Rate of energy transfer by conduction through the window?

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SUMMARY

The rate of energy transfer by conduction through a thermal window consisting of two layers of glass and an air space is calculated using the formula H=kA[(Thot-Tcold)/L]. Given the thermal conductivity of glass as 0.8 W/m°C, an area of 6.00 m², and a temperature difference of 58°C, the effective lengths for each layer must be considered. The discussion emphasizes the need to account for energy flow through each layer and suggests using a resistance variable "R" for simplification. This approach allows for simultaneous solving of the thermal conduction equations across the layers.

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A thermal window with an area of 6.00 m2 is constructed of two layers of glass, each 4.00 mm thick and separated from each other by an air space of 3.00 mm. If the inside is at 20.0°C and the outside is at -38.0°C, what is the rate of energy transfer by conduction through the window?

i am using the equation:

H=kA[(Thot-Tcold)/L]

where:
k=Thermal Conductivity constant of glass (0.8 W/m*C)
A=6.0 m^2
change in temp=58 C
and i have tried the lengths at 0.011 m, 0.008 m, and 0.003 m
 
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First, the change in Temperature is zero ... they're asking about a
static situation. There is a Temperature difference (gradient).

Have you done simple thermal conduction problems, with only one layer? If so, you know that you're computing the flow of Energy through the layer.

Now, what happens to the Energy that flows through the first layer?
How much of it later flows through the second layer?
How much of it eventually flows through the third layer?

You have the same H in each layer:
k_I A \frac{(T_1 - T_2)}{L_I} = k_M A \frac{(T_2 - T_3)}{L_M} = k_O A \frac{(T_3 - T_4)}{L_O}
this looks a lot neater with Tempaerature diferences across each layer:
k_I A \frac{\delta T_I }{L_I} = k_M A \frac{ \delta T_M}{L_M} = k_O A \frac{ \delta T_O}{L_O} , where \delta T_I + \delta T_M + \delta T_O = 58 C
Now "just" solve these simultaneously.
It might help to make a new variable "R" = L / kA for each layer ...
 

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