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Earth's Energy balance including a convective component

  • #1
Hey guys, i'm really stuck right now

1. Homework Statement

Habitable exoplanet:
Now we will include a convective heat flux (H), with H being proportional to the difference between the atmospheric and ground surface temperature, $$H=c_p(T_a-T_p)$$ with c = 1.2 Wm−2K−1. Calculate the temperature of the planet. How does the convective heat flux respond to an increase in the atmospheric emissivity?

Homework Equations


An overview of the symbols
  • S0/4 is the incoming heat flux (W/m2)
  • A_s is planet's ground surface albedo
  • Epsilon is the emissivity of atmosphere
  • Alpha is the absorptivity of atmosphere
  • Sigma is radiative constant
  • 30% of incoming flux reaches the atmosphere
  • 70% goes through to the planet
  • 10% is reflected by atmosphere
  • everything is known except the emissivity of the atmosphere and T_a and T_p
An energy balance has been set up for the case without the convection:

fwGruyv.jpg



The Attempt at a Solution



I've set up the energy balance above and calculated it completely (to get at least partial points), but now i need to add convection in here somehow and i don't really know how. For starters, this is my derivation to solve the problem if convection wouldn't play a role. The first balance is an energy balance for the entire planet (the striped box):

$$\frac{S_0}{4}=0.7\frac{a_s S_0}{4}+\epsilon\sigma T_a^4+(1-\alpha)\sigma T_p^4 + 0.1\frac{S_0}{4}$$
And for the atmosphere the energy balance is:
$$\alpha\sigma T_p^4+0.3\frac{S_0}{4}=2\epsilon\sigma T_a^4 + 0.1 \frac{S_0}{4}$$
Since we are interested in the temperature of the atmosphere, we can rewrite the equation into:
$$T_a^4=\frac{\alpha\sigma T_p^4+0.2\frac{S_0}{4}}{2\epsilon\sigma}$$
Filling this into equation (\ref{eq:1d1}) we get
$$\frac{S_0}{4}=0.7\frac{a_s S_0}{4}+\epsilon\sigma \left( \frac{\alpha\sigma T_p^4+0.2\frac{S_0}{4}}{2\epsilon\sigma} \right)+(1-\alpha)\sigma T_p^4 + 0.1\frac{S_0}{4}$$
Rewriting this finally gives us the final equation
$$T_p=\sqrt[4]{\frac{(0.8-0.7a_s)S_0}{4\sigma(1-\frac{\alpha}{2})}} \approx 248.43K$$

I think there should be a term of
$$H=c_p(T_a-T_p)$$
Included on the left-hand side of the energy balance of the atmosphere, but if i do that i don't know how to calculate the temperatures anymore.

Given from examples in the lecture-slides i assume my derivation is correct, but i'm not sure. For starters i have assumed for now that alpha ~= epsilon, but it is given in the exercise that "Kirchhof’s law may be applied to broadband radiation" which would mean they can be taken as the same. I'm really stuck here. (Even is they're not the same, i think the derivation without convection is still on the right track)

I hope it's clear what the question is, if anything is unclear please let me know. :-)
 

Answers and Replies

  • #2
Twigg
Gold Member
196
35
I appreciate you including the diagram, it made it a lot easier to follow.

Intuitively, the atmosphere loses/gains a quantity of heat ## H = c_{p} (T_{p} - T_{a}) ## to/from the surface. I believe that the way you've set up your balance equations, you put heat flows into the atmosphere on the left-hand side. If I understand your convention correctly, then you should be able to add the term I've given to the left hand side (as it's a flow into the atmosphere). I.e.,

## \alpha\sigma T^{4}_{p} + 0.3*0.25*S_{0} + c_{p} (T_{p} - T_{a}) = 2\epsilon\sigma T^{4}_{a} + 0.1*0.25*S_{0}##


If you do that, you'll have two equations in two unknowns. If it proves to be algebraically yucky, just look for a numerical solution. If you have MATLAB, you can use fzero. Alternatively, I'm sure there are online calculators for finding roots of a quartic equation.

Did this answer your question?
 
  • #3
Ok then my reasoning was correct. I indeed was thinking that the professor wouldn't have made such a yucky analytic solution, but i'll just try to uncover the quartic roots with MATLAB, thanks for the help! :-)
 

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