Calculating Resistances of a Cavity Wall

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This discussion focuses on calculating the thermal resistances of a cavity wall with a U value of 0.48. The wall consists of 103mm brickwork, 50mm insulation, 100mm lightweight concrete block, and 13mm plasterboard. The U value is defined as the inverse of the sum of the resistances, which includes the thermal conductivities of each material. To accurately calculate the resistances, the thermal conductivity values for each material are essential, with specific values provided: brickwork (0.69), insulation (0.035), concrete block (0.42), and plasterboard (0.17).

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laheer
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Hi,

I am trying to work out the resistances for the following cavity wall. I am having to work backwards. I have the U value, which is 0.48. The brickwork is 103mm, the insulation in cavity is 50mm, lightweight concrete block is 100mm and plaster board is 13mm.

I know

total thickness (0.103+ 0.050 +.100+.013) / total resistance = a

1/a = .048.

Thanks
 
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Simply stated, you cannot find what you are looking for if this is all the information that you have.

The U value is the overall heat transfer coefficient, and it means just that. It accounts for not only the resistance values of the materials that apply for conductance and a given temperature difference, but also convection coefficents on the inside and outside faces of the wall. Generally speaking, the U value will be defined under typical conditions for a given application, but the convection is still accounted for.

If you chose to neglect this and simplify the problem to a temperature difference from one face to the other, you are still short of information needed to solve your problem. You would have one equation, but four unknowns. The U value is the inverse of the sum of the resistances in the series, i.e. convection inside surface, brickwork, insulation, concrete block, plaster board, and convection on the outside surface. That's six but since we will neglect convection, we'll say four. So:

U=1/sum(R)

or

1=U*(R1+R2+R3+R4)

where R=l/k

l being the thickness and k being the thermal conductivity

so:

1=U*[(l1/k1)+(l2/k2)+(l3/k3)+(l4/k4)]

You have the U value, the thickness for each, but no thermal conductivity values. Finding the resistance values you want given the thicknesses you have provided, requires the respective thermal conductivity values. You would need the other three to solve for one of them.

Taking the thicknesses and materials you have specified and using this page as a reference:
http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

Brick work: 0.69
Insulation: 0.035
Concrete, light: 0.42
Gypsum or plaster board: 0.17

0.48*[(0.103/0.69)+(0.05/0.035)+(0.1/0.42)+(0.013/0.17)=0.908

This value being lower than one is not surprising because the resistance for the inside and outside surface convection has been neglected. Also of course, the values from this reference source and another are likely to vary some (especially the insulation which in this case accounts for over 75% of the total resistance). Hopefully this helps.
 

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