A motor is 5 5 kilowatts, the voltage is 380 volts, the frame is 50 meters, and the cross sectional

Calculation of conductor cross-sectional area and current carrying capacity.

1. The safe ampacity of the general copper conductor is determined according to the maximum temperature, cooling conditions and laying conditions of the allowable core. Generally, the safe current carrying capacity of copper wire is 5 8a mm2, and the safe current carrying capacity of aluminum wire is 3 5a mm2. 《Key Points》 Generally, the safe ampacity of copper wire is 5 8A mm2, and the safe current carrying capacity of aluminum wire is 3 5A mm2.

2. Calculate the cross-sectional area of the copper conductor using the recommended value of the safe ampacity of the copper conductor 5 8A mm2, and calculate the upper and lower ranges of the selected copper conductor cross-sectional area S: s=< i (5 8)>= i i(mm2) s--- copper conductor cross-sectional area (mm2) i--- load current (a).

3. Power calculation: General loads (which can also be used as electrical appliances, such as lights, refrigerators, etc.) are divided into two types, one is resistive load, and the other is inductive load. Formula for resistive load: p=ui Formula for fluorescent load:

p = uicos, where the power factor cos = of the fluorescent lamp load. The power factor of different inductive loads is different, and the power factor cos can be taken when the household appliances are calculated in a unified manner. That is to say, if all the appliances in a household plus the total power is 6000 watts, the maximum current is i=p ucos =6000 220* However, under normal circumstances, it is impossible to use all the electrical appliances in the home at the same time, so add a common factor, and the common coefficient is average.

So, the above calculation should be rewritten as i=p*commonity ucos =6000*, i.e., the total current value of this home is 17a. Then the main gate air switch can not use 16A, should use greater than 17A.

5500/380=

From a safety point of view, the current of a squared line is 23(a).That's it.

Summary. According to the title, it can be known that the output power of the motor:

110kw voltage: 380v power supply distance: 300 meters According to Max Planck's law, the magnitude of the current is equal to the magnitude of the power divided by the voltage, that is:

I = p u = 110KW 380V = to get the required current is according to the power supply distance of 300 meters, can be known 200V voltage drop cable, according to IEC227-1-CDDV cable selection standard in the corresponding wire diameter, according to the table, I 185A, with low-voltage cable mm2; Thus, the cross-section of the copper core cable is obtained, and the voltage drop is 200V.

The power supply voltage of the 110kw motor is 380V, and the power supply distance is 300 meters.

According to the title, it can be known that the output power of the motor is 110kw voltage:

380V power supply distance: 300 meters by Max Planck's law, the size of the current is equal to the power divided by the size of the voltage, that is: I = p u = 110KW 380V = to get the required current is according to the power supply distance of 300 meters, it can be known that the 200V voltage drop cable bench envy, according to the IEC227-1-CDDV cable selection standard in the corresponding wire diameter, according to the table, I 185A, with low-voltage cable mm2; Thus, the cross-section of the copper wheel punch cable is obtained, and the voltage drop is 200V.

Dear, are you dissatisfied?

A 15kw three-phase asynchronous motor, the power factor is, the line voltage is 380V, and the line current when working is.

According to the formula of electric power, the formula of three-phase electric power is: p = 3 u i cos where p is the power, the vertical mill u is the line voltage, i is the line current, and the power factor angle is the power factor.

Substituting the known parameters into the formula yields: 15kw = 3 380v i The solution current i is: i = 15kw 3 380v Therefore, the line current at work is about .

5 sets of 54 kW motors should be equipped with how large the conductor and the distance is 150 meters.

How big a wire should be matched with 5 sets of 54 kilowatt motors, the distance is 150 meters, <> 54 kilowatt equipment needs to use 25 square wires. 54 kilowatts with a cable of 380, 25 square meters. First, the current of 54 kilowatts is calculated by the formula of a three-phase motor:

p = so the rated current of power 54kw i = p , cos is. The voltage is 380V. According to the current of 102A, check the cable manual YJV, you need to use 32mm aluminum cable, 25mm copper cable.

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The motor power is 315kw, the voltage is 380v, and the distance is 100 meters, so a 300mm copper core cable should be selected.

From the formula of three-phase motor: p=

Therefore, the rated current of 315kW is i=p, COS is, and the voltage is 380V.

According to the current 594A, check the cable manual YJV, you need to use a 300mm copper core cable.

Since the BAI voltage is below 10kV, the cross-section is selected according to the heating strip

Therefore, there is a cross-section to choose the DAO option:

pc=pn*kd/η=16*4*

qc=pc*tan(arccosφ)=

sc=root(pc*pc+qc*qc)version=

ic=sc (root number(3)*.)

Therefore, the conductor with a cross-section A=25mm2 is selected, and its IOC=102A meets the requirements, so the BLX-500-1*25 type wire is selected.

Verification of mechanical strength:

Its minimum allowable cross-section a=10mm2, so it meets the requirements of the verification voltage loss:

Look up the table r0= x0=

The equivalent line length is 35m = so the voltage loss = (pc*r + qc * x) u = (so the voltage loss % = <5% meets the requirements.

Rated working current = 16*4, considering the starting current, you should choose 50mm2 BLV wire.