# How to choose the thickness of the electric heating tube

- Apr 29, 2019-

The general customer will always ask, how to choose the thickness of a heating tube cable, ask more, explain it to everyone, to solve the customer's problem, in addition to explain how to choose the thickness of the electric heating tube wire, also prepared for the customer Some questions about the selection process. I hope to help customers. The safe current carrying capacity of a typical copper wire current carrying wire is determined according to the maximum core temperature allowed, cooling conditions, and laying conditions. Generally, the safe current carrying capacity of copper wires is 5~8A/mm2, and the safe current carrying capacity of aluminum wires is 3~5A/mm2.

Generally, the safe current carrying capacity of copper wires is 5~8A/mm2, and the safe current carrying capacity of aluminum wires is 3~5A/mm2. Such as: 2.5 mm2 BVV copper wire safe current carrying capacity recommended value 2.5 × 8A / mm2 = 20A 4 mm2 BVV copper wire safe current carrying capacity recommended value 4 × 8A / mm2 = 32A

Calculate the cross-sectional area of the copper wire. Use the recommended value of the safe current carrying capacity of the copper wire to 5~8A/mm2, and calculate the upper and lower ranges of the cross-sectional area S of the selected copper wire: S=< I /(5~8)>=0.125 I ~0.2 I (mm2) S—– copper conductor cross-sectional area (mm2) I—–load current (A)

Power calculation General load (can also be used as electrical appliances, such as lighting, refrigerators, etc.) is divided into two types, one type of resistive load and one type of inductive load. For the calculation of the resistive load: P = UI For the calculation of the fluorescent lamp load: P = UIcos ф, where the power factor of the fluorescent lamp load is cos ф = 0.5. The power factor of different inductive loads is different. When the household electrical appliance is uniformly calculated, the power factor cosφ can be taken as 0.8. That is to say, if the total power of a household appliance is 6000 watts, the maximum current is I=P/Ucosф=6000/220*0.8=34(A). However, in general, it is impossible for home appliances to be used at the same time. , so add a common coefficient, the common coefficient is generally 0.5. Therefore, the above calculation should be rewritten as I = P * common coefficient / Ucos ф = 6000 * 0.5 / 220 * 0.8 = 17 (A) That is, the total current value of this family is 17A. The total brake air switch cannot use 16A, it should be greater than 17A.