Chapter 5, Problem 5.1P

Determine the thermal conductivity of 4 in. (100 mm) of insulation with a unit conductance of $\text{0}{\text{.2Btu/(hr-ft}}^{\text{2}}\cdot \text{F)[1}{\text{.14W/(m}}^{\text{2}}\cdot \text{C)]}$ in (a) English units and (b) SI units.

To determine

The thermal conductivity of insulation in English unit.

Answer to Problem 5.1P

The thermal conductivity of insulation in English unit is $k=\text{0}\text{.80}\text{Btu}\cdot \text{inch}/\left(\text{hr}\cdot {\text{ft}}^{\text{2}}\cdot \text{F}\right)$ .

Explanation of Solution

Given:

The unit conductance of insulation is $C=0.2\text{Btu}/\left(\text{hr}\cdot {\text{ft}}^2\cdot \text{F}\right)$ .

The thickness of insulation is $dx=4\text{inch}$ .

Formula used:

The thermal conductivity is given as,

  $k=Cdx$

Here, $k$ is thermal conductivity.

Calculations:

The thermal conductivity can be calculated as,

  $\begin{array}{c}k=Cdx\\ =0.2\text{Btu}/\left(\text{hr}\cdot {\text{ft}}^2\cdot \text{F}\right)\times 4\text{inch}\\ \text{=0}\text{.80}\text{Btu}\cdot \text{inch}/\left(\text{hr}\cdot {\text{ft}}^{\text{2}}\cdot \text{F}\right)\end{array}$

Conclusion:

Therefore, the thermal conductivity of insulation in English unit is $k=\text{0}\text{.80}\text{Btu}\cdot \text{inch}/\left(\text{hr}\cdot {\text{ft}}^{\text{2}}\cdot \text{F}\right)$ .

To determine

The thermal conductivity of insulation in SI unit.

Answer to Problem 5.1P

The thermal conductivity of insulation in SI unit is $k=\text{0}.1140\text{W}/\text{m}\cdot \text{°C}$ .

Explanation of Solution

Given:

The unit conductance of insulation is $C=1.14\text{W}/{\text{m}}^{\text{2}}\cdot \text{°C}$ .

The thickness of insulation is $dx=100\text{mm}$ .

Formula used:

The thermal conductivity is given as,

  $k=Cdx$

Here, $k$ is thermal conductivity.

Calculations:

The thermal conductivity can be calculated as,

  $\begin{array}{c}k=Cdx\\ =1.14\text{W}/{\text{m}}^{\text{2}}\cdot \text{°C}\times \left(100\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\\ =\text{0}.1140\text{W}/\text{m}\cdot \text{°C}\end{array}$

Conclusion:

Therefore, the thermal conductivity of insulation in SI unit is $k=\text{0}.1140\text{W}/\text{m}\cdot \text{°C}$ .