# Psychorometric equation : calculation of dew point

I would like to calculate a dew-point temperature having wet and dry bulb temperatures and atmospheric pressure. I have the equation, but this equation uses following coefficients:

• A: 6.116441
• K: 0.000662 ( Psychrometer constant )
• m: 7.591386
• Tn: 240.7263 (Triple point temperature )

To calculate water vapor saturation pressure I use simplified formula: $$\mathrm{Pws} = A\times\cfrac{m \cdot T}{T+Tn}$$ T is temperature

The given values are good for temperatures around 20 Celsius, but they fail around -5 or even lower. The accuracy around -20 degree ( wet bulb temperature) are terrible ( error around 3 : 5 degree ). What are the names of m and A coefficients? Where can I find these values for low temperatures?

First figure out how you should solve it with a psychrometric diagram $(p_w;T)$:

1. Find the wet bulb temperature on the saturation curve;
2. Follow isenthalpic curve until reaching the dry bulb temperature: you have found the mixture point;
3. Then follow the isobaric curve until reaching the saturation curve: you reach the dew point temperature.

Now, knowing the curve equations, you may try it to solve it analytically if you are looking for a formulae or numerically if you just care about an answer that can be automatized.

Correct thermodynamic data should be found in a Engineering or Chemical Handbook.

Saturation curve can be modelized by an exponential using Clapeyron-Clausius relation:

$p_\mathrm{sat}(T) = p_0 \exp \left\{ -\frac{ \Delta_\mathrm{vap}H}{R} \left(\frac{1}{T} - \frac{1}{T_0}\right) \right\}$

Where $p_0$ and $T_0$ is your reference pressure and temperature.

Recalling mass and amount balances you should be able to find that mass ratio $x$ (mass of water on mass of dry air) is linked to water pressure by:

$p_w(x) = p_0\frac{x}{x+\delta}$

Where $\delta = 0.62196$ is the ratio of molar mass (water/dry air). Isobaric curve then mean $x = k$.

Recalling the energy balance, isenthalpic curve have the following equation:

$h(T) = c_{p,a}(T-T_0) + x(c_{p,w}(T-T_0) + \lambda_\mathrm{vap}) = k$

Where subscript $a$ stands for dry air and $w$ stands for water. Quantities $c_p$ are isobaric specific heat and $\lambda$ is defined as follow:

$\lambda = \frac{\Delta H_\mathrm{vap}}{M_w}$

Those identities are easily demonstrable, you just need to write down balances. Then you have almost everything to plot a psychrometric diagram and you can solve every mixture you want.

If you want more explanation, you should consult some Chemical Engineering books, look for dry operation unit. You will find everything you need.