16 square cables can carry about 80A of current? The 18.5KW motor's rated current is only 37A. How can you see that 16 square cables are enough to drive an 18.5KW motor?

But no one thought that the motor burned out one after another, and finally the electrician found that the voltage of the motor was only 340V due to the cable length of 1000 meters and the cable distance caused by the voltage reduction, thus burning the motor.

Cause of voltage reduction of motor

The voltage drop is due to the cable distance being too long, the wire diameter being too small, and the cable resistance increasing, resulting in a voltage reduction. The normal voltage range of the motor is 380V, and the allowed voltage fluctuation range is lower than 5% of the normal voltage and higher than 10% of the normal voltage. (380*5% = 19V)

The voltage is lower than the normal voltage, but it can be used normally if it is not more than 19V; if it is more than 19V, it will cause damage to the motor or even burn.

The effect of low voltage on the motor

When the voltage is reduced, the current of the motor will increase, and the winding of the motor will heat. When the voltage is reduced by more than 10%, the winding temperature is too high for a long time, which will affect the service life of the motor, and the motor may be burned out if it is serious.

How to calculate voltage drop?

Voltage drop = current through the wire x resistance of the wire

The current through the wire is the current required to drive the load. The rated current of the 18.5KW motor is 37A, so the wire needs to pass the 37A current.

Resistance of wire = length of wire *resistivity of wire ÷diameter of wire

(The resistivity of the wire is 0.0175 for copper and 0.0283 for aluminum.)

For example:

18.5 KW motor with 16 cables, cable distance of 1000 meters, how much voltage will be reduced?

Resistance of wire = length of wire 1000 meters * resistivity of wire Copper wire 0.0175÷ Wire diameter 16=1000*0.0175÷16 = 1.1Ω

Current on the wire = 18.5*2 = 37A (Calculate the load current according to 1 KW of power for 2A quick calculation).

Voltage drop = current on wire 37A x resistance of wire 1.1Ω= 37 x 1.1 = 40.7V

That is, a 18.5 KW motor with 16 cables, a cable distance of 1000 meters, the voltage will be reduced by 41 volts from the normal voltage of 380 volts to 41 volts, which is far beyond the range of the normal voltage, so the electric machine will burn out, so the configuration of the cable needs to increase the wire diameter to reduce the voltage drop.

How to configure the cable considering the voltage drop?

Cross-sectional area of copper wire = (current on wire * length of wire) ÷ (54.4*allowable voltage fluctuation range 19V)

Cross-sectional area of aluminum wire = (current on wire * length of wire) ÷ (34*allowable voltage fluctuation range 19V)

For example:

18.5 KW motor, cable length 1000 meters, configuration of how big cable?

Copper wire cross-sectional area = (the current on the wire 37A*the length of the wire 1000 meters) ÷ (54.4*19V) = (37*1000) ÷ (54.4*19) = 35.8 wire, that is, we have to choose not less than 35.8 copper wire cable.

Aluminum wire cross-sectional area = (the current on the wire 37A*the length of the wire 1000 meters) ÷ (34*19V) = (37*1000) ÷ (34*19) = 57.3 wire, that is, we have to choose not less than 57.3 aluminum wire cable.

Therefore, dry electrical, known load power, depends on the configuration of the cable, not only to take into account the current of the load but also to take into account the load too far away and the impact of the voltage drop!