Calculation of motor power and wiring diameter, and how to choose circuit breakers, etc.

Calculation of motor power and wiring diameter, and how to choose circuit breakers, etc.

1. Calculation of motor power and wiring diameter

First, calculate the line current of the 100KW load.

For the three-phase balanced circuit, the formula for calculating the power of the three-phase circuit is: P=1.732IUcosφ.

It can be deduced from the three-phase circuit power formula:

Line current formula: I=P/1.732Ucosφ

In the formula: P is the circuit power, U is the line voltage, three-phase is 380V, cosφ is the power factor of the inductive load, generally take 0.8 of the line current of your 100KW load:

The current value should also be corrected according to the nature and quantity of the load.

If there are many large motors in the load, since the starting current of the motor is very large, which is 4 to 7 times of the working current, the starting current of the motor should also be considered, but the starting current time is not very long. Factors of 1.3 to 1.7 are considered.

If you take 1.5, then the current is 285A. If the number of 60KW loads is large, and everyone does not use them at the same time, the use factor can be taken as 0.5 to 0.8, where 0.8 is taken, and the current is 228A. You can select wires, air switches, contactors, thermal relays and other equipment according to this current. Therefore, the step of calculating the current cannot be omitted.

Wire selection:

According to the wire allowable ampacity meter of a wire manufacturer, choose 50 square copper core rubber wire, or choose 70 square copper core plastic wire.

Transformer selection:

There are also many conditions for transformer selection. Here, we simply divide the total capacity by the power factor and then round it. S=P/cosφ=100/0.8=125KVA

Just choose a transformer larger than 125KVA.

How much current a 50 square copper core cable can withstand depends on the laying method and ambient temperature, as well as the structure type of the cable and other factors.

50 square 10/35KV XLPE insulated cable Long-term allowable current carrying capacity Air laying Long-term allowable current-carrying capacity

(10KV three-core cable) 231A (35KV single-core cable) 260A long-term allowable current-carrying capacity for direct burial (soil thermal resistance coefficient 100° (10KV three-core cable) 217A (35KV single-core cable) 213A

2. Simple calculation according to the power distribution cable

It is known that the rated power of the motor is 22KW, and the rated voltage is 380V. The transformer is 400 meters away from the well site. How big is the cable with a large cross-sectional area?

(The resistivity P of copper is taken as 0.0175) (1) Calculate the rated current of the motor under rated power with rated capacity

Solution: From P=S×COSφ, S=P/COSφ=22/0.8=27.5KVA, where P is the rated power, COSφ is the power factor, which is 0.8 according to the motor nameplate

There is S=I×U to calculate the rated current at rated power I=S/U=27500/380=73A is obtained from the calculation formula

Estimation formula:

Multiply it by nine under 2.5, subtract one straight number up and go.

Thirty-five times three-and-a-half, and subtracting one-point five in pairs.

Conditions are subject to change plus discount, high temperature 10% off copper upgrade.

The number of pipes passing through is two, three or four, and the full load is 876%.


(1) 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 expresses it by multiplying the cross section by a certain multiple, which is obtained by mental calculation.

The multiplier decreases as the cross section increases.

“Multiply 2.5 times by 9, subtract 1 and go up” refers to aluminum core insulated wires with various cross-sections of 2.5mm’ and below, and the current carrying capacity is about 9 times the number of cross-sections. Such as 2.5mm’ wire, the current carrying capacity is 2.5×9=22.5(A). The relationship between the current carrying capacity and the number of sections of wires of 4mm’ and above is the upward row along the line number, and the multiple is successively reduced by l, that is, 4 × 8, 6 × 7, 10 × 6, 16 × 5, 25 × 4.

“Thirty-five times 3.5, minus 5.5″ in pairs means that the current carrying capacity of the 35mm” wire is 3.5 times the number of 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 sections becomes a group of two line numbers, and the multiple is reduced by 0.5 in turn. That is, 50, 70mm’

The current-carrying capacity of the wire is 3 times the number of sections; the current-carrying capacity of the 95 and 120mm” wires is 2.5 times the number of the cross-sectional area, and so on.

“Conditions are changed and discounted, and high-temperature copper upgrades are 10% off.”

The above formulas are determined by the aluminum core insulated wire and the open lay at the ambient temperature of 25 ℃.

If the aluminum core insulated wire is laid openly in an area where the ambient temperature is higher than 25 °C for a long time, the current carrying capacity of the wire can be calculated according to the above formula, and then 10% off;

When using not aluminum wire but copper core insulated wire, its current carrying capacity is slightly larger than that of aluminum wire of the same specification.

For example, the current carrying capacity of 16mm’ copper wire can be calculated according to 25mm2 aluminum wire.

Then calculate the cross-sectional area of ​​the wire from the allowable voltage drop. The minimum allowable working voltage of the motor is 360V, the secondary voltage of the transformer is 380V, the maximum allowable voltage drop under rated power is △U is 20V, and the allowable resistance under rated power is 20V. for

R line=△U/I=20/73=0.27Ω

From R line = ΡL/S, calculate the cross-sectional area of ​​the wire: S = ΡL/R line

=0.0175×400/0.27Ω=24.62mm square

Conclusion: 25 square copper cables should be selected

3. How to choose circuit breakers and thermal relays

How to choose a cable with a large cross-sectional area according to the current, the cable we choose is a copper core cable.

Let’s give an example, we want to wire a 18.5KW motor, and we can calculate that its rated current is 37A. It is also based on experience that 1 square millimeter of copper wire can pass 4~6A of current. We take the middle value of 5A, then the cable The cross-sectional area of ​​the line should be 37/5 = 6.4 square millimeters. Our standard cables are 6mm square and a total of 10 square millimeters, in order to ensure reliability, we choose 10 square cables. In fact, in the specific selection, we may also choose 6 square meters. This should be considered comprehensively. How much power is consumed when the load is working? If it is only less than 60% of the rated power, you can choose this way. If you basically want to work near the rated power , that can only choose 10 square cables.

How to consider the rated voltage of the motor, current wiring, selection of circuit breakers and thermal relays according to the power of the motor

Three-phase 220 motor, kW 3.5 amps.

Three hundred and eighty motors are commonly used, one kilowatt and two amps.

Low voltage 660 motor, kW 1.2 amps.

High voltage three kilovolt motor, four kilowatts one amp.

High voltage six kilovolt motor, eight kilowatts one amp.

A three-phase motor, in addition to knowing its rated voltage, must also know its rated power and rated current, such as: a three-phase asynchronous motor, 7.5KW, 4 poles (commonly used are generally 2, 4, 6, the number of stages Different, its rated current is also different), its rated circuit is about 15A.

1. Circuit breaker:

Generally choose its rated current 1.5-2.5 times, commonly used DZ47-6032A,

2. Wire:

According to the rated current of the motor 15A, select the wire with the appropriate current carrying capacity. If the motor starts frequently, choose a relatively thick wire, otherwise, it can be relatively thin.

3. AC contactor:

Just choose the appropriate size according to the motor power, 1.5-2.5 times. Generally, there are models in the selection manual. Here we choose Chint CJX2-2510. We must also pay attention to the matching of auxiliary contacts. Don’t buy them when the auxiliary contacts are not enough.

4. Thermal relay

The setting current can be adjusted, generally adjusted to 1-1.2 times the rated current of the motor.

(1) Multiple motors with wires: add up the total power of the motors and multiply by 2 to get their total current.

(2) The cross section of the wire within 50 meters of the line is: the total current divided by 4. (put a little margin appropriately)

(3) If the line length exceeds 50 meters, the cross section of the conductor: divide the total current by 3. (add a little more appropriately)

(4) The current density of large cables over 120 square meters is lower

4. Summary:

Circuit breaker selection:

The rated current of the motor is multiplied by 2.5 times, and the set current is 1.5 times that of the motor, so as to ensure frequent startup and sensitive short-circuit action.

The setting value of the thermal relay is 1.1 times the rated current of the motor.

AC Contactor:

The AC contactor selection is 2.5 times the motor current.

This ensures long-term frequent work

The Links:   AT25010A-10TU-27 LM64P806 LCD-SOURCE

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