# Calculating (Glass) thermal conductivity

I read that glass has a thermal coefficient of 0.8-1 W/mK. Given that my window window thickness is around $5\ mm$, then I would calculate my heat loss per area being $160-200\ Wm^{-2}K^{-1}$.

Yet standard glass has a U value of $5.6\ Wm^{-2}K^{-1}$.

What is going on here? I'm missing something huge...

• Interesting that someone added a homework tag. I'm working on understanding calculations for my custom ventilation and heat transfer system in my house. I guess it really is home-work - in the truest sense of the word. – Stephen Feb 8 '13 at 0:00

## 2 Answers

I guess window glass' temperature is somewhere between the temperature of the air inside and the air outside (far from the window). Air has low thermal conductivity. So you take into account the thickness of your window glass, but do not take into account the thickness of air layers near the glass. These layers' temperatures differ from the temperatures of air far from the window (inside and outside, respectively).

• So this would mean that high air flow on both sides would theoretically produce this transfer of $200 Wm^{-2}K$. An interesting idea, but I don't completely buy it... I haven't found any references for standard window panes yet, but commercialwindows.org/ufactor.php has the general theory. – Stephen Feb 4 '13 at 0:23
• So I agree with what they say, but that does not mean that my answer is wrong. If there were a medium with high thermal conductivity at both sides of the glass (case B), rather than air (case A), you would get those $L=200W/m^{-2}K$ heat losses. However, thermal conductivity of air is very low, and air convection's effect is limited, so the difference of temperatures at the outside and inside surfaces of the window glass is much lower in case A than in case B. – akhmeteli Feb 4 '13 at 1:41
• I could rephrase my explanation as follows. If you measure the actual difference of temperatures at the outside and inside surfaces of the window glass $\triangle T_a$ and multiply it by $L$ from my previous comment, you would get the actual heat losses. However, these losses are much lower than what you obtain by multiplying $L$ by $\triangle T_c$, where $\triangle T_c$ is the difference of temperatures in the center of your room and outside far from your window, whereas I guess they use $\triangle T_c$ to calculate the U-value, as $\triangle T_c$ is what matters to us. – akhmeteli Feb 4 '13 at 1:50
• Another thing. Are you talking about single glazing or double glazing? – akhmeteli Feb 4 '13 at 2:17
• My apologies - I said "standard" where I meant "single". I really am interested in this theory, but I'd like a reference to something to indicate that a 30x factor can be explained by air's (s)low conduction/convection... – Stephen Feb 5 '13 at 3:11

There's a big difference between a piece of glass and a window structure that is made of glass. For the structure, the thermal U-factor is the sum of the U-factors of the glass itself, plus other structural parameters, such as air infiltration which is very specific to the type of window you are considering. Suggest that you look at window manufacturers' websites for tech details (Andersen Corp, for example). You have properly identified your glass thickness, but note that there is a 2/1 difference in U factor between 6mm and 3mm.
Sorry about the lateness of the reply.

• Yes, but air infiltration would make it worse though wouldn't it? If I just count the size of the window, "glass-physical-material" is 30-40 times more conductive than "installed-glass-pane". I'm looking for reasons why that is so (and so far have come up with the idea that the air touching the glass on each side actually has a thermal delta that's 30-40 times less than the main room/outside body of air) – Stephen Sep 21 '15 at 0:00