# Power transformer working point

Hello,
In one article I read that power transformer works very near saturation area beause of better energy efficiency. It is well known
that magnetic flux - current characteristics is nonlinear, but there is one area that can be considered linear. I wonder why is that, and how working point near saturation area can provide better efficiency.
Thank you
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It doesn't. Running high flux density as close to satruation as possible allows a transformer to be as small and cheap as possibe, but it's efficiency is not the best. The closer to saturation it runs, the higher the magnetizing current, the higher the core losses, the higher the winding losses and the hotter it runs. That's not the condition for high efficiency.
The B_H curve of the core material is the curve of the amount of flux density you get, B for the amount of magnetizing force, H. This curve has a relatively straight line portion at low values of H but curves off flattening the flux density, B as you approach saturation. When fully saturated, B = H and the core material acts like air. You don't want to run there! The straight line portion is relatively linear.
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| It doesn't. Running high flux density as close to satruation as possible | allows a transformer to be as small and cheap as possibe, but it's | efficiency is not the best. The closer to saturation it runs, the higher the | magnetizing current, the higher the core losses, the higher the winding | losses and the hotter it runs. That's not the condition for high efficiency.
What is the condition for highest efficiency?
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|WARNING: Due to extreme spam, I no longer see any articles originating from |
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wrote:

the
efficiency.
Highest efficiency occurs when the core losses and the winding losses are as low as possible. Core losses increase with core weight and flux density. But, flux density decreases with larger core area. Therefore these things are in opposition to each other.
Winding losses are due to I^2*R in the wire so the wire wants to be as large in diameter as possible and as short in length as possible. High flux density means a shorter wire but fatter wire means large core with more space for wire. Again, these things oppose each other.
It can be shown mathematically that the highest efficiency occurs when the core losses equal the winding losses at the maximum current. That's where these opposing parameters balance one another.
In general, with modern silicon steel core materials, a flux density in the region of 1.2 to 1.3 Teslas will result the highest efficiency. Exact calculations based on core configuration, allowed temperature rise, desired regulation and so on are required to get an exact answer.
If you are a glutton for punishment and want to get into a lot of mathematics see: Transformer and Inductor Design Handbook by Col. Wm. T. McLyman.
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snipped-for-privacy@ipal.net says...

The cooler it runs?
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Keith

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wrote:

To supplement what Bob Eld has said: Decreasing flux density by enlarging the core cross sectionwill reduce core loss per unit volume of core. However, tthat also increases the core volume so this tactic is a wash-and can increase the length of a turn of the winding so the resistance loss rises. Increasing the number of turns will reduce core loss and magnetising current but again the conductor I^2R loss will rise. This can be countered by larger conductors but that will result in some increase in core length.
For a given manufacturing cost then there is a balance between core and copper losses and the maximum efficiency is where these are equal. For something like a distribution transformer the average load is in the 50% range so the balance should give maximum efficiency in this range. For a large transformer the maximum efficiency should be near full load. Large station transformers will have maximum efficiency in excess of 99% while distribution transformers have max efficiency about 97-98% -- tis a matter of \$ Efficiencies can be increased by throwing more money at the problem but there is a point where the increase in capital cost exceeds the decrease in losses- and there isn't much room to increase. Going from 99% to 99.9% efficiency would be very expensive.
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Don Kelly snipped-for-privacy@shawcross.ca
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The company I worked for a long time ago now, specified a \$ value for no-load losses (core and continuously energized pumps and/or fans) and a separate \$ value for load losses. The idea being that the designer could make tradeoffs in the design to favor one or the other. For example if fans and pumps were staged on/off at different loading that would affect the load loss. A little more or less copper might determine at what load a cooler would be brought on line at some given standardized ambient temperature and allow an evaluation to be made between more or less copper vs running and extra pump or fan.
I don't have a copy of the old transformer spec or I'd look up the wording but I seem to recall some sort of present worth of a kW of loss over the life of the transformer adjusted somehow for the load factor in the case of load losses.
Transformers were tested and money changed hands depending on whether the loss guaranties were met or exceeded. This let the designer improve the efficiency if he could make money doing so.
This was done on large generator step-ups and transmission system transformers. I don't think we did heat runs on station auxiliary and substation transformers at medium voltages. I suspect that distribution transformers were type tested or sampled somehow.
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Fred Lotte
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Thanky you for the replies... I have read that almost all power transformers work near saturation area. From your reples, I think that is not true. Based on your arguments, I agree that working point near saturation does not provide good effeciency. Does it mean that power transformers (say 200 MVA) don't have working points on linear piece near saturation?