A thermal window consists of two panes of glass separated by...

A thermal window consists of two panes of glass separated by an air gap. Each pane of glass is $3.00 \mathrm{~mm}$ thick, and the air gap is $1.00 \mathrm{~cm}$ thick. Window glass has a thermal conductivity of $1.00 \mathrm{~W} /(\mathrm{m} \mathrm{K})$, and air has a thermal conductivity of $0.0260 \mathrm{~W} /(\mathrm{m} \mathrm{K})$. Suppose a thermal window separates a room at temperature $20.00{ }^{\circ} \mathrm{C}$ from the outside at $0.00^{\circ} \mathrm{C}$. a) What is the temperature at each of the four air-glass interfaces? b) At what rate is heat lost from the room, per square meter of window? c) Suppose the window had no air gap but consisted of a single layer of glass $6.00 \mathrm{~mm}$ thick. What would the rate of heat loss per square meter be then, under the same temperature conditions? d) Heat conduction through the thermal window could be reduced essentially to zero by evacuating the space between the glass panes. Why is this not done?

A thermal window consists of two panes of glass separated by an air gap. Each pane of glass is $3.00 \mathrm{~mm}$ thick, and the air gap is $1.00 \mathrm{~cm}$ thick. Window glass has a thermal conductivity of $1.00 \mathrm{~W} /(\mathrm{m} \mathrm{K})$, and air has a thermal conductivity of $0.0260 \mathrm{~W} /(\mathrm{m} \mathrm{K})$. Suppose a thermal window separates a room at temperature $20.00{ }^{\circ} \mathrm{C}$ from the outside at $0.00^{\circ} \mathrm{C}$.
a) What is the temperature at each of the four air-glass interfaces?
b) At what rate is heat lost from the room, per square meter of window?
c) Suppose the window had no air gap but consisted of a single layer of glass $6.00 \mathrm{~mm}$ thick. What would the rate of heat loss per square meter be then, under the same temperature conditions?
d) Heat conduction through the thermal window could be reduced essentially to zero by evacuating the space between the glass panes. Why is this not done?

Key Concepts

Heat Transfer Rate through Conduction

The rate of heat transfer through a material layer is determined by the temperature difference across the material, the material's thermal conductivity, its thickness, and the area through which heat is transferred. This concept is fundamental to understanding how much heat is lost or gained in systems such as insulated windows.

Thermal Resistance in Series

When multiple layers of different materials are used, each layer offers a thermal resistance to heat flow. The overall thermal resistance is the sum of the individual resistances when the layers are arranged in series. This concept is used to analyze and design composite structures where precise control over heat transfer is required.

Practical Limitations of Vacuum Insulation

While evacuating the space between insulating layers can greatly reduce heat conduction by eliminating the medium for energy transfer, practical issues such as maintaining a vacuum, structural integrity, and cost make it less feasible for everyday applications like windows in residential buildings.

Thermal Conduction

Thermal conduction is the process by which heat energy is transferred within a material or between materials in direct contact. It involves the transfer of energy from high-temperature regions to lower-temperature regions through the vibration and interaction of particles within the material.

Thermal Conductivity

Thermal conductivity is a material property that indicates how well a material can conduct heat. High conductivity materials transfer heat efficiently, whereas low conductivity materials, like air, act as insulators. This property is critical in designing systems like thermal windows where controlling heat flow is essential.

Transcript

00:01 So here the conduction rate per square meter, so per unit area, would be equalling the thermal conductivity of glass, multiplied the t .0, in this case, room temperature, t0 minus t sub 1.

00:24 My apologies.

00:27 And this would be divided by l sub g, the thickness of the glass, essentially.

00:33 And so the heat flow through all three layers must be the same.

00:39 So we can say that then the conduction rate should be equal to k sub g a t -not minus t -s -sub -1.

00:51 This would be divided by l -s -g, and this is equaling then k -s -g, multiplied by t -1 minus t -sup -2 divided by l -s -g, and this is equalling then k -s -s -g, multiplied by t sub 2 minus t sub 3 divided by then l sub g.

01:27 And this, my apologies, this would be, we have an air layer in the middle.

01:33 So this, we can kind of change our subscripts here.

01:37 My apologies.

01:41 So this would be the thermal conductivity of air, the thickness of the air layer.

01:47 And so we can use these equations to find values for t where we can then say that k sub a a t sub 1 minus t sub 2 this will be divided by the thickness of the air layer equalling k sub g a t sub 2 minus t sub 3 divided by the thickness of the glass layer and at this point we can solve and we can say that k sub a, l sub g, t sub 1 plus k sub g, l sub a, t sub 3.

02:47 This will equal then t sub 2 multiplied by k sub a l sub g plus k sub g, l sub a.

03:01 And we then find that we can say k sub g a so this is from the previous equation once again t not minus t some 1 this would be divided by l sub g this would be equaling then k sub g a t sub 2 minus t sub 3 divided by the thickness of the class layer and here we can say that then t sub g a t sub 2 minus t sub 3 divided by the thickness of the class layer and here we can say that then t0 minus t sub 1 equals t sub 2 minus t sub 3 or t sub 1 equals t not minus t sub 2 plus t sub 3 so plugging one into the next and we can say that solving for t sub 2 this will be equal to then thermal conductivity of air times the thickness of the glass layer multiplied by t not plus t 3 plus thermoconductivity of glass thickness of the air layer t sub 3 this would all be divided by two times the thermoconductivity of the air times the thickness of the glass plus the thermoconductivity of the glass times the thickness of the air layer and at this point we can solve for part a.

04:48 So t sub 2 would be equalling then.

04:52 0 .026.

04:54 I'll lose the units for the sake of space, but this would be 0 .26 watts per meter per kelvin, multiplied by 0 .03 meters or 3 millimeters, multiplied by then 293 kelvin or room temperature, plus t sub 3, we know it to be 273, this would be plus than 1 .00, again, watts per meter per kelvin, multiplied by 0 .01 meters or 10 millimeters, multiplied by 273 kelvin.

05:39 This will all be divided by 2, multiplied by, again, 0 .026 watts per meter per kelvin.

05:48 This will be multiplied by the thickness of the glass layer .00.

05:53 3 plus 1 .00 watts per meter per kelvin multiplied by .01 meters.

06:03 And so t sub 2 would be equaling then to .154 degrees celsius or 273.

06:18 .154 kelvin.

06:23 Now, we can substitute this value in, and t sub 1 would then be equaling to 293 kelvin minus 273 .154 kelvin plus 273 kelvin.

06:44 T71 is found to be equal to 292 .846 kelvin.

06:54 Or we can say approximately 19 .8 degrees celsius...