Hello everyone, I have an 800Kva transformer in my unit in a hurry

There should be two reasons, the sampling current of the reactive power compensation meter of the general low-voltage distribution room is only one of the phases, most of them are phase B, so the power factor displayed is phase B, of course, the power factor of phase A and C may be lower than phase B, or higher than phase B, when the power factor of phase A and C is lower than phase B, then the total reactive power will be larger, on the contrary, phase A and C may be overcompensated, and the reactive power will also increase; In another case, as mentioned on the first floor, the capacitance of several channels of reactive power compensation can be selected to be smaller. In short, in order to achieve a truly comprehensive reactive power compensation, the instrument should choose three-phase current detection, and the compensation capacitor connection method adopts a mixture of star and triangle, and the capacity of each channel is selected according to the actual situation.

The capacitance chosen is too large. If you don't vote, it's too low, and if you don't vote, it's too big. Use a smaller capacitor.

A transformer with a capacity of 800 kVA can load 640 kW of electrical appliances.

On the one hand, the transformer itself has losses, and on the other hand, part of the margin should be reserved for safety reasons. Therefore, the transformer generally takes 80% of its rated capacity as its load power, that is, 800x80% = 640 kilowatts.

A transformer is a device that uses the principle of electromagnetic induction to change the alternating voltage, and the main components are the primary coil, the secondary coil and the iron core (magnetic core). The main functions are: voltage conversion, current conversion, impedance conversion, isolation, voltage regulation (magnetic saturation transformer), etc.

According to the use, it can be divided into: power transformer and special transformer (electric furnace transformer, rectifier transformer, power frequency test transformer, voltage regulator, mining transformer, audio transformer, intermediate frequency transformer, high frequency transformer, impact transformer, instrument transformer, electronic transformer, reactor, transformer, etc.). Circuit symbols often start with t.

Examples: T01, T201, etc.

Transformer load 70 80 is the most economical, in addition, pay attention to the power factor to be high, low will also affect the load capacity, can be the best, the lowest also.

It depends on the nature of your load, if the inductive load is large (e.g. electric motor, lamp with rectifier), then there is a power factor problem, (con).p= 3 (root number) * v * i * con * efficiency). Generally speaking, the power carried by the transformer is about half of the rated power, which belongs to economic operation.

It is necessary to determine the nature of electricity production, if the metallurgical textile cement plant industry is 800kw, if the mechanical processing is twice the calculation, 1600kw. If it is a light industry, it is inverted seven folds, that is, about 1140kw, this is an empirical algorithm, maximize the use of transformer capacity, don't worry about the transformer capacity is not enough.

There is a general formula for your reference:

1. Single-phase transformer: the input current on the high-voltage side (the line current is also the phase current) I1=PN U1N (A), wherein: PN is the rated capacity of the transformer (volt-ampere), and the rated voltage (volt) on the high-voltage side of the U1N.

The output current of the low-voltage side is i2n=pn u2n (ampere), where: the rated voltage of the low-voltage side of u2n (volts).

2. Three-phase transformer: the rated input line current on the high-voltage side is i1=PN U1N 3 (A), wherein: PN is the rated capacity of the transformer (volt-ampere), and U1N is the rated line voltage (volt) on the high-voltage side.

The output line current on the low-voltage side is i2n=pn u2n 3 (ampere), where: u2n is the rated line voltage (volts) on the low-voltage side.

3. Take 800kva as an example, if the high-voltage rated line voltage is 10000 volts. Three-phase transformer with a low-voltage rated line voltage of 400 volts: the input line current on the high-voltage side is i1=800*1000 10000 amperes; The output line current on the low voltage side is i2=800*1000 400 amps.

You can do the rest.

To add, the rated power of the transformer refers to the apparent power, generally newly installed, without considering the reserved power capacity, the load inductance accounts for more of the load rate, not much load rate can be the most economical load rate, the load rate based on lighting can be 1, the maximum current in actual use should not exceed the rated current.

It is wrong to calculate by kilowatts, 800kva transformer is safe to carry current, if you are a motor load, no kw is calculated according to 2a, it can carry 570kw load.

800kva transformer:

1. The rated current of the primary side of the high voltage is: i=p (3 u)=800 (. (Rated voltage 10kv).

2. The rated current of the secondary side of the low voltage is: i=p ( 3 u)=800 (。 (Rated voltage.)

This is related to the power factor, if it is 1 it is 800 kilowatts, 0,9 is 720 kilowatts, and 0,8 is 640 kilowatts.

The load relationship of the transformer is related to the power factor, the optimal load of the transformer is between 60% and 80%, nothing is considered, according to the current rating.

800x transformer efficiency x load power factor.

800kva* is the kw you want, and then 65-75% of the rated current of the transformer is the load that the transformer can carry. You are 800*08*075 here, and the rated capacity of the general transformer is multiplied by the secondary current, and your secondary current is 1200A

800kVA is the rated apparent power of the transformer, divided by the rated voltage of both sides is the rated current on both sides.

The apparent power multiplied by the power factor of the load equals the active power.

Generally, the transformer works at 70% load.