The thermal insulated pipe problem - TU/e

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1 The thermal insulated pipe problem In order to calculate heat flow passing through a pipe with or without insulation, it is necessary to calculate the thermal resistance for each material layer. For cylinders this thermal resistance varies not only with the insulation thickness and thermal conductivity, but also with the pipe outer diameter and the average material temperature.

2 The thermal resistance of each (insulation) layer is calculated using the following equation: ln ( ) = [m.K/W] 2 Where: re = the outer radius of that layer [m] ri = the inner radius of that layer [m] k = the thermal conductivity of that layer [W/m.K] The total thermal resistance of a pipe, with or without insulation layer(s) and surface heat transfer resistances to inner fluid and extern air, can be calculated with following equation: (+1) ln ( ) = + 1 =1 + [m.K/W] 2 1 2 2 Where: Ri = inner surface heat transfer resistance between fluid and material (1/hi) [m2.K/W] Re = outer surface heat transfer resistance between fluid and material (1/hi) [m2.K/W] Suppose there is a metal pipe with thermal conductivity of 52 W/m.K and an insulated layer of 5 cm thick with thermal conductivity of 0,03 W/m.K. The inner radius of the pipe is 1,60 cm and outdoor radius is 2,13 cm. The inner surface heat transfer resistance coefficient to inner flowing liquid is 1/2000 m2.K/W and outer resistance coefficient to the air is 1/8 m2.K/W. The thermal resistance over the pipe without insulation is: 0,0213 1/2000 ln ( ) 1/8 0,016 = 20,016 + 252 + 20,0213 = 0,94 m.K/W The total resistance with 5 cm thick insulation is: 0,0213 0,0713 1/2000 ln ( ) ln ( ) 1/8 0,016 0,0213 = 20,016 + 252 + 20,03 + 20,0713 = 6,69 m.K/W The heat flow passing through the pipe surface is calculated using the following equation: = [W] When the temperature of the inner liquid is 60C and the extern air has a temperature of 20C, the heat flow through the pipe surface without insulation is: 6020 = 0,94 = 42,6 [W/m] When the pipe is insulated the heat flow passing through the pipe is: 6020 = 6,77 = 6,0 [W/m] When the temperatures of inner and outer surfaces are known, the temperature at any distance in the material layer(s) can be calculated with the following equation: 1 2 = 1 [K or C] ln 2 1 1 2

3 Numerical simulation with Comsol Multiphysics First we have calculated the heat loss through the pipe surface with or without insulation. Now we simulate this heat flow problem by using the software Comsol Multiphysics v3.5. In Fig. 1 the two-dimensional model of the pipe with (left) and without (right) insulation layer is given, the colors in this figure indicate the thermal conductivity of the materials. Fig. 1 The two-dimensional model of the pipe with and without diameter, indicated by their thermal conductivities In this simulation model the same temperatures, conductivities and surface heat transfer coefficients are used as in the calculations before. Under these stationary conditions the temperature distribution though the pipe and insulation are shown in Fig. 2: Fig. 2 Temperature distribution through the pipe an insulation material 3

4 The heat flux through the material is given by the colors and arrows in Fig. 3: Fig. 3 Heat flux to outside indicated by colors and arrow length By multiplying the maximum heat flux, which accorded to the inner surface with the circumference of the inner surface, we get the total lost of heat flux. This is for the pipe without insulation: = 423,6 (2 0,016 ) = 42,6 [W/m] The heat loss of the insulated is: = 59,5 (2 0,016 ) = 6,0 [W/m] So the results out the numerical simulation are the same than the results we have obtained out of the calculations. Fig. 4 shows that the total thermal resistance is not proportional to the thickness of the insulation layer. These graphs are based on the same material thermal conductivities and pipe dimensions as before. Fig. 4 Equivalent thermal resistance of an insulated pipe depending on the insulation thickness in meter. 4

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