91Ó°ÊÓ

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

$R=\frac{\mathrm{Δ}x}{k_t}$

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 $k_t=0.026\mathrm{J}â‹\dots {\mathrm{s}}^{−1}â‹\dots {\mathrm{m}}^{−1}â‹\dots {\mathrm{K}}^{−1}$, we have:

$R_{air}=\frac{\mathrm{Δ}x}{k_t}=\frac{0.001}{0.026}=0.0385{\mathrm{J}}^{−1}â‹\dots \mathrm{s}â‹\dots {\mathrm{m}}^2â‹\dots \mathrm{K}$

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

$R_{\text{glass}}=\frac{\mathrm{Δ}x}{k_t}=\frac{0.0032}{0.8}=0.004{\mathrm{J}}^{−1}â‹\dots \mathrm{s}â‹\dots {\mathrm{m}}^2â‹\dots \mathrm{K}$

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 $\mathrm{F}â‹\dots {\mathrm{ft}}^2â‹\dots \mathrm{hr}â‹\dots {\mathrm{Btu}}^{−1}$where a British thermal unit $\text{Btu}$is the amount of energy necessary to elevate one pound of water by $1^{∘}\mathrm{F}$. 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 $1^{∘}\mathrm{K}$. Therefore:

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

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

$1^{∘}\mathrm{F}×{\mathrm{ft}}^2×\mathrm{hr}×{\mathrm{Btu}}^{−1}={\frac{5}{9}}^{∘}\mathrm{K}â‹\dots (0.3048{)}^2{\mathrm{m}}^2×(3600)\mathrm{s}×(1054.85{)}^{−1}{\mathrm{J}}^{−1}$

$1^{∘}\mathrm{F}â‹\dots {\mathrm{ft}}^2â‹\dots \mathrm{hr}â‹\dots {\mathrm{Btu}}^{−1}={0.176}^{∘}\mathrm{K}â‹\dots {\mathrm{m}}^2â‹\dots \mathrm{s}â‹\dots \mathrm{J}$

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

$R_{air}=\frac{1}{0.176}×0.0385={0.2187}^{∘}\mathrm{F}â‹\dots {\mathrm{ft}}^2â‹\dots \mathrm{hr}â‹\dots {\mathrm{Btu}}^{−1}$

$R_{\text{glass}}=\frac{1}{0.176}×0.004={0.0227}^{∘}\mathrm{F}â‹\dots {\mathrm{ft}}^2â‹\dots \mathrm{hr}â‹\dots {\mathrm{Btu}}^{−1}$

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 $\frac{Q}{\mathrm{Δ}t}=−k_{t1}A\frac{\mathrm{Δ}T}{\mathrm{Δ}{x}_1}$, so from the material $1$we have:

Let be Equation ($1$)

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

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

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

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

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

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

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

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 $T_a$:

$\frac{T_2−T_a}{R_2}=\frac{T_a−T_1}{R_1}→R_1{T}_2−R_1{T}_a=R_2{T}_a−R_2{T}_1$

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

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

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

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

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

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

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

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

$R_c=R_1+R_2$

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" $R_c=R_{\text{glass}}+2×R_{\text{air}}$

using the results from step 1,

localid="1650266585531" $R_{\text{air}}=0.0385{\mathrm{J}}^{−1}â‹\dots \mathrm{s}â‹\dots {\mathrm{m}}^2â‹\dots \mathrm{K}$

localid="1650266588974" $R_{\text{glass}}=0.004{\mathrm{J}}^{−1}â‹\dots \mathrm{s}â‹\dots {\mathrm{m}}^2â‹\dots \mathrm{K}$

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

localid="1650266597424" $R_c=0.004+2×0.0385=0.081{\mathrm{J}}^{−1}â‹\dots \mathrm{s}â‹\dots {\mathrm{m}}^2â‹\dots \mathrm{K}$

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

$\frac{Q}{\mathrm{Δ}t}=−k_{t}A\frac{\mathrm{Δ}T}{\mathrm{Δ}x}$

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

localid="1650266613094" $\frac{Q}{\mathrm{Δ}t}=−A\frac{\mathrm{Δ}T}{R_c}$

substitute,

localid="1650266651710" $\frac{Q}{\mathrm{Δ}t}=−1×\frac{20}{0.081}=246.91$localid="1650266657465" $watts$

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

Therefore, it is concluded that,

localid="1650266679151" $R_{air}=0.0385{\mathrm{J}}^{−1}â‹\dots \mathrm{s}â‹\dots {\mathrm{m}}^2â‹\dots \mathrm{K}$

localid="1650266683604" $R_{\text{glass}}=0.004{\mathrm{J}}^{−1}â‹\dots \mathrm{s}â‹\dots {\mathrm{m}}^2â‹\dots \mathrm{K}$

localid="1650266688380" $1^{∘}\mathrm{F}â‹\dots {\mathrm{ft}}^2â‹\dots \mathrm{hr}â‹\dots {\mathrm{Btu}}^{−1}={0.176}^{∘}\mathrm{K}â‹\dots {\mathrm{m}}^2â‹\dots \mathrm{s}â‹\dots \mathrm{J}$,

localid="1650266693304" $R_{\text{air}}={0.2187}^{∘}\mathrm{F}â‹\dots {\mathrm{ft}}^2â‹\dots \mathrm{hr}â‹\dots {\mathrm{Btu}}^{−1}$

localid="1650266702329" $R_{\text{glass}}={0.0227}^{∘}\mathrm{F}â‹\dots {\mathrm{ft}}^2â‹\dots \mathrm{hr}â‹\dots {\mathrm{Btu}}^{−1}$

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

localid="1650266713776" $R_c=0.081{\mathrm{J}}^{−1}â‹\dots \mathrm{s}â‹\dots {\mathrm{m}}^2â‹\dots \mathrm{K},\frac{\mathrm{Q}}{\mathrm{Δ}t}=246.91\text{watts}$