$1.$ Window Parallel Heat Transfer

In the last homework, we treated windows as a collection of heat transfer rates in series. Now with our new knowledge of the effects that framing can have on window heat loss, we can apply the parallel heat transfer method we learned for walls to windows and their frames. You are worried that your frame is affecting the total window heat loss because you still feel cold next to it after changing the glazing from single to double.

The window glazing, now that you put in double$-$pane glazing with a low$-$e coating, has an R$-$value of conduction of $3($not including $\backslash (\backslash$mathrm$\{$R$\}\_\{\backslash$mathrm$\{$cv$\}\}\backslash )$ and $\backslash (\backslash$mathrm$\{$R$\}\_\{\backslash$mathrm$\{$cvo$\}\}\backslash )).$ The frame is aluminum with an R $-$value of conduction of $0\mathrm{.}2($not including $\backslash ($ R$\_\{\backslash$mathrm$\{$cv$\}\}\backslash )$ and $\backslash ($ R$\_\{\backslash$mathrm$\{$cvo$\}\}\backslash )).$ The total window opening is $\backslash (3\backslash$mathrm$\{$ft$\}$ x $4\backslash$mathrm$\{$ft$\}\backslash )$; of that $\backslash (7\mathrm{.}8\backslash$mathrm$\{$ft$\}^\{2\}\backslash )$ is center$-$of$-$glass area, $\backslash (2\mathrm{.}5\backslash$mathrm$\{$ft$\}^\{2\}\backslash )$ is edge$-$of$-$glass area, and the rest is frame area. You want to find out the total R$-$value of the entire window assembly $($glazing plus frame$).$

a$.$ While you know the center$-$of$-$glass $($COG$)$ R$-$value and frame R$-$value from the manufacturer you do not know the edge$-$of$-$glass R$-$value. So you borrow the Infrared $($IR$)$ Imager from the CEE$176$A Renewable Energy Lab and find that the interior surface temperature of the edge$-$of$-$ glass $($the $2\mathrm{.}5$ inches around the perimeter of the glazing$)$ is $\backslash (60^\{\backslash$circ$\}\backslash$mathrm$\{$F$\}\backslash ),$ while it is $\backslash (46^\{\backslash$circ$\}\backslash$mathrm$\{$F$\}\backslash )$ outside and $\backslash (70^\{\backslash$circ$\}\backslash$mathrm$\{$F$\}\backslash )$ inside. Assume $\backslash (\backslash$mathrm$\{$R$\}\_\{\backslash$mathrm$\{$cVI$\}\}=0\mathrm{.}68\backslash )$ and $\backslash (\backslash$mathrm$\{$R$\}\_\{\backslash$mathrm$\{$cvo$\}\}=0\mathrm{.}17\backslash ).$ What is the total edge$-$of$-$glass R $-$value $($including $\backslash (\backslash$mathrm$\{$R$\}\_\{\backslash$mathrm$\{$cv$\}\}\backslash )$ and $\backslash (\backslash$mathrm$\{$R$\}\_\{\backslash$mathrm$\{$cvo$\}\}\backslash ))?$

b$.$ Now using the areas provided, use the parallel heat transfer method for the three areas to find the total average U $-$value of the window assembly. What is the total average R $-$value?

c$.$ You are upset about your recent findings and decide to call back the window company and ask them to replace your aluminum frame too. They say they can give you a new aluminum frame but this time with a thermal break inside the frame, for an R$-$value of conduction of $1.$ After the retrofit, you borrow the IR Imager again to see if you get an R$-$value of the total window closer to the COG R$-$value. You find the interior surface temperature of the edge$-$of$-$glass area to be $\backslash (63^\{\backslash$circ$\}\backslash$mathrm$\{$F$\}\backslash ),$ while it is a similar $\backslash (46^\{\backslash$circ$\}\backslash$mathrm$\{$F$\}\backslash )$ outside and $\backslash (70^\{\backslash$circ$\}\backslash$mathrm$\{$F$\}\backslash )$ inside. What is the edge$-$of$-$glass R $-$value now?

d$.$ What is the total average U$-$value of the window now with the new frame with a thermal break? The total average R$-$value?