First, the general copper wire carrying current The safety of the conductor is based on the maximum allowable core temperature, cooling conditions, laying conditions to determine. Generally, the safe current carrying capacity of copper wire is 5 ~ 8A / mm2, and the safe current of aluminum wire is 3 ~ 5A / mm2. <Key points> General copper wire safety current carrying capacity of 5 ~ 8A / mm2, aluminum wire safety current carrying capacity of 3 ~ 5A / mm2. Such as: 2.5mm2BVV copper wire recommended safe carrying capacity of 2.5 × 8A / mm2 = 20A4mm2BVV copper wire recommended current carrying capacity of 4 × 8A / mm2 = 32A

Second, calculate the copper conductor cross-sectional area of ​​the use of copper wire safety carrying capacity of the recommended value of 5 ~ 8A / mm2, calculate the selected copper wire cross-sectional area S range: S = <I / (5 ~ 8)> = 0.125I ~ 0.2I (mm2) S ----- copper wire cross-sectional area (mm2) I ----- load current (A)

Third, the power calculation of the general load (can also be used as electrical appliances, such as lighting, refrigerator, etc.) is divided into two kinds, a resistive load, one is the inductive load. For the resistive load calculation formula: P = UI for the fluorescent lamp load calculation formula: P = UIcosф, where the fluorescent lamp power factor cosф = 0.5. Different inductive load power factor is different, unified calculation of household appliances can be used when the power factor cosф take 0.8. That is, if a household with all the appliances with a total power of 6000 watts, the maximum current is I = P / Ucosф = 6000/220 * 0.8 = 34 (A) However, under normal circumstances, the home appliances can not 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 / Ucosф=6000*0.5/220*0.8=17(A) That is, the total current value of this family is 17A. The total gate air switch can not use 16A, should be greater than 17A.

Estimated formula:

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 53, 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 the group of five points," said 35mm "wire carrying capacity of 3.5 times the number of 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 of two, followed by 0.5 times, that is, 50,70 mm 'conductor carrying capacity of 3 times the number of crossings; 95,120 mm " The flow rate is 2.5 times the cross-sectional area, 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 line carrying capacity, according to 25mm2 aluminum line calculation

Optimization of Transmission Cable Section

In the past, when selecting a power distribution cable, the cable type is usually determined according to the laying conditions, and then the cable section is selected according to the heating conditions. Finally, the cable section meets the requirements of the current carrying capacity and meets the requirements of voltage loss and thermal stability.

If the economic benefits are taken into account, the optimal cross-section of the cable should be the minimum cross-section for the initial investment and the cost of the entire cable's economic life. From this point of view to select the cable cross-section, the need for heat conditions selected in the cross-section basis, and then artificially increase the 4 to 5 section, called the section for the best cross-section.

As the cross-section increases, the line resistance is reduced, so that the line pressure drop is reduced, thus greatly improving the power supply quality, power loss is reduced, so that the operating costs of the cable to reduce the capacity of the cable, So that the total cost of the entire cable can be guaranteed to be the lowest.

The following will be used to prove the total cost of ownership method, the cable should be the best cross-section in accordance with conventional methods based on the selected, and then increase the 4 to 5 level.

To a potter dryer, for example, the three-phase power of 70kW, the supply voltage of 400V, current 101A, line length of 100m. 2 Select the cable cross section according to the heating conditions

According to the laying requirements of the selection of YJLV type, 1kV three-core power cable, pipe straight buried laying, according to the heat conditions selected cable cross-section S is 25mm2, this section allows the closure of 125A.

3 Select the cable section by total cost of ownership

The total cost of ownership method is a common method for comparing the economic benefits of various schemes. The current investment of the comparative scheme and the future cost of the scheme are expressed by the current value. The future cost of the scheme is multiplied by the present value coefficient Q, and the total cost of ownership is calculated after the calculation.

Total cost of ownership C = initial investment + PV value

PV value is called the present value PV = Q × annual energy loss

The initial investment in this equipment, including cable prices plus laying the integrated cost. A variety of cross-section of the power cable, the length of 100m when the initial investment in Table 1.

Table 1 the initial investment of various cross-section power cables

Cable cross-section cable unit price (yuan / m) cable price (yuan) equipment comprehensive cost (× 105 yuan) initial investment C

257.757750.1616775

359.179170.1616917

Cable initial investment C = cable unit price × cable length + laying the integrated cost. Total cost of ownership:

Power loss P = 3I2r0l × 10-3 (kW), where I = 101A, l = 0.1km.

Annual power loss A = Pτ (kWh), where τ is the annual maximum load loss hours, take τ = 4500h.

Annual energy loss costs Cf = A × electricity price (yuan), take the Northeast industrial electricity price (0.398 yuan / kWh).

PV value (present value) = Q × Cf (yuan), Q (present value coefficient)

Q = {1 - [(1 + a) / (1 + i)] n} / (i-a)

Where i - annual interest rate, i = 7%;

A - year inflation rate, a = 0;

N - years of use, n = 20 years. Substituting Q-style

Q = {1- [1 / (1 + 0.07)] 20} /0.07=10.59

The optimal economic cross section of the distribution cable is 120 mm2, with the lowest total cost of ownership. As the price increases, the optimal cross section of the distribution cable will become larger.

Calculation of conductor carrying capacity

1, use: a variety of wire current carrying capacity (safe current) can usually be found from the manual. But with the formulas and then with some simple mental arithmetic, can be directly calculated, do not look up the table. (Aluminum or copper), type (insulated wire or bare wire, etc.), laying method (Ming or pipe, etc.) and the ambient temperature (25 degrees or so Greater) and so on, the impact of more factors, the calculation is more complex.

10 on the fifth, 100 on the second.

25,35, four or three circles.

70,95, twice and a half.

Penetration temperature, eighty-nine fold.

Bare plus half.

Copper wire.

4. Description: formula is aluminum core insulated wire, Ming Fu in the ambient temperature of 25 degrees prevail. If the conditions are different, there is another statement. Insulation lines include various types of rubber insulated wire or plastic insulated wire. Formulas for a variety of cross-section of the current (current, safety) is not directly pointed out, but "with a certain number of crossings" to express. To this end, should be familiar with the wire cross-section, (square mm) arrangement:

11.52.54610162535507O95l20150185 ...

The cross-sectional area of ​​the aluminum core insulated wire of the manufacturing plant usually starts at 2.5, and the copper insulated wire starts from 1; the bare aluminum line starts from 16; the bare copper wire starts from 10

① This formula pointed out: aluminum core insulation line carrying capacity, security, can be calculated by the number of cross the number of times. In the formula, the Arabic numerals indicate the wire cross section (square millimeters), and the Chinese characters indicate the multiple. The arrangement of the cross section of the formula and the multiples are as follows:

..1016-2535-5070-95120 ....

Five times twice as much as twice as twice as twice

Now and then the formula is even more clear. The original "10 next five" refers to the cross-section from 10 below, carrying capacity are five times the number of cross-section. "100 on the two" (read the first two), refers to more than 100 cross-section, carrying capacity is twice the number of cross-section. Section 25 and 35 are four times and three times the boundaries. This is the "tricks 25,35 four three circles." While the cross-section 70,95 was 2.5 times. From the above arrangement, it can be seen: in addition to 10 below and 100 or more, the middle of the wire cross-section is the same for each of the two specifications.

Below to cover the aluminum core insulated wire, the ambient temperature of 25 degrees, for example:

[Example 1] 6 square millimeters, according to 10 five, calculate the load flow of 30 An.

[Example 2] 150 square millimeters, according to 100 on the second, calculate the flow of 300 amps.

[Example 3] 70 square millimeters, according to 70,95 two and a half times, calculate the load flow of 175 am.

It can also be seen from the above arrangement that the multiple decreases with increasing cross-section. At the junction of multiple transformations, the error is slightly larger. For example, sections 25 and 35 are four times and three times the boundary, 25 is four times the range, but close to the three times the side of the change, it is four times the tone, that is 100A. But actually less than four times (according to the manual for 97). And 35 on the contrary, according to the formula is three times, that is, 105 An, is actually 117 An. But the impact on the use of this is not large. Of course, if the number of chest, in the choice of wire cross-section, the 25's not let it over 100 An, 35 can be slightly more than 105 A more accurate. Similarly, the 2.5 mm square wire position at five times the initial (left) end, actually more than five times <up to 20 or more>, but in order to reduce the power loss within the wire, usually do not have to be so big, manual In the general only standard 12 An.

② from below, the formula is to change the conditions of treatment. (Including the trough plate and other laying, that is, with a protective sheath layer, not exposed) by ① calculation, and then hit 20% (by 0.8) if the environment Temperature over 25 degrees, should be calculated by ①, then hit 10% off. (By 0.9).

On the ambient temperature, according to the provisions of the summer is the hottest month the average maximum temperature. In fact, the temperature is variable, under normal circumstances, it affects the conductor current is not very large. Therefore, only for some high-temperature workshop or more hot areas more than 25 degrees, only consider the discount.

There is also a situation where both conditions change (higher in pipe and temperature). According to ① calculated after 20% discount, hit 10% off. Or simply a dozen odds (ie 0.8 × 0.9 = 0.72, about 0.7). This can also be said that the tube temperature, eighty-nine fold the meaning.

For example: (aluminum core insulated wire) 10 square millimeters, through the tube (20% off) 40A (10 × 5 × 0.8 = 40)

Tube and high temperature (30%) 35A (1O × 5 × 0.7 = 35)

95 square millimeters, through tube (20% off) 190 Ann (95 x 2.5 x 0.8 = 190)

High temperature (10% off), 214 am (95 x 2.5 x 0.9 = 213.8)

Tube and high temperature (Qizhe). 166A (95 x 2.5 x 0.7 = 166.3)

② for bare aluminum current carrying capacity, mouth code that, plus half of the bare line, that is, by ① after the calculation of half (by l.5). This refers to the same section of the aluminum core insulated wire compared with aluminum bare wire, carrying capacity can be increased by half.

[Example 1] 16 mm square bare aluminum wire, 96 amp (16 x 4 x 1.5 = 96). High temperature, 86A (16 × 4 × 1.5 × 0.9 = 86.4)

[Example 2] 35 square millimeters bare aluminum wire, 150A (35 × 3 × 1.5 = 157.5)

[Example 3] 120 square millimeters bare aluminum wire, 360 amp (120 × 2 × 1.5 = 360)

③ for copper wire current carrying capacity, formulas that copper line operator count. That is, the copper wire cross-section in order to enhance the order of a row, and then according to the corresponding aluminum conditions.

[Example 1] 35 square bare copper wire 25 degrees, upgrade to 50 square millimeters, and then by 50 square millimeters bare aluminum wire, 25 degrees calculated for 225 An (50 × 3 × 1.5)

[Example 2] 16 square mm copper insulated wire 25 degrees, according to the same conditions of 25 square millimeters of aluminum insulation, calculated as 100A (25 × 4)

[Example 3] 95 square millimeters of copper insulated wire 25 degrees, through the 120 square millimeters of aluminum insulated wire of the same conditions, calculated as 192 An (120 × 2 × 0.8).