Atmospheric Physics: Max Env. Lapse Rate of Unsaturated Air Layer

TobyDarkeness
Messages
38
Reaction score
0

Homework Statement


Consider a layer of unsaturated air on Earth, 2000 m thick, whose base is at a height of 4000 m above sea level. The layer sinks and is compressed till its base is at 350 m and its top is at 1650 m. If the layer now provides a subsidence inversion, calculate the maximum possible environmental lapse rate for the layer in its original position (assuming the layer to have a uniform environmental lapse rate).


Homework Equations


\Gammad=9.8 Kkm^-1 adiabatic lapse rate
this must be smaller than the environmental lapse rate for stability
dT/dz=lapse rate T is temp in Kelvin, z is altitude in km

The Attempt at a Solution


To my thinking, the environmental lapse rate must be smaller than adiabatic for subsidence inversion to remain stable, so using 9.8 Kkm^-1 I can obtain the dT of the air parcels.
so for the top air parcel 2km 19.6K and for 1.3km(bottom parcel) 12.74K.
using these temperatures and the total altitude change of 6km for the large parcel and 1.65km for the smaller parcel I find that the max is 7.72K km^-1, assuming uniform distribution. I obtain 7.72 from 12.74/1.65km. Both answers are smaller than 9.8 so are consistant, i have simply picked the larger number as the maximum is required.

Is this correct?
 
Physics news on Phys.org
TobyDarkeness said:

Homework Statement


Consider a layer of unsaturated air on Earth, 2000 m thick, whose base is at a height of 4000 m above sea level. The layer sinks and is compressed till its base is at 350 m and its top is at 1650 m. If the layer now provides a subsidence inversion, calculate the maximum possible environmental lapse rate for the layer in its original position (assuming the layer to have a uniform environmental lapse rate).

Homework Equations


\Gammad=9.8 Kkm^-1 adiabatic lapse rate
this must be smaller than the environmental lapse rate for stability
dT/dz=lapse rate T is temp in Kelvin, z is altitude in km

The Attempt at a Solution


To my thinking, the environmental lapse rate must be smaller than adiabatic for subsidence inversion to remain stable, so using 9.8 Kkm^-1 I can obtain the dT of the air parcels.
so for the top air parcel 2km 19.6K and for 1.3km(bottom parcel) 12.74K.
using these temperatures and the total altitude change of 6km for the large parcel and 1.65km for the smaller parcel I find that the max is 7.72K km^-1, assuming uniform distribution. I obtain 7.72 from 12.74/1.65km. Both answers are smaller than 9.8 so are consistant, i have simply picked the larger number as the maximum is required.

Is this correct?
Hi Toby, did you manage to solve this as I have the same question :)
 

Thread 'Please tell me where I am going wrong in this integral'

##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
View full post »

Thread 'Direction of the magnetic field of an oscillating charge'

Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
View full post »

Thread 'Initial conditions for orbits around a wormhole'

Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
View full post »

Similar threads

Matlab code using ode45 difficulty
Replies
3
Views
3K
Strong negative feedback in Climate
Replies
1
Views
6K

Hot Threads

Recent Insights

Back
Top