Chapter 11, Problem 9P

Calculate the R-value for the following materials:

a. 4 in. thick brick

b. 10 cm thick brick

c. 12 in. thick concrete slab

d. 20 cm thick concrete slab

e. 1 cm thick human fat layer

To determine

Find the $\text{R-value}$ for $4\text{in}$ thick brick.

Answer to Problem 9P

The $\text{R-value}$ for $4\text{in}$ thick brick is $0.10\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

Explanation of Solution

Given data:

The thickness of the brick, $L=4\text{in}$.

From table 11.3 in the textbook, the thermal conductivity of the brick, $k=1.0\frac{\text{W}}{\text{m}\cdot \text{k}}$.

Formula used:

The formula to find $\text{R-value}$ is,

$R=\frac{L}{k}$ (1)

Here,

$L$ is the thickness of the material, and

$k$ is the thermal conductivity of the material.

Calculation:

Substitute $4\text{in}$ for $L$, and $1.0\frac{\text{W}}{\text{m}\cdot \text{k}}$ for $k$ in equation (1) to find $R$,

$\begin{array}{c}R=\frac{4\text{in}}{1.0\frac{\text{W}}{\text{m}\cdot \text{k}}}\left[\therefore 1\text{in}=0.0254\text{m}\right]\\ =\frac{\left(4\text{in}\right)\left(\frac{0.0254\text{m}}{1\text{in}}\right)}{1.0\frac{\text{W}}{\text{m}\cdot \text{k}}}\\ =\frac{\left(4\times 0.0254\right)\text{m}}{1.0\frac{\text{W}}{\text{m}\cdot \text{k}}}\\ R=0.10\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}\end{array}$

Therefore, $\text{R-value}$ for $4\text{in}$ thick brick is $0.10\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

Conclusion:

Hence, the $\text{R-value}$ for $4\text{in}$ thick brick is $0.10\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

To determine

Find the $\text{R-value}$ for $10\text{cm}$ thick brick.

Answer to Problem 9P

The $\text{R-value}$ for $10\text{cm}$ thick brick is $0.10\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

Explanation of Solution

Given data:

The thickness of the brick, $L=10\text{cm}$.

Calculation:

Substitute $10\text{cm}$ for $L$, and $1.0\frac{\text{W}}{\text{m}\cdot \text{k}}$ for $k$ in equation (1) to find $R$,

$\begin{array}{c}R=\frac{10\text{cm}}{1.0\frac{\text{W}}{\text{m}\cdot \text{k}}}\left[\therefore 1\text{m}=100\text{cm}\right]\\ =\frac{\left(10\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}{1.0\frac{\text{W}}{\text{m}\cdot \text{k}}}\\ =\frac{0.1\text{m}}{1.0\frac{\text{W}}{\text{m}\cdot \text{k}}}\\ R=0.10\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}\end{array}$

Therefore, $\text{R-value}$ for $10\text{cm}$ thick brick is $0.10\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

Conclusion:

Hence, the $\text{R-value}$ for $10\text{cm}$ thick brick is $0.10\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

To determine

Find the $\text{R-value}$ for $12\text{in}$ thick concrete slab.

Answer to Problem 9P

The $\text{R-value}$ for $12\text{in}$ thick concrete slab is $0.22\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

Explanation of Solution

Given data:

The thickness of the concrete slab, $L=12\text{in}$.

From Table 11.3 in the textbook, the thermal conductivity of the concrete slab, $k=1.4\frac{\text{W}}{\text{m}\cdot \text{k}}$.

Calculation:

Substitute $12\text{in}$ for $L$, and $1.4\frac{\text{W}}{\text{m}\cdot \text{k}}$ for $k$ in equation (1) to find $R$,

$\begin{array}{c}R=\frac{12\text{in}}{1.4\frac{\text{W}}{\text{m}\cdot \text{k}}}\left[\therefore 1\text{in}=0.0254\text{m}\right]\\ =\frac{\left(12\text{in}\right)\left(\frac{0.0254\text{m}}{1\text{in}}\right)}{1.4\frac{\text{W}}{\text{m}\cdot \text{k}}}\\ =\frac{\left(12\times 0.0254\right)\text{m}}{1.4\frac{\text{W}}{\text{m}\cdot \text{k}}}\\ R=0.22\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}\end{array}$

Therefore, $\text{R-value}$ for $12\text{in}$ thick concrete slab is $0.22\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

Conclusion:

Hence, the $\text{R-value}$ for $12\text{in}$ thick concrete slab is $0.22\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

To determine

Find the $\text{R-value}$ for $20\text{cm}$ thick concrete slab.

Answer to Problem 9P

The $\text{R-value}$ for $20\text{cm}$ thick concrete slab is $0.14\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

Explanation of Solution

Given data:

The thickness of the concrete slab, $L=20\text{cm}$.

Calculation:

Substitute $20\text{cm}$ for $L$, and $1.4\frac{\text{W}}{\text{m}\cdot \text{k}}$ for $k$ in equation (1) to find $R$,

$\begin{array}{c}R=\frac{20\text{cm}}{1.4\frac{\text{W}}{\text{m}\cdot \text{k}}}\left[\therefore 1\text{m}=100\text{cm}\right]\\ =\frac{20\text{cm}\times \frac{1\text{m}}{100\text{cm}}}{1.4\frac{\text{W}}{\text{m}\cdot \text{k}}}\\ =\frac{0.2\text{m}}{1.4\frac{\text{W}}{\text{m}\cdot \text{k}}}\\ R=0.14\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}\end{array}$

Therefore, $\text{R-value}$ for $20\text{cm}$ thick concrete slab is $0.14\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

Conclusion:

Hence, the $\text{R-value}$ for $20\text{cm}$ thick concrete slab is $0.14\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

To determine

Find the $\text{R-value}$ for $1\text{cm}$ thick human fat layer.

Answer to Problem 9P

The $\text{R-value}$ for $1\text{cm}$ thick human fat layer is $0.05\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

Explanation of Solution

Given data:

The thickness of the human fat layer, $L=1\text{cm}$.

From table 11.3 in the textbook, the thermal conductivity of the human fat layer, $k=0.2\frac{\text{W}}{\text{m}\cdot \text{k}}$.

Calculation:

Substitute $1\text{cm}$ for $L$, and $0.2\frac{\text{W}}{\text{m}\cdot \text{k}}$ for $k$ in equation (1) to find $R$,

$\begin{array}{c}R=\frac{1\text{cm}}{0.2\frac{\text{W}}{\text{m}\cdot \text{k}}}\left[\therefore 1\text{m}=100\text{cm}\right]\\ =\frac{\left(1\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}{0.2\frac{\text{W}}{\text{m}\cdot \text{k}}}\\ =\frac{0.01\text{m}}{0.2\frac{\text{W}}{\text{m}\cdot \text{k}}}\\ R=0.05\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}\end{array}$

Therefore, $\text{R-value}$ for $1\text{cm}$ thick human fat layer is $0.05\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.

Conclusion:

Hence, the $\text{R-value}$ for $1\text{cm}$ thick human fat layer is $0.05\frac{{\text{m}}^{\text{2}}\cdot \text{K}}{\text{W}}$.