HW$08$ Wall Temperature gradient: OPTION B $($chose Option A or Option B$)$

Complete the table and draw the steady$-$state temperature gradient as illustrated on page $1.$

The wall assembly is one where XPS $($extruded polystyrene$)$ is sandwiched permanently between a

interior structural concrete wall $($#$1)$ and an exterior $($self$-$supporting, $!$

$($#$2).$ To simplify thing, we disregard the resistance of the air layers insi $T_d=(\frac{R_c}{R_t}$:$)*T$

al$)$ concrete wall

Total temperature difference: $70$ F

Reasoning: The structural layer of concrete $($#$1)$ has $R=6.$ In relation with the total resistance of the whole wall

made of three layers, which is $R=30,$ the resistance of the concrete layer is $\frac{6}{30},$ that is$,$ one fifth of the total

resistance. From that, we can conclude that the temperature drop due to this layer is also on$-$fifth of the total

temperature difference between inside and outside, i$.$e$.70$ F $.$ One$-$fifth of $70$ is $14.$ The temperature drop in the

concrete layer # $1$ is $14$ F $.$ Consequently the temperature on the outside$-$facing face $($point $"$B$")$ of concrete wall #$1$

is $70F-14F=56F.$

You can$/$should proof check your

answer! Did the low conductivity $($i$.$e$.$

high R$-$value$)$ material, the XPS$,$ come

with a big temperature difference

between its two faces? Yes? Then this

must have translated into a steep

slope of the gradient line within the

insulation layer! Similarly, did the high

conductivity $($i$.$e$.$ low R$-$value$)$

materials such as concrete walls come

with a small temperature difference

between their faces of that material,

which translated into a gentle slope of

the gradient line? If so$,$ you are

definitely in the ballpark!