380V voltage 70 aluminum wire, load 40 kilowatts, distance 400 meters, what is the line loss 20

The current of 40KW of three-phase load is about 80A, the line resistance of each line: r = ohm, the voltage drop of each line loss: u =, the three-phase line loss power: p = 80 * 80 *

A distance of 400 meters, a voltage of 380V, and 160 kilowatts of electricity require a large aluminum wire.

400 meters of distance, 380V voltage, 160 kilowatts of electricity, need to return to guess and large aluminum Zhaoxiao line Hello dear, 4 square aluminum wire open line laying safety interception flow of about 30A. If the power factor of the three-phase electricity is calculated, the maximum power that the aluminum wire can withstand is: p= 3 u i cos =.

The power supply distance of 400 meters is calculated as: taking the starting voltage of 400V and the total current of the 25KW welding machine of 50A as an example, the voltage required for normal operation at the end is 380V, then the voltage drop between the lines is 20V, the voltage drop of each wire is 10V, and the line resistance value of each wire: r u i 10 50 The cross-section of each three-phase aluminum wire (the resistivity of the aluminum wire is:

Summary. The issue is not available.

In addition to line loss, there is also reactive power loss.

If you are using an inductive load, there is no reactive power compensation at the terminal, and the reactive power loss is much greater than the line loss.

The daily electricity consumption is about 400 degrees, and if you use electricity 24 hours a day, the loss is negligible.

35 square aluminum wire, three-phase four-wire, line length of 1000 meters, daily electricity consumption of about 400 degrees, how much is the line loss?

The issue is not available. In addition to line loss, there is also reactive power loss. If you are using an inductive load, the terminal does not have the state to shoot reactive power compensation, and the reactive power loss is far greater than the line loss spike.

The daily electricity consumption is about 400 degrees, and if you use electricity 24 hours a day, the loss is negligible.

Does negligible mean that the loss is small?

Yes. That's not 24 hours of electricity consumption.

The daily electricity consumption is about 400 kWh.

Three-phase four-wire power supply, the distance is 1600 meters, the load current is about 50A, with 35 square aluminum wire, ask what the voltage drop is.

Three-phase four-wire power supply, the distance is 1600 meters, the load current is about 50A, with 35 square aluminum wire, ask how much the voltage drop is Hello pro<>, first, the line voltage drop calculation formula; △u=2*i*r；where i is the line current, r is the resistance, and l is the line length. Resistivity 1, Copper is Euro*2 m2, Aluminum is Euro*2m. Second, the current calculation formula, i=p.

Third, the resistance calculation formula, r= *l s (cable cross-section mm2). Fourth, the voltage drop U=IR 5%U is enough to meet the requirements. Example:

There is a load of 30kw at a distance of 800 meters, and the 70 2 copper core cable is used to see if it meets the requirements? Untie; i=p/ 。r= l cable cross-section = ohms.

U=2*IR=2*>19V (5%U= Not compliant.) Set of 95mm2 copper core cables. 2. The single-phase power supply is zero, and the live wire (2 wires) can constitute the voltage difference, and the three-phase power supply is based on the line voltage, so it is also 2 wires.

The voltage drop can be the loss of a single wire conductor, but the front line voltage is 380V (the line and line voltage is 2 wires), for example, the voltage at the end is the voltage drop between the front end line and the line voltage minus the end line and line (2 wires), so whether it is single-phase or three-phase, the voltage drop calculation is 2 wires Ohm's law: u r*i But the load current data and the wire resistance value must be calculated in order to calculate. Resistivity of copper wire: Resistivity of aluminum wire: Example: The length of the single-phase power supply line is 100 meters, the load power of the copper core 10 square wire is 10kw, the current is about 46A, and the terminal voltage drop is obtained.

Untie; 1. Find the resistance of a single line: r l s, and find the voltage drop at the end of a single line: u ri.

3. The single-phase power supply is zero, and the fire is 2 wires, and the total voltage drop at the end is .

Summary. 1. Under normal circumstances, a motor with a 380V circuit power of 300kW needs to use more than 290 square meters of aluminum wire.

The distance is 300 meters, and the 380V power is 300 kilowatts.

1. Under normal circumstances, a motor with a 380V circuit power of 300kW needs to use more than 290 square meters of aluminum wire.

It is a single-wire aluminum wire.

300 kilowatts with 290 square aluminum cable 1, under normal circumstances, 380V circuit power 300kw motor needs to use more than 290 square aluminum wire. 2. Current = 300000 380, consider some surplus, take more than 290 square aluminum wire. 3. If the three-phase is slightly unbalanced, the current of a certain phase may exceed 570A, and the cross-sectional area of the aluminum wire is half of the current value.

Calculation: Missing Belt Through calculation, it can be seen that the distance of 70 square meters of aluminum wire, 330 meters, according to the voltage drop of about 5 percent, can drive 41 kilowatts, according to the limit of 10 percent of the voltage drop calculation, the limit load power is 82 kilowatts hail.

380V three-phase 150kw power load, capacitance compensation 60 thousand (less, should be 80 thousand short), the line disturbs only the power factor is about nearby, and the load current is back to the fierce current

i p 300m70 square aluminum wire 30m50 square aluminum wire, resistance value of each series wire:

r＝ρ×l/s＝

Voltage drop for each series wire:

When the full load leakage bridge at the end of the u ri is running, the voltage drop between the lines is

u＝37×2＝74(v)

The terminal voltage is only around 310V, so it is certain that it cannot operate normally.

380V three-phase 150kw electricity load, line length 330m, with how much overhead aluminum wire, currently with 70 square overhead aluminum stranded wire, 1 75kw motor (compensation for 60 thousand deficiency) The problem description is unknown, the motor is not described before the working conditions, starting the opening of the pin, etc., the line length of 330m is not less than LGJ-120 overhead aluminum wire, and the motor is heavy-duty with soft start or inverter start.

It is recommended to choose 240 square aluminum-core steel strands, and the compensation capacitor is installed on site, that is, the capacitor is installed at the electrical equipment. Otherwise, the line will be lost.

Underwater 380v150kw waterproof Yongxue brother magnetic synchronous servo motor manufacturer Ganzhou Shengfu Motor **** has been focusing on underwater stepper motors, waterproof DC servo motors, underwater waterproof AC servo motors, large ridge friends hole power DC brushless motors, AC permanent magnet synchronous motors, diving waterproof DC brushless motors, brushless DC motors, hub brushless motors, hub servo motors, waterproof DC hub motors, hybrid stepper progression, DC servo motors, AC servo motors, AC permanent magnet synchronous servo motor series products, planetary reducer research and development, sales and service as one of the manufacturers. Now the company has advanced production equipment, testing equipment, perfect production process, has a strong and powerful product development, design of professional and technical team, with the spirit of Chinese craftsmen, intelligent manufacturing of domestic and foreign underwater waterproof stepper motors, waterproof motors, underwater motors, submersible motors, seawater motors, deep-water motors, marine motors, stainless steel underwater motors.

380V aluminum wire 240, how many degrees of electricity loss per meter per hour for one kilometer.

Hello, the following answers to your question about the key to be identified are: Explanatory examples; Known; l = 1000m s = 240mm2 (aluminum) if the voltage u = 380v power p = 100kw (three symmetrical load) cos= acacia land resistance = find resistance r = euro current i = 100 * 1000 calculate loss p = 3i2r = 3 * answer: the line load 100kw 1 hour loss degree), each kilowatt-hour loss of wire.

After the answer, I hope it can help you.