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

[TobyDarkeness]

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?

 

[K.QMUL]

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 :)