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Unit-Substation Transformers Efficiency
- Thread starter Hertz
- Start date
ozmosis
Electrical
- Oct 12, 2003
- 1,794
I'm not sure if this will cover your needs but there is a useful report from the Dutch market about "Energy saving with industrial transformers" that includes a lot of useful information. It's quite a read but some good data.
Click on the above download within this link:
Click on the above download within this link:
If you are considering single ending your sub and shutting down 1 transformer you need to put that transformer in "layup" to prevent moisture absorbtion of the insulation, that means there will be some time required to get that transformer ready to energize if you need it after a loss of the other side. One of my customers (automotive) toyed with this idea to save a little money and determined it was a overall bad idea.
Scott Peterson
Training Manager
Power Plus Engineering
Scott Peterson
Training Manager
Power Plus Engineering
Transformer losses neatly divide into two categories: no-load and load. The no-load losses are primarily iron losses related to just having the core energized. These are essentially constant regardless of load. The load losses vary with the square of the load current.
So with two transformer at 50% load versus one at 100% load, the load losses are reduced, but the core losses double.
So with two transformer at 50% load versus one at 100% load, the load losses are reduced, but the core losses double.
- Thread starter
- #5
Thanks all for your excellant comments. I remember from my college days in the last century that the maximum efficiency of the transformer is at a point where the
no-load(iron), and load (copper) losses are equal. I want to know where that point is? Is it at 100%, 90%, 50% or what?
no-load(iron), and load (copper) losses are equal. I want to know where that point is? Is it at 100%, 90%, 50% or what?
Clearly the efficiency depends upon the load and no-load loss characteristics of the transformer.
As an estimation a 2.5 MVA 11/0.433kV transformer will have approximately 2.8 kW no-load loss & 21 kW load loss.
Assuming these figues the maximum efficiency will be approximately 99.4% at unity power factor and will be obtained between 30% and 50% loading.
To be honest there is only a small a variation in efficiency over a wide loading. Between 10% and 100% load the efficiency variation approximately 98.8% to 99.4%
You should also note that the power factor of the load will also reduce the efficiency.
Maybe you should also consider the loss capitalisation costs to help you decide at what sort of level to operate your transformer at.
As an estimation a 2.5 MVA 11/0.433kV transformer will have approximately 2.8 kW no-load loss & 21 kW load loss.
Assuming these figues the maximum efficiency will be approximately 99.4% at unity power factor and will be obtained between 30% and 50% loading.
To be honest there is only a small a variation in efficiency over a wide loading. Between 10% and 100% load the efficiency variation approximately 98.8% to 99.4%
You should also note that the power factor of the load will also reduce the efficiency.
Maybe you should also consider the loss capitalisation costs to help you decide at what sort of level to operate your transformer at.
bacon4life
Electrical
The efficency depends heavily on the particular design for the transformer. For a recent 15 MVA Transformer we went out for bids on, noload/load losses ranged from 16.50/39.00=2.36 to 11.00/65.00=5.91. However, for a mobile substation application where size is a much more important factor than efficiency, the noload/load losses are 197.0/12.9=15.3.
Also, check out this Guide for the Evaluation of Large Power Transformer Losses
Also, check out this Guide for the Evaluation of Large Power Transformer Losses
Referring to Cuky2000's reply I would like to desagree with the efficiency formula in the picture. According to my opinion and to some (at least!) litarature, in the denominator there is a fault! If the efficiency is calculated as per ANSI practice - i.e. the rated power is the output -, all the losses (n2.Psc+Poc) must be added to the rated power and a term is missing; for IEC - the rated power is the input - no losses are included in the denominator, so Poc is to exclude.
There is somebody with me about this?
There is somebody with me about this?
There are differences between IEEE Transformer efficiency calculations and IEC. I have a paper "somewhere" that describes the differences, but finding it might take some time. I do recall that IEEE efficiencies are generally lower than IEC version.
I think IEEE requires that all parasitic loads, such as cooling fans and pumps be included in the input power, while IEC does not include these. But this difference has nothing to do with the subject at hand.
I think IEEE requires that all parasitic loads, such as cooling fans and pumps be included in the input power, while IEC does not include these. But this difference has nothing to do with the subject at hand.
Hertz
The graph above (Good graph Cuky) shows the effeciency well, at very light loads the no-load losses (iron) bring down the effeciency drastically.
The load losses are I2R losses therefore load dependant, iron losses are considered constant losses and are typically 0.5%.
For a 2.5MVA transformer, you could expect thye manufacturer to quote load losses of 25-28kW at 75C
You dont mention the transformer type, if you have an indoor compact substation with AN/AF cast resin transformers then you could load the transformers up to 70% of their nominal rating, that way you can support the full substation load from one transformer by force cooling the unit, otherwise if you have ONAN or AN type without forced cooling then you can load to 50% as you noted.
The graph above (Good graph Cuky) shows the effeciency well, at very light loads the no-load losses (iron) bring down the effeciency drastically.
The load losses are I2R losses therefore load dependant, iron losses are considered constant losses and are typically 0.5%.
For a 2.5MVA transformer, you could expect thye manufacturer to quote load losses of 25-28kW at 75C
You dont mention the transformer type, if you have an indoor compact substation with AN/AF cast resin transformers then you could load the transformers up to 70% of their nominal rating, that way you can support the full substation load from one transformer by force cooling the unit, otherwise if you have ONAN or AN type without forced cooling then you can load to 50% as you noted.
Griseo,
Thanks for the sanity check. However,the simplified formula gives sufficient accurate values for practical purposes.
For instance, in the example above for loading factor range up to full load and PF=1, the worst deviation between the two formulas is less than 2x10-5%.
Therefore, the term n2Psc can be neglected.
NOTE: If an accurate formula is required, the efficiency may be calculate with more precision as follow:
n= {1 - (Poc +n^2Psc)/[n.S.PF(1-U[φ]/100)+Poc+n^2.Psc]}100%
Where: U[φ] = The voltage regulation determined as the difference between the rated voltage of a winding and the voltage which occurs at particular values of load and power factor when rated voltage is applied to the other winding or windings.
Thanks for the sanity check. However,the simplified formula gives sufficient accurate values for practical purposes.
For instance, in the example above for loading factor range up to full load and PF=1, the worst deviation between the two formulas is less than 2x10-5%.
Therefore, the term n2Psc can be neglected.
NOTE: If an accurate formula is required, the efficiency may be calculate with more precision as follow:
n= {1 - (Poc +n^2Psc)/[n.S.PF(1-U[φ]/100)+Poc+n^2.Psc]}100%
Where: U[φ] = The voltage regulation determined as the difference between the rated voltage of a winding and the voltage which occurs at particular values of load and power factor when rated voltage is applied to the other winding or windings.
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