Animals in cold climates often depend on two layers of...

Animals in cold climates often depend on two layers of insulation: a layer of body fat (of thermal conductivity $0.20 \mathrm{~W} /(\mathrm{m} \cdot \mathrm{K})$ ) surrounded by a layer of air trapped inside fur or down. We can model a black bear as a sphere $1.5 \mathrm{~m}$ in diameter having a layer of fat $4.0 \mathrm{~cm}$ 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 $2.7^{\circ} \mathrm{C}$ during hibernation, while the inner surface of the fat layer is at $31.0^{\circ} \mathrm{C}$. (a) What is the temperature at the fat-inner fur boundary, and (b) how thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of $50.0 \mathrm{~W} ?$

Key Concepts

Heat Transfer in Spherical Geometry

When dealing with spherical objects, heat transfer analysis must account for the curvature of the object. The geometry affects the area through which heat is conducted, as it varies with the radius. Specialized formulas are used to calculate the thermal resistance of spherical shells, making it possible to determine temperature gradients and heat flow in objects like animals or engineered systems with spherical symmetry.

Steady-State Heat Transfer

In steady-state conditions, the temperature distribution and heat flow remain constant over time. This assumption simplifies calculations because it allows the heat flux to be equated across every element in the system. The same heat transfer rate applies through each insulating layer, which is essential for determining temperature distributions and designing insulation with desired thermal properties.

Thermal Conduction

This concept involves the transfer of heat through a material due to a temperature gradient. It is governed by Fourier’s law, which states that the rate of heat transfer is proportional to the temperature difference and the cross?sectional area, and inversely proportional to the material thickness. This fundamental principle applies across various scenarios where heat is transmitted through solids, liquids, or gases.

Thermal Resistance in Composite Layers

This concept is used to analyze heat transfer through materials composed of several layers, each with its own thermal conductivity and thickness. In such systems, each layer presents a thermal resistance to heat flow. By modeling these layers as resistors in series, the overall resistance can be calculated, allowing the determination of temperature drops across each layer given a known heat flux.

Transcript

00:01 Here's an example of heat conduction through two layers.

00:05 What's a little bit different about this is that our layers are spherical, one layer being a sphere of fat around the hibernating bear and the other, the fur slash air that is on top of the bear.

00:26 So if you have watched animals hibernate, they do ball up.

00:31 But here, the heat conduction equation is q.

00:36 Dot is equal to, at least in magnitude, q.

00:41 Dot is equal to temperature difference between one side of the layer and the other.

00:48 So we'll call that t high and t low divided by the resistance to thermal heat flow.

00:59 So this is very similar to an electricity where you have a current.

01:05 The q .dot is similar to current, except it's in watts.

01:09 It's a heat flow.

01:10 It is driven by a temperature difference, and the resistance to thermal heat flow gets bigger.

01:21 The thicker the layer is delta x, we'll call that.

01:25 It gets smaller as the area grows, and then there is a constant kappa, which is defined to be the thermal conductivity and is a property of the material that you are using to impede the flow of heat.

01:49 So here we have two layers, and what do we know? we know that the inside temperature of the bear, the hottest part where all the metabolism is going on is 31 degrees centigrade.

02:05 We know on the outside we have roughly 2 .7 degrees centigrade.

02:12 So we assume the bear is in some sort of warmer area that's a little bit warmer than the outside or the ice outside.

02:25 And the surface area, we're going to have to use the area as the surface area of a sphere.

02:34 And that's a little bit problematic because there isn't just one sphere, but the sphere kind of grows.

02:42 But we'll do an approximation there.

02:45 And the rate of metabolism, we will assume as q .d.

02:52 Equal to 50 watts.

03:00 Okay, so there are a couple things that we can do, but realize that the layers are in series, the two layers of fat and air.

03:11 All right, so we've got fat.

03:13 That's four centimeters thick, and we will assume that the thermal conductivity for the fat is 0 .2 watts per, meter per degree centigrade.

03:34 There is a layer of air, what we'll just call the fur, an air.

03:41 We'll just call that an air layer to simplify.

03:46 And that has a thermal conductivity, a tenth of what the fat is.

03:55 But these are in series.

03:59 So what we know is that the conductivity through both layers is the same.

04:06 Same, the current, quote unquote, the thermal current is equal.

04:18 So 50 watts of energy through the fat from the inside and then through the air into the outside.

04:27 We also know that the resistance to thermal heat flow from both layers just adds in series.

04:42 So notice that we do not know the thickness of the air layer.

04:52 So that is one of the things that we can find is the thickness of the air layer.

04:58 We can also find the temperature after heat has passed through the fat, but before it goes through the hair.

05:08 So we'll call that, let's see, we'll call that t inner, and it should be in between 31 and 2 .5.

05:17 0 .7 degrees centigrade.

05:19 Let's start with that.

05:22 Find tn...

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