# Single Phase 380v Ac Transformer

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Dear all,

I have a single phase transformer (used for resistance spot welding) with following ratings:

Rated input U: 380V AC

Freq.: : 50 Hz

Rated output capacity :150 KVA (at 50% duty cycle)

Secondary noload voltage : 24 V

Turn ratio: : 16

2. How to calculate L (inductance) of the primary side and secondary side?

3. 150kVA at 50% duty cycle means what?

Rated current = S/ U = 150 000/380 = 394.7A (why circuit breaker rated current is 225A only)

Does it mean the transformer never reach 100% of its capacity?

As far as I understand

Duty cycle = number of cycle (n) / Total number of cycles (N) * 100%.

For example, with f =50Hz frequency then N = 50

At 50% duty cycle means n =25

Am I right or wrong?

4. Power saving?

Currently, sec. current is set at 10 000 A, weld time = 20 cycles = t (ms)

S = 380V x 625A (10 000A/16 (turn ratio)) = 131 kVA

P = S x cosφ (kW)

A = P x t x 1/3 600 000 (kWh)

If I reduce current to 9000 A

S = 380V x 562.5A (9000A / 16 (turn ratio)) = 213.75 kVA

then i will save 10% of energy?

please give me some feedbacks and guides.

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Hello.

Only some short notes. Outputs are always kW not KVA. When calculating the apparent current you seems to have forgotten to divide by sq root of 2 (1.414): 150 000/380/1.414 = 279 A (but still, too big for the circuit breaker).

Yuri

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Hello yuri

Outputs are always kW not KVA.

Not where I come from, transformers are always rated in KVA, not KW. You can fully load a transformer at 10% KW if the power factor is 0.1 This is what the whole thing is about power factor correction. It allows more load on transformers and reduces losses.

Best regards,

Mark.

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Hello

1. I do not believe that it is possible to calculate the no load current from the information given. The no load current is essentially the magnetizing current of the transformer. If the full load power factor was quoted, an estimation could be made from that.

2. Same as 1. Insufficient data.

3. 50% duty cycle means ON for 50% of the time. 150KVA at 50% duty cycle will give you an average load of 75KVA. Provided that the length of the ON time is short relative to the thermal curve of the circuit breaker, you could use a smaller breaker.

4. I assume that the transformer is designed to deliver the correct voltage and current for the spot weld so you can not reduce these values. If you can establish that the voltage and/or current are higher than required for your application, you could then look at reducing the output voltage to reduce the load and energy.

Best regards,

Mark.

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Hello yuri

Not where I come from, transformers are always rated in KVA, not KW. You can fully load a transformer at 10% KW if the power factor is 0.1 This is what the whole thing is about power factor correction. It allows more load on transformers and reduces losses.

Best regards,

Mark.

Hello Mark,

- Max input capacity: 380KVA (rated input voltage 380VAC)

- Max short circuit current: 16kA

- Max secondary output voltage: 23.75 (at 7.8 duty cycle)

- PF = 0.55

When secondary side is noload (open circuit), I measure the current at the primary side so I can have no load current?

2. Same question how can I know L=? (power factor = 0.55)

3. The length of the time is only max 20 cycles.

4. The equipment use constant current control (loop control by means of primary current feedback). Current and time can be reduce, even turn ratio also can be changed. At secondary side we can change current setting via manual programming device. I think we also can reduce secondary voltage (output voltage) by changing the turn ratio.

Pls correct me iF I am wrong at anything.

Kind regards,

Hung

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• 4 weeks later...
Hello Mark,

- Max input capacity: 380KVA (rated input voltage 380VAC)

- Max short circuit current: 16kA

- Max secondary output voltage: 23.75 (at 7.8 duty cycle)

- PF = 0.55

When secondary side is noload (open circuit), I measure the current at the primary side so I can have no load current?

2. Same question how can I know L=? (power factor = 0.55)

3. The length of the time is only max 20 cycles.

4. The equipment use constant current control (loop control by means of primary current feedback). Current and time can be reduce, even turn ratio also can be changed. At secondary side we can change current setting via manual programming device. I think we also can reduce secondary voltage (output voltage) by changing the turn ratio.

Pls correct me iF I am wrong at anything.

Kind regards,

Hung

Hello Mark,

Could you pls answer remaining questions with above information.

1. How much no load current?

2. L value?

Kind regards,

Hung

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