Wires and cables refer to materials used for power, communication, and related transmission purposes. There is no strict boundary between "wires" and "cables". Generally, products with a small number of cores, small outer diameter, and simple structure are called wires; those without insulation are called bare wires, and the rest are called cables. Wires with a larger conductor cross-sectional area (greater than 6 mm²) are referred to as large wires, while those with a smaller cross-sectional area (less than or equal to 6 mm²) are called small wires; insulated wires are also known as wiring cables.

In a broad sense, wires and cables are also simply called "cables". In a narrow sense, a cable specifically refers to an insulated cable. It can be defined as an assembly consisting of the following parts: one or more insulated cores, along with their respective possible covering layers, an overall protective layer, and an outer sheath. Cables may also include additional uninsulated conductors.

In China, wire and cable products are classified into the following five major categories based on their applications:

Basic structure of wires and cables:

The safe current-carrying capacity of a conductor is determined by the maximum allowable temperature of the conductor core, cooling conditions, and laying conditions. Generally, the safe current-carrying capacity of copper conductors ranges from 5 to 8 A/mm², and that of aluminum conductors ranges from 3 to 5 A/mm².

Using the recommended safe current-carrying capacity of copper conductors (5–8 A/mm²), the upper and lower ranges of the selected copper conductor cross-sectional area (S) can be calculated as follows:S = < I / (5–8) > = 0.125I ~ 0.2I (mm²)Where:

Different inductive loads have different power factors. When calculating household electrical appliances uniformly, the power factor (cosφ) can be taken as 0.8. For example, if the total power of all electrical appliances in a household is 6000 W, the maximum current is calculated as:I = P / (Ucosφ) = 6000 / (220 × 0.8) = 34 (A)

However, under normal circumstances, household appliances cannot be used simultaneously, so a demand factor (usually 0.5) is applied. Therefore, the above calculation should be revised to:I = (P × Demand Factor) / (Ucosφ) = (6000 × 0.5) / (220 × 0.8) = 17 (A)

In other words, the total current of this household is 17 A. Therefore, a 16 A main air switch cannot be used; instead, a switch with a rating greater than 17 A should be adopted.

The mnemonic "For sizes below 2.5 mm², multiply by 9; for larger sizes, decrease the multiple by 1 in sequence" means that for aluminum-core insulated wires with a cross-sectional area of 2.5 mm² or smaller, the current-carrying capacity is approximately 9 times the cross-sectional area. For example, the current-carrying capacity of a 2.5 mm² conductor is 2.5 × 9 = 22.5 (A). For conductors with a cross-sectional area of 4 mm² or larger, the multiple relationship between current-carrying capacity and cross-sectional area follows the conductor size in ascending order, with the multiple decreasing by 1 each time (i.e., 4×8, 6×7, 10×6, 16×5, 25×4).

The mnemonic "35 mm² multiplies by 3.5; for larger sizes, group in pairs and decrease the multiple by 0.5" indicates that the current-carrying capacity of a 35 mm² conductor is 3.5 times its cross-sectional area (i.e., 35 × 3.5 = 122.5 A). For conductors with a cross-sectional area of 50 mm² or larger, the multiple relationship between current-carrying capacity and cross-sectional area changes to grouping two conductor sizes as a pair, with the multiple decreasing by 0.5 sequentially. Specifically, the current-carrying capacity of 50 mm² and 70 mm² conductors is 3 times their cross-sectional area; the current-carrying capacity of 95 mm² and 120 mm² conductors is 2.5 times their cross-sectional area, and so on.

The mnemonic "Adjust for changing conditions with conversion; 10% discount for high temperature, and upgrade copper conductors by one size" is based on aluminum-core insulated wires laid in the open at an ambient temperature of 25°C. If aluminum-core insulated wires are laid in areas where the ambient temperature is consistently higher than 25°C, the current-carrying capacity of the conductor can be calculated using the above mnemonic and then a 10% discount (multiplied by 0.9) is applied. When copper-core insulated wires are used instead of aluminum wires, their current-carrying capacity is slightly higher than that of aluminum wires of the same specification. In this case, the current-carrying capacity can be calculated by upgrading the copper wire size by one level and then using the above mnemonic for aluminum wires. For example, the current-carrying capacity of a 16 mm² copper wire is calculated as that of a 25 mm² aluminum wire.

Notes:

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Power cables are suitable for power transmission and distribution on AC 50 Hz power lines with a rated voltage of 0.6/1 kV or below. At an ambient temperature of 25°C, the operating temperature of the cable conductor shall not exceed 70°C.

Specifications are expressed using the number of cores, nominal cross-sectional area, and voltage level.

(1) In the past, when selecting distribution cables, the cable model was usually determined based on laying conditions first, then the cable cross-sectional area was chosen according to heat generation conditions, and finally a cross-sectional area that meets the requirements of current-carrying capacity, voltage drop, and thermal stability was selected.

If economic benefits are taken into account, the optimal cross-sectional area of the cable should be the one that minimizes the sum of the initial investment and the loss cost during the entire economic life of the cable. From this perspective, when selecting the cable cross-sectional area, it is necessary to artificially increase the cross-sectional area by 4 to 5 grades on the basis of the area selected according to heat generation conditions; this area is called the optimal cross-sectional area.

Increasing the cable cross-sectional area improves the current-carrying capacity, thereby extending the service life of the cable; as the cross-sectional area increases, the line resistance decreases, which reduces the line voltage drop, thus greatly improving the power supply quality, reducing power loss and lowering operating costs. In this way, the minimum total cost during the entire economic life of the cable can be ensured.

The following uses the total cost of ownership method to demonstrate that the optimal cross-sectional area of the cable should be increased by 4 to 5 grades on the basis of the area selected by conventional methods.

Take a ceramic dryer as an example: its three-phase power is 70 kW, the supply voltage is 400 V, the current is 101 A, and the line length is 100 m.

(2) Selection of Cable Cross-Sectional Area According to Heat Generation ConditionsAccording to the laying requirements, a YJLV-type 1 kV three-core power cable is selected for pipe-buried laying. The cable cross-sectional area (S) selected according to heat generation conditions is 25 mm², and the allowable current-carrying capacity of this cross-section is 125 A.

Total Cost of Ownership (C) = Initial Equipment Investment + Present Value (PV)PV is called the present value, and PV = Q × Annual Power Loss Cost. In this example, the initial equipment investment includes the cable price plus the comprehensive laying cost. The initial investment for power cables of various cross-sections with a length of 100 m is shown in Table 1.

Initial Cable Investment (C) = Cable Unit Price × Cable Length + Comprehensive Laying Cost.

Calculation of Total Cost of Ownership:

Substituting into the formula for Q, we get Q = {1 - [1/(1+0.07)]²⁰}/0.07 = 10.59.The optimal economic cross-sectional area (S) of the distribution cable is 120 mm², at which the total cost of ownership is the lowest. As the electricity price rises, the optimal cross-sectional area of the distribution cable will become larger.

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① Purpose: The current-carrying capacity (safe current) of various conductors can usually be found in manuals. However, with the help of this mnemonic and simple mental calculation, it can be calculated directly without looking up the manual. The current-carrying capacity of a conductor is related to its cross-sectional area, as well as the conductor material (aluminum or copper), type (insulated wire or bare wire, etc.), laying method (open laying or pipe laying, etc.) and ambient temperature (around 25°C or higher). There are many influencing factors, and the calculation is relatively complex.

Mnemonic: "Below 10 mm², multiply by 5; above 100 mm², multiply by 2; 25 mm² and 35 mm² follow the 4-3 rule; 70 mm² and 95 mm², multiply by 2.5; for pipe laying and temperature adjustment, take 80% and 90% respectively; bare wires add half; copper wires are calculated by upgrading one grade."

② Explanation: This mnemonic is based on aluminum-core insulated wires laid in the open at an ambient temperature of 25°C. If the conditions are different, additional explanations for the mnemonic apply. Insulated wires include various types of rubber-insulated wires or plastic-insulated wires. The mnemonic does not directly indicate the current-carrying capacity (in amperes) for various cross-sections, but instead expresses it as "multiplying the cross-sectional area by a certain multiple". For this reason, one should first be familiar with the sequence of conductor cross-sections (in mm²): 1, 1.5, 2.5, 4, 6, 10, 16, 25, 35, 50, 70, 95, 120, 150, 185...

Manufacturers usually start producing aluminum-core insulated wires with a cross-sectional area of 2.5 mm², while copper-core insulated wires start from 1 mm²; bare aluminum wires start from 16 mm², and bare copper wires start from 10 mm².

The following are examples of aluminum-core insulated wires laid in the open at an ambient temperature of 25°C:[Example 1] For a 6 mm² conductor, according to "below 10 mm², multiply by 5", the current-carrying capacity is calculated as 30 A.[Example 2] For a 150 mm² conductor, according to "above 100 mm², multiply by 2", the current-carrying capacity is calculated as 300 A.[Example 3] For a 70 mm² conductor, according to "70 mm² and 95 mm², multiply by 2.5", the current-carrying capacity is calculated as 175 A.

Cross-sectional area (mm²): 10; 16-25; 35-50; 70-95; 120...Corresponding multiple: 5; 4; 3; 2.5; 2...

The following are examples of aluminum-core insulated wires laid in the open at an ambient temperature of 25°C:[Example 1] For a 6 mm² conductor, according to "below 10 mm², multiply by 5", the current-carrying capacity is calculated as 30 A.[Example 2] For a 150 mm² conductor, according to "above 100 mm², multiply by 2", the current-carrying capacity is calculated as 300 A.[Example 3] For a 70 mm² conductor, according to "70 mm² and 95 mm², multiply by 2.5", the current-carrying capacity is calculated as 175 A.

It can also be seen from the above sequence that the multiple decreases as the cross-sectional area increases. At the boundary where the multiple changes, the error is slightly larger. For example, 25 mm² and 35 mm² are the boundary between the multiple of 4 and 3. 25 mm² belongs to the range of the multiple of 4, but it is close to the side where the multiple changes to 3. According to the mnemonic, it is multiplied by 4, resulting in a current-carrying capacity of 100 A, but the actual value is less than 4 times (97 A according to the manual). On the contrary, for 35 mm², according to the mnemonic, it is multiplied by 3, resulting in a current-carrying capacity of 105 A, but the actual value is 117 A. However, this has little impact on practical use. Of course, if one has a clear understanding, when selecting the conductor cross-sectional area, the current-carrying capacity of a 25 mm² conductor will not be fully utilized up to 100 A (the maximum can reach more than 20 A), but to reduce power loss in the conductor, it is usually not used to such an extent, and the manual generally only marks 12 A.

From this point on, the mnemonic deals with adjustments for changing conditions. The phrase "for pipe laying and temperature adjustment, take 80% and 90% respectively" means that if the conductor is laid in a pipe (including laying in a trough, i.e., the conductor is covered with a protective layer and not exposed), after calculating according to ①, multiply by 0.8 (take 80%); if the ambient temperature exceeds 25°C, after calculating according to ①, multiply by 0.9 (take 90%).

According to regulations, the ambient temperature refers to the average maximum temperature in the hottest month of summer. In fact, the temperature is variable, and under normal circumstances, it has little impact on the current-carrying capacity of the conductor. Therefore, a discount is only considered for certain high-temperature workshops or areas where the temperature is much higher than 25°C. There is also a case where both conditions change (pipe laying and high temperature). In this case, after calculating according to ①, multiply by 0.8 first, then by 0.9, or simply multiply by 0.7 (approximately 0.72 from 0.8 × 0.9) for calculation. This is also the meaning of "for pipe laying and temperature adjustment, take 80% and 90% respectively".

For the current-carrying capacity of bare aluminum wires, the mnemonic states "bare wires add half", which means that after calculating according to ①, multiply by 1.5 (add half). This means that compared with aluminum-core insulated wires of the same cross-sectional area, the current-carrying capacity of bare aluminum wires can be increased by half.

[Example 1] For a 16 mm² bare aluminum wire, the current-carrying capacity is 96 A (16 × 4 × 1.5 = 96); at high temperature, it is 86 A (16 × 4 × 1.5 × 0.9 = 86.4).[Example 2] For a 35 mm² bare aluminum wire, the current-carrying capacity is 150 A (35 × 3 × 1.5 = 157.5).[Example 3] For a 120 mm² bare aluminum wire, the current-carrying capacity is 360 A (120 × 2 × 1.5 = 360).

For the current-carrying capacity of copper wires, the mnemonic states "copper wires are calculated by upgrading one grade", which means that the cross-sectional area of the copper wire is upgraded by one grade according to the cross-sectional area sequence, and then calculated according to the corresponding conditions for aluminum wires.

[Example 1] For a 35 mm² bare copper wire at 25°C, upgrade it to 50 mm², then calculate according to the 50 mm² bare aluminum wire at 25°C, resulting in a current-carrying capacity of 225 A (50 × 3 × 1.5).[Example 2] For a 16 mm² copper-insulated wire at 25°C, calculate according to the same conditions as the 25 mm² aluminum-insulated wire, resulting in a current-carrying capacity of 100 A (25 × 4).[Example 3] For a 95 mm² copper-insulated wire laid in a pipe at 25°C, calculate according to the same conditions as the 120 mm² aluminum-insulated wire laid in a pipe, resulting in a current-carrying capacity of 192 A (120 × 2 × 0.8).