How big a line should be used for 5000W, how much A is open in space, how to calculate...

Assuming the voltage is 220V) 5000W uses 4 square millimeters.

The copper core wire, the empty open.

The calculation is as follows:

p=ui，i=p÷u=5000÷220=

According to the wire ampacity.

Choose 4 square millimeters of copper conductors.

Although the square millimeter copper core wire also meets the requirements, but the surplus is small, so the 4 square millimeter copper core wire is selected).

The current carrying capacity of the square copper core wire: 16a-25a

4 square copper core wire current carrying capacity: 25a-32a

5 kilowatts divided by 220V to obtain the current, according to the current line selection and air opening (greater than the rated value 2-3 times).

Are you a civilian point of 220v, or a 380? If you don't explain, how to choose a thread for you.

If 150 kilowatts is the normal operation of the full load, the current can reach 400 amperes, taking into account the starting impulse current and other factors, the air opening of 630 amperes should be selected, and the unit current carrying capacity of the aluminum wire is 2 after the cross-sectional area of 100 square millimeters, that is to say, the aluminum wire of 240 square millimeters can meet the requirements, and the copper wire can be reduced to 180 square millimeters.

However, if the 150 kW you mentioned is not running at full load, that is, the combined equipment is not running at the same time, then the standard should be appropriately lowered as needed.

In the same time, the work done by the current through different electrical appliances is generally not the same. For example, the work done by an electric current passing through the motor of an electric locomotive is significantly greater than that done by the motor passing through an electric fan in the same amount of time. In order to represent how fast or slow an electric current does work, the concept of electrical power is introduced into physics.

The work done by an electric current per unit of time is called electrical power. Electrical power is expressed by p, p=w t, and w = u i t (i.e., voltage multiplied by current multiplied by time), so p=ui.

Assuming that the voltage is 220V) 5000W to use 4 square millimeters of copper core wire, the empty opening.

The calculation is as follows:

p=ui，i=p÷u=5000÷220=

According to the current carrying capacity of the wire, a copper conductor of 4 square millimeters is selected.

Although the square millimeter copper core wire also meets the requirements, but the surplus is small, so the 4 square millimeter copper core wire is selected).

The current carrying capacity of the square copper core wire: 16a-25a

4 square copper core wire current carrying capacity: 25a-32a

When there is a short circuit or overload phenomenon in the household electricity, the air opening will automatically trip and cut off the power, which can protect the electrical appliances from damage and power accidents, so how much air is used for 3000w?

1. How much space does 3000w use?

It is recommended to use 20A to open empty, 3000W 220V =, that is, the current is around the right, but because the electrical appliance has a large current in the early stage of opening, to prevent it from being burned out, it is necessary to use 20A. If the air switch is too small, the working wattage of the appliance is too large, and the switch will automatically trip. The air opening option is too large, in case of current overload of the electrical appliance, the air opening cannot automatically trip.

In severe cases, the bridge can also lead to fire.

Second, the purchase skills of air switches.

1. Before purchasing the empty opening, the user should figure out the entrance switch and the loop opening and closing in the air, look at the total fixed current, and select the minimum rated current value; When choosing an air opening, it is recommended to adhere to the principle of choosing from large to small, and if it is used at home, it is necessary to consider whether it can carry the current as a whole, such as the total rated current of the living room, kitchen, and bathroom.

2. The main purpose of the design of air opening is to ensure the safety of electricity, if the current is too large, once it can not be beared, the air opening will trip, so when you buy, you should check whether the air opening can be tripped in a short time when the current is too large, so as to avoid the occurrence of electric shock accidents.

3. When you choose the air open, see if the model matches it, there are several models of air switches, among which the smallest model is DZ5, which is suitable for circuits with a rated current of 50A, and the largest model is DZ47, which is suitable for a rated current of 60A.

Conclusion of the article: The relevant knowledge about the empty opening will be introduced to you here, I hope it can help you. If you want to know more about related knowledge, please continue to pay attention to the Qeeka Home information platform, and more exciting content will be presented to you in the future.

For 6000 watts of single-phase 220V AC, according to the power calculation formula p ui (p is power, u is voltage, i is current), its passing current is 25A. There are certain differences in circuit application requirements, so it is necessary to choose an air switch with 220V 30A.

The initial level of 380VAC is 6000W, and the current flowing is about 13A, and the air switch of 380V 16A should be selectedAir switch is a very important electrical appliance in the system of low voltage switch network system. It combines control and multiple protection functions into one.

Not all electrical appliances in the home need to install sockets, and it is more dangerous to install them, and the old electrician will teach you step by step, practical electrician technology!

If it is a household single-phase water heater (rated voltage 220V), 6000W 220V, use a 30A circuit breaker. If it is an industrial three-phase water heater, 6000W 380V, use a 25A circuit breaker.

Method 1: 250A three-phase empty open at home.

Method 2: There are two water heaters, then buy two empty ones, one is 60a.

63a's is sufficient, or two 32a's respectively.

13000w with a few square lines, how much space opens.

Dear, I have received your problem, the solution to the related problem is as follows, I hope to be able to help you 13000w with a few square wires, how much empty open answer is as follows: 13000 watt induction cooker needs square wires 13000 watts = 13 kilowatts First calculate the current of 13 kilowatts of power cable current i = p (calculated with the standard three-phase mains voltage, cos is the load power factor, the general motor takes ). The rated current that can be borne by each square copper wire is 6A, so the square number of copper wire = 26 6 = flat high square. Conclusion:

Air open 50Hz, 400V6000A inlet wire with how big the wire.

According to the current current information of this town, with 50Hz current, 400V voltage, 6000A incoming wire, you can use a 4 m bottom line for power supply. According to the standard "Installation of Feeding Power Wires and Cables", the absolute standard current of the cross-linked copper wire of the lowest low-voltage wire is 6000A, and its minimum wire area should be more than 4 m. Travelers therefore supply power in the above case in Senga, and you can use a 4 m bottom line for power supply.

Make sure you use the right grade of wire and install it correctly to ensure safe use.