Problem 17.104 Animals in cold climates often depend on two layers of insulation
ID: 2309989 • Letter: P
Question
Problem 17.104
Animals in cold climates often depend on twolayers of insulation: a layer of body fat [of thermal conductivity 0.200 W/(mK) ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1.40 m in diameter having a layer of fat 3.90 cm thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at 2.80 C during hibernation, while the inner surface of the fat layer is at 31.3 C.
Assume the surface area of each layer is constant and given by the surface area of the spherical model constructed for the black bear.
Part A
What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 50.2 W ?
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Part B
How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 50.2 W ?
Problem 17.104
Animals in cold climates often depend on twolayers of insulation: a layer of body fat [of thermal conductivity 0.200 W/(mK) ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1.40 m in diameter having a layer of fat 3.90 cm thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at 2.80 C during hibernation, while the inner surface of the fat layer is at 31.3 C.
Assume the surface area of each layer is constant and given by the surface area of the spherical model constructed for the black bear.
Part A
What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 50.2 W ?
T = CSubmitMy AnswersGive Up
Part B
How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 50.2 W ?
Lair = cmExplanation / Answer
rate of heat loss=conductivity*area*change in temperature/change in length
let temperature at fat-iiner fur boundary be T degree celcius.
surface area of a sphere=4*pi*radius^2
then 50.2 =0.2*4*pi*(1.4/2)^2*(31.3-T)/0.039
=>50.2= 31.57*(31.3-T)
=>31.3-T=1.59
=>T= 31.3 - 1.59 = 29.71 degree celcius
hence temperature at fat-inner fur boundary is 29.71 degree celcius
part B:
temperature at inner surface of air layer is 29.71 degree celcius
temperature at outer surface is 2.8 degree celcius
let thick ness be x m
thermal conductivity of air =0.025 W/(m.K
then 50.2=0.025*4*pi*(1.4/2)^2*(29.71-2.8)/x
=>x=0.025*4*pi*(1.4/2)^2*(29.71-2.8)/50.2 = 0.08251 m