Re: At what speed does wind chill become fri ction heat

The subject of aerodynamics, including heating by air friction, is a very complicated one that experts study using complicated methods, including sophisticated numerical simulations using high-powered computers. My attempt to simplify the answer is not going to be exact, nor will it be satisfactory to someone wanting much precision, but I think we can obtain a rough order-of-magnitude answer to your question. You are asking specifically about wind chill, which is calculated on the basis of the perceived temperature of air around a human face (bare skin), so I will limit my discussion to that small part of our bodies.

I am going to make some assumptions (Which is almost always the case in any physics problem!!) to simplify things to where some calculations can be made. First, I make the assumption that the "face" being studied is the shape of a sphere. This simplification is needed to obtain an estimate of the drag coefficient. The right thing to do would be to take a "typical" person to a wind tunnel and measure the drag coefficient at various wind speeds, but I don't have access to a wind tunnel, and I have been unable to find wind-velocity-dependent coefficient tables anywhere, for a human body.

Second, I assume that there are only three sources (or sinks) of energy. The first source is the wind itself, and friction and/or compression will do some heating. The second "source" is conductional cooling by the air passing the "face" (or heating, if the air temperature is higher than the body temperature!). The third "source" is radiational (largely infrared) cooling by the "face".

The second and third are easy to compute, assuming we know something about the "face". The emissivity of bare skin is generally taken to be 0.97 (The emissivity of a perfect blackbody is 1.0.). Further assume that the "face" has a diameter of 0.10 meters. The thermal conductivity of air at 273K is 0.0243 and at 293K is 0.0257. The density of the air is 1.29 kg m^-3 while air's molecular viscosity is 17x10^-6 kg m^-1 s^-1, both at an air temperature of 273K. These values are pretty commonly accepted.

Cooling (or heating!) by conduction roughly follows the equation
P_c = -k A' / d (T_b - T_a)
where k is the thermal conductivity, A' is the surface area, T_b is the body's temperature, T_a is the air temperature, and d is the thickness of the boundary layer. The limiting layer is the boundary layer, whose thermal conductivity we will use (k of air). The boundary layer thickness is computed using the equation
d = L1 / sqrt(r v L1 / m)
where L1 is the face's diameter, "sqrt" is the squareroot, r is the density of the air, v is the velocity of the air, and m is the molecular viscosity.

(You can find all these equations at Internet pages dealing with air drag.)

For radiational cooling we will use the equation
P_r = -e s A' (T_b⁴ - T_a⁴)
where e is the emissivity, s is the "Stefan-Boltzman" constant having the value 5.67x10^-8, and A' is the surface area of the "spherical face".

Now, lastly, for the first source of energy, being the frictional heating of the moving air. Because power is the product of force and velocity, if we can compute the force due to the air we can compute the power that is (largely) dissipated as heat of friction (and/or compression) to the "face". To compute the force we need to know the coefficient of drag, and that is not a constant, but rather depends on the air velocity, as mentioned in early comments in this answer. Fortunately much study has gone into this subject, and it can be found on many Internet sites that spheres have been pretty thoroughly studied. The drag coefficients are determined empirically for the most part, and vary from 0.48 to 0.41 for the velocities we will consider (1 to 60 meters per second). The drag force is
F_f = 0.5 C_d r A v²
where C_d is the (tabulated) drag coefficient, r is again the air density, A is the frontal surface area (cross sectional area), and v is the velocity of the air. To obtain the power we multiply the force times the velocity, so we obtain, for the "frictional" heating,
P_f = 0.5 C_d r A v³
I will tabulate six wind speeds: 1, 5, 10, 20, 40, and 60 meters per second. I stop at 60 m s^-1 because that is the top wind speed for which wind chills are typically calculated. (For those who want to know what these are in miles per hour, they correspond to 2.2, 11, 22, 45, 90, and 134 miles per hour.)

The calculations can be done in a spreadsheet or some other software tool.

Putting all this together, let's first use a body temperature of 307K (which is 3 Kelvin (or Celcius) degrees below what we usually use as "body temperature"; the surface of our outer body is not quite at "body temperature"), and an air temperature of 273K (which is the freezing point of water). The second table shows the results for an air temperature of 293K (which is a typical value for "room temperature"):

Air Temperature 273K

Notice several qualitative aspects of the results. First, radiational cooling is very small compared to the other ways in which heat energy is transferred. Second, although the conducted energy is less for the higher air temperature, it is still a cooling effect for the lower wind speeds.

The main characteristic that I notice is that there actually is a net heating effect at higher wind speeds, according to my simplistic calculations. The crossover from cooling to heating is between 40 and 60 m s^-1 at an air temperature of 273K, but is between 20 and 40 m s^-1 at an air temperature of 293K.

One final note. Many resources will call my "conductional" effect "convection", because convection is the transport of heat energy by bulk movement of the medium. I use the term "conduction" because the actual mechanism of heat energy transport is the conduction between the face and the air, and I have included the effect of the thickness of the boundary layer.

John Link, MadSci Physicist