How Does Wind Speed Affect Perceived Temperature and Heat Transfer?

[gfd43tg]

Homework Statement

It is well known that wind makes the cold air feel much colder as a result of the windchill effect that is due to the increase in heat transfer coefficient with increasing air velocity. The windchill effect is usually expressed in terms of the windchill factor, which is the difference between the actual air temperature and the equivalent calm-air temperature. For example, a windchill factor of 20 °C for an actual air temperature of 5 °C means that the windy air at 5 °C feels as cold as the still air at -15 °C.

Estimate the wind-chill temperature on a day when the air temperature is -10 °C and the wind speed is 10 m/s, giving a convective heat transfer coefficient of 50 W/m2 K. A radiation heat transfer coefficient of 5 W/m2 K can be used, and under calm conditions the convective heat transfer coefficient can be taken to be 5 W/m2 K. Assume a 3 mm layer of skin (k=0.35 W/m K), clothing equivalent to 8 mm-thick wool (k=0.05 W/m K), and a temperature of 35 °C below the skin. Also calculate the skin outer temperature.

Homework Equations

The Attempt at a Solution

First off, this problem is worded so strangely that I am having trouble making sense out of it. First they want to know the wind chill temperature, and then they give a bunch of information about heat transfer coefficients, then start to ask about the temperature on the skin.
I just assumed in this problem that they meant that the person is wearing a coat over their skin (its -10 C after all) and went from there. Since the PF phone app appears to not be working, I made a sketch on paint and posted my interpretation of the problem.

So I see its simultaneous heat transfer, so I figure I would add up all of the resistances to heat transfer,
R_{total} = \sum_{i} {R_{i}}
So I go on to calculate the Resistance to heat transfer,
R_{total} = \frac{\Delta x_{1}}{k_{1}A} + \frac{\Delta x_{2}}{k_{2}A} + \frac {1}{h_{conv}A} + \frac {1}{h_{rad}A}

Then I realize that I am unsure if there is such thing as resistance to radiation heat transfer??
Either way, I proceed knowing that

\frac {1}{UA} = R_{total}

where $U$ is the overall heat transfer coefficient.

This simplifies to

\frac {1}{U} = \frac{\Delta x_{1}}{k_{1}} + \frac {\Delta x_{2}}{k_{2}} + \frac {1}{h_{conv}} + \frac {1}{h_{rad}}

And I plug in the numbers given in the problem statement, yielding

$U = 2.57 \frac {W}{m^2 K}$

I know that the heat flow rate due to conduction, $Q_{cond}$ is constant through the skin and wool coat, so my inclination to calculate the temperature at the surface of the skin is the following

\dot{Q_{cond}} = \frac {35 - T_{s}}{R_{1}}

knowing when calculating the heat flow rate

\dot{Q} = \frac {\Delta T}{R_{total}}

I wonder if my equation is legitimate for the conduction through the skin only? Either way, Area is unknown so I wouldn't be able to calculate this. The second equation, is that for all the heat flows added up? conduction, convection, and radiation?

I'm definitely missing something here, or the question is lacking pertinent information.

I then wonder, do I have to add the radiation of the wool coat to the surroundings, as well as the radiation of the skin to the surroundings? And from where? Do I do the radiation from the outside of the skin to the air, inside to air, outside of wool to air? There are so many possibilities.

So now I am boggled by this question because of the lack of clarity and seemingly lack of information. How can I calculate the temperature on the outside of the coat without skin temperature, etc.

 

[Chestermiller]

Maybe this will help. The convective and radiative resistances are in parallel, and are in series with both the skin layer resistance and the clothing resistance. Let
T_air = air temperature (-10C)
T_ca = the temperature at the interface between the clothing and the air
T_sc = the temperature at the interface between the surface skin and the clothing
T_b = the temperature at the interface between the surface skin and the body (35C)
The heat flux through all the layers and through the convective/radiative layer between the air and the clothing will be the same.

You are going to need to find the temperatures at the interfaces. Once you know the heat flux based on the overall combined resistance, you can look at each layer individually.

The only thing you did incorrectly so far was to assume that the convective and radiative resistances are in series, rather than in parallel.

Chet

 

[gfd43tg]

I redid it with parallel convective and radiative resistances, and after some algebra the expression simplified to

\frac {1}{UA} = \frac {\Delta x_{1}}{k_{1}A} + \frac {\Delta x_{2}}{k_{2}A} + \frac {1}{A(h_{conv}+h_{rad})}

and I calculate $U$ to be $5.35 \frac {W}{m^2 K}$

I then calculate the heat flux rate, $\dot{q}$, as $\dot{q} = U \Delta T$, which comes out to be $5.35(35 C - (-10 C)) = 240.75\frac {W}{m^2}$

Then I look at the skin clothing interface,
\dot{q}A = \frac {(35 - T_{sc})}{\frac {\Delta x_{1}}{k_{1}A}}

\dot{q}A = \frac {(35 - T_{sc})k_{1}A}{\Delta x_{1}}

And I calculate $T_{sc} = 32.7 \hspace{0.05 in} C$. Similar analysis (except it will be $(32.7 - T_{ca})$ in the numerator) yields $T_{ca} = -5.8 \hspace{0.05 in} C$.

So I solved one part of the problem, which was to find the temperature of the skin. I can redo the same exact calculations on a calm day when the heat transfer coefficient is 5 $\frac {W}{m^2 K}$ and get the skin temperature on a calm day. However, after all that the question first asked to find the wind chill factor, so what do I do for that?

I redid the calculations for calm conditions, and I get

$T_{sc} = 33.6 \hspace{0.05 in} C$ and $T_{ca} = 6.8 \hspace{0.05 in} C$. So the temperature of the skin/clothing interface is only slightly warmer, but the temperature of the clothing/air interface is substantially warmer. But where does the wind chill factor come into play here?

Should it just be the outer skin temperature on the calm day minus the outer skin temperature on the windy day? Or the clothing air interface temperature difference between windy and calm?

 

[Chestermiller]

The problem statement isn't clear on how the wind chill should be calculated. I would tend to calculate it as if the person wasn't wearing any insulating clothing, but that may not be what they want. In any event, you did the calculations correctly, which is the important part. If you really want to know how wind chill is determined, look it up on google. They must have a specific prescription that they use.

Chet