<p><strong>My results:</strong></p>

<p>If an <span class="math-container">$\alpha$</span> sized chunk was moved up, then what remains is <span class="math-container">$(1-\alpha)$</span>.  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:</p>

<p><span class="math-container">$$350K(1-\alpha)+325K\alpha = 325K(1-\alpha)+375K\alpha$$</span>
<span class="math-container">$$\alpha = \frac{1}{3}$$</span></p>

<p>Further,
<span class="math-container">$\frac{1}{3}$</span> mass movements equates to <span class="math-container">$\frac{1}{3}$</span> of a <span class="math-container">$300mb$</span> layer, which is <span class="math-container">$100mb$</span> subsidence. </p>

CC BY-SA 4.0
2018-10-30T04:37:40.900Z
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. :)
JeopardyTempest
CC BY-SA 4.0
2018-10-30T04:40:21.580Z
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
JeopardyTempest
CC BY-SA 4.0
2018-10-30T05:10:46.673Z
@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.
PattyWatty27
CC BY-SA 4.0
2018-10-30T07:42:49.060Z
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.
JeopardyTempest
CC BY-SA 4.0
2018-10-30T07:58:46.003Z
Further considered the feasibility of the 100 mb. [A 50 m/s is possible in strong thunderstorms](http://www.theweatherprediction.com/habyhints/319/)... and based the [lowest 300 mb being about 3 km deep](https://www.spc.noaa.gov/exper/soundings/18103000_OBS/), 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
JeopardyTempest
CC BY-SA 4.0
2018-10-30T08:01:02.047Z
And so that may well be a key factor in equalizing the air at the different levels as well.  But that upper air working out, around, and back into the updraft is definitely a factor too.  Certainly found the atmosphere tough to visualize, and questions like yours challenging when taking meteorology courses.  But amazing how it all starts to click eventually.  Hope your studies go well :)
JeopardyTempest
CC BY-SA 4.0
2018-10-30T01:48:49.477Z
2018-10-30T01:48:49.477Z
PattyWatty27
<hr>

<p>P = 300mb, Pot. Equiv. Temp = 375K </p>

<hr>

<p>P = 600mb, Pot. Equiv. Temp = 325K</p>

<hr>

<p>P = 900mb, Pot. Equiv. Temp = 350K  </p>

<hr>

<p>(surface)</p>

<p>Given that a deep convective plume sends a portion of the bottom layer to the top:</p>

<p>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?</p>

<p>I need direction on how to begin this.  Which equation will assist with this answer.</p>

<p><strong>My attempt</strong>:</p>

<p>If I remove a chuck sized <span class="math-container">$\alpha$</span> from the bottom layer, it will contain <span class="math-container">$\frac{3m}{2}$</span> the mass of the middle layer of same size.  So, is the answer simply <span class="math-container">$\frac{2}{3}$</span> 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.  </p>

CC BY-SA 4.0
2018-10-29T04:17:25.770Z
:)  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?
JeopardyTempest
CC BY-SA 4.0
2018-10-29T04:27:44.310Z
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 :)
JeopardyTempest
<atmosphere><temperature>
Given a 3 box model of the atmosphere,determine fraction of mass transport from lower layer
PattyWatty27
# Given a 3 box model of the atmosphere,determine fraction of mass transport from lower layer

(Asked by: username_0 on 2018-10-29)

---

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.

**Tags:** atmosphere, temperature

> **Comments:**
> **username_1**: :)  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?  
> **username_1**: 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 :)  

---

## 1 answers

### Answer by username_0 on 2018-10-30. Score: 1 | Accepted answer

**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.

> **Comments:**
> **username_1**: 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. :)  
> **username_1**: 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  
> **username_0**: @username_1 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.  
> **username_1**: 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.  
> **username_1**: Further considered the feasibility of the 100 mb. [A 50 m/s is possible in strong thunderstorms](http://www.theweatherprediction.com/habyhints/319/)... and based the [lowest 300 mb being about 3 km deep](https://www.spc.noaa.gov/exper/soundings/18103000_OBS/), 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  
> **username_1**: And so that may well be a key factor in equalizing the air at the different levels as well.  But that upper air working out, around, and back into the updraft is definitely a factor too.  Certainly found the atmosphere tough to visualize, and questions like yours challenging when taking meteorology courses.  But amazing how it all starts to click eventually.  Hope your studies go well :)  

---

