KWH Reading Difference

The difference between the two readings will be the heat that is dissipated within the transformer.

Note that the efficiency of a transformer will generally be at its highest (typically ≈95%) when fully loaded. At 50% loading, it may be down to ≈75%.

So to answer your question, there will almost certainly be a significant difference.

It is not a problem to get accurate metering on the low side of the transformers. The method is to use metering that can be programmed with "compensation" factors to account for the transformer losses.

There are several good reasons to use compensated metering, primarily cost of equipment and safety. No meter will be handling 20kV, so there will be an expensive voltage transformer with it's own factors needed on the high side. Both sides will require CT's (Current Transformers), but the high side version are quite a bit more expensive.

The details required to make the compensation are listed on the transformer nameplate, and the metering will be quite accurate. A metering engineer, a qualified meter technician and the meter-man's handbook will be required, but they easily pay for themselves.

Regards, CJM

there is something interesting about transformers. If you see the efficiency curves, the peak is somewhat abot at 50% load (and not 100%)

Then the efficiency slowly tapers off till 100% (and sharply on the 0% side)

This is about your second portion

The first portion is already answered- the secondary side power reading will be load power, assuming the transfromer efficiency appox 98-99% you may get the primary side power.

http://nptel.iitm.ac.in/courses/IIT-MADRAS/Electrical_Machines_I/pdfs/1_10.pdf

http://www.bicronusa.com/tpt_efficiency.html

The most probable reasson being that these are not supposed to work at full load (since for most - especially the distribution ones) the load is much lower that the full load value.

Just to give you an example- in my shed we have one 1MVA sub-station - a few days back I was surprised to find that the connected load (machines) is 1.5 MVA approx

However as we know that the machines do not run at full load eg a VTL drive motor of 75KW will rarely draw this power, and then all the machines at full load will not synchronise.

WSimilarly, all the domestic loads do not.

My bad - I was looking at ferroresonant transformers:

[Source]

Yes on Ferro resonant transformers you are correct.

I think this is what you meant to show:

This is a good graph; it breaks out iron and copper losses then adds them together to make the "total loss" curve. It also shows peak efficiency at around 75-80% (which is what I have always been told working in the trade). Here's a link if you want to buy the book:

http://books.google.com/books?id=yKMJAAAAIAAJ&pg=PA309&lpg=PA309&dq=transformer+efficiency+curves&source=bl&ots=m4DF1f_ulf&sig=C4-Aa4Uurb8vxSaXSTefJfTnk88&hl=en&ei=fUhCSt_ALc3ktgeE6L2ZCQ&sa=X&oi=book_result&ct=result&resnum=10

Anyway, the efficiency of a transformer is very good, but it is not perfect, so it is most accurate to meter on the primary side of the transformer. However, if this is not practical, then metering on the secondary side will yield a value that is close to actual power used - provided that the transformer is not severely "under-loaded."

If you have a continuous load on the transformer, then it would not be unreasonable to treat the power loss in the transformer as a constant.

The efficiency rating of the transformer should be provided by the manufacturer. In the US it would be found on the nameplate of the transformer - it is usually abbreviated eff.

Use the eff rating to estimate "total loss" (which should be ~1-5%) by subtracting from 100%:

100% - eff = "total loss"