Chapter 1, Problem 1.1P

The thermal conductivity of a sheet of rigid, extrudedinsulationis reported to be $k=0.029\text{W/m}\cdot \text{K}$ . Themeasured temperature difference across a 20-mm-thicksheet of the material is $T_1-T_2=10°\text{C}$ .
(a) What is the heat flux through a $2\text{m}\times 2\text{m}$ sheet ofthe insulation?
(b) What is the rate of heat transfer through the sheet of insulation?

To determine

Heat Flux through a Sheet of insulation of dimension $2\text{ m}\times 2\text{ m}$ .

Answer to Problem 1.1P

Heat Flux through a Sheet of insulation of dimension $2\text{ m}\times 2\text{ m}$ is $\text{14}\text{.5 }\text{W}/{\text{m}}^{\text{2}}$ .

Explanation of Solution

Given:

Thermal conductivity of insulation is $0.029\text{W}/\text{m}\cdot \text{K}$ .

Thickness of Sheet is $20\text{ mm}$ .

Temperature difference through sheet is $10°\text{C}$ .

Dimension of sheet is $2\text{m}\times \text{2}\text{m}$ .

Concept used:

Write the expression for heat flux passing through the sheet of insulation.

  $\frac{Q}{A}=k\frac{\Delta T}{\Delta x}$ ........ (1)

Here, $Q$ is the rate of heat transfer, $A$ is the cross-sectional area of sheet, $k$ is thermal conductivity of specimen, $\Delta T$ is the temperature difference through the sheet and $\Delta x$ is the thickness of sheet.

Calculation:

Substitute $10°\text{C}$ for $\Delta T$ , $0.029\text{W}/\text{m}\cdot \text{K}$ for $k$ and $20\text{mm}$ for $\Delta x$ in equation (1).

  $\begin{array}{c}\frac{Q}{A}=\left(0.029\right)\frac{10}{\left(20\text{ mm}\left(\frac{1\text{ m}}{1000\text{mm}}\right)\right)}\\ =14.5\text{W}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Thus, heat Flux through a Sheet of insulation of dimension $2\text{ m}\times 2\text{ m}$ is $\text{14}\text{.5 }\text{W}/{\text{m}}^{\text{2}}$ .

To determine

Rate of heat transfer through sheet of insulation.

Answer to Problem 1.1P

Rate of heat transfer through sheet of insulation is $\text{58}\text{W}$ .

Explanation of Solution

Concept used:

Calculate the area of rectangular sheet as follows:

  $\begin{array}{c}A=2\text{ m}\times 2\text{ m}\\ {\text{=4 m}}^2\end{array}$

Write the expression for rate of heat transfer.

  $Q=kA\frac{\Delta T}{\Delta x}$ ........ (2)

Calculation:

Substitute $10°\text{C}$ for $\Delta T$ , $0.029\text{W}/\text{m}\cdot \text{K}$ for $k$ , $4{\text{m}}^2$ for $A$ and $20\text{mm}$ for $\Delta x$ in equation (1).

  $\begin{array}{c}Q=\left(4\right)\left(0.029\right)\frac{10}{\left(20\text{ mm}\left(\frac{1\text{ m}}{1000\text{mm}}\right)\right)}\\ =58\text{W}\end{array}$

Conclusion:

Thus, rate of heat transfer through sheet of insulation is $\text{58}\text{W}$ .