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How many kilowatts can a 1250kva transformer carry, do you know?

# How many kilowatts can a 1250kva transformer carry, do you know?

• Categories:Industry News
• Author:
• Origin:
• Time of issue:2018-11-28
• Views:45

(Summary description)Even if the 1250KVA transformer provides all active power, that is, KW, it is only 1250KW. Of course it is impossible to afford it.

# How many kilowatts can a 1250kva transformer carry, do you know?

(Summary description)Even if the 1250KVA transformer provides all active power, that is, KW, it is only 1250KW. Of course it is impossible to afford it.

• Categories:Industry News
• Author:
• Origin:
• Time of issue:2018-11-28
• Views:45
Information

Even if the 1250KVA transformer provides all active power, that is, KW, it is only 1250KW. Of course it is impossible to afford it.

If the equipment is running at the same time, the active power is 3115KW, all of which are inverters, then the power factor is estimated to be quite low, assuming 0.8 is good, then at least the transformer capacity is required to be at least 3115/0.8

KVA, generally speaking, the transformer needs a margin, about 80%, so if the power factor is 0.8, the transformer efficiency is 80%.

The load of a 1250Kva transformer is 1000 kilowatts.

The load of a transformer is generally 0.8 times the capacity of the transformer, 1250x0.8=1000kw, so the load of a 1250Kva transformer is 1000kw.

The transformer is used as a load, and it runs smoothly and economically under small load conditions. When the load factor of the transformer is between 63% and 67%, the efficiency of the transformer is the highest.

Transformer: It is a device that uses the principle of electromagnetic induction to change the AC voltage. The main components are the primary coil, the secondary coil and the iron core. The main functions are: voltage transformation, current transformation, impedance transformation, isolation, voltage stabilization, etc.

Its rated current is the capacity divided by the voltage multiplied by six divided by ten. The specific load needs to be P=root number 3 multiplied by line voltage multiplied by line current multiplied by power factor. The 400V side current is about 1800A. On the other side, you have to calculate it yourself, and no voltage is provided.

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