Effect of a Window in a Door. A carpenter builds a solid...

Effect of a Window in a Door. A carpenter builds a solid wood door with dimensions 2.00 $\mathrm{m} \times 0.95 \mathrm{m} \times 5.0 \mathrm{cm} .$ Its thermal conductivity is $k=0.120 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}$ . The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional 1.8 -cm thickness of solid wood. The inside air temperature is $20.0^{\circ} \mathrm{C},$ and the outside air temperature is $-8.0^{\circ} \mathrm{C}$ (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window 0.500 $\mathrm{m}$ on a side is inserted in the door? The glass is 0.450 $\mathrm{cm}$ thick, and the glass has a thermal conductivity of 0.80 $\mathrm{W} / \mathrm{m} \cdot \mathrm{K}$ . The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 $\mathrm{cm}$ of glass.

Key Concepts

Composite Thermal Conduction

Composite thermal conduction involves the analysis of heat transfer through systems made up of multiple layers or materials, each with its own thermal conductivity and thickness, as well as possibly additional resistances such as those due to convective films. Understanding composite conduction is essential for evaluating how different components interact, particularly when materials with very different thermal properties are combined, as the overall thermal performance is influenced by the contribution of each layer.

Thermal Resistance

Thermal resistance is a concept used to quantify the opposition to heat flow through a material or an assembly. It is analogous to electrical resistance in circuits. In thermal systems, different layers or components (such as solids and convective air films) each contribute a resistance to heat transfer, and these resistances can be summed when they are arranged in series. This approach simplifies the analysis of heat flow through complex systems by treating them as thermal circuits.

Temperature Difference as a Driving Force

The temperature difference between two regions is the driving force for heat conduction. In any steady-state conduction situation, the rate of heat transfer is directly proportional to this difference. A larger temperature gradient increases the thermal energy transfer rate, following the principles established by Fourier's law, making it a critical parameter in the design and analysis of thermal systems.

Steady-State Heat Conduction

Steady-state heat conduction refers to the transfer of thermal energy through a material or assembly when the temperature distribution does not change with time. In such conditions, the rate at which heat enters any region is exactly balanced by the rate at which it leaves, allowing for the use of simplified formulas and principles to calculate the heat flow based on material properties, geometry, and temperature differences.

Thermal Conductivity

Thermal conductivity is a fundamental material property that measures a material's ability to conduct heat. A higher thermal conductivity indicates that the material can transfer heat more efficiently. It appears as a proportionality constant in Fourier's law of heat conduction, where the rate of heat transfer through a material is directly proportional to the thermal conductivity, the area through which heat flows, and the temperature difference, and inversely proportional to the thickness of the material.

Transcript

00:01 Okay, so for part a of this question, we want to find the rate of heat flow through a solid wooden door.

00:08 And to keep in mind that there are air films on the inner and outer surfaces of the door, that the same combined thermal resistance as an additional 1 .8 centimeters of solid wood.

00:20 So when it's asking for the heat flow, we want to think of heat current.

00:24 There's a flow of heat from outside, inside, inside, inside, depending on the temperatures.

00:29 So in this case, we have, that's given to us that the temperature of outside is minus 8 celsius, temperature of inside is 20 celsius.

00:37 So really all we have to do here is use the equation for heat current, which is h equals, h equals k, which is thermal resistance or thermal conductivity, excuse me, thermal conductivity, times a, which would be the area of our door, and times th, so the hotter temperature, minus tc the colder temperature and that difference is divided by l or like the width or the length at which the heat will travel from since we're talking about a door that length is actually going to be the thickness of the door if you can imagine kind of the cold air on the left side of the door outside of the door and that hot air on the inside it's going to travel through that door's thickness so the thing i want to keep in mind is what is a and l because we already have what k is we know the hot temperature is 20 celsius and the cold temperature is 8 celsius.

01:43 But what is l and a? so a is just going to be the area of the door, which is 2 meters times 0 .95 meters, the length times the height of the door.

01:56 So that's just going to be 1 .9 meters squared.

02:04 Okay.

02:05 Okay.

02:05 And the length, another width of the door is five centimeters, right? that's given.

02:10 But they give us this extra information about these air films.

02:13 And that's why we have to make sure we read all the given information because this adds an extra 1 .8 centimeters of thickness.

02:21 So these air films act as about two centimeters of extra wood.

02:25 That's on the door.

02:26 That's not actually there.

02:27 Just the filters that construction people put on the doors to help trap the heat better and kind of help the transfer of heat in your home.

02:35 So 5 centimeters plus this 1 .8 centimeters and that's going to be that length through which the heat's going to try to move through so we have to do now plug in our values and solve for this h we'll do h we call this h sub door that was not very well written we do it for simplicity sake h sub d that's equal to times k which is 0 .12 watts and while you're writing this out keep in mind like what you want our units to be.

03:19 So we're talking about heat flow and heats is obviously energy.

03:24 So that'll be in jules, but talking about a current or a flow, that's over time.

03:29 So we want our units to be watts.

03:35 In times the area, 1 .9 meter squared.

03:44 In times a temperature difference, again, we're doing a temperature difference so you can use celsius or kelvin, but since our thermal conductivity's in kelvin, we should use kelvin.

03:52 So that temperature difference is actually going to be 28 because we're doing 20 minus a positive 8.

03:58 I'm sorry, 28 minus a negative 8.

04:01 So 28 kelvin divided by our length in meters.

04:05 Again, keep si units.

04:07 So 6 .8 centimeters is 0 .068 meters.

04:18 If you plug and chug that out, we'll find that h sub d or this heat flow or heat current through the door is 93, 0 .83 .88 or 93 .9 .9.

04:43 Okay, great.

04:45 There's part a done.

04:51 So for part b, we would now want to find by what factor is the heat flow increased if a window is inserted into the door.

05:03 So essentially we're doing the same type of equation, but now we have two materials that make up our door, the new glass window and then the remaining area of the wooden door.

05:12 So essentially we're going to have two heat current values that i'll app together to get a new total heat current value for this new door.

05:20 So let's first find heat current to the glass.

05:27 Let's say hg.

05:30 Okay, so again, they give us a new thermal conductivity, which you can also look up where it's in the book.

05:38 We'll call that case of g.

05:41 And then what's the area of the window? it says it's 0 .5 by 0 .5...

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