Chapter 20, Problem 21P

To determine

To Calculate:The rate of heat conduction. The ratio of the heat that escapes through the door knob to the heat that escapes through the door. The overall R-factor for the door.

Answer to Problem 21P

The total thermal current is $\text{0}\text{.3kW}$

The ratio of the heat that escapes through the door knob to the heat that escapes through the door is $5.0$ .

The overall R-factor for the door is $\text{85W/(m}\text{.K)}$

Explanation of Solution

Given:

The thickness of the door = $\text{2}\text{.0 inch}$

The thermal conductivity of brass = $\text{85 W/(m}\text{.K)}$

Formula used:

For thermal current

  ${\text{I}}_{\text{x}}{\text{=k}}_{\text{o}}{\text{A}}_{{}_{\text{x}}}\frac{\text{dT}}{\text{dx}}$

  ${\text{I}}_{\text{x}}$ is the thermal current of material x

  ${\text{k}}_{\text{o}}$ is the thermal conductivity constant

  ${\text{A}}_{{}_{\text{x}}}$ is the area of material x

T is the temperature

Calculation:

The thermal current through the door is

  ${\text{I}}_{\text{door}}{\text{=k}}_{\text{o}}{\text{A}}_{{}_{\text{door}}}\frac{\text{dT}}{\text{dx}}$

The thermal current through the knob is

  ${\text{I}}_{\text{knob}}{\text{=k}}_{\text{B}}{\text{A}}_{{}_{\text{knob}}}\frac{\text{dT}}{\text{dx}}$

The total thermal current is

  ${\text{I = k}}_{\text{o}}{\text{A}}_{\text{door}}\frac{\text{dT}}{\text{dx}}{\text{+k}}_{\text{B}}{\text{A}}_{\text{knob}}\frac{\text{dT}}{\text{dx}}$

Assume that, ${\text{Δx}}_{\text{door}}{\text{=Δx}}_{\text{knob}}\text{=Δx}$

  ${\text{I = k}}_{\text{o}}{\text{A}}_{\text{door}}\frac{\text{dT}}{\text{dx}}{\text{+k}}_{\text{B}}{\text{A}}_{\text{knob}}\frac{\text{dT}}{\text{dx}}$

Considering the thickness of door and the length of the knob as same.

  $\text{I=}\frac{\text{dT}}{\text{dx}}\left({\text{k}}_{\text{o}}{\text{A}}_{\text{door}}{\text{+k}}_{\text{B}}{\text{A}}_{\text{knob}}\right)$

  $\text{I=}\frac{{\text{20}}^{\text{o}}\text{C-}\left({\text{-20}}^{\text{o}}\text{C}\right)}{\left(\text{2}\text{.0in}\right)\left(\frac{\text{1m}}{\text{39}\text{.37in}}\right)}\left[\left(\frac{\text{0}\text{.15W}}{\text{m}}\text{×K}\right)\left(\text{2}{\text{.0m}}^{\text{2}}\right)\text{+}\left(\frac{\text{85W}}{\text{m}}\text{×K}\right)\left(\frac{\text{π}}{\text{4}}\right)\left(\text{3}{\text{.0×10}}^{\text{-2}}{\text{m}}^{\text{2}}\right)\right]$

  $\text{I=1812}\text{.3975W=1}\text{.812kW}$

The heat escaped through the knob is divided with the heat escaped through the door.

  $\frac{{\text{I}}_{\text{knob}}}{{\text{I}}_{\text{door}}}\text{=}\frac{{\text{k}}_{\text{B}}{\text{A}}_{\text{knob}}}{{\text{k}}_{\text{o}}{\text{A}}_{\text{door}}}$

  $\text{=}\frac{\text{0}\text{.15W/m×K)}\left(\text{2}{\text{.0m}}^{\text{2}}\right)}{\left(\text{85}\frac{\text{W}}{\text{m}}\text{×K}\right)\left(\text{3}{\text{.0×10}}^{\text{-2}}{\text{m}}^{\text{2}}\right)}\text{=0}\text{.1175}$

As the door and the knob are parallel, the equivalent R-factor is,

  $\frac{\text{1}}{{\text{R}}_{\text{eq}}}\text{=}\frac{\text{1}}{{\text{R}}_{\text{knob}}}\text{+}\frac{\text{1}}{{\text{R}}_{\text{door}}}\mathrm{......}\left(\text{3}\right)$

R-factor of the knob is

  ${\text{R}}_{\text{knob}}\text{=}\frac{\left|\text{dx}\right|}{{\text{k}}_{\text{B}}{\text{A}}_{{}_{\text{knob}}}}$

R-factor of the door is

  ${\text{R}}_{\text{door}}\text{=}\frac{\left|\text{dx}\right|}{{\text{k}}_{\text{o}}{\text{A}}_{{}_{\text{door}}}}$

From equation (3)

The overall R-factor for the door is

  $\begin{array}{l}\frac{\text{1}}{{\text{R}}_{\text{eq}}}\text{=}\frac{\text{1}}{{\text{R}}_{\text{knob}}}\text{+}\frac{\text{1}}{{\text{R}}_{\text{door}}}\\ \text{=}\frac{\text{1}}{\frac{\left|\text{dx}\right|}{{\text{k}}_{\text{B}}{\text{A}}_{\text{knob}}}}\text{+}\frac{\text{1}}{\frac{\left|\text{dx}\right|}{{\text{k}}_{\text{o}}{\text{A}}_{\text{door}}}}\\ \text{=}\frac{{\text{k}}_{\text{B}}{\text{A}}_{\text{knob}}{\text{+k}}_{\text{o}}{\text{A}}_{\text{door}}}{\text{dx}}\end{array}$

  $\begin{array}{l}{\text{R}}_{\text{eq}}\text{=}\frac{\text{dx}}{{\text{k}}_{\text{B}}{\text{A}}_{\text{knob}}{\text{+k}}_{\text{o}}{\text{A}}_{\text{door}}}\\ \text{=}\frac{\text{0}\text{.051m}}{\left(\text{85W/m×K}\right)\left(\text{7}{\text{.0×10}}^{\text{-6}}{\text{m}}^{\text{2}}\right)\text{+}\left(\text{0}\text{.15W/m×K}\right)\left(\text{2}{\text{.5m}}^{\text{2}}\right)}\\ \text{=0}\text{.135K/W}\\ \approx \text{0}\text{.14K/W}\end{array}$

Conclusion:

  $\text{0}\text{.14K/W}$ is the overall R-factor for the door.

  $0.1175$ is the ratio of the heat that escapes through the door knob to the heat that escapes through the door.

  $\text{1}\text{.812}\text{kW}$ is the total thermal current.