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No load condition of energized tranformwer

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HamidEle

Electrical
Feb 20, 2007
309
We are shutting down the major loads on 5MVA transformer (25kV/4.16KV), which will be slightly loaded eventually(almost no-loaded). The power factor at the primary feeder will be higher because of the no-load condition. Do we need to correct the power factor for this feeder because of the shedding the loads? To me, the power factor for 25kV system will be higher than before. Are there any other negative impact because of this unloaded conditions?

Do we need to reduce the no load current in an egergized transformer?

Thanks in advance for any inputs.
 
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If you have existing PF correction, you may have to make an adjustment since you are subtracting some inductive load, or you may be over-compensating.
 
The transformer no-load current is going to be whatever it is - there is no way to change that unless you change the excitation voltage.

Check the load power factor at your new loading condition and see if it too low or too high.

 
We got a call from a customer regarding a 90% penalty for a 10% power factor. It turned out to be a building that they had stopped using. They turned off all the loads. The total draw on the primary metering was the transformer losses and the exciting current. The good news was that it was a large percentage penalty on a very small power bill.
respectfully
 
The problem we are having is, we would have to maintain the minimum small loads. There are no ways to shut down the main transformer.
 
I am looking for the typical data of power factor of unloaded(or slightly loaded) 5MVA transformer. Is it 10%? Is there any curve for this?
 
The power factor will probably be close to zero percent. If you measure the no-load current you can determine the no-load KVA. This will be almost pure KVARs. Use this figure to calculate size of a capacitor bank to improve the power factor. Don't go over 100% correction.
respectfully
 
It may not be as bad as you're thinking. The core losses require real power input.

If you have the transformer test data, you can probably calculate the no-load power factor.
 
The no-load current of such large trafos is around 1% of FLC. Your original test certificate should contain this info.

You convert the primary no-load current to secondary current (acc to volt ratio) and use a capacitor on the secondary to compensate this current.

Doesn't your primary side have metering where you can read the pf directly ?

*Why a man thinks he outrun a chasing dog when it has twice as many legs?*
 
There is a misconception that no-load power factor of transformers is very low nearly zero.But actually no-load power factor of transformers is pretty high 0.8-0.9 for both distribution and large power transformers.No-load current of Power Transformers is quite low 0.05-o.1% of full load current.For small power transformers below 10 MVA this willl be 0.1-1 %.For pole mounted trnsformers and small distribution transformers it will be 2-3 %. From this we can understand the load KVA (as % of rated kVA ) on primary feeder by an unloaded transformer.It will be same % as of no-load current.

The above figures can be verified from transformer test reports.Net pf on the grid due to transformer depends on load pf plus no-load pf.But the effect of load pf alone will be significant as the no-load kVA is negligible.So load pf improving capacitors can be switched off during no-load operation.
 
PPC,
The power factor is very low because the only loss is from the core loss of the transformer, which is almost purely inductive.

To reduce no-load current in an energized tranformer, a two terminal control and power swicthing circuit for reducing electric current flow in the primary winding when no load is connected can be utilized, which is called Triac, in series with the power transformer primary and utility. But I am not sure about the price range and practicablity about this.
 
HamidEle,

Any actual energy loss in the transformer must result in in-phase current, not inductive.

I believe prc is correct in stating that the transformer no-load current will not be highly inductive.

In reviewing transformer test data for a 12/16/20 MVA 110 kV transformer, I come up with a power factor of about 0.85 for no-load current, which is 0.18% of the rated current.

The main concern with a lightly loaded transformer is generally the core losses, since these are constant regardless of transformer load.
 
The only power factor correction I'd consider is whether or not I had to remove a capacitor from the feeder due to the load reduction and a possible leading power factor condition at the feeder level. Otherwise, if the 1% of rated capacity no-load losses on a 5 MVA power transformer result in 80% power factor at this point you're talking 30 kVAr - hard to even notice on a 25kV feeder.
 
If my understanding is correct. Core loss is purely resistive, exciting curerent causes reactive loss.
 
It's all exciting current, in reality.

Any true loss (watts) has to be related to current in phase with the voltage. The term "reactive loss" is not accurate. For a hypothetical pure reactance, there would be no loss.
 
Thanks,DPC,
But there is still a power consumption due to excitation current.
 
I agree with dpc. The no load loss is real, even though it is not caused by the current flowing through the winding. The current still has to have the phase relationship to the voltage as indicated by the power factor (no-load loss in % of rated divided by exciting current in % of rated)

Reactive loss is a bad term because it is generally used for the load current flowing through the transformer reactance.

In any case, as far as feeder overall power factor is concerned, the current may be exciting, but it not significant. [rofl]
 
The no-load current(Ie) is not purely real. It consists of two components: the magnetizing current(Im)and the core loss(Ih). Im lags the applied voltage by 90Degree, while Ih is in phase with the applied voltage. Ih is very small in comparison with Im, and Im is nearly equal to Ie.
 
HamidEle,

As mentioned a couple of times previously, your assumption that Ih is very small compared with Im is not supported by the facts as shown in transformer test data.

But more importantly, what is your point? No one said that Ie was purely real.
 
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