Q. 1.57

Question

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

$R \equiv \frac{Δ x}{k_{t}}$

(a) Calculate the $R$ value of a $1 / 8 -inch$ ( $3.2 mm$ ) piece of plate glass, and then of a $1 mm$ 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, $^{\circ} F \cdot {ft}^{2} \cdot hr / Btu$ . A Btu, or British thermal unit, is the energy needed to raise the temperature of a pound of water $1^{\circ} F$ . 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 $1.0$ $mm$ layer of still air on each side. (The effective thickness of the air layer will depend on how much wind is blowing; $1 mm$ 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 $1 - m^{2}$ single-pane window when the temperature in the room is $20^{\circ} C$ higher than the outdoor temperature.

Step-by-Step Solution

Verified

Answer

From the following

a. The value of $R$ both glass and air has been found.

$R_{a i r} = 0.0385 J^{- 1} \cdot s \cdot m^{2} \cdot K$

$R_{glass} = 0.004 J^{- 1} \cdot s \cdot m^{2} \cdot K$

b. By conversion of units

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

$R_{air} = {0.2187}^{\circ} F \cdot {ft}^{2} \cdot hr \cdot {Btu}^{- 1}$

$R_{glass} = {0.0227}^{\circ} F \cdot {ft}^{2} \cdot hr \cdot {Btu}^{- 1}$

c. The effective value of $R$

$R_{c} = R_{1} + R_{2}$

d. Therefore using the effective $R$ value we get,

$R_{c} = 0.081 J^{- 1} \cdot s \cdot m^{2} \cdot K$

$\frac{Q}{Δ t} = 246.91 watts$

1Step 1: Thermal conductivity (part a)

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

$R = \frac{Δ x}{k_{t}}$

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

$R_{a i r} = \frac{Δ x}{k_{t}} = \frac{0.001}{0.026} = 0.0385 J^{- 1} \cdot s \cdot m^{2} \cdot K$

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

$R_{glass} = \frac{Δ x}{k_{t}} = \frac{0.0032}{0.8} = 0.004 J^{- 1} \cdot s \cdot m^{2} \cdot K$

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

2Step 2: Conversion of Units (part b)

In the US, the units of $R$ are $F \cdot {ft}^{2} \cdot hr \cdot {Btu}^{- 1}$ where a British thermal unit $Btu$ is the amount of energy necessary to elevate one pound of water by $1^{\circ} F$ . To convert to SI units of $R$ , we need to convert $1 Btu$ 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 $1^{\circ} K$ . Therefore:

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

One foot is $0.3048 m$ and $1$ hour is $3600$ seconds, so

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

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

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

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

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

3Step 3: Pictorial Representation

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 $T_{a}$ .

4Step 4: At a steady flow rate of flow of heat

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 $\frac{Q}{Δ t} = - k_{t 1} A \frac{Δ T}{Δ x_{1}}$ , so from the material $1$ we have:

Let be Equation ( $1$ )

$\frac{Q}{Δ t} = - k_{t 1} A \frac{Δ T}{Δ x_{1}} = - k_{t 1} A \frac{T_{a} - T_{1}}{Δ x_{1}}$

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

$\frac{Q}{Δ t} = - k_{t 2} A \frac{Δ T}{Δ x_{2}} = - k_{t 2} A \frac{T_{2} - T_{a}}{Δ x_{2}}$

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

$k_{t 2} \frac{T_{2} - T_{a}}{Δ x_{2}} = k_{t 1} \frac{T_{a} - T_{1}}{Δ x_{1}}$

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

$R_{1} = \frac{Δ x_{1}}{k_{t 1}} R_{2} = \frac{Δ x_{2}}{k_{t 2}}$

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 ( $6$ ).

$\frac{T_{2} - T_{a}}{R_{2}} = \frac{T_{a} - T_{1}}{R_{1}} \frac{T_{a} - T_{1}}{R_{1}} = \frac{T_{2} - T_{1}}{R_{c}}$

5Step 5: Solving T a to find R c

now we need to solve these two equations for $R_{c}$ . From the first equation, we can solve for $T_{a}$ :

$\frac{T_{2} - T_{a}}{R_{2}} = \frac{T_{a} - T_{1}}{R_{1}} \to R_{1} T_{2} - R_{1} T_{a} = R_{2} T_{a} - R_{2} T_{1}$

$R_{1} T_{2} + R_{2} T_{1} = R_{2} T_{a} + R_{1} T_{a} \to R_{1} T_{2} + R_{2} T_{1} = T_{a} (R_{2} + R_{1})$

$T_{a} = \frac{R_{1} T_{2} + R_{2} T_{1}}{(R_{2} + R_{1})}$

Substituting into the second equation in ( $6$ ) we can solve for $R_{c}$ :

$\frac{T_{a} - T_{1}}{R_{1}} = \frac{T_{2} - T_{1}}{R_{c}} \to \frac{1}{R_{1}} [T_{a}] - \frac{T_{1}}{R_{1}} = \frac{T_{2} - T_{1}}{R_{c}}$

$\frac{1}{R_{1}} [\frac{R_{1} T_{2} + R_{2} T_{1}}{(R_{2} + R_{1})}] - \frac{T_{1}}{R_{1}} = \frac{T_{2} - T_{1}}{R_{c}}$

$[\frac{R_{1} T_{2} + R_{2} T_{1}}{R_{1} (R_{2} + R_{1})}] - \frac{(R_{2} + R_{1}) T_{1}}{(R_{2} + R_{1}) R_{1}} = \frac{T_{2} - T_{1}}{R_{c}}$

$\frac{R_{1} T_{2} + R_{2} T_{1} - R_{2} T_{1} - R_{1} T_{1}}{R_{1} (R_{2} + R_{1})} = \frac{T_{2} - T_{1}}{R_{c}}$

$\frac{R_{1} (T_{2} - T_{1})}{R_{1} (R_{2} + R_{1})} = \frac{T_{2} - T_{1}}{R_{c}}$

$R_{c} = R_{1} + R_{2}$

6Step 6: To find the effective R value

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

$R_{c} = R_{glass} + 2 \times R_{air}$

using the results from step 1,

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

$R_{glass} = 0.004 J^{- 1} \cdot s \cdot m^{2} \cdot K$

so, the effective $R$ value of this system is, therefore:

$R_{c} = 0.004 + 2 \times 0.0385 = 0.081 J^{- 1} \cdot s \cdot m^{2} \cdot K$

When the temperature difference is $Δ T = 20^{\circ} K$ , the rate of heat is:

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

with, $R_{c} = \frac{Δ x}{k_{t}}$ so:

$\frac{Q}{Δ t} = - A \frac{Δ T}{R_{c}}$

substitute,

$\frac{Q}{Δ t} = - 1 \times \frac{20}{0.081} = 246.91$ $w a t t s$

This compares with the heat loss through the glass on its own $\frac{A Δ T}{R_{g} l a s s} = 5000$ $w a t t s$ . Thus the air layer provides most of the insulation.