Chapter 11, Problem 44P

The thermal conductivities of human tissues vary greatly. Fat and skin have conductivities of about 0.20 W/m · K and 0.020 W/m · K, respectively, while other tissues inside the body have conductivities of about 0.50 W/m · K. Assume that between the core region of the body and the skin sin face lies a skin layer of 1.0 mm, fat layer of 0.50 cm, and 3.2 cm of other tissues. (a) Find the R-factor for each of these layers, and the equivalent R-factor for all layers taken together, retaining two digits. (b) Find the rate of energy loss when the core temperature is 37°C and the exterior temperature is 0°C. Assume that both a protective layer of clothing and an insulating layer of unmoving air a absent, and a body area of 2.0 m².

To determine

The R factor for skin, fat, and tissue.

Answer to Problem 44P

The R factor for skin, fat, and tissue is $5.0\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$, $2.5\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$, and $6.4\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$ respectively.

Explanation of Solution

Section1:

To determine: TheR factor for skin.

Answer: TheR factor for skin is $5.0\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$

Explanation:

Given info: Thickness of the skin is 1.0 mm and thermal conductivity is $0.20\text{W}/\text{m}\cdot \text{K}$.

Formula to calculate R factor  for skin is,

$R_{\text{skin}}=\frac{L_{\text{skin}}}{k_{\text{skin}}}$

• $R_{\text{skin}}$ is the R factor for skin,
• $L_{\text{skin}}$ is the thickness of the skin,
• $k_{\text{skin}}$ is the thermal conductivity of the skin,

Substitute 1.0 mm for $L_{\text{skin}}$ and  $0.20\text{W}/\text{m}\cdot \text{K}$ for $k_{\text{skin}}$ to find $R_{\text{skin}}$.

$\begin{array}{c}{R}_{\text{skin}}=\frac{1.0\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{0.20\text{W}/\text{m}\cdot \text{K}}\\ =5.0\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}\end{array}$

Therefore, the R factor for the skin is $5.0\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$

Section2:

To determine: TheR factor for fat layer.

Answer: TheR factor for fat layer is $2.5\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$

Explanation:

Given info: Thickness of the fat layer is 0.50 cm and thermal conductivity of fat layer is $0.02\text{W}/\text{m}\cdot \text{K}$.

Formula to calculate R factor for fat layer is,

$R_{\text{fat}}=\frac{L_{\text{fat}}}{k_{\text{fat}}}$

• $R_{\text{fat}}$ is the R factor for fat layer,
• $L_{\text{fat}}$ is the thickness of the fat layer,
• $k_{\text{fat}}$ is the thermal conductivity of the fat layer,

Substitute 0.50 cm for $L_{\text{fat}}$ and  $0.20\text{W}/\text{m}\cdot \text{K}$ for $k_{\text{fat}}$ to find $R_{\text{skin}}$.

$\begin{array}{c}{R}_{\text{skin}}=\frac{0.50\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}{0.02\text{W}/\text{m}\cdot \text{K}}\\ =2.5\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}\end{array}$

Therefore, the R factor for the fat layer is $2.5\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$

Section3:

To determine: TheR factor for tissue.

Answer: TheR factor for tissue is $6.4\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$

Explanation:

Given info: Thickness of the tissue is 3.2 cm and thermal conductivity of tissue is

$0.50\text{W}/\text{m}\cdot \text{K}$.

Formula to calculate R factor for tissue is,

$R_{\text{tissue}}=\frac{L_{\text{tissue}}}{k_{\text{tissue}}}$

• $R_{\text{tissue}}$ is the R factor for tissue,
• $L_{\text{tissue}}$ is the thickness of the tissue,
• $k_{\text{tissue}}$ is the thermal conductivity of the tissue,

Substitute 3.2 cm for $L_{\text{tissue}}$ and  $0.50\text{W}/\text{m}\cdot \text{K}$ for $k_{\text{tissue}}$ to find $R_{\text{tissue}}$.

$\begin{array}{c}{R}_{\text{tissue}}=\frac{3.2\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}{0.50\text{W}/\text{m}\cdot \text{K}}\\ =6.4\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}\end{array}$

Therefore, the R factor for the tissue is $6.4\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$

Conclusion:

Therefore, the R factor for skin, fat, and tissue is $5.0\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$, $2.5\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$, and $6.4\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$ respectively.

To determine

The rate of energy loss from the body.

Answer to Problem 44P

The rate of energy loss from the body is $5.3\times {10}^2\text{W}$

Explanation of Solution

Given info:  The surface area of the body is $2.0{\text{m}}^{\text{2}}$, temperature of the core is $37°\text{C}$, and exterior temperature is $0°\text{C}$.

The R factor of the body is the sum of R factor of skin, fat layer and tissue.

$R_{\text{body}}=R_{\text{skin}}+R_{\text{fat}}+R_{\text{tissue}}$

Formula to calculate the rate at which energy is transferred by conduction from the body is,

$P=\frac{A\left(T_h-T_c\right)}{R_{\text{body}}}$

• $P$ is the rate at which energy is transferred by conduction from the body,
• $R_{\text{body}}$ is the R factor of the body,
• $A$ is the surface area of the body,
• $T_h$ is core temperature,
• $T_c$ is exterior temperature,

Use $R_{\text{skin}}+R_{\text{fat}}+R_{\text{tissue}}$ for $R_{\text{body}}$ in $P=\frac{A\left(T_h-T_c\right)}{R_{\text{body}}}$ to rewrite P.

$P=\frac{A\left(T_h-T_c\right)}{R_{\text{skin}}+R_{\text{fat}}+R_{\text{tissue}}}$

Substitute $2.0{\text{m}}^{\text{2}}$ for A, $0°\text{C}$ for $T_c$ , $5.0\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$ for $R_{\text{skin}}$, $2.5\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$ for $R_{\text{fat}}$, and $6.4\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}$ for $R_{\text{tissue}}$ and $37°\text{C}$ for $T_h$ to find P.

$\begin{array}{c}P=\frac{\left(2.0{\text{m}}^{\text{2}}\right)\left[\left(\text{37+273}\right)\text{K}-\left(0\text{+273}\right)\text{K}\right]}{5.0\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}+2.5\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}+6.4\times {10}^{-2}{\text{m}}^{\text{2}}\cdot \text{K}/\text{W}}\\ =5.3\times {10}^2\text{W}\end{array}$

Conclusion:

Therefore, the rate at which energy is transferred through body is $5.3\times {10}^2\text{W}$.