Fiberglass Covered Insulated rectangular Aluminum Wire
Glass fiber covered rectangular aluminum wire
Fiberglass Covered Insulated rectangular aluminum wire is in accordance with customer requirements in aluminum conductor uniform whole winding package 1-2 layers of alkali-free fiberglass, with the necessary temperature resistance index and compatible adhesive paint dipping and baking process, glass fiber and conductor between bonded into a whole. The insulating layer has a certain wear resistance, adhesion, flexibility and resistance to voltage, Fiberglass Covered Insulated rectangular aluminum wire has good thermal shock resistance and overload performance. Fiberglass Covered Insulated rectangular aluminum wire is generally applied to high-voltage electrical appliances, electrical appliances, electrical appliances, medium capacity of electrical appliances, electrical appliances, large capacity of electrical products, electrical products, the current temperature index of 155, 180, three levels, 20, 30, 40, 50, four grade.
The bare electrical aluminum wire from extrusion process is the most ideal materials to produce Fiberglass Covered Insulated rectangular aluminum wire.
＊Product name and type :
Diameter range(mm) : : 3.0-7.0 rectangular : a: 1-10 b: 3-35
Temp Class : 155, 180
Standard : IEC,NEMA,GB,JIS,GB/T7672,IEC60317
Resistivity : (copper) 20℃<=0.017241Ω(mm)^2/m Aluminum 20℃<=0.028000Ω(mm)^2/m
Package : PT-25, PT-30,30kg/50kg/150kg wooden Spool according to customers requirements.
Feature :
Wrapped smoothly and tightly, without missing layers, bright and clean.
Material :
Drawing or extruding bare wire confirm to National Standard. smooth surface, no defects
Applicaton of Fiber Glass covered rectangular aluminum wire :
The product is widely used in transformers, electromagnets, welders or other similar products, medium and large electrical appliances winding.
＊Product name and type :
R20 during 20℃ unit length DC resistance shall comply with the following calculation of the value of the R20=P20/S min type : Smin conductor minimum cross-sectional area, in accordance with the minimum size of a and b, and r ‘s maximum value calculation: Smin=Amin × Bmin- 0.8584 r2 max × P20= 20℃conductor resistivity , mm2 / m for Aluminum Conductor: P20=1/35.7 mm2 / m fiberglass should be closely evenly a on the conductor of rectangular aluminum, double Fiberglass covered aluminum wire with two layers of glass wire package directions should be opposite, and double Fiberglass covered aluminum wire insulation thickness should be in conformity with the following as specified in the table (narrow on the edge of the insulation thickness (B-b) not to do assessment.)
