The ceiling of a single-family dwelling in a cold climate should have an R-value of$\mathbf{30}$.To give such insulation, (a) how thick would a layer of polyurethane foam and (b) how thick would a layer of silver have to be?

1. R value of the ceiling$R=30$
2. Thermal conductivity of polyurethane foam$k_p=0.024\text{\hspace{0.17em}W/m.K}$
3. Thermal conductivity of silver$k_s=428\text{\hspace{0.17em}W/m.K}$

Thermal radiation is the process of transferring heat through electromagnetic radiation that is produced by the thermal motion of matter particles. We use the concept of thermal resistance to conduction. The lower the thermal conductivity of the material; then the higher is the R-value. So, something that has a higher R-value is a poor thermal conductor and thus a good thermal insulator.

Formula:

The thickness of the layer, $L=kR$ …(i)

The thickness of the layer of polyurethane foam using equation (i) is given as:

$\begin{array}{c}{L}_p=\left(0.024\text{\hspace{0.17em}W/m.K}\right)\left(30{\text{\hspace{0.17em}ft}}^{\text{2}}\text{.F°.h/Btu}\right){(1\text{\hspace{0.17em}m}/3.281\text{\hspace{0.17em}ft})}^2\left(3600\text{s/h}\right)(5°\text{C}/9\text{F°})(1\text{\hspace{0.17em}Btu}/1055\text{J})\\ =0.13\text{m}\end{array}$

Hence, the thickness of the layer of polyurethane is$0.13\text{m}$

Using equation (i), the thickness of the silver layer with respect to the polyurethane foam is given as:

$\begin{array}{c}{L}_s=\frac{k_s{R}_s}{k_p{R}_p}.L_p\\ =\frac{428\text{\hspace{0.17em}W/m.K}\times 30}{0.024\text{\hspace{0.17em}W/m.K}\times 30}\times 0.13\text{\hspace{0.17em}m}\\ =2.3\times {10}^3\text{m}\end{array}$

Hence, the thickness of the layer is$2.3\times {10}^3\text{m}$