The T.M.Haja Sahib Transformer efficiency thread

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jusme123

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NY
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JW
......your not allowed to argue here, your post will be censored or deleted. Stay on topic please.
 
T

T.M.Haja Sahib

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If I bid for a project, the overwhelming probability is that I will be in competition. Possibly with competitors and/or other projects for funding.
If I price for larger than required transformer ratings for the application the chances are that Joe Bloggs or John Doe, will be able to undercut my offer and maybe win the business that we could otherwise have had.
You would be quoting to a set of specs prepared in advance taking into account present load,required spare capacity,maximum efficiency operating point etc.,and so you may quote confidently expecting the award of work to you.So be ready.
 

Besoeker

Senior Member
Location
UK
You would be quoting to a set of specs prepared in advance taking into account present load,required spare capacity,maximum efficiency operating point etc.,
Actually, that's not generally the case. Many of specifications, particularly those drawn up by consultants, are pretty vague on ratings and much else.
An actual example:
"The earth wire must be adequately rated for the use intended taking into account potential fault levels."
If they had omitted that requirement and I chose to make it inadequately rated, would I then be still be spec. compliant?
Anyway, since I'd be supplying the the transformer and can make a reasonable guess at the supply fault level, I come up with a rating for that.

Transformer rating is covered by this:
"All equipment supplied under this contract shall be suitably rated for this application."
So it's up to the bidder to determine the rating. Same with efficiency. The spec doesn't tell you what it has to nor at what duty point. But you have to provide guaranteed efficiency figures at the bid stage. And harmonics consideration of which may well determine the most suitable transformer configuration. Again, that's not specified as a rule. It's for the bidder to decide.
 

RichB

Senior Member
Location
Tacoma, Wa
Occupation
Electrician/Electrical Inspector
HUH??????? Isn't an "oversized" transformer the same as a "higher sized"---oversized is bigger than needed and higher sized is also bigger than needed--kinda like "Teacher I don't understand--My house burn down--or My house burn up--still no house right?"
 
T

T.M.Haja Sahib

Guest
Actually, that's not generally the case. Many of specifications, particularly those drawn up by consultants, are pretty vague on ratings and much else.

Then I can not help you to win this bid.
 
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Besoeker

Senior Member
Location
UK
Then I can not help you to win this bid.
I really had no such expectation and, anyway, my post wasn't specifically about any particular bid.

The thread is about transformer efficiency. Yes, there is a specific operating point at which a transformer runs at it's highest efficiency.
But efficiency isn't the whole story. Efficiency is a ratio. For least cost operation, you need the lowest losses in absolute terms.
 

topgone

Senior Member
I didn't think there was. I said it was only new to me. I was, and am still, sure it would be old hat to someone such as yourself, that fully understands the physics. The reason I brought it up was no one was discussing the math models.


At my level of understanding, this is not readily apparent to me by inspection. I'm sure it is a simple derivation - its nothing I would ask you to do here. I'll see if I can work this one out.

.... You will only achieve maximum efficiency when the Cu losses equal the Fe losses. Per example, the total losses = 3960.7 X 2 = 7,921.4 watts; maximum efficiency = 400/(400+7.9214) = 0.98.058 or 98.06%. ...
I don't get this one at all. The efficiency calc you use is for full load - 400kw. But the coil losses you are using are less than that listed for full load. ???


Again, at my level of understanding, this is not readily apparent to me by inspection. This may be a simple, but I suspect, tedious derivation. Perhaps you could list a reference.

I'm a bit surprised the maximum efficiency is not influenced by the ambient. Is that a third order effect and not part of your model? Or not even relavent?

ice

It would be easy to grasp if you read some ideas on "maximum efficiency of transformers".

Google is my friend. It can be yours too. Read here.
 
T

T.M.Haja Sahib

Guest
But efficiency isn't the whole story. Efficiency is a ratio. For least cost operation, you need the lowest losses in absolute terms.
Surely,higher the efficiency lower is the loss in the transformer.
So I think you mean net cost to the owner of the transformer?
 

Besoeker

Senior Member
Location
UK
Surely,higher the efficiency lower is the loss in the transformer.
Efficiency isn't the whole story. A larger transformer might have a higher efficiency but losses can be greater than those of a smaller transformer depending on loading. As an extreme example to illustrate the point, 50kVA of loading on say, a 75kVA transformer is likely to result in significantly lower losses than if that 50kVA load was fed from a 7,500kVA transformer
 

jaymack

Member
For maximum efficiency of a transformer, the iron losses equals the iron losses. this won't occur at full load. It is usual however to select a distribution transformer that has varying loads, based upon the restrictions of transformer regulation on load, allowance for future capacity and the fault level etc.

Transformers can be run overload condition for a period of time, guidelines are available; Montsinger's law states that the rate of deterioration of mechanical properties for paper insulation, doubles for each 5 -10?C, although this doubling factor is usually rounded off to 6?C.

Regards
 
T

T.M.Haja Sahib

Guest
Efficiency isn't the whole story. A larger transformer might have a higher efficiency but losses can be greater than those of a smaller transformer depending on loading. As an extreme example to illustrate the point, 50kVA of loading on say, a 75kVA transformer is likely to result in significantly lower losses than if that 50kVA load was fed from a 7,500kVA transformer
The trick is to use a higher size transformer so that the net cost to the owner is lower (taking into consideration future load expansion and maximum efficiency point not lower than that for present load) than using a lower size transformer catering to the present load only.
 

Besoeker

Senior Member
Location
UK
The trick is to use a higher size transformer so that the net cost to the owner is lower (taking into consideration future load expansion and maximum efficiency point not lower than that for present load) than using a lower size transformer catering to the present load only.
What if the transformer is designed for purpose and future expansion isn't a consideration?
 

Besoeker

Senior Member
Location
UK
Then it is not of modern electrical design.
You seem to have completely missed my point:

"What if the transformer is designed for purpose and future expansion isn't a consideration?"

Here's a selection of a few of the lines* of my design data for one of the Static Scherbius systems we built:

ISK05.jpg


The system parameters are defined which in turn sets the transformer rating, voltage, and configuration. The transformer is an integral part of the system and thus designed for purpose.

What's not modern about this transformer design? How would you personally have done it differently?
 
T

T.M.Haja Sahib

Guest
Bes:
How did you arrive at maximum KVA of the transformer i.e 471 KVA in the wind mill example of your last post in this thread?Please put up the principles here.
 

Besoeker

Senior Member
Location
UK
Bes:
How did you arrive at maximum KVA of the transformer i.e 471 KVA in the wind mill example of your last post in this thread?Please put up the principles here.
Wind mill?? Wot wind mill, guv??
Sometimes Google is not your friend. Using it as a search tool can make you appear knowledgeable sometimes and foolish at other times.

This application was a wound rotor variable speed drive on a centrifugal fan. Now, although you have previously told me that I don't understand motor speed control and may still be of that opinion, I'll provide the explanation of where the 471kVA comes from anyway.

First a few fundamental points.
  • This type of variable speed drive is sometimes known as a slip recovery system and the speed control is on the rotor side.
  • For a centrifugal load, torque is proportional to the square of the speed and power to the cube of speed.
  • Rotor voltage is proportional to slip.
  • Motor torque is proportional to rotor current*.

From this you can deduce that, at the bottom end of the controlled speed range, rotor voltage is maximum and rotor current is minimum.
As the speed increases the rotor voltages reduces and the rotor current increases. For any given speed you can thus calculate rotor power. This peaks at 15% of rated drive power.
The rotor power is fed back (recovered) to the supply via a mains commutated inverter operating at fixed firing angle. This firing angle, which is set to give a safe commutation margin, determines the power factor of recovered power and hence the required kVA of the recovery transformer. That, in a nutshell, is how I got the 471kVA.

Some curves:

ISK06.jpg


*It's a fairly good approximation but I usually use the value calculated from the motor equivalent circuit if I have it. In this particular case the approximation would have resulted in an error of rather less than 1%.
 
T

T.M.Haja Sahib

Guest
Bes:

Thanks for your reply.I thought,since doubly fed induction motor is also used in wind mill application,your post related to it.

But you still did not explain why maximum capacity i.e 471 KVA for the transformer is required?What about the minimum capacity of the transformer in this case?
 

Besoeker

Senior Member
Location
UK
Bes:

Thanks for your reply.I thought,since doubly fed induction motor is also used in wind mill application,your post related to it.
You assumed.

But you still did not explain why maximum capacity i.e 471 KVA for the transformer is required?
I did.
"As the speed increases the rotor voltages reduces and the rotor current increases. For any given speed you can thus calculate rotor power. This peaks at 15% of rated drive power.
The rotor power is fed back (recovered) to the supply via a mains commutated inverter operating at fixed firing angle. This firing angle, which is set to give a safe commutation margin, determines the power factor of recovered power and hence the required kVA of the recovery transformer."
That works for any rating of this configuration.

What about the minimum capacity of the transformer in this case?
If you mean what is the lowest operational duty, then it is zero. You can clearly see that from the curves.
 
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