# Atmospheric convection

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.40.

The barometric equation was derived under the assumption that the atmosphere is stable, so that at a given height ${z}$ the pressure is equal to the weight of the column of air of unit cross-sectional area above that altitude. The barometric equation is

$\displaystyle \frac{dP}{dz}=-\frac{mg}{kT}P \ \ \ \ \ (1)$

Earlier, we assumed that ${T}$ was constant, which allowed us to integrate the equation. In reality, the temperature decreases with increasing ${z}$. If the temperature gradient ${\left|dT/dz\right|}$ passes a critical value, the density of the warmer air near the surface drops to a point where it starts to rise, resulting in convection, or the mass movement of air. Similarly, the density of the cooler air higher up is large enough that it falls towards the surface, so that a cycle is set up.

If we assume that the velocity of these air masses is high enough that little heat is lost or gained as the air moves vertically, we can use the adiabatic formula to analyze the situation. For adiabatic expansion, the pressure, volume and temperature are related by

 $\displaystyle VT^{f/2}$ $\displaystyle =$ $\displaystyle K\ \ \ \ \ (2)$ $\displaystyle V^{\gamma}P$ $\displaystyle =$ $\displaystyle A \ \ \ \ \ (3)$

where ${K}$ and ${A}$ are constants and ${\gamma=\left(f+2\right)/f}$. Taking differentials we get

 $\displaystyle T^{f/2}dV+\frac{f}{2}VT^{\frac{f}{2}-1}dT$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (4)$ $\displaystyle \gamma V^{\gamma-1}PdV+V^{\gamma}dP$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (5)$

Simplifying, we get

 $\displaystyle TdV+\frac{f}{2}VdT$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \gamma PdV+VdP$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle dT$ $\displaystyle =$ $\displaystyle -\frac{2T}{fV}dV\ \ \ \ \ (8)$ $\displaystyle dP$ $\displaystyle =$ $\displaystyle -\frac{\gamma P}{V}dV\ \ \ \ \ (9)$ $\displaystyle \frac{dT}{dP}$ $\displaystyle =$ $\displaystyle \frac{2T}{f\gamma P}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{f+2}\frac{T}{P} \ \ \ \ \ (11)$

Now if ${\frac{dT}{dz}}$ is at the critical point, then 1 still applies and, since adiabatic expansion is just about to start, 11 should apply as well. From 1 we get

 $\displaystyle \frac{dP}{P}$ $\displaystyle =$ $\displaystyle -\frac{mg}{kT}dz\ \ \ \ \ (12)$ $\displaystyle dT$ $\displaystyle =$ $\displaystyle \frac{2T}{f+2}\frac{dP}{P}\ \ \ \ \ (13)$ $\displaystyle \frac{dT}{dz}$ $\displaystyle =$ $\displaystyle -\frac{2mg}{\left(f+2\right)k} \ \ \ \ \ (14)$

The average molecular mass ${m}$ can be found from the mass of a mole of dry air at room temperature and 1 atm pressure, which is ${0.0289697\mbox{ kg}}$:

$\displaystyle m=\frac{0.0289697}{6.02\times10^{23}}=4.81\times10^{-26}\mbox{ kg} \ \ \ \ \ (15)$

so the critical temperature gradient is

$\displaystyle \frac{dT}{dz}=-\frac{2\times4.81\times10^{-26}\times9.8}{\left(5+2\right)1.38\times10^{-23}}=-0.00976\mbox{ deg m}^{-1}=-9.76\mbox{ deg km}^{-1} \ \ \ \ \ (16)$

If the temperature gradient reaches around 10 degrees per kilometre, we can expect convection to occur. This is known as the dry adiabatic lapse rate.

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