Question

Use the conditions provided in the previous problem. Atmosphere statically stable at the base of the mountain, where pressure = 1000 hPa, potential temperature lapse rate of 3.1◦C per 50 hPa. Atmosphere statically unstable at the top, where pressure = 850 hPa, potential temperature lapse rate of -0.61◦C per 50 hPa. Please find the change in relative vorticity of the column of air.

Answer 1

change in relative vorticity 0.0590

Step-by-step explanation:

Given data

pressure = 1000 hPa

temperature lapse rate q1 = 3.1◦C per 50 hPa

pressure = 850 hPa

temperature lapse rate q2= -0.61◦C per 50 hPa

to find out

change in relative vorticity

solution

we will apply here formula that is

N = (g / potential temperature ) × (potential vertical temperature) × exp^1/2 ............................1

here we know g = 9.8 m/s

and q1 = potential temperature=3.3 degree celsius

potential vertical temperature gradient = 3.1 - 0.61 / 1000 -850

potential vertical temperature gradient = 0.0166 degree celsius/hpa

so

N = 9.8 / 2.75 × 0.0166 × exp^1/2

N = 0.0590