Question

As the result of a thermal inversion the prevailing air temperature profile increases 1°C/100m above the ground level. To what maximum height will polluted air rise if it is released at ground level 15°C warmer than the atmosphere? The adiabatic lapse rate is −0.65°C/100m.

Answer 1

909.1 m

Step-by-step explanation:

Rate of temperature increase with 100 m elevation = 1°C

h = Maximum Height

Adiabatic lapse rate = -0.65°C/100 m

We have the relation

`(1^(\circ)C)/(100)=(15^(\circ)C)/(h)+\left(-(0.65)/(100)\right)\\\Rightarrow (1^(\circ)C)/(100)+\left(-(0.65)/(100)\right)=(15^(\circ)C)/(h)\\\Rightarrow h=(15^(\circ)C)/(1.65^(\circ)C)* 100\\\Rightarrow h=909.1\ m`

The maximum height is 909.1 m