# Why are wet bulb temperatures so warm compared to a naive calculation?

Wikipedia defines wet bulb temperature as:

the temperature of a parcel of air cooled to saturation (100% relative humidity) by the evaporation of water into it, with the latent heat supplied by the parcel.

I would like to calculate this temperature for:

\begin{align*} T & = 300\;\mathrm{K} \\ RH & = 0.5 \\ P & = 1\;\mathrm{atm} \approx 10^5\;\mathrm{Pa} \end{align*}

where $$RH$$ is the relative humidity.

The vapor pressure under these conditions is about $$P_V \approx 3.5 \times 10^3\;\mathrm{Pa}$$ (source)

So I estimate

$$\Delta T = \dfrac{Q}{n_{air}c_P}$$

where $$\Delta T$$ is the difference between wet bulb temperature and ambient temperature, $$Q$$ is the heat absorbed by the water as it evaporates, $$n_{air}$$ is the number of moles of air, and $$c_P$$ is the heat capacity at constant pressure of air.

Using $$Q = n_{water} L,$$ where $$n_{water}$$ is the number of moles of water that evaporate and $$L$$ is the latent heat of fusion of water,

$$\Delta T = \dfrac{n_{water}L}{n_{air}c_P}$$

The ideal gas law says

$$\dfrac{n_{water}}{n_{air}} = \dfrac{\Delta P_{vapor}}{P_{air}}$$

where $$\Delta P_{vapor}$$ is the change in the water vapor pressure as the water evaporates.

I have

\begin{align*} \Delta P_{vapor} & = .5\cdot 3.5\times 10^3 \;\mathrm{Pa} \\ L & = 40.8 \;\mathrm{kJ/mol} \\ c_P & = 30 \;\mathrm{J/(K\;mol)} \end{align*}

$$c_P$$ is calculated from $$c_P = \dfrac72 R$$ for a diatomic gas. The source on $$L$$, the latent heat of vaporization of water, is here.

Putting in these numbers, I find

$$\Delta T \approx 24 K$$

However, calculators like this one give $$\Delta T \approx 7\;\mathrm{K}$$.

My calculation leaves off some effects such as the heat needed to cool the vapor, but because the partial pressure of the vapor is small compared to that of air, this should be a small effect. So where is the factor of $$\approx 3-4$$ error coming from?

• It's for any size parcel of air. I'm at a loss to understand your comment; my question didn't have anything to do with boundary layers and wet cloths. It was about the theoretical concept of wet bulb temperature. Is your commenting supposing that the website I looked at is not a calculator for wet bulb temperature, but instead a calculator for real temperatures measured by swinging around wet thermometers? – Mark Eichenlaub Mar 12 at 21:11
• Well, that could be, but in that case there's a factor of 3 or 4 between the theoretical temperature and what that calculator finds, and Wikipedia at least doesn't acknowledge that this factor exists, and in fact claims that it is close to one. – Mark Eichenlaub Mar 12 at 22:05
• I removed my muddled comments. I wrote an answer below. Thank you for asking this question. – Pieter Mar 12 at 22:59

## 1 Answer

The problem in your calculation is that it is not possible to evaporate the amount of water that you assumed. You took 50 % of the vapor pressure at $$27\ ^\circ$$C ($$300$$ K). But that many moles would supersaturate at the calculated $$\Delta T = 24$$ K. The amount of water evaporated in a parcel of air is much less.

The psychrometric chart below shows the evaporative cooling line from $$27\ ^\circ$$C at 50 % RH intersecting with the vapor pressure curve at a wet bulb temperature of $$20\ ^\circ$$C. This is the limit for evaporative cooling, in agreement with the calculator that you quoted. (from http://www.coolbreeze.co.za/psyevap.htm )