Thermal Conductivity of a Cylindrical Wall

Consider infinitely long cylindrical wall (tube) with internal radius r1, external radius r2, having constant coefficient of thermal conductivity λ. Internal surface of the tube is held at temperature T1. Inside the wall there are uniformly distributed sources of heat qν. The heat generated in the wall is dissipated to the ambient environment through the external surface of the tube (see figure).

General solution of this problem has the form:

T = C1 ln (r) + C2 - qν r2/ 4λ.

The constants C1 and C2 are determined from conditions prescribed on the internal (r = r1) and external ( r = r2 )surfaces of the tube:

T |r=r1 = T1 ,

λ (dT / dr) |r=r2 = q .

Thus,

Let us use the following data: internal radius of the tube r1 = 100 mm, external radius of the tube r2 = 250 mm, length of the tube L = 1000 mm.
Coefficient of the thermal conductivity λ, of material of the tube is equal to 43 W/m.K.
Energy Q of sources of heat located inside the tube is equal to 4500 W.
Since the sources of heat qν are uniformly distributed over the volume of the tube,

Specific heat flux on the external surface of the tube q = -15000 W / m2. Temperature T1 on the internal surface of the tube is equal to 373.15 K (or 100 oC).

Thermal Conductivity of a Cylindrical Wall, the finite element model with applied thermal loads

The finite element model with applied thermal loads

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	3920	15107

Table 2. Result "Temperature" at r = r2 = 0.250 m *

Numerical solution Temperature T*, К	Analytical solution Temperature T, К	Error δ = 100* \| T* - T \| / \| T \|, %
3.03125183E+002	3.03081513E+002	0.01

*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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