Recent content by bohire

Yes, all the other ones in that chapter did.

I have no idea but I doubble checked and 40.1W is the answer given. All that is given in the question is in my original post. I guess Carl-Eric Hagentoft (author of the book) is to blame for coming up with a nonsense exercise.

Well that's pretty irritating.. I also see now that all the other exercises in the same chapter do give us interior and exterior temperatures. Anyway thank you for your time and have a good day.

No and that is exactly why I'm struggling to solve it I also tried this from Q=0.33A(TL-TM)=1.4A(TM-TR) I get TM=TL-Q/0.33A And TM=Q/(1.4A)-TR ⇒TL-Q/0.33A=Q/(1.4A)-TR ... ⇒ Q=(A/3.74)(TL-TR) But the problem with the missing temperatures still remains.. Do you think that the question might be...

Thank you for your answer Chet, this is how far I get K1=λ1*A/d1=0.033A/0.1=0.33A K2=λ2*A/d2=0.14A/0.1=1.4A Q=0.33A(TL-TM)=1.4A(TM-TR) Not sure if I'm supposed to do the rest or how I'm supposed to eliminate TM 1.4/0.33≈4.24 TL-TM=4.24(TM-TR) ... TM=(TL+4.24TR)/5.4 Then I'm pretty much stuck..

Thank you Svein but do you know where I go from there? I still need to find the heat flow and temperature between the materials.

Homework Statement Consider the following multi-layer wall consisting of an interior insulation layer and an external light-weight concrete layer. Calculate the heat flow thorugh the structure and the temperature at the interface between the two materials. Layer 1: Thermal insulation...