Transformer Efficency Curves

[elecguy]

I'm looking for old transformer efficiency curves. Building was built in 1970 so basically between 1970-1990. trying to make a comparision between the old transformers and the new. Anyone know where I can get some? Looking for all types of kVA ratings.

thanks

 

[templdl]

I think what you should request are the losses which should be published values from the transformer manufaturer for that transformer. There should be NL losses, FL, 3/4, 1/2, 1/4 load and total losses.

 

[gar]

120104-2138 EST

Look in "Alternating-Current Machinery" by Bailey and Gault, McGraw-Hill, 1951, p16 thru 22 for a theoretical discussion. They provide a theoretical discussion and show a graph of a 10-kva unit with 98% efficiency at 60% load of unity PF. In the page 20 example at 10-kva load the copper loss is 181 watts and the core loss is 61 W producing 97.6% efficiency at full load.

To be more efficient you need less copper, and lower core losses. This means more expensive core material.

.

 

[jim dungar]

Everything depends on the loading.

Prior to the latest government regulations, transformers were designed to be efficient at about 70-80% loading.
Then someone noticed that nobody actually loads transformers much past about 30-40% (think offices and schools after hours), so enter the TP-1 regulations which forced manufacturers to design transformers to be efficient at 35% for about 75% of the time.

 

[T.M.Haja Sahib]

I'm looking for old transformer efficiency curves. Building was built in 1970 so basically between 1970-1990. trying to make a comparision between the old transformers and the new. Anyone know where I can get some? Looking for all types of kVA ratings.

thanks

A modern distribution transformer has maximum efficiency between 50% to 75% of its loading.So it is beneficial to load it in that range.

 

[kingpb]

A modern distribution transformer has maximum efficiency between 50% to 75% of its loading.So it is beneficial to load it in that range.

Could you please furnish the technical reference(s) for which you are referring.

 

[gar]

120105-1904 EST

You can create your own efficiency curves.

efficiency = power output / power input = power output / (power output + transformer losses)

Transformer losses = I^2*R losses + core losses

Core losses at a fixed input voltage can be considered constant independent of load as a first approximation. Whereas load current losses are a function of load.

If you want lower core losses there are several things to do:
Reduce the amount of iron,
Reduce the flux density,
Reduce lamination thickness (raise its resistance),
Reduce the frequency, and
Change the core material.

Some might require more turns, meaning more copper resistance.

If without changing the copper losses you could lower the core losses, then the peak efficiency point can be at a lower % of full load. If core losses could be made zero, then where is the peak efficiency point?

.

 

[rcwilson]

The old Westinghouse Consulting Engineering Application Guide had a table of transformer losses, no load and full load. I have a 1999 version (Eaton /Cutler-Hamer) that has a couple pages of data. PM me with your e-mail and I'll scan and send it.

It will be better than nothing.

 

[templdl]

rcwilson,

This is what you are refering to:

http://www.eaton.com/Electrical/Consultants/ConsultingApplicationGuide/index.htm

Then click on section 19 which will be a document that includes examples of efficiencies.

 

[T.M.Haja Sahib]

Could you please furnish the technical reference(s) for which you are referring.

I took it from

'Energy conservation in steel rolling mills' from ABI & ABI Publishers,India.

The following example also.

Suppose you have two 1000kva transformers and load is 500kw.It is better efficiency to operate the two transformers in parallel than using one transformer to supply the load.

 

[T.M.Haja Sahib]

I took it from

'Energy conservation in steel rolling mills' from ABI & ABI Publishers,India.

The following example also.

Suppose you have two 1000kva transformers and load is 500kw.It is better efficiency to operate the two transformers in parallel than using one transformer to supply the load.

Sorry,there is a mistake.The load is 1000kw and not 500kw.

 

[jim dungar]

I took it from

'Energy conservation in steel rolling mills' from ABI & ABI Publishers,India.

The rules, regulations and standard practices in the United States are certainly different than industry practices in India.

In 2005 the US passed an Energy Policy Act that required all general purpose transformers sold after 2007 to be designed to a target efficiency of a '24-hour average loading of 35%'.
There are also provisions, for non-general purpose transformers as well as large (i.e. >500kVA) units.

 

[ron]

In 2005 the US passed an Energy Policy Act that required all general purpose transformers sold after 2007 to be designed to a target efficiency of a '24-hour average loading of 35%'.
There are also provisions, for non-general purpose transformers as well as large (i.e. >500kVA) units.

Isn't it unfortunate that the actual loading of an average transformer is likely 50-60% during the day and 10-20% at night, but the standard requires 35% which is not a value that the transformer actually runs at. I'm not sure what the standard should have been, but picking 35% seems wrong.

 

[T.M.Haja Sahib]

The rules, regulations and standard practices in the United States are certainly different than industry practices in India.

In 2005 the US passed an Energy Policy Act that required all general purpose transformers sold after 2007 to be designed to a target efficiency of a '24-hour average loading of 35%'.
There are also provisions, for non-general purpose transformers as well as large (i.e. >500kVA) units.

Globalization process is on throughout the world.So The rules, regulations and standard practices etc., are bound to converge.

 

[kingpb]

Isn't it unfortunate that the actual loading of an average transformer is likely 50-60% during the day and 10-20% at night, but the standard requires 35% which is not a value that the transformer actually runs at. I'm not sure what the standard should have been, but picking 35% seems wrong.

That 35% is a 24-Hr average. If you take 60% for 8 hrs, and 20% for 16 Hrs, you get a 24Hr avg of 33.3%. Pretty darn close.

 

[ron]

That 35% is a 24-Hr average. If you take 60% for 8 hrs, and 20% for 16 Hrs, you get a 24Hr avg of 33.3%. Pretty darn close.

I don't know how linear the transformer efficiency curve is to know if at 10% for a majority of the day it is really terrible, to result in the real measured efficiency over a day would average to 35% (or significantly less) as compared to only measuring the efficiency at 35% for 24 hours.

 

[zog]

Globalization process is on throughout the world.So The rules, regulations and standard practices etc., are bound to converge.

I hope so, then maybe we can get some manufacturing jobs back in the USA.

 

[gar]

ron:

My post at #7 tells you how to calculate the points on a transformer efficiency curve.

In a somewhat different form this is:

efficiency = K*i / (K*i + i^2*R + Cl)

where
K is a constant
i is the load current variable (varies from 0 at no load to some maximum values at full load)
K*i is the power output and the power input excluding losses
i^2*R is the copper loss
Cl is the fixed core losses that are obtained by measuring the power input at no load,
core losses are assumed constant independent of load

If you plot this curve for a transformer that has some core losses, then efficiency starts a 0 with no load increases to a maximum at some load current and then starts to drop off with increasing load.

Given that you know the no load loss, and full load copper loss, then you can plot the curve for a particular transformer.

The reference provided by templdl has this loss information. If you use some of this data you will see rather high efficiency and that there is not a lot of drop from peak efficiency to full load. One I looked at was 98.5 at full load and something near 98.9 for the peak.

.