Frost On My Car
So, with the brunt of winter upon us I thought that I would share some of my experiences. Living in Oklahoma, our winter hovers on the edge of freezing quite often. One thing that I noticed right away was that on many nights that were near freezing (35°F or so), I would have to scrape ice off of one side of my car, but not the other. A few questions came to mind, First: Why is my car frosting when the low temperature was not below freezing (32°F)?, Second: Why was the ice only forming on one side of my car and not the other?
So this is approximately the setup of my car sitting in the drive way (garage door in front of it):
Question 1: Temperature at Which the Window Should FrostNow, to do a quick thought experiment to determine lowest approximate ambient temperature (T∞) that might lead to the window frosting over.
RadiationThe amount of heat radiated from your car window to space can be described using the Stephan Boltzmann Law (assuming your car window is a black body):
q = σ (Twindow - Tspace)^4
Where q is the heat flux (per unit area), σ is the Stefan Boltzmann constant (5.67E-8 W/m^2/K^4), Twindow is the temperature of your car in Kelvin, and Tspace is the temperature of outer space.
However, the story is not so simple because the car window is actually receiving radiation from all around (the adjacent buildings, walls, etc). Here is a schematic (and actually picture) from the car of the adjacent building that shows the "view factor" of the window to space.
And the actual picture from the car:
In this case, the "view factor", accounts for what percentage of the window surface can radiate to another object (in this case, we'll consider outer space and the adjacent house). From this picture, I'll estimate the view factor of the window to space as ~0.1 (i.e., 10% of the radiation from the window is radiated to space while 90% is radiated to the adjacent house/ground/etc.). Also, the window is at a slightly lower temperature (let's say the freezing temperature of water, Twindow=32°F=0°C=273.15°K) than the adjacent house (which is warmer--possibly around or slightly higher than the ambient temperature, T∞).
The net heat loss from the window due to radiation can be computed by:
Where we've assumed that space is approximately 0°K and that the adjacent house temperature is slightly higher than the ambient temperature (Thouse = T∞+ΔT). With this, we've developed a function for the amount of radiative heat transfer out of the window given the ambient temperature (T∞) and an assumption for the house temperature relative to ambient (say ΔT = 4°C or so--as in the house hasn't quite cooled down to ambient yet--which is reasonable).
Natural ConvectionNow that we've considered the radiative component of heat transfer, let's consider the counteracting force of convective heat transfer. In this case, because the window is at the frosting temperature (Twindow=0°C=273.15K) and the ambient air is possibly a higher temperature, the ambient air will actually act to warm the window. This will be what is known as natural convection (i.e., cooling by air that moves over a surface due to the fact that it's being heated or cooling, as opposed to force convection which might happen if it is windy).
The relevant equations that allow us to compute the convective heat transfer coefficient, h is that for the Nusselt number (Nu):
Where kair is the thermal conductivity of air at 0°C (http://www.engineeringtoolbox.com/air-properties-d_156.html) and Lwindow is the characteristic dimension of the window (assumed to be ~1 meter). The remaining parameter is the Rayleigh number (Ra), which is given by:
The Rayleigh number is a dimensionless number that is the product of two other dimensionless numbers: the Grashof number (Gr) and the Prandtl number.
The Grashoff number takes the ratio of the fluid buoyancy to the viscous driving forces. Here, we use the thermal expansivity (β = 1/T for an ideal gas) and the kinematic viscosity of air (νair).
We can compute the Ra number for a given ambient temperature (T∞).
Which can be used to compute the convective heat transfer coefficient (h) as a function of T∞.
By balancing the convective (qconv = h*(T∞-Twindow)) and radiative heat fluxes, we can come up with a final answer to the question of what temperature might the window on that side of the car freeze:
The ambient temperature (T∞) which gives a net heat flux of zero (i.e., which can sustain the window at Twindow = 273.15K), for the given assumptions is about 275°K (~35.3°F). This seems like a reasonable estimate, and explains why your car window will frost at above 32°F.
Question 2: Why doesn't the other side of the car frost over?
So to answer question two: why didn't the window on the side of my house ice over? The answer is simple: those windows didn't radiate directly out to space. Their view factor to the adjacent house was too high, and therefore we had significantly less radiative heat loss (the house was actually radiating its heat to the car as well since they are close). The neighbors house, on the other hand, was too far away to provide the same effect (as it had a lower view factor).
In the case of the smaller view factor to space (in the limit of zero), the window will not freeze until the ambient temperature reaches freezing (in reality, it will have to be below freezing to cool the window). This is why it's much more difficult for a car to frost when sitting in a garage or under a car port.
Interestingly, previous questions have been asked before: