Deriving the precipitable water

Consider a column of liquid water with cross-sectional area $A$, and height $h$. This height ($h$) is the precipitable water. The mass of this column of liquid water is:

$$m_l = \rho_l \ A \ h \quad ,$$

where $\rho_l$ is the density of liquid water ($\rho_l = 1$ g/cm$^3$).

Now, the mass of the water vapor in an atmospheric column with cross-sectional area $A$ is:

$$m_v = A \ m_w \ \int_0^\infty \ n_w(z) \ dz \quad ,$$

where $m_w$ is the mass of each water molecule ($m_w = 18$ amu), $n_w$ is the number density of water molecules, and the integration is done over altitude $z$. This integration over altitude is why it is strictly necessary to have the full vertical distribution of water vapor in order to calculate the precipitable water. However, if the water vapor is distributed exponentially (similar to the bulk of the lower atmosphere) like:

$$n_w(z) = n_0 \ e^{-(z - z_0)/H} \quad ,$$

where $n_0$ is the number density of water vapor at $z_0$ (the surface, practically speaking), and $H$ is the scale height of the water vapor distribution (the e-folding distance), then the integral can be done analytically, resulting in:

$$m_v = A \ m_w \ n_0 \ H \quad .$$

To find the precipitable water, equate the mass of the vapor to that of the liquid (conservation of mass):

$$m_l = m_v \qquad \Rightarrow \qquad \rho_l \ A \ h = A \ m_w \ n_0 \ H \quad ,$$

therefore,

$$h = {{m_w \ n_0 \ H} \over \rho_l} \quad .$$

So, given a measurement of $n_0$, and an estimate for $H$, the precipitable water can be estimated. From the ideal gas law, the number density of water molecules is related to the water vapor partial pressure ($P_0$) and the temperature ($T_0$) via:

$$n_0 = {P_0 \over {k \ T_0}} \quad ,$$

Making this substitution, the precipitable water is then:

$$h = {{m_w \ P_0 \ H} \over {\rho_l \ k \ T_0}} \quad .$$

The surface temperature, $T_0$ is measured and recorded regularly in the observing logs. In fact, it is measured and recorded on the visibility archive tapes as well, but getting at that data is logistically harder, and the accuracy and time resolution gain is not really needed. The surface water vapor partial pressure can be derived from the surface dew point ($D$) via (Clark 1987):

$$P_0 = e^{\left(1.81 + {{17.27 \, D} \over {D + 237.3}}\right)}\quad ,$$

where the dew point is in degrees C, and $P_0$ is in millibar. The dew point is also measured and recorded regularly in the observing logs. Again, electronic versions of these logs exist, and these can be parsed for the temperature and dew point quantities.

What to use for the scale height $H$? Formally, for an isothermal atmosphere in hydrostatic equilibrium the scale height is given by:

$$H = {{k \ T} \over {m_v \ g}} \quad ,$$

where $g$ is the gravitational acceleration. If this were the right value for $H$, then substituting this would yield the following very simple equation for the precipitable water:

$$h = {P_0 \over {\rho_l \ g}} \quad .$$

Unfortunately, it turns out that the above formal expression for the scale height is not correct for water vapor. Given a typical surface temperature ($T = 10^\circ$C), that expression would give a scale height of about 13 km. Observationally, the scale height of water vapor in the Earth's atmosphere is between 1.5 and 2 km (e.g. Ulich 1980). Therefore, the slightly more complicated expression (with scale height and surface temperature explicitly included) must be used. In this memo, a scale height of 1.5 km will be assumed. Since the derived precipitable water is linearly proportional to the assumed scale height, the results can be scaled as desired with little effort.