This, in response to a question from Ben Silver, is just for fun.

WHY DO T HEY DO T HE DEW T HE WAY T HEY DO?

A modest contribution to the annals of weather by Fred Bisshopp

Note 1: When first I started to look for ways to find the dewpoint I was somewhat irritated to learn that it is seldom mentioned in fluid mechanics texts. Bad news, because books about the weather can be really hard to read - too much jargon (but do read on). Later, I found that I own a pretty good reference that has given me enough information to put together an algorithm. The book is: A short course in Cloud Physics by R. R, Rogers and M. K. Yau, Butterworth-Heineman, 1989.

Note 2: The three measurable quantities are what we have on our weather stations, namely pressure (p), temperature (T) and relative humidity (h - to be defined presently).

Note 3: Subscripts (e.g. p_{s}(T) and ρ_{s}(T)) and/or diacritical marks (e.g. ρ́, ρ̀ and ρ̂) will be used for quantities that are not directly measured.

Note 4: The acute accent refers to the dry component, composed of O_{2}, N_{2} and other gases, and the grave accent refers to water vapor.

Note 5: Saturation levels for water vapor are p_{s}(T) and ρ_{s}(T), and the dewpoint is T_{s}(p, T, h), where p_{s}(T_{s}) = p̀(p, T, h).

Now here we go with a mixture of two ideal gases in equilibrium, so

ρ̀ = \frac{M̀p̀}{RT} ( density - wet component - M̀ ≈ 18 ),

ρ́ = \frac{Ḿṕ}{RT} ( density - dry component - Ḿ ≈ 28.5 ),

p = ṕ + p̀ ( Dalton’s law ) ,

ρ̂ = ρ́ + ρ̀ ( Gertrude’s observation ),

and R is the universal gas constant. From these, the density of the mixture is

ρ̂ = \frac{M̀p̀ + Ḿ(p − p̀)}{RT} = \frac{p}{ŔT} (1 − (1 − \frac{M̀}{Ḿ} ) \frac{p̀}{p}) ≈ \frac{p}{ŔT} (1 − 0.378\frac{p̀}{p})

(Note the use of the dry-air gas constant, Ŕ = R/Ḿ ≈ 287 J/kg/^{∘}K.) The concentrations are ć = ρ́/ρ̂ and c̀ = ρ̀/ρ̂ (with ć + c̀ = 1), and the mixing ratio is

x̂ = \frac{ρ̀}{ρ́} = \begin{matrix}M̀ & p̀ \\ Ḿ & ṕ \end{matrix} ≈ 0.622\frac{p̀}{p − p̀}.

In terms of pressure and vapor pressure, the specific humidity is

q̀(p, p̀) (= c̀) ≈ \frac{M̀p̀}{Ḿ(p − 0.378p̀)} ≈ 0.622\frac{p̀}{p − 0.378p̀}

Now at the temperature T, water vapor and the other phase (solid or liquid) are in equilibrium at the saturation pressure

p_{s}(T) ≈ 6.112 exp (\frac{17.67 T}{T + 243.5}) ,

where p is in mb and T is in ^{∘}C. (The empirical formula is cited by Rogers & Yau.) Given p_{s}(T) the saturation mixing ratio and specific humidity are

x_{s}(p, T) ≈ 0.622\frac{p_{s}(T)}{p − p_{s}(T)} and q_{s}(p, T) ≈ 0.622\frac{p_{s}(T)}{p − 0.378p_{s}(T)},

and, at last (sigh), the relative humidity is h = q̀/q_{s}(p, T) (times 100 for %).

The rest is pretty easy, but not very pretty. First comes the evaluation,

p_{s}(T) ≈ 6.112 exp (\frac{T}{13.78(1 + T/243.5)}) .

(This has been rewritten to indicate the reasonable approximation when |T/243.5| is small.) Then q̀ = hq_{s} gives the evaluation of p̀(p, T, h),

\frac{p̀}{1 − 0.378 p̀/p} = \frac{hp_{s}}{1 − 0.378 p_{s}/p} → p̀ = \frac{hp_{s}(T)}{1 − 0.378(1 − h)p_{s}(T)/p}.

In the definition of T_{s} the pressure p and the mixing ratio x̂ are fixed, so

\frac{p_{s}(T_{s})}{1 − p_{s}(T_{s})/p} = \frac{p̀}{1 − p̀/p} → p_{s}(T_{s}) = p̀.

Finally, from the inversion of the empirical formula for p_{s}(T_{s}) comes

ℓ_{s} = ln \frac{p̀}{6.112} and T_{s} = \frac{13.78 ℓ_{s}}{1 − ℓ_{s}/17.67}.

So how do they do the dew? Well, according to Rogers & Yau: Back in the days before computers began their destruction of the quality of our lives, it was done graphically, with clever diagrams called tephigrams, on which contours of constant pressure, temperature, entropy and mixing ratio were conveniently displayed. (T emperature, Entropy, Pressure, Humidity? - nobody said meteorological jargon couldn’t be fun.) I suspect that these days it is more often done with a numerical algorithm that looks something like this:

θ_{s} = \frac{T}{13.78(1 + T/243.5)} ;

ℓ_{s} = θ_{s} + ln h − ln (1 − \frac{2.31 (1 − h) exp θ_{s}}{p} ) ;

T_{s} = \frac{13.78 ℓ_{s}}{1 − ℓ_{s}/17.67} ;

and I think I prefer tephigrams. 4