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Finding the temperature change of Earth's surface over 12 hours

  1. Apr 27, 2013 #1
    Ok, this question goes on a little further. I want to find the velocity of air moving from the equator to the poles due to the change in pressure caused by differing temperatures.

    I've found the flux density from the sun, so I have a value of flux (hope im using the right term here - whatever W/m^2 is) for the equator and the N pole and I factored in different albedos too :approve:

    But now I'm stuck.

    I have a flux on the equator, so energy/s for 1m^2 essentially. I want to find how much the surface heats up and use that to find how much that heats the column of air above it.

    If we consider at sunrise the angle between the surface and the sunlight is 0°, at midday 90°, and sunset 180°. So we are going from minimum>maximum>minimum. I've drawn myself a graph, the maximum flux is 1270 W/m^2

    As the sun moves across the sky, the 1m^2 of ground we are considering has a differing amount of flux on it as F = (Fmax) sinθ so we wind up with a graph that looks like the sin curve. What I want to know is how can I work out the total energy on this 1m^2 of ground over 12 hours (43200 seconds if thats easier)? I guess it's an integral but so far nothing has worked for me :cry:
     
  2. jcsd
  3. Apr 27, 2013 #2

    BruceW

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    well, you've got the equation F = (Fmax) sinθ which is the power on this 1m^2, right. So yes, to get the total incident energy, it is going to be an integral over time. So now, you need to find out what is θ as a function of time? hint: it is fairly simple, think about the rotation of the earth.

    edit: also, I'm guessing you are making the assumption that it is the longest day on the equator, so that the equator is the point closest to the sun. (or equivalently, just ignoring the effect of the seasons).
     
  4. Apr 27, 2013 #3
    Check out Introduction to Dynamic Meteorology by Holton
     
  5. Apr 28, 2013 #4
    Ok, so I subbed 180*t/43200 as a value for θ into the integral:

    E = ∫ Fmax*sin(t/240) dt and used the limits 0 - 43200

    I found an answer to be 609600J.. seems like a good answer! Thanks :biggrin:

    Just trying to figure out the change in air temperature because of that... need some help with this bit!

    I jumped a step and rearranged bernoulli's theorem to find the change in velocity due to the change in pressure
     
  6. Apr 28, 2013 #5

    BruceW

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    hmm. your method looks good. But I get a different answer than you did. You are using Fmax=1270, right? Maybe you accidentally used pi=180 instead of the radians answer?
     
  7. Apr 28, 2013 #6
    yeah, i used 1270.

    I worked the whole thing through with degrees, shouldn't make a difference?

    E = ∫ Fmax*sin(t/240) dt
    E = 240*Fmax*(-cos(t/240)

    once you stick in limits of 0 and 43200 i get:

    E = [240*Fmax*(-cos(180))] - [240*Fmax*(-cos(0))]
    E = [240*Fmax*1] - [240*Fmax*-1] = 480*Fmax = 609600J

    How did yours differ?
     
  8. Apr 29, 2013 #7

    BruceW

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    It is not OK to use degrees when the angle comes out of the sin function (when you did the integration). If the angle was only inside the sin function, then you can use degrees or radians, it doesn't matter. But when you're using an angle outside of a sinusoid function, you have to use radians.

    For example, with radians the arc length = radius * angle But this equation only works when the angle is in degrees. And other similar formulas require radians, not degrees.
     
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