8.1.3 Hydrostatic equilibrium

At large scales, the air pressure and density depend essentially on the massive and slowly varying dry component and are well described by hydrostatic equilibrium. The air temperature, as we have seen, depends significantly on the abundance and distribution of water, CO₂ (and O₃ for the stratosphere).

At equilibrium:

  

$$-g\rho$$ dp/dh = (8.6)

$$\frac{\rho R T'}{M_a}$$ p = (8.7)

where $\rho$ is the density at an altitude h, p is the pressure, $T'\simeq T$the air ``virtual'' temperature. $M_a\simeq{29}$ is the average molecular weight, and g the local gravitational field.

$$\frac{dp}{p}=\frac{-g M_a}{R T'} dh$$ (8.8)

In the ``standard atmosphere'' model, T', the temperature of the air varies linearly with altitude and is given throughout the troposphere (i.e. between h=0 and 11 km) by:

 

T'=T'_o-b(h-h_o) (8.9)

where b= 6.5 Kkm^-1 is a constant.

Let us first consider a relatively small change in altitude: $h-h_o\leq 1$ km, $T'\simeq T'_{ave}= (T'(h)+T'(h_o))/2$; we find Laplace's hydrostatic formula:

$$\frac{\rho}{\rho_0} \simeq \frac{p}{p_0} \simeq e^{-\frac{g M_a h}{R T'_{ave}}} = e^{-\frac{h}{h_o}}$$ (8.10)

where $\rho_o$ is the density at sea level and h_o=RT/M_ag=8.4 (T/288) km, the scale height. The gas column density (expressed in g.cm^-2) along the vertical above a point at sea level is:

$$N_o M_a = \int \rho dh = \rho_o h_o %$$ (8.11)

and that above a point at an altitude h:

 

N = N_o e^-h/h_o (8.12)

For larger altitudes, from Eq.8.8 dh = -dT'/b, then Eq.8.9 yields

$$\frac{dp}{p}=\frac{-g M_a dh}{b R}\frac{dT'}{T'}$$ (8.13)

$$\rho=\rho_o (\frac{T'}{T'_o})^{s-1}$$ (8.14)

with s = -g/(b R_a).

Although the above equations represent fairly well the density and pressure throughout the troposphere, the temperature distribution can depart significantly from Eq.8.9 near the ground. This latter heats up faster than the transparent air during the day, and cools off more rapidly during the night. The temperature gradient at low altitudes (up to 1-2 km) can be thus steeper or shallower than shown in Eq.8.9. Occasionally, it can be inverted, the temperature increasing with altitude. The inversion layer usually stops briskly at 1 or 2 km altitude and the normal temperature gradient resumes above. Inversion layers are common during the night over bare land. They can also be caused by hot winds blowing from the sea.

The local temperature gradient determines stability of the air to vertical motions. A rising bubble of wet air expands adiabatically as long as the water vapor it contains does not condense. Expanding, it cools almost as an ideal gas with:

$$T \propto p^{m}\, \; {\rm and}\; m= (1-0.23 q)m_a \simeq m_a$$ (8.15)

The pressure is set by the surrounding air (Eq.8.10), and the bubble seen to cool down with an ``adiabatic'' gradient of

$$\frac{dT}{dh} = -g \frac{M_a}{C_p} = 9.8 \mathrm{K km}^{-1}$$

If the actual temperature gradient is smaller than the adiabatic gradient, the bubble becomes cooler, hence denser than the surrounding air and its ascent stops. The air is stable. If the local gradient is larger than the adiabatic gradient, the bubble becomes less dense than the surrounding air; the air is unstable and a thick convection layer develops, a situation likely to happen in a hot summer afternoon.