A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and

A carpenter builds an exterior house wall with a layer of...

A carpenter builds an exterior house wall with a layer of wood $3.0 \mathrm{~cm}$ thick on the outside and a layer of Styrofoam insulation $2.2 \mathrm{~cm}$ thick on the inside wall surface. The wood has a thermal conductivity of $0.080 \mathrm{~W} /(\mathrm{m} \cdot \mathrm{K}),$ and the Styrofoam has a thermal conductivity of $0.010 \mathrm{~W} /(\mathrm{m} \cdot \mathrm{K})$. The interior surface temperature is $19.0^{\circ} \mathrm{C}$. and the exterior surface temperature is $-10.0^{\circ} \mathrm{C}$. (a) What is the temperature at the plane where the wood meets the Styrofoam? (b) What is the rate of heat flow per square meter through this wall?

Key Concepts

Thermal Resistance

Thermal resistance represents the opposition to heat flow through a material and is calculated as the thickness of the material divided by its thermal conductivity. It is an essential concept in analyzing and designing systems for thermal management and insulation.

Composite Wall Conduction

Composite wall conduction involves analyzing heat transfer through a structure made up of layers of different materials, each with its own thermal conductivity and thickness. In such systems, the overall heat transfer is determined by adding the thermal resistances of each individual layer in series, allowing for the calculation of temperature drops at interfaces and the overall rate of heat flow.

Fourier's Law

Fourier's Law of heat conduction states that the rate of heat transfer through a material is proportional to the negative gradient of temperatures and the area through which the heat flows. It is fundamental in determining how temperature differences drive heat flow in materials.

Thermal Conductivity

Thermal conductivity is a material property that indicates its ability to conduct heat. It quantifies how easily heat is transferred through a substance, with higher values representing better conductivity and lower values indicating insulating properties.

Transcript

00:01 Here we have an example of heat conduction through two layers in series.

00:07 The heat conduction equation for a single slab of area a, we'll call it, which is a cross -sectional area, is an equation for the heat loss, i'll call that q dot, is equal to a thermal conductivity constant.

00:29 That's a property of the material times the cross -sectional area, times the difference in temperature between one side and the other, divided by the thickness of the layer.

00:47 And it's helpful to think about this equation as very similar to v equals ir in electronics, with i being similar to the key, q.

01:01 Dot, so it's thermal current.

01:07 The change in temperature is the driver, or the potential difference, if you will.

01:14 It's like delta v in electronics.

01:18 And the thermal resistance is equal to delta x over the area times the property, thermal conductivity.

01:32 Yes, thermal resistance is related to the inverse of the conductivity.

01:39 So the thicker the layer, the more resistance to heat flow or heat loss, the larger the area, of course, the more heat loss that will be.

01:49 And just like in electronics, if you have multiple layers, each layer will have its thermal resistance.

02:00 Here we've got two layers, a wood layer and a styrofoam layer.

02:08 And the total resistance is going to just be the two of those added up.

02:17 You can also have a parallel situation with the layers side by side, but that's not usually how you build things.

02:29 Okay, so there's some things we know about the two layers.

02:34 We do not know the heat loss rate, which is going to be the same, through both, just like in series circuits, we have the same qdot through both layers and through the individual layers.

02:52 But let's write down some of the parameters so we can figure out the thermal resistance.

02:58 The styrofoam is a couple centimeters thick, 2 .022 meters, and it has a lower thermal conductivity.

03:11 It is 0 .010.

03:16 Let's see, in units of watts per meter squared kelvin.

03:32 For the wood, we have a slightly different set, three centimeters thick, and thermal conductivity a little bit higher.

03:58 Okay.

03:59 And we don't know the area.

04:00 So what will be finding is the power per unit area.

04:06 So we'll go ahead and work that out.

04:14 And since i've got everything in si units, i will not be dealing with the units as i figure this out.

04:24 Furthermore, we know that the temperature, high temperature, we're expecting to be on the inside layer.

04:34 And let's see, that is 19 degrees centigrade and the temperature on the cold side is equal to, let me go look that up, minus 10 degrees centigrade.

05:07 So lots of things to keep track of.

05:10 Let's make a little table just so we have these down where we can see them.

05:16 So the r1 is basically 0 .022 divided by the area by point 010...

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