Calculating Heat Transfer in a Cavity Block Wall with Windows

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Homework Help Overview

The problem involves calculating heat transfer through a cavity block wall that includes windows. The wall consists of multiple layers, including concrete, air, polystyrene, and glass, with specified thermal conductivities and temperature differences.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of heat flow through different components of the wall and windows, questioning the representation of the second concrete wall and the area calculations. There are attempts to clarify the structure of the wall and the implications of the window areas on the total heat flow.

Discussion Status

Some participants express confusion regarding the setup of the problem and the calculations presented. There is an ongoing effort to clarify the interpretation of the wall structure and the areas involved in the heat transfer calculations. Guidance is being offered to reconsider certain terms and assumptions in the equations.

Contextual Notes

Participants note potential misinterpretations of the problem statement, particularly regarding the areas of the windows and the surrounding concrete. There is also mention of specific thermal conductivity values that may have been incorrectly applied in the calculations.

Darth Frodo
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Homework Statement


A cavity block wall of area 10m2 constructed of 100 mm concrete, 50 mm air gap, 50 mm of polystyrene and a second 100 mm wall of concrete. The wall contains three windows with dimensions of 0.5 m x 0.7 m with glass 4 mm thick. The outside temperature is 5 °c and the inside temperature is 20°C.

Calculate
(A) the flow of heat through the glass, and
(B) the total heat flow through the wall.

Average thermal conductivities are
Concrete 0.6 W m·' K-'.
Polystyrene 0.03 W m-' K-'.
Air 0.024 Wm-' K-'.
Glass 0.8 W m-' K-'.

Homework Equations



H = kA\frac{ΔT}{L}

H = \frac{A(ΔT)}{\sum\frac{l}{k}}

The Attempt at a Solution



(A) H = (0.8)(0.35)(\frac{15}{0.004})

H = 1050 J/K

Total Heat through Glass = 3150(B): Total heat = Heat though Cavity wall + Heat though windows + Heat through the concrete surrounding the windows

\frac{10(15)}{\frac{0.1}{0.6}+ \frac{0.05}{0.024}+ \frac{0.05}{0.03}} + 3150 + \frac{(8.95)(15)(0.6)}{0.1}

total = 3993.8 J/K

Is this correct?
 
Last edited:
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Darth Frodo said:
(B): Total heat = Heat though Cavity wall + Heat though windows + Heat through the concrete surrounding the windows
I don't understand the third term there. That would suggest there's an area of concrete surrounding the windows which is not cavity wall.
\frac{10(15)}{\frac{0.1}{0.6}+ \frac{0.05}{0.03}+ \frac{0.05}{0.03}} + 3150 + \frac{(8.95)(15)(0.6)}{0.1}
Some problems there.
You have 0.03 instead of 0.24 for air conductivity. The second concrete wall should be represented by a fourth term below the line in the first expression, not as a third expression added on the end.
 
Sorry for the confusion. By "Heat though Cavity wall" I meant those sections of the cavity wall that were 10 square metres in area. Then the outer most wall was split into 2 parts, The windows and the surrounding concrete?"The second concrete wall should be represented by a fourth term below the line in the first expression" But the area of the second outermost wall isn't 10 square metres
 
Darth Frodo said:
Sorry for the confusion. By "Heat though Cavity wall" I meant those sections of the cavity wall that were 10 square metres in area. Then the outer most wall was split into 2 parts, The windows and the surrounding concrete?
I think you are misreading the problem:
A cavity block wall constructed of ... and a second wall of concrete. The wall contains three windows​
The 'cavity block wall' consists of four layers: concrete-air-polystyrene-concrete.
The 'wall' that contains three windows is the cavity block wall as a whole, not just the second layer of concrete. (Otherwise you wouldn't be able to see through the windows.)
And I forgot to mention (as follows from this) that the 10(15) in your first expression should have the window area subtracted.
 
Oh god! I feel like a fool! I really misread that bloody question! I assumed I had x-ray vision. Thanks!
 

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