# Power transformer working point

• posted

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

• posted

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.

• posted

| 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?

• posted

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.

• posted

The cooler it runs?

• posted

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.

• posted

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.

• posted

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?

• posted

---------------------------- "Pandiani" wrote in message news: snipped-for-privacy@d1g2000hsg.googlegroups.com...

-------------------- I never indicated that working the iron near (and below) saturation is inefficient and I didn't get that impression from others. Certainly one doesn't want it operating above the knee (in the saturation region) as then there are problems.

The working point at maximum voltage is set a bit below the knee of the saturation curve- i.e. near the top of the "linear" section. At this point, with the steels used, the exciting current is small and core losses are small- this is part of the metallurgical design of the steel and core laminations. This is making efficient use of the iron considering all factors.

Yes, losses per unit volume of core can be reduced by reducing the flux density. If the flux density is halved, there are three choices a) double the cross section of the core resulting in a decrease in core loss by a factor of nearly 2 (nearly because core will need to be longer)but also will increase the length of a turn so that the copper losses will increase (and the voltage regulation would be worse). b)leave the core cross section alone and double the number of turns . This will double the copper losses and increase the reactance and worsen voltage regulation. c) something in between-there is some optimum balance depending on the use of the transformer and the money that one wants to spend on it (One could reduce the losses to a very low level but the associated costs (monetary and ecological) of doing this would exceed the cost of losses over the lifetime of the transformer). The copper losses can be reduced by using larger conductors but this will require a larger core window so the core volume will increase- increasing core losses. However it will worsen the voltage regulation as the winding inductance (which is typically much larger than the resistance) will be increased . A 200MVA transformer would be designed to have maximum efficiency near full load so the emphasis would be on reducing copper loss and voltage regulation and it would be worked near (and below) the knee of the saturation curve. Such a transformer often has a peak efficiency over 99% so a significant improvement in efficiency will greatly increase the cost. --

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

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