Steady-State Temperature of a Multilayer Wall

Let us consider a problem of steady-state flow of heat in a plate of thickness h with thermal conductivity k, the surface of which is held at temperatures t1 and t2 (see figure).

The change in temperature along the thickness of the plate h is defined by relation:

Thus, the heat flux at any point is equal to:

Now, let us assume that the plate is a composite, that is, consists of n layers with thicknesses h1,h2,...,hn and coefficients of thermal conductivity k1,k2,...,kn, respectively. Then, the heat flux for each layer f1,f2,...,fn can be found from the formula:

fi = - ki (ti+1 - ti) / hi = (ti - ti+1) / Ri ,
Ri = hi / ki ,
ti < ti+1 , i = 1,2,...,n

Let the layers have ideal thermal contact across the interfaces; then the heat flux will be continuous when going from one layer to another, and for this particular problem, it will be the same at any point (that is, f1 = f2 = ... = fn = f). The change in temperature between the opposite external surfaces of the whole composite plate is equal to the sum of temperature changes in each single layer:

(t1 - t2) + (t2 - t3) + ... + (ti - ti+1) + ...+ (tn - tn+1) = t1 - tn+1

Then:

t1 - tn+1 = f1 R1 + f2 R2 + ...+fn Rn = f ( R1 + R2 + ...+ Rn ),

f = ( t1 - tn+1 ) / (R1 + R2 + ...+ Rn) .

Let us use the following data: number of layers n = 3, length and width of each layer are 0.5 m and 0.3 m respectively, thicknesses of layers h1,h2,h3 are equal to 0.007 m, 0.01 m and 0.003 m . Applied temperatures t1 and t4 are equal to 273.15 K (or 0 oC) and 373.15 K (or 100 oC) respectively.

Coefficients of thermal conductivity: k1 = 200 W / m.K, k2 = 390 W / m.K, k3 = 43 W / m.K .

Steady-State Temperature of a Multilayer Wall, the finite element model with applied temperatures

The finite element model with applied temperatures

Thus, f = - 7.6682E+005 W / m2, t2 = -R1 f + t1 = 299.9887 K, t3 = - (R1 + R2) f + t1 = 319.6508 K.

After carrying out calculation with the help of AutoFEM, the following results are obtained:

Table 1. Parameters of finite element mesh

Finite element type	Number of nodes	Number of finite elements
linear tetrahedron	6324	27000

Table 2. Result "Temperature"

Surface Sij of separation of layers i and j	Numerical solution Temperature T*, K	Analytical solution Temperature T, К	Error δ = 100%* \|T* - T\| / \|T\|
S12	2.999887E+02	2.999887E+02	0.01
S23	3.196508E+02	3.196508E+02	0.01

Steady-State Temperature of the Multilayer Wall, temperature distribution

Table 3. Result "Heat flux"

Numerical solution Heat flux f *, W/m2	Analytical solution Heat flux f , W/m2	Error δ = 100%* \| f* - f \| / \| f \|
7.66821375E+005	7.66821372E+005	3.91E-007

Steady-State Temperature of a Multilayer Wall, Heat Flux

*The results of numerical tests depend on the finite element mesh and may differ slightly from those given in the table.

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