Suggested Response - Q3

** Calculate the overall heat transfer coefficient through a multi-layer wall. From the outside moving in, the wall contains a 100 mm brick section, a 50 mm insulation section, a 15 mm cement section, and a 5 mm plaster finish section. The film heat transfer coefficients for the inside and outside walls are, respectively, 8 and 34 Wm-2 K-1. The thermal conductivities of the materials are listed in the table below.

Material

Thermal Conductivity

Wm-1 K-1

Brick

0.84

Insulation

0.035

Cement

0.5

Plaster

0.5

 

Response:

We first need to determine the equation for the overall heat transfer coefficient, U. The equation needs to have a term for each layer of resistance to heat transfer. In this scenario we have convective heat loss into the wall, conductive heat loss through each layer of the wall and convective again into the air. Therefore, the equation for U is:

1 over U equals 1 over h subscript 1 plus x over k plus x over k plus x over k plus x over k plus 1 over h subscript 2

It is important to keep track of the material thickness and thermal conductivity for each term. Entering the values, we have:

1 over U equals 1 over 8 plus fraction numerator 0.1 over denominator 0.84 end fraction plus fraction numerator 0.05 over denominator 0.035 end fraction plus fraction numerator 0.015 over denominator 0.5 end fraction plus fraction numerator 0.005 over denominator 0.5 end fraction plus 1 over 34

1 over U space equals space 0.125 space plus space 0.120 space plus space 1.43 space plus space 0.03 space plus space 0.01 space plus space 0.029

1 over U space equals space 1.74

U space equals space 0.573 space W m to the power of negative 2 end exponent space K to the power of negative 1 end exponent

At this point, it would be good practice to alter the problem on your own to see the change in the overall heat transfer coefficient. Perhaps you add another layer of insulation or increase a layer thickness? What would happen to the rate of heat transfer?

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