How to calculate the current carrying capacity of refrigeration equipment wires

How to calculate the current carrying capacity of refrigeration equipment wires

The working temperature is 30℃ and the current carrying capacity under continuous 90% load for a long time is as follows:

　　 1.5 square millimeters-14A,

　　2.5mm2――26A,

　　 4 square millimeters-32A,

　　 6 square millimeters-47A

　　 16 square millimeters-92A

　　 25 square millimeters-120A

　　 30 square millimeters-150A

　　 Current converted power:v

　　 1A=220W, 10A=2200W, and so on.

　　 If the current carrying capacity is 14A copper wire, that is: 220W×14=3080W, then the power of 1.5 square copper wire is 3.08 kilowatts.

　　Copper core wire allows long-term current

　　2.5 square millimeters (16A～25A)

　　 4 square millimeters (25A～32A)

　　 6 square millimeters (32A～40A)

　　Aluminum core wire allows long-term current

　　2.5 square millimeters (13A～20A)

　　 4 square millimeters (20A～25A)

　　 6 square millimeters (25A～32A)

for example

　　 1. Each computer consumes about 200 to 300W (about 1 to 1.5A), so 10 computers need a 2.5 square millimeter copper core wire for power supply, otherwise a fire may occur.

　　2, the power consumption of a large 3-hp air conditioner is about 3000W (about 14A), so one air conditioner needs a single 2.5 square millimeter copper core wire for power supply.

3. The incoming wires of houses are generally 4 square millimeters of copper wire. Therefore, the household appliances that are turned on at the same time must not exceed 25A (5500 watts). It is useless to replace the wires in the house with 6 square millimeters of copper wire. , Because the wire going into the meter is 4 square millimeters.

　　4. In the early housing (15 years ago), the incoming line was generally 2.5 square millimeters of aluminum wire. Therefore, the household appliances that were turned on at the same time should not exceed 13A (2800 watts).

　　5. Household appliances with relatively large power consumption are: air conditioner 5A (1.2hp), electric water heater 10A, microwave oven 4A, rice cooker 4A, dishwasher 8A, washing machine with drying function 10A, electric water heater 4A.

　　 90% of the fires caused by the power supply are caused by the heat of the joints, so all joints must be welded, and the contact devices that cannot be welded must be replaced in 5-10 years (such as sockets, air switches, etc.).

　　 Long-term current allowed by national standard

　　 4 square is 25-32A

　　 6 square is 32-40A

　　 In fact, these are theoretically safe values, and the limit values ​​are even greater than these.

　　 2,5 square copper wire allows the use of z* high power: 5500W. 4 square 8000W, 6 square 9000W is no problem. A 40A digital meter is absolutely no problem with a normal 9000W. The mechanical 12000W will not burn.

　　Copper core wire and cable ampacity standard cable ampacity port decision

　　 Estimated formula:

　　 Two and five times down and multiply by nine, and go up and down by one.

　　 Thirty-five times three and five points, both groups minus five points.

　　 The conditions are changed and the conversion is added, and the high temperature 10% copper upgrade.

　　 The number of piercing pipes is two, three, four, and the full load current is eight or seventy percent off.

　　Description:

　　 The formula in this section does not directly indicate the current carrying capacity (safe current) of various insulated wires (rubber and plastic insulated wires), but is expressed by "cross section multiplied by a certain multiple", which is obtained through mental calculation.

　　 "Multiply two and five times by nine, and go up and down by one" refers to various cross-section aluminum core insulated wires of 2.5mm2 and below, and their current carrying capacity is about 9 times the number of cross-sections. Such as 2.5mm2 wire, the current carrying capacity is 2.5 × 9 = 22.5 (A).

　　The current-carrying capacity of 4mm’ and above conductors are multiplied by the number of cross-sections, and the multiples are successively reduced by 1, namely 4×8, 6×7, 10×6, 16×5, 25×4. "Thirty-five times 3.5, doubles in groups minus five", it is said that the current carrying capacity of 35mm2 wire is 3.5 times the number of cross sections, that is, 35×3.5=122.5(A).

　　From 50mm’ and above, the multiple relationship between the current-carrying capacity and the number of cross-sections becomes a set of two wire numbers, and the multiples are sequentially reduced by 0.5.

　　 means that the current carrying capacity of 50 and 70mm2' wires is 3 times the number of cross sections; the current carrying capacity of 95 and 120mm2 wires is 2.5 times the number of cross sections, and so on. "Conditions have changed, plus conversion, high temperature 10% copper upgrade."

　　The above formula is determined by the aluminum core insulated wire and the open coating at an ambient temperature of 25℃. If the aluminum core insulated wire is exposed to the area where the ambient temperature is higher than 25℃ for a long time, the current carrying capacity of the wire can be calculated according to the above formula calculation method, and then a 10% discount is enough; when the copper core insulated wire is not used, Its current-carrying capacity is slightly larger than that of the aluminum wire of the same specification. According to the above formula method, the current-carrying capacity of one wire larger than the aluminum wire can be calculated. For example, the current carrying capacity of 16mm2’ copper wire can be calculated as 25mm2 aluminum wire.