How Do I Calculate Heat Transfer Through a Multi-Layered Ceiling?

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To calculate heat transfer through a multi-layered ceiling, the indoor temperature is 21°C and outdoor temperature is 0°C, with specific thicknesses and thermal conductivities for concrete, air, and Armstrong plates provided. The formula Q = U S (T1-T2) is used, where U is derived from the sum of d/k and 1/alpha values. The challenge lies in determining the convection coefficients (alphas) for the air layers, particularly due to the natural air movement and temperature differences. There is confusion regarding the significant difference in calculated heat flux values, with conduction yielding 165.5W compared to 42,253W for concrete alone, raising questions about the role of convection. The discussion emphasizes the importance of accurately accounting for convection in heat transfer calculations to understand energy efficiency in ceiling design.
Ziga Casar
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Hello guys,

I have a problem and I don't know how to solve it properly ...

So I have this situation:
armstrongplosca.png

And I must calculate the heat transfer (Q) through it ...

I know that the Indoor Temperature is 21C and the outdoor Temperature is 0C, I also know the following data for all layers:
Concrete: d (thickes) = 0.4m, k (Thermal conductivity) = 1.10 W/mK
Air: d = 2.25m, k = 0.0244 W/mK
Armstrong plate: d = 0.015m, k = 0.06 W/mK

And the surface is S = 731.9 m^2

I know I should calculate through this expression: Q = U S (T1-T2) and 1/U = SUM(d/k) = SUM(1/alpha)

the sum of d/k is no problem but for sum of 1/alpha, I do not know hot to calculate the alphas (Convection coefficient), because there is one alpha from air to Armstrong and one alpha from concrete to air (Or are there 2 alphas for each because the temperature under Armstrong plate is higher that the temperature above the Armstrong plate ...?)

So here is the problem, because there is a natural motion of air and a big horizontal plate I do not know hot to calculate the alphas... Can somebody please help me?
 
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You need to concentrate on what the variables represent - and think: resistors in series.
 
I do not exactly understand what you mean. Can you explain it a little bit better?
 
How would you find the current through resistors in series?
Now: heat flow = current, temperature difference voltage.
 
I completely understand the resistance circuits. But when I calculate the heat flux through the concrete only I get 42 253W, but when I calculate the heat flux through concrete, air and plate I get 165.5W. But this is only conduction, in real there is there also convection, and for convection I need alphas ... or is it not important to consider convection by calculating the heat flow?

I mean the heat flux difference is huge. I can not imagine that if you lower a ceiling with this plates you safe so much of energy? Air is a good isolator, but so good?
 

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