How Thick Should the Air Layer Be to Limit Heat Loss in Winter Clothing?

[BhavinP]

Homework Statement

A man of 75 kg (1.77 m^2 surface area) in weight, dressed in appropriate winter clothing stands on a winter pond. The temperature outside is -5 degrees Celsius. Assume that the surface layer of his skin is 5 mm thicker and is not perfused with blood (behaves as an insulating layer). The temperature of blood in the vasculature just below this layer is 20 degree. Compute the thickness of air layer that needs to be trapped in the clothing such that heat loss by conduction is 100 W/m^2.

My only problem is the equation. I cannot figure out what to use. The class has no textbook and the lecture slides have the equation. I think I missed an equation when taking notes. I asked the prof but not reply yet. Any information is key! What is throwing me off is the skin layer.

Homework Equations

I do have one equation and its Watts/Area = k(T1 - T2) / d

T1 and T2 are temperatures of the skin and ambient area, respectively.
d is the distance from skin to the ambient temperature outside.
k is the conductance constant

No info is given about the jacket's material etc. The question is stated above.

 

[jbsteele]

The Attempt at a SolutionI am trying to find the distance d which is the thickness of the air layer. W/A = k (T1 - T2) / d 100 W/m^2 = k (20 - (-5)) / d k = 8.33 x 10-4 W/mK 100 W/m^2 = 8.33 x 10-4 W/mK (20 - (-5)) / d d = 0.00374 m The thickness of the air layer needed to be trapped in the clothing is 0.00374 m.

 

[kerimek]

The equation you have is the correct one to use for this problem. To solve for the thickness of the air layer needed, you will need to rearrange the equation to solve for d. The equation can be rewritten as:

d = k(T1 - T2) / (Watts/Area)

Substituting in the values given in the problem, we get:

d = k(20 - (-5)) / 100

Since the man's skin layer is 5 mm thick, we need to subtract this from the total distance d to get the thickness of the air layer (since the air layer is trapped between the skin layer and the ambient temperature outside). So the final equation would be:

d = k(20 - (-5)) / 100 - 0.005

The value of k will depend on the material of the clothing, but it is usually given in the problem or can be looked up in a table. For example, if we assume the clothing has a k value of 0.03 W/mK, then the final equation would be:

d = (0.03)(20 - (-5)) / 100 - 0.005 = 0.0155 m = 15.5 mm

Therefore, the thickness of the air layer that needs to be trapped in the clothing is 15.5 mm.