Design of Armoured Electric Heating MI Armored

Published on:

2023-03-14 16:42

The method of calculating the heat required by the indirect heating method using the MI heating cables is as follows: calculate the heat release of the heated liquid while maintaining a set temperature, and use mineral insulated heating cables with MI to supply the heat corresponding to this value. However, if the mineral insulated heating cable emits unlimited heat, overheating will result in aging, shutting down, etc. D., so choose a mineral insulated heating cable with the appropriate characteristics.

The design sequence is as follows:

1. Calculate the heat release of pipes, pipelines, etc., transporting and storing heated liquids

Heat dissipated by pipelines and cylindrical pipes (i. E. Required heat), equal to:

W = W * L

In the formula: w is heat per unit length, W/m;

L -- Pipeline length, m.

λ-thermal conductivity coefficient of insulation material, kcal/mch℃;

T-maintenance temperature, ° C;

T0-ambient temperature, ° C;

α-surface heat dissipation coefficient, standard value 20, kcal/m2h °C;

r0 is the outer radius of the insulation layer, m;

r is the outer radius of the pipe, m;

μ is the heating efficiency, generally 0.7-1.0

2. Determine the number of circuits by the total electrical power and voltage of the system. Calculate the approximate length of the cable depending on the temperature of the heating cable sheath.

3 Depending on the heating power and the applied voltage, the resistance of the primary circuit conductor can be obtained, and then the cable used can be selected according to the cable length and the resistance of the conductor.

In the event that the type and power of the MI heating cable cannot be obtained, it is necessary to study and analyze the parameters such as the voltage used, the number of circuits, the number of veins, the thickness of the insulation layer.

4. After selecting different parameters, the cable sheath temperature should be checked not to exceed the allowable value

Pipeline heating design example

asphalt 100 -5 Calcium Silicate 50mm 218
liquid type

maintain the temperature (℃)

Capacity Tube (A)

Ambient Temperature (℃)

Insulation materials

required heat (W/m)

Heavy Oil

55

125

0

Calcium Silicate 50mm

84

water

5

50

-10

glass and vitriol 25mm

9

160

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