3
$\begingroup$

P = 300mb, Pot. Equiv. Temp = 375K


P = 600mb, Pot. Equiv. Temp = 325K


P = 900mb, Pot. Equiv. Temp = 350K


(surface)

Given that a deep convective plume sends a portion of the bottom layer to the top:

What fraction of the mass of the lower layer must rise in order to just barely stabilize the lower layer with respect to the middle layer? How many millibars of environmental subsidence does this imply?

I need direction on how to begin this. Which equation will assist with this answer.

My attempt:

If I remove a chuck sized $\alpha$ from the bottom layer, it will contain $\frac{3m}{2}$ the mass of the middle layer of same size. So, is the answer simply $\frac{2}{3}$ of the mass must move out of the lower layer? I believe I am wrong, it seems to be more involved than this, but I cannot grasp this concept.

$\endgroup$
  • $\begingroup$ :) Remember stability is all about how temperature (and more concisely, θ$_e$) changes with height... so the answer is very much to do with the temperatures. Any thoughts from that, maybe remind yourself more about the definitions of stability from your notes? $\endgroup$ – JeopardyTempest Oct 29 '18 at 4:17
  • $\begingroup$ The definition I think you're looking for to figure out the starting part, how the air masses need to change... is actually quite a simple looking one :) $\endgroup$ – JeopardyTempest Oct 29 '18 at 4:27
1
$\begingroup$

My results:

If an $\alpha$ sized chunk was moved up, then what remains is $(1-\alpha)$. Because each layer is 300mb thick, they have equal parts mass. What goes up must be compensated, and due to conservation of mass, there is no gain or loss of mass in my system. Potential temperature is a conserved quantity, so I can say:

$$350K(1-\alpha)+325K\alpha = 325K(1-\alpha)+375K\alpha$$ $$\alpha = \frac{1}{3}$$

Further, $\frac{1}{3}$ mass movements equates to $\frac{1}{3}$ of a $300mb$ layer, which is $100mb$ subsidence.

$\endgroup$
  • $\begingroup$ Definitely seems you're more on top of it... because yes, I agree with the thoughts they'd be equal mass regions, and that $θ_e$ would be conserved. You don't explain how you got to your equation... but it's the same one I come up with in basic thinking (probably want to explain it in the HW answer you turn in, but leaving it unexplained here is useful in case others in the class go looking for a shortcut later). And so I get the same answer if it's valid. :) $\endgroup$ – JeopardyTempest Oct 30 '18 at 4:37
  • $\begingroup$ 100 mb seems like a shocking amount of mass turnover, so I can't 100% we aren't missing something. Also, real life would have the subsidence displaced from the convection, so the subsidence wouldn't immediately shut it down, but instead lead to going beyond that value until the subsided air finally gets entrained into the convective area. Plus there's other complications like evaporative cooling (if the rising air caused condensation/precipitation). But overall, I think your answer seems reasonable to me, and glad to see you got latched onto the right ideas by sticking with it :D :D $\endgroup$ – JeopardyTempest Oct 30 '18 at 4:40
  • 1
    $\begingroup$ @JeopardyTempest We did ignore effects of condensation, etc. You said once the subsided air finally gets entrained in the convective area, shut down occurs. I never thought about that. So, it's the subsidence of this now warm, dry air from upper troposphere that stops this convective area ultimately? Pretty cool. $\endgroup$ – PattyWatty27 Oct 30 '18 at 5:10
  • $\begingroup$ In some circumstances, certainly could be. Out here in the eastern US, particularly Florida, I think more of moist convection, where it's usually the cold precipitation (that doesn't subscribe to adiabatic warming!) that tends to shuts things down. But yes, confident that entrainment of drier air can definitely be a cause to some convective shutdowns. $\endgroup$ – JeopardyTempest Oct 30 '18 at 7:42
  • $\begingroup$ Further considered the feasibility of the 100 mb. A 50 m/s is possible in strong thunderstorms... and based the lowest 300 mb being about 3 km deep, I believe that's then a rate of about 300 mb worth/minute (60*50/3000). If that's all valid, never realized how much an insane engine a supercell can be. Of course that quickly points to the fact that any updraft is drawing in A LOT of air around it. And while large-scale subsidence will be displaced from the updraft, mixing still occurs $\endgroup$ – JeopardyTempest Oct 30 '18 at 7:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.