  2. Wind speed? (Replies: 1)

  3. Wind speed (Replies: 4)

Loading...