Worked exercise — Double glazing with air gap

Problem

A window is made of two panes of glass $d_v = 4 \ \text{mm}$ thick ($k_v = 1{,}0 \ \text{W/(m·K)}$), separated by a gap of still air $d_a = 12 \ \text{mm}$ thick ($k_a = 0{,}025 \ \text{W/(m·K)}$). The area is $A = 2 \ \text{m}^2$, with $T_1 = 20 \ ^{\circ}\text{C}$ and $T_2 = -5 \ ^{\circ}\text{C}$. Calculate the flow $\dot Q$ and the temperature of the two inner faces, assuming a steady-state flow.

Solution

The three layers (glass, air, glass) are crossed in series by the same flow, so their thermal resistances $R = \dfrac{d}{k A}$ add up: $R = \frac{d_v}{k_v A} + \frac{d_a}{k_a A} + \frac{d_v}{k_v A} = \frac{2 \cdot 0{,}004}{1 \cdot 2} + \frac{0{,}012}{0{,}025 \cdot 2} = 0{,}004 + 0{,}24 = 0{,}244 \ \text{K/W}$ The heat flow is $\dot Q = \frac{\Delta T}{R} = \frac{25}{0{,}244} \approx 102 \ \text{W}$ about $120$ times less than the $12500 \ \text{W}$ that a single pane of glass would disperse. The first inner face, on the warm side, is obtained by subtracting the temperature drop across the first pane: $T_{i1} = 20 - \dot Q \, \frac{d_v}{k_v A} = 20 - 102 \cdot \frac{0{,}004}{2} \approx 19{,}8 \ ^{\circ}\text{C}$ The second inner face, on the cold side, is obtained by adding the drop across the last pane: $T_{i2} = -5 + \dot Q \, \frac{d_v}{k_v A} = -5 + 102 \cdot \frac{0{,}004}{2} \approx -4{,}8 \ ^{\circ}\text{C}$ Almost the entire temperature drop occurs in the air gap: it is the still air that insulates, not the glass. $\ev{\dot Q \approx 102 \ \text{W}}$

Collegamenti

Argomenti: Termologia Concetti: Trasmissione del calore