Which one has, know the power of electrical appliances, and choose the mantra of cable. And explain

There are eight specifications of domestic cloth wires of 25mm and below.

That is, 1. The relationship between the wire section and the straight warp.

Cross-section s s 1 4 6 10 16 25 (mm).

Diameter d 3 7 7

Calculation of the traverse section and the straight warp: s r (the square of the radius

Estimation of load current:

Three-phase motor 1kw 2a

Single-phase 220 uses 1kw of electricity

Single-phase 380 uses 1kw of electricity

The conductor is selected according to the load current, laying method, and laying environment.

Select the wire formula:

10 under 5; 100 2;

World; Two and a half times, through the tube, temperature.

Eight, nine-fold; Copper wire upgrade calculation; Bare wires are added in half.

That is, every 1mm of wire below 10mm is calculated as 5A; Every 1mm of wire above 100mm is calculated as 2A; 25mm wire is calculated as 4A per 1mm; 35mm wire is calculated as 3A per 1mm; 70 95mm wire per 1mm is calculated;

Threading, temperature.

Eight, nine-fold; When the pipe is laid darkly, the current carrying capacity of the wire is discounted by 20%; When the ambient temperature is greater than 35, the current carrying capacity of the wire is discounted;

Copper wire upgrade calculation; Bare wires are added in half. Because the formula is calculated by aluminum wire, when using insulated copper wire, it is calculated according to the insulated aluminum wire with an increased cross-section; When using bare wires, multiply the ampacity of insulated wires of the same cross-section.

Select an example. Example 1: The load current is 33A, the copper wire is required to be laid darkly, and the ambient temperature is 35. Trial calculation:

If 6mm rubber copper wire (e.g., BX 6) is used, according to the formula, the current carrying capacity of 10mm insulated aluminum wire can be calculated according to the formula, which is 10 5 = 50A; Dark laying, 50; The ambient temperature is pressed at 35, 40. Therefore, it can be used.

Example 2: The load current is 66A. Aluminum wire is required to be laid darkly, and the ambient temperature is calculated according to 35

Design: 16mm plastic aluminum wire (such as BLV 16). According to the formula, 16 4 = 64a.

Dark laying, 64. Use 25 mm2 plastic aluminum wire (e.g. BLV 25) instead. According to the formula, 25 4 = 100a.

Dark laying, 100. The ambient temperature is pressed at 35, 80. Can be used.

Several types of wire cross-sections for fixed requirements:

1.Insulated wire for pipe penetration, the minimum cross-section of copper wire is 1mm; The minimum cross-section of the aluminum wire is.

2.The secondary circuit of various electrical equipment (except for the secondary circuit of current transformer), although the current is very small, in order to ensure the mechanical strength of the secondary line, the insulated copper wire with a cross-section of not less than is often used.

Note! Selecting a line according to the formula is only suitable for wiring the device. Because the current density shown in the formula only ensures the safety of the wire itself, and how much voltage drop and line loss at the end of the wire are not considered.

3.The conductor used in the secondary circuit of the current transformer often uses insulated copper wire with a cross-section.

Note! Selecting a line according to the formula is only suitable for wiring the device. Because the current density shown in the formula only ensures the safety of the wire itself, and how much voltage drop and line loss at the end of the wire are not considered.

Cable current carrying port:

Estimation formula: 2.5 times 9, subtract one straight up and go.

Thirty-five times 3.5, double in a group minus five.

The conditions have been changed and converted, and the high-temperature nine-fold copper has been upgraded.

The number of pipe roots is two, three, four, eight, seven and six folds full of load current.

Note: The current carrying capacity (safety current) of various insulated wires (rubber and plastic insulated wires) is not directly indicated in this formula, but is expressed by "multiplying the cross-section by a certain multiple", which is obtained by mental arithmetic.

The multiplier decreases as the cross-section increases.

2.5 times 9, up minus 1 number "refers to 2 5mm' and below various cross-section aluminum core insulated wires, its current carrying capacity is about 9 times the number of cross-sections. For example, a 2 5mm' wire has a current carrying capacity of 2 5 9 22 5 (a). The relationship between the current carrying capacity and the number of cross-sections of the wire from 4mm' and above is to go up along the wire number, and the multiple is reduced by l, that is, 5, 25 4.

"Thirty-five times three .five, double and double groups minus five", which means that the current carrying capacity of the wire of 35mm is 3 5 times the number of cross-sections, that is, 35 3 5 122 5 (a). From 50mm' and above wires, the multiple relationship between the current carrying capacity and the number of cross-sections becomes a group of two two wire numbers, and the multiples are reduced by 0 5 in turn. i.e. the current carrying capacity of the mm' wire is 3 times the number of cross-sections; 95, 120mm" wire current carrying capacity is 2 5 times of its cross-sectional area, and so on.

The conditions have been changed and converted, and the high-temperature nine-fold copper has been upgraded.