7
$\begingroup$

It is well known that temperature sensors are often shaded, partially to mitigate the variability due to varying cloud cover. However, most people experience temperature within the sunlight. Other measurements of temperature that describe how temperature is felt are the heat index and wind chill, but both neglect the effect of the sun. For example, a very low wind chill can partially be overcome by a clear, sunny day.

Is there such a formula that perhaps gives an upper echelon of experiential temperature? I envision such a formula would involve the solar angle, albedo and perhaps the radiative balance.

$\endgroup$
  • 1
    $\begingroup$ Are you interested in the temperature of the ground? The temperature as perceived by a person? Or the temperature of an instrument under direct sunlight? All of those are pretty different cases. Please clarify. $\endgroup$ – Camilo Rada Mar 19 '18 at 16:41
  • 1
    $\begingroup$ When you're standing in sunlight, you're being heated by the sun's radiation. Google "insolation" for more details. The hotter you get, the more energy you lose to the surrounding environment (if the air temperature is roughly constant), until you're receiving as much radiation heat as you're giving off conductive/convective heat. That thermal equilibrium point would be the "in-sun temperature". As @CamiloRada notes, it would be different for a living creature, because we create heat internally as well. $\endgroup$ – Barry Carter Mar 19 '18 at 21:26
  • 1
    $\begingroup$ I think this would also broadly fall under biometeorology and personal head exposure research; personal experience of temperature is going to vary with factors such as humidity, hydration, clothing, ability to perspire, overall health, and light and heat reflection from nearby surfaces. $\endgroup$ – jeffronicus Mar 19 '18 at 22:35
  • $\begingroup$ @BaroclinicCplusplus - do you want the 2 m temperature value output by models ? $\endgroup$ – gansub Mar 20 '18 at 14:30
  • $\begingroup$ @CamiloRada I'd look more for either the instrument or the person. The ground is too complicated since it requires a variable heat source. People's skin temperature is weakly 84 degrees Fahrenheit (29 Celsius). The instrument has negligible heat source (assuming the thermoelectric effect). The person-perceived may be more practical yet uncertain, but the instrument-perceived may be more useful. Preferably instrument temperature then. $\endgroup$ – BarocliniCplusplus Mar 20 '18 at 14:39
5
$\begingroup$

It is not an easy problem even for a simple object. The temperature will vary as the conditions change, but the usual approach to such problems is to find the equilibrium temperature for a set of conditions. This is, to find the final temperature that the object will reach if exposed to a given solar irradiation, with air at a given temperature, for given pressure, humidity, wind, etc.

Such equilibrium state will allow you to find the temperature by assuming energy balance. That can be stated as requiring that the incoming energy into the body equals the outgoing energy, or

$Ẹ_{in} = E_{out}$

To simplify the explanation let's assume the object was in equilibrium with air temperature during the night and we want to calculate its temperature onces the sun is shining on it. In this case the incoming energy will be function of the absorbed solar energy, therefore it will be function of the radiation intensity ($R_{sun}$) that depends on solar elevation, cloud cover, etc. (and can be obtain as a direct output of climate models), the albedo of the object ($\alpha$) and the area exposed to sunlight (i.e. coross-section) ($A$), that will be given by the shape of the object, solar elevation and azimuth. So, that is equivalent to:

$E_{in}=R_{sun} (1-\alpha) A$

Now, the energy leaving the object will have two components.

  1. Radiation: The object will cool down emitting electromagnetic radiation (mostly infrared) according to the Stefan-Boltzmann law.
  2. Conduction: As the object get hotter than the surrounding air, it will loose heat by thermal conduction into it.

So

$E_{out}=\kappa (T - T_{air}) + \varepsilon \sigma T^4 S$

Where $\kappa$ is the thermal conductivity of air, $T$ is the temperature of the object and $S$ its surface area. $\varepsilon$ is the emissivity (that is 1 for a black-body), and $\sigma$ the Stefan-Boltzmann constant.

The three equations above are sufficient to solve for the temperature of the object, but as you can see there are many parameters that depends on the object shape, and material, so it is impossible to generalize.

Also, in the absence of wind, the temperature of the air in contact with the object won't be the same as the surrounding air, further complicating the problem.

Furthermore, in addition to $A$, $\varepsilon$ and $\alpha$ can as well change with solar elevation and azimuth (if the object is made out different materials).

And to further complicate the problem, the thermal conductivity of the air $\kappa$ is a function of temperature, pressure and humidity. And it is not an easy relationship, it is usually tabulated or calculated using empirical formulas.

At engineeringtoolbox.com you can find tables and graphs of $\kappa$ for dry air, here is one example: enter image description here

But the humidity have also an impact much more difficult to quantify, and whose importance grow with temperature. There is an article at electronics-cooling.com that explain the effect, summarized in the following graph:

enter image description here

So, that is hard enough for an object, let alone for a person. In such case there would a lot more parameters that come to play. Like the amount of insulation provided by clothing, surface and moisture content of the skin exposed to the wind, body heat production, etc.

Apparent temperature

That been said, there are some attempts to estimate the perceived temperature for people including humidity, wind and solar radiation. The best example is perhaps the Wet-bulb globe temperature ($\mathbf{WBGT}$), that consist on the weighted mean of the readings of three different types of thermometers: A normal one, a wet-bulb one (affected by wind and humidity) and a very uncommon black-globe thermometer (affected by solar radiation).

The formula is:

$\mathbf{WBGT} = 0.7 \, T_{wet-bulb} + 0.2 \, T_{black-globe} + 0.1 \, T_{air}$

However, this value is not often used because is very location specific (depend on clouds for example). Nevertheless, perhaps a good approach you can take is to use actual solar irradiation from weather models to compute the temperature that a wet-bulb and a black-globe thermometer would produce, something that I think would be easy to achieve, and then compute the $\mathbf{WBGT}$ with the above formula.

Be aware that the $\mathbf{WBGT}$ value could be useful for people to know how cold/hot it will feel, but won't tell you anything about the actual temperature of different objects exposed to those conditions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.