# A double-pane window uses two 6-mm thick pieces of plate glass separated by 6 mm of dead air. The...

## Question:

A double-pane window uses two 6-mm thick pieces of plate glass separated by 6 mm of dead air. The surface of the window inside the house is exposed to air at T{eq}_{in} {/eq} = 20{eq}^\circ {/eq}C having h{eq}_{in} {/eq} = 8 W/m{eq}^2 {/eq}.K, and the surface of the window outside the house is exposed to air at T{eq}_{out} {/eq} = -10{eq}^\circ {/eq}C having h{eq}_{out} {/eq} = 35 W/m{eq}^2 {/eq}{eq}\cdot {/eq}K. Assume k{eq}_{glass} {/eq} = 1.1 W/ mK for the plate glass and k = 0.025 W/m. K for the dead air between the pieces of glass.

a) Find the heat flux through the window.

b) Find the temperature on the inside surface of the window.

## Heat transfer:

In thermodynamics, the study about the transfer or the conversion of the heat energy between two bodies is known as the heat transfer. The heat transfer process can be takes place by conduction, convection and radiation.

## Answer and Explanation:

**Given Data**

- Inside temperature is: {eq}{T_i} = 20^\circ {\rm{C}} = 293\;{\rm{K}} {/eq}

- Outside temperature is: {eq}{T_o} = - 10^\circ {\rm{C}} = 263\;{\rm{K}} {/eq}

- Thickness of the plate is: {eq}{t_p} = 6\;{\rm{mm}} = 6 \times {10^{ - 3}}\;{\rm{m}} {/eq}

- Thickness of the gap is: {eq}{t_g} = 6\;{\rm{mm}} = 6 \times {10^{ - 3}}\;{\rm{m}} {/eq}

- Internal thermal convection is: {eq}{h_{in}} = 8\;{\rm{W/}}{{\rm{m}}^2} \cdot {\rm{K}} {/eq}

- Outer thermal convection is: {eq}{h_{out}} = 35\;{\rm{W/}}{{\rm{m}}^2} \cdot {\rm{K}} {/eq}

- Thermal conductivity of the glass is: {eq}{k_g} = 1.1\;{\rm{W/m}} \cdot {\rm{K}} {/eq}

- Thermal conductivity of the dead air between the glass is: {eq}{k_a} = 0.025\;{\rm{W/m}} \cdot {\rm{K}} {/eq}

**(a)**

Expression to calculate the thermal resistance is

{eq}R = \dfrac{1}{{{h_{in}}}} + \dfrac{{{t_p}}}{{{k_g}}} + \dfrac{{{t_g}}}{{{k_a}}} + \dfrac{{{t_p}}}{{{k_g}}} + \dfrac{1}{{{h_{out}}}} {/eq}

Substitute the value in above expression

{eq}\begin{align*} R& = \dfrac{1}{{8\;{\rm{W/}}{{\rm{m}}^2} \cdot {\rm{K}}}} + \dfrac{{6 \times {{10}^{ - 3}}\;{\rm{m}}}}{{1.1\;{\rm{W/m}} \cdot {\rm{K}}}} + \dfrac{{6 \times {{10}^{ - 3}}\;{\rm{m}}}}{{0.025\;{\rm{W/m}} \cdot {\rm{K}}}} + \dfrac{{6 \times {{10}^{ - 3}}\;{\rm{m}}}}{{1.1\;{\rm{W/m}} \cdot {\rm{K}}}} + \dfrac{1}{{35\;{\rm{W/}}{{\rm{m}}^2} \cdot {\rm{K}}}}\\ R &= 0.404\;{\rm{k/W}} \end{align*} {/eq}

Expression to calculate the heat flux is

{eq}{H_f} = \dfrac{{{T_i} - {T_o}}}{R} {/eq}

Substitute the value in above expression

{eq}\begin{align*} {H_f} &= \dfrac{{293\;{\rm{K}} - 263\;{\rm{K}}}}{{0.404}}\\ {H_f} &= 74.25\;{\rm{W/}}{{\rm{m}}^2} \end{align*} {/eq}

**(b)**

Expression to calculate the temperature inside surface of the window is

{eq}\begin{align*} {H_f}& = \dfrac{{{T_i} - {T_w}}}{{\dfrac{1}{{{h_{in}}}}}}\\ {H_f} &= {h_{in}}\left( {{T_i} - {T_w}} \right) \end{align*} {/eq}

Substitute the value in above expression

{eq}\begin{align*} 74.25\;{\rm{W/}}{{\rm{m}}^2} &= 8\;{\rm{W/}}{{\rm{m}}^2} \cdot {\rm{K}}\left( {293\;{\rm{K}} - {T_w}} \right)\\ {T_w} &= 226.75\;{\rm{K}} \end{align*} {/eq}

Thus the inside temperature of the window is {eq}226.75\;{\rm{K}}{/eq}.

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from High School Physics: Help and Review

Chapter 17 / Lesson 12