thermal conductivity


by Vanessa23
Tags: conductivity, thermal
Vanessa23
Vanessa23 is offline
#1
Nov17-07, 12:05 AM
P: 41
1. The problem statement, all variables and given/known data
A mountain climber wears a goose down jacket 3.28 cm thick with total surface area 1.10 m2. The temperature at the surface of the clothing is -19.3C and at the skin is 36.0C. Determine the rate of heat flow by conduction through the jacket assuming it is dry and the thermal conductivity, k, is that of down.
Part 2:
Determine the rate of heat flow by conduction through the jacket assuming the jacket is wet, so k is that of water and the jacket has matted down to 0.462 cm thickness.


2. Relevant equations
thermal conductivity of goose down is .025 J/(mKs)
thermal conductivity of water is .561W/(mKs)

phi dot = KA (delta T/delta X)

K is goose down (.025 j/smk)
A is area
T is temperature
x is distance heat flows

Part 2
The rate of heat flow= (the rate of heat flow you found in
part 1) x (k water/ k goose down) x (thickness of goosedown/
thickness of wet jacket)

3. The attempt at a solution
.025*(.0328*1.10)*(36+19.3/.0328)

I get the rate is 1.52 W, yet it is the wrong answer. If I had that answer I could use it to get part 2 where I would do the following
rate=rate1*(.561/.025)*(.0328/.00462)

Please let me know what I am doing wrong! Thank you!
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Shooting Star
Shooting Star is offline
#2
Nov17-07, 01:24 AM
HW Helper
P: 1,986
Why is the distance between the two surfaces in the numerator? Delta_x should be in the bottom.


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