The current carrying capacity of the wire is related to the cross section of the conductor. It is also related to the material, type, laying method and ambient temperature of the wire. The factors affecting the calculation are more complicated. The carrying capacity of various conductors can usually be found in the manual. But with the formulas and then with some simple mental arithmetic, can be directly calculated, do not look up the table.
Two hundred and fifty times multiplied by nine, go up by a straight to go.
Thirty-five by 3.5, both groups of five points.
Conditions have changed, high temperature Jiujiang copper upgrade.
Piercing the number of two hundred thirty-four, eight seven six fold full load.
Description:
(Safety current) is not directly pointed out, but "cross section multiplied by a certain number of times" to express, through the mental arithmetic derived from the heart of the line (rubber and plastic insulated wire) As can be seen from Table 5 3:
The multiple decreases with increasing cross-section.
"2.5 points multiplied by nine, go up by a straight run" that is 2.5mm 'and below the various sections of aluminum core insulated wire, the carrying capacity of about 9 times the number of cross-section. Such as 2.5mm 'wire, carrying capacity of 2.5 × 9 = 22.5 (a).
From the 4mm 'and above the conductor of the current and the number of cross-sectional number of the relationship is the number of lines along the line row, multiple times by 1, that is 4 × 8,6 × 7,10 × 6,16 × 5,25 × 4.
"35 by 3.5, double group of five points," said 35mm "wire carrying capacity of 3.5 times the cross-section, that is, 35 × 3.5 = 122.5 (a). From 50mm 'and above the wire, the carrying capacity and the number of crossings between the multiple relationship between the two lines into a group, the number of times in turn 0.5. That is, 50,70 mm' conductor carrying capacity of the number of cross-section 3 Times; 95,120mm "wire carrying capacity is the number of its cross-sectional area of 2.5 times, and so on.
"Conditions are variable conversion, high temperature Jiujiang copper upgrade." The above formula is aluminum core insulated wire, the application of the ambient temperature of 25 ℃ depending on the conditions. If the aluminum wire insulation line in the ambient temperature long-term higher than 25 ℃ in the region, the line carrying capacity can be calculated according to the formula formula, and then hit the nine fold can be; when the use of aluminum wire is not copper wire, It is slightly larger than the capacity of the same specifications of the aluminum line, according to the above formulas to calculate a line than the aluminum line to increase the current carrying capacity. Such as 16mm 'copper current carrying capacity, according to 25mm2 aluminum line calculation, for the bare aluminum line current capacity, formulas that "half plus half of the line" that is calculated after the addition of half. This refers to the same section of bare aluminum wire and aluminum core insulation compared to the carrying capacity can be increased by half.
For example, the calculation of bare aluminum line carrying capacity:
When the cross-section of 16 square millimeters, the carrying capacity of 16 × 4 × 1.5 ═ 96 An, if at high temperatures, the carrying capacity of 16 × 4 × 1.5 × 0.9 = 86.4 Ann.
For the copper conductor current carrying capacity, the formula pointed out that the "copper line operator", that is, the copper wire cross-section order to enhance the level, and then according to the corresponding aluminum conditions.
For example, the cross-section of 35 square millimeters bare copper wire ambient temperature of 25 ℃, carrying capacity calculation is: according to the upgrade to 50 square millimeters bare aluminum line that was 50 × 3 × 1.5 = 225 amps.
For the cable, the formulas are not introduced. Generally directly buried high-voltage cable, in general, can be directly used in the first sentence formulas in the multiple calculation. Such as 35 square millimeters high pressure armored aluminum cable buried laying capacity of 35 × 3 = 105A. 95 square millimeters of about 95 × 2.5 ≈ 238 Ann.
Three-phase four-wire system in the zero line, usually selected for phase line section of about 1/2. Of course, not less than the minimum cross-section required by mechanical strength requirements. In a single-phase line, the zero line and the phase line through the same load current, so the zero line cross-section should be the same phase.
