How many lamps can be connected to a circuit of street lamps at most 250W sodium lamps 5

The maximum number of lights connected to the street light circuit needs to be calculated according to the rated power of the circuit and the power of each lamp.

Suppose the rated power of the street light circuit is 250W, and the power of each lamp is 250W sodium lamp. First, we need to determine the current rating of the loop.

According to the relationship between power and current, power is equal to current multiplied by voltage. Assuming that the voltage of the loop is 220V, then the rated current of the loop can be calculated by the following formula:

Rated current = rated power voltage.

Rated current = 250W 220V

Next, we need to determine the current of each lamp. According to the relationship between power and current, power is equal to current multiplied by voltage. Assuming that the voltage of each lamp is also 220V, then the current per lamp can be calculated by the following formula:

Current per lamp = power per lamp voltage.

Current per lamp = 250W 220V

Since the current of each lamp and the rated current of the circuit are equal, it means that the power of each lamp and the rated power of the circuit are equal. Therefore, a maximum of 1 lamp can be connected to the street light circuit.

To sum up, the street light circuit can be connected to up to 1 250W sodium lamp.

Inadequate. According to the national standard, a circuit can access a maximum of 25 300-watt street lights, and access 29 will exceed the rated load, which is easy to cause a series of serious consequences such as line overload, heat loss, and short circuit. Therefore, 29 street lights one circuit is not enough, the correct way is to divide the 29 street lights into at least 3 circuits, and each circuit is connected to 25 street lights.

Summary. Depending on your problem, there is a light for every 100 meters. The current of each lamp is about right.

Considering the cost problem, of course, the smaller the transformer, the better. It would be best if a single-phase transformer could be used. This makes it possible to put one every two kilometers.

Using 16 square copper wire, the transformer is in the middle, put one kilometer on both sides, the voltage loss is about 10V, and it can be used normally. Every four kilometers, it is necessary to use more than 35 copper wires to use normally, but the cost of this method may be expensive. It is recommended that you use a 10kw single-phase transformer and put 12 pcs.

How to wiring 24 kilometers of street lights with several transformers, a total of 240 250W lights.

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Depending on your auspiciousness problem, there is a light per 100 meters. The current of each lamp is about right. Considering the cost problem, of course, the smaller the transformer, the better.

It would be best if a single-phase transformer could be used. This makes it possible to put one every two kilometers. Using 16 square copper wire, the transformer is in the middle, put one kilometer on both sides, the voltage loss is about 10V, and it can be used normally.

Every four kilometers of the bridge to hold one, it is necessary to use more than 35 copper wires to use normally, but the cost of this rolling method may be expensive. It is recommended that you use a 10kw single-phase transformer and put 12 pcs.

The length of the street lamp circuit is 1000 meters, and every 20 meters there is a lamp (600W), and the pressure drop at the end is more than the beard and less limbs?

Answer: With 220 volt pants Lushi power supply, the voltage drop cannot be calculated, taking into account the voltage drop of the wire.

40 street lamps, each 150W sodium lamp, with a total length of 1 kilometer, requires 35 square copper cables.

According to the power formula p=ui there is:

i=p/u40x150/220

27A (calculated according to the three-phase mains, the power factor is taken.)

Check the cable ampacity table below, consider the distance, you can try to choose a 25 square two-core copper cable.

From the table, it is found that the resistance of 25 square copper wire for 1 km (1000 meters) is ohm (at room temperature 20), the corresponding voltage loss is volts, and the voltage loss rate is in line with the requirements of the specification. Considering that the resistance of the wire will increase when the temperature increases, it is better to choose 35 square copper wire.

Put your power distribution box in the middle of the lamp, 500 meters on one side, calculate, 150w*20=3000w In addition to the high-pressure sodium lamp, the real return power is generally 20% larger than the answer So your lamp is calculated according to 200w, that is, 4 kilowatts, 4 kilowatts are calculated as the line, 4000 220= current.

Then add factors such as voltage drop, 6 square lines, and use three-phase electricity, that is, 5*6 cables, it's OK, pay attention to three-phase balance.

It is recommended to use 16 square copper wire, because the line is far away, the voltage of the sodium lamp is low and cannot be lit.

200 watt lamp destroyer brigade bulb filament resistance local union: p=ui=u r, then r=u p=220 200 242 ( ).

2 stools 220 volts 200 watt incandescent lamp series filament resistance: r=484 ( ) connected to a 380 volt line, current: i=

u/r=380/484=

After plugging, the voltage of each Yuntenxun bulb is 110V

The resistance is 220 2 100 = 484 ohms next to this.

The actual selling power is 2 * 110 2 484 = 50W, which is definitely right

If you don't have a requirement for contrast, 4 square is enough!

If the illuminance of the lamp needs to meet the requirements of the version, it will not be affected.

The service life of the lamp requires a copper core of 35 square meters. This satisfies the voltage drop of less than 10%. Except for the installation of a pressure regulating device!

Input Parameters]:

Line operating voltage u = (kv).

Line type: general.

Line cross-section s = 35 (mm2).

Calculate the operating current ig = (a).

Line length l = (km).

Power factor cos =

Circuit Material: Copper.

Intermediate Parameter]:

Resistance r = (km).

Reactance x = (km).

Calculation Formula and Result]:

The general line voltage loss is:

u% = (173 / u ) ig * l * r * cosφ +x * sinφ)

4 cores are sufficient. Because the two cores are connected in parallel up to 5mm2, the voltage drop of 2500 meters is only less than 10%, which is no problem.

If it is only an LED light, it will be powered by 300V DC after rectification.

200 watt bulb filament resistance: p=ui=u r, so roll tease shouting r=u p=220 200 242 ( ) finger.

2 x 220 volt 200 watt incandescent lamps in series filament resistance: r=484 ( ).

Connected to a 380 volt line, through Ohno overcurrent: i= u r=380 484=