Home owners and builders discuss thermal conductivities in terms of the value ($\mathbf{R}$for resistance) of a material, defined as the thickness divided by the thermal conductivity:

$R\equiv \frac{\Delta x}{k_t}$

(a) Calculate the $R$value of a $1/8\text{-inch}$($3.2\mathrm{mm}$) piece of plate glass, and then of a $1\mathrm{mm}$layer of still air. Express both answers in SI units.

(b) In the United States, $R$values of building materials are normally given in English units,${}^{\circ }\mathrm{F}\cdot {\mathrm{ft}}^2\cdot \mathrm{hr}/\mathrm{Btu}$. A Btu, or British thermal unit, is the energy needed to raise the temperature of a pound of water $1^{\circ }\mathrm{F}$. Work out the conversion factor between the SI and English units for values. Convert your answers from part (a) to English units.

(c) Prove that for a compound layer of two different materials sandwiched together (such as air and glass, or brick and wood), the effective total $R$value is the sum of the individual $R$values.

(d) Calculate the effective $R$value of a single piece of plate glass with a $1.0$$\mathrm{mm}$layer of still air on each side. (The effective thickness of the air layer will depend on how much wind is blowing; $1\mathrm{mm}$is of the right order of magnitude under most conditions.) Using this effective $R$value, make a revised estimate of the heat loss through a $1-{\mathrm{m}}^2$single-pane window when the temperature in the room is ${20}^{\circ }\mathrm{C}$higher than the outdoor temperature.

In the construction industry, thermal conductivity is frequently expressed in terms of the $R$value, which is defined as:

$R=\frac{\mathrm{\Delta }x}{k_t}$

Since $R$depends on the inverse of the thermal conductivity and directly on the thickness of the insulating material, a larger $R$means a better insulator.

For the $1\mathrm{mm}$layer of still air with $k_t=0.026\mathrm{J}\cdot {\mathrm{s}}^{-1}\cdot {\mathrm{m}}^{-1}\cdot {\mathrm{K}}^{-1}$, we have:

$R_{air}=\frac{\mathrm{\Delta }x}{k_t}=\frac{0.001}{0.026}=0.0385{\mathrm{J}}^{-1}\cdot \mathrm{s}\cdot {\mathrm{m}}^2\cdot \mathrm{K}$

Given that $k_t=0.8\mathrm{J}\cdot {\mathrm{s}}^{-1}\cdot {\mathrm{m}}^{-1}\cdot {\mathrm{K}}^{-1}$for glass, the $R$value of a $3.2\mathrm{mm}$thick sheet of glass is:

$R_{\text{glass}}=\frac{\mathrm{\Delta }x}{k_t}=\frac{0.0032}{0.8}=0.004{\mathrm{J}}^{-1}\cdot \mathrm{s}\cdot {\mathrm{m}}^2\cdot \mathrm{K}$

Thus if there is a $1\mathrm{mm}$layer of still air next to a window, it actually provides more insulation than the window glass itself.

In the US, the units of $R$are $\mathrm{F}\cdot {\mathrm{ft}}^2\cdot \mathrm{hr}\cdot {\mathrm{Btu}}^{-1}$where a British thermal unit $\text{Btu}$is the amount of energy necessary to elevate one pound of water by $1^{\circ }\mathrm{F}$. To convert to SI units of $R$, we need to convert $1\text{Btu}$to Joules. One Fahrenheit degree is$\frac{5}{9}$of a Kelvin, there are $453.592$grams per pound, and $4.186$joules are required to raise $1$a gram of water by $1^{\circ }\mathrm{K}$. Therefore:

$1\mathrm{Btu}={\frac{5}{9}}^{\circ }\mathrm{K}\times \left(4.186\right)\mathrm{J}\times \left(453.592\right)\mathrm{g}=1054.85\mathrm{J}$

One foot is $0.3048\mathrm{m}$and $1$hour is $3600$seconds, so

$1^{\circ }\mathrm{F}\times {\mathrm{ft}}^2\times \mathrm{hr}\times {\mathrm{Btu}}^{-1}={\frac{5}{9}}^{\circ }\mathrm{K}\cdot (0.3048{)}^2{\mathrm{m}}^2\times (3600)\mathrm{s}\times (1054.85{)}^{-1}{\mathrm{J}}^{-1}$

$1^{\circ }\mathrm{F}\cdot {\mathrm{ft}}^2\cdot \mathrm{hr}\cdot {\mathrm{Btu}}^{-1}={0.176}^{\circ }\mathrm{K}\cdot {\mathrm{m}}^2\cdot \mathrm{s}\cdot \mathrm{J}$

The previous $R$values in English units are therefore:

$R_{air}=\frac{1}{0.176}\times 0.0385={0.2187}^{\circ }\mathrm{F}\cdot {\mathrm{ft}}^2\cdot \mathrm{hr}\cdot {\mathrm{Btu}}^{-1}$

$R_{\text{glass}}=\frac{1}{0.176}\times 0.004={0.0227}^{\circ }\mathrm{F}\cdot {\mathrm{ft}}^2\cdot \mathrm{hr}\cdot {\mathrm{Btu}}^{-1}$

The $R$values of a compound layer of two different materials are the sum of the individual $R$values. This can be seen as follows. Assume we have a composite layer made up of two components.: material $1$and material $2$. The temperature on the exposed side of the material $2$is $T_2$and on the material $1$side is $T_1$. At the place where the two materials meet, the temperature is$T_a$.

The rate of heat flow across the two layers must be the same in the steady-state (otherwise, heat would build up elsewhere), and employing $\frac{Q}{\mathrm{\Delta }t}=-k_{t1}A\frac{\mathrm{\Delta }T}{\mathrm{\Delta }{x}_1}$, so from the material $1$we have:

Let be Equation ($1$)

$\frac{Q}{\mathrm{\Delta }t}=-k_{t1}A\frac{\mathrm{\Delta }T}{\mathrm{\Delta }{x}_1}=-k_{t1}A\frac{T_a-T_1}{\mathrm{\Delta }{x}_1}$

and for material $2$, we have equation ($2$):

$\frac{Q}{\mathrm{\Delta }t}=-k_{t2}A\frac{\mathrm{\Delta }T}{\mathrm{\Delta }{x}_2}=-k_{t2}A\frac{T_2-T_a}{\mathrm{\Delta }{x}_2}$

by equating equation ($1$) and equation ($2$), we have equation ($3$):

$k_{t2}\frac{T_2-T_a}{\mathrm{\Delta }{x}_2}=k_{t1}\frac{T_a-T_1}{\mathrm{\Delta }{x}_1}$

where $A$(area) is dropping out since in both layers the area is the same. we have equation ($4$):

$R_1=\frac{\mathrm{\Delta }{x}_1}{k_{t1}}{R}_2=\frac{\mathrm{\Delta }{x}_2}{k_{t2}}$

substitute from ($4$) into ($3$), we get equation ($5$),

$\frac{T_2-T_a}{R_2}=\frac{T_a-T_1}{R_1}$

The overall rate of heat transfer across the compound layer must now be the same as well.. If we define $R_c$it to be the effective $R$value of the compound layer, then:

$\frac{T_2-T_a}{R_2}=\frac{T_a-T_1}{R_1}=\frac{T_2-T_1}{R_c}$

so we have two equations, let's be equation ($6$).

$\frac{T_2-T_a}{R_2}=\frac{T_a-T_1}{R_1}\frac{T_a-T_1}{R_1}=\frac{T_2-T_1}{R_c}$

now we need to solve these two equations for $R_c$. From the first equation, we can solve for $T_a$:

$\frac{T_2-T_a}{R_2}=\frac{T_a-T_1}{R_1}\to R_1{T}_2-R_1{T}_a=R_2{T}_a-R_2{T}_1$

$R_1{T}_2+R_2{T}_1=R_2{T}_a+R_1{T}_a\to R_1{T}_2+R_2{T}_1=T_a\left(R_2+R_1\right)$

$T_a=\frac{R_1{T}_2+R_2{T}_1}{\left(R_2+R_1\right)}$

Substituting into the second equation in ($6$) we can solve for $R_c$:

$\frac{T_a-T_1}{R_1}=\frac{T_2-T_1}{R_c}\to \frac{1}{R_1}\left[T_a\right]-\frac{T_1}{R_1}=\frac{T_2-T_1}{R_c}$

$\frac{1}{R_1}\left[\frac{R_1{T}_2+R_2{T}_1}{\left(R_2+R_1\right)}\right]-\frac{T_1}{R_1}=\frac{T_2-T_1}{R_c}$

$\left[\frac{R_1{T}_2+R_2{T}_1}{R_1\left(R_2+R_1\right)}\right]-\frac{\left(R_2+R_1\right)T_1}{\left(R_2+R_1\right)R_1}=\frac{T_2-T_1}{R_c}$

$\frac{R_1{T}_2+R_2{T}_1-R_2{T}_1-R_1{T}_1}{R_1\left(R_2+R_1\right)}=\frac{T_2-T_1}{R_c}$

$\frac{R_1\left(T_2-T_1\right)}{R_1\left(R_2+R_1\right)}=\frac{T_2-T_1}{R_c}$

$R_c=R_1+R_2$

Using this compound $R$value, we can estimate the rate of heat loss from a $1{\mathrm{m}}^2$single-pane window of thickness $3.2\mathrm{mm}$, but with a $1$$mm$layer of still air on each side. The effective $R$value of this system is:

localid="1650266582013" $R_c=R_{\text{glass}}+2\times R_{\text{air}}$

using the results from step 1,

localid="1650266585531" $R_{\text{air}}=0.0385{\mathrm{J}}^{-1}\cdot \mathrm{s}\cdot {\mathrm{m}}^2\cdot \mathrm{K}$

localid="1650266588974" $R_{\text{glass}}=0.004{\mathrm{J}}^{-1}\cdot \mathrm{s}\cdot {\mathrm{m}}^2\cdot \mathrm{K}$

so, the effective localid="1650266593411" $R$value of this system is, therefore:

localid="1650266597424" $R_c=0.004+2\times 0.0385=0.081{\mathrm{J}}^{-1}\cdot \mathrm{s}\cdot {\mathrm{m}}^2\cdot \mathrm{K}$

When the temperature difference is localid="1650266601163" $\mathrm{\Delta }T={20}^{\circ }\mathrm{K}$, the rate of heat is:

$\frac{Q}{\mathrm{\Delta }t}=-k_{t}A\frac{\mathrm{\Delta }T}{\mathrm{\Delta }x}$

with, localid="1650266605707" $R_c=\frac{\mathrm{\Delta }x}{k_t}$so:

localid="1650266613094" $\frac{Q}{\mathrm{\Delta }t}=-A\frac{\mathrm{\Delta }T}{R_c}$

substitute,

localid="1650266651710" $\frac{Q}{\mathrm{\Delta }t}=-1\times \frac{20}{0.081}=246.91$localid="1650266657465" $watts$

This compares with the heat loss through the glass on its own localid="1650266667372" $\frac{A\mathrm{\Delta }T}{R_{g}lass}=5000$localid="1650266673006" $watts$. Thus the air layer provides most of the insulation.

Therefore, it is concluded that,

localid="1650266679151" $R_{air}=0.0385{\mathrm{J}}^{-1}\cdot \mathrm{s}\cdot {\mathrm{m}}^2\cdot \mathrm{K}$

localid="1650266683604" $R_{\text{glass}}=0.004{\mathrm{J}}^{-1}\cdot \mathrm{s}\cdot {\mathrm{m}}^2\cdot \mathrm{K}$

localid="1650266688380" $1^{\circ }\mathrm{F}\cdot {\mathrm{ft}}^2\cdot \mathrm{hr}\cdot {\mathrm{Btu}}^{-1}={0.176}^{\circ }\mathrm{K}\cdot {\mathrm{m}}^2\cdot \mathrm{s}\cdot \mathrm{J}$,

localid="1650266693304" $R_{\text{air}}={0.2187}^{\circ }\mathrm{F}\cdot {\mathrm{ft}}^2\cdot \mathrm{hr}\cdot {\mathrm{Btu}}^{-1}$

localid="1650266702329" $R_{\text{glass}}={0.0227}^{\circ }\mathrm{F}\cdot {\mathrm{ft}}^2\cdot \mathrm{hr}\cdot {\mathrm{Btu}}^{-1}$

localid="1650266707608" $R_c=R_1+R_2$

localid="1650266713776" $R_c=0.081{\mathrm{J}}^{-1}\cdot \mathrm{s}\cdot {\mathrm{m}}^2\cdot \mathrm{K},\frac{\mathrm{Q}}{\mathrm{\Delta }t}=246.91\text{watts}$