It is lower, but certainly not worthy of the "low" label it was given. Even "medium" would have been a poor characterization.DaveC426913 said:Compared to maximum radiative cooling in a vacuum, the loss in Antarctica is low(er).
True, but considering the OP's knowledge of the subject, I think numbers don't tell the whole story and need some framing.Gokul43201 said:PS: When we can actually get quantitative comparisons for cheap, why resort to vague, qualitative descriptions?
sneez said:Now that I am thinking about it, is the wind actually relevant? The wind acts only to slow down the process of equilibriation between body and the air, by keeping theT gradient between the body and air by blowing off the near skin molecules of air that my body heated. The body, even in height wind, cannot cool bellow the ambient air temperature, right?
Stop and think about that for a sec - you've felt wind before, right? You've worn heavy clothes, right? The purpose of heavy clothes (insulation) is to create a shallow temperature gradient. With no clothes and a heavy wind, you have the steepest possible temperature gradient and since heat loss is proportional to that gradient, you have the maximum possible heat loss.sneez said:Now that I am thinking about it, is the wind actually relevant? The wind acts only to slow down the process of equilibriation between body and the air, by keeping theT gradient between the body and air by blowing off the near skin molecules of air that my body heated.
Of course. But that's the steady-state condition where you are frozen solid and in a normal winter, that takes hours. We're talking about the heat loss in the beginning, when your body temperature is near normal and you are not dead.The body, even in height wind, cannot cool bellow the ambient air temperature, right?
I don't think that's Integral's point. The point is that "wind chill" is a measure of convective heat loss. It is basically how cold still air would have to be to provide the same convective heat loss as windy air at a higher temperature.I sure did, is it than reasonalbe to assume the windchill temperature rather than the air temperature in this example?
I'm not sure where you are getting the 1/2 number, and in the "other work" section, others who have calculated the ratio got .7 instead of 1/3 (2nd and fourth paragraphs). With wind, the first saw a 40% increase in convective rate with a 1mph wind (very, very low wind) and the fourth saw a 178% increase (10W/m^2) in a 3 m/s (2 mph) wind. So as that shows, the effect of wind is huge. With a very light wind (ie, just by walking), the convective heat loss rate swamps the radiative heat loss rate.sneez said:Well, quick back of envelope results given convection heat loss is 1/2 of radiative loss in wind from http://www.drphysics.com/convection/convection.html,where 1/3 of heat loss is due to convection without wind condition.
Stop and think about that for a sec - you've felt wind before, right? You've worn heavy clothes, right? The purpose of heavy clothes (insulation) is to create a shallow temperature gradient. With no clothes and a heavy wind, you have the steepest possible temperature gradient and since heat loss is proportional to that gradient, you have the maximum possible heat loss.
Of course. But that's the steady-state condition where you are frozen solid and in a normal winter, that takes hours. We're talking about the heat loss in the beginning, when your body temperature is near normal and you are not dead.
Thats exactly why I posed it as a question , do you have an answer? And due to its definition I think its reasonable to use the concept in the wind condition. Basically, I use wind chill temperature in calculation as a temperature surrounding air would "have" given no wind.I don't think that's Integral's point. The point is that "wind chill" is a measure of convective heat loss. It is basically how cold still air would have to be to provide the same convective heat loss as windy air at a higher temperature.
I sort of arbitrarily assumed it. The article claims 1/3 and other sources also cite that, but that's for no wind condition, I thought with wind convection gets more and more important as you say, so I gave it more importance .I'm not sure where you are getting the 1/2 number, and in the "other work" section, others who have calculated the ratio got .7 instead of 1/3 (2nd and fourth paragraphs). With wind, the first saw a 40% increase in convective rate with a 1mph wind (very, very low wind) and the fourth saw a 178% increase (10W/m^2) in a 3 m/s (2 mph) wind. So as that shows, the effect of wind is huge. With a very light wind (ie, just by walking), the convective heat loss rate swamps the radiative heat loss rate.
You said the wind slows the body reaching equilibrium with the air - it speeds it up.sneez said:Thats exactly what I thought I said.
Certainly not.1st point:Are you implying that it is possible for anybody temperature to cool below the T of the air due to wind effects at certain conditions?
The body isn't frozen solid, so talking about how heat would be removed when it is frozen solid is irrelevant.2nd: explain how that is not applicable in the beginning
?? An answer to what? If we should assume wind chill temp? I guess if your equation doesn't have a term for wind speed, you can use the wind chill temp instead.Thats exactly why I posed it as a question , do you have an answer?
Yes! That is why we sweat. Otherwise everyone in Phoenix would die.sneez said:1st point:Are you implying that it is possible for anybody temperature to cool below the T of the air due to wind effects at certain conditions?
So neglecting evaporation just considering convection? No, I don't think that convection can cool things below the air temperature. I think the only way that would even be theoretically possible is for the the flow to cause some unusually low pressure region which would experience an adiabatic drop in temperature. But surely such effects would be negligably small and very difficult to model for a human. The wind-tunnel tests might be fun though!sneez said:DaeSpam, sweat is cooling by evaporation, I was trying to see if object will cool below the air temperature just due to wind alone.