# 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...

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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

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