Calculating Transformer losses part II

The current showing in the primary side is in addition to the capacitive load of cables, and the low IR values of cable.

Better to isolate the main O/G breaker so that you can measure the actual ccurrent on primary.

Anand.mandarapu

How did you get the avg current? These reading are with the secondary load removed?

I calculated the average of Phase A and C currents. Phase B current was listed as zero on the meter when we did this. I know this particular calculation is correct because I have access to the utility website for billing analysis and my calculated number correlates directly to what the system has for that moment in time.

LTU student

You didn't specify the size and distance of cable lenght and size. A difficult calculation indeed. Incorporate three phase interaction, plus impedence level difference between buildings or structures with power off and on. often you may find zero ohms of resistance, but the circuit of a ground wire may carry 2 or 2.5 amps of current due to different earth resistance with all power off. Oh, may I add...all dirt conduction aint the same Good luck. If you figure it out... You will be rich.

The transformers are directly outside the building with less than 20 feet of direct busbar connecting them to the switchgear. I am trying to find the losses to justify removing the two large transformers and replacing them with a single, much smaller one.

LTU Student

Surely the answer is to remove one transformer and feed the whole facility from just one unit? Is there some reason that cannot be done?

Replacing the transformers will take ages to recuperate the costs, as one of the responders to your previous post pointed out.

Chas

That makes a lot of sense, but is not that easy. These transformers are separated by several hundred feet along the side of the building. They are bus-barred into the building's switch gear. Therefore, removing one would require replacing the busbars with conduit and adding several hundred feet of cable in conduit.

I agree that it makes sense and I was trying to put some numbers around the task to show the business owners what the payback would be. It began as an exercise for a class project and turned into a real-life cost-benefit analysis.

I appreciate all the answers from this website. I have learned a lot from the exercise.

LTU student.

You will have to do that cabling if you replace both anyway!

Chas

How did you get the avg current? How do you get it?

These reading are with the secondary load removed?

This is a follow up thread to the one I posted last month. I performed the "experiment" of turning off all the loads in the building at the switchgear and taking note of the readings on the meter. The meter is on the primary side of the transformers. The arrangement they have with the meter is that only two legs of the 4800 volt delta circuit are recorded. At the point that no loads were on the transformers, the data read:...

Go through the post.

OP:

Ideally well designed transformer (or other machines too) should have almost equal iron losses (Fixed losses) and copper losses (variable losses) at rated load.

The iron losses will be almost constant across the working range. Ideally totally constant, but a few parts of it varies just slightly.

With these assumptions and also assuming that at the condition (of no load) the copper losses are negligible Your full load copper loss (rated transformer value) will be equal to current loss (at no load)

Further assuming you have taken the readings, and the multiplication factors (ie the readings are post-multiplication factor- some times, at least in my factory, these panel meters are marked taking into account the CTs and PTs ie 100A on the cable shows 100A on ammeter (though the CT has brought it down to 5A only, but the sceles are re-graduated)

The FL Cu loss will be another 7KW or so so total loss at full rated load will be about 14KW. At lower load these will proportionately go down.

Assumptions validity check up (not too preposterous I hope)

What are the ratings of the transformer ?

_{BTW: do you want us to solve total problem here or leave something to you ?}

When the tranformer is on idle mode the pf is 0.58

that is a bit high power factor (unless you have some fixed capacitor banks)

Why is it contradicting with the current pf of 0.164 ?

The idle means partially loaded ? ie some resistive loads on?

This type of pf (0.5-0.6) we have seen in our installation when the lighting loads are on but the main loads (the machines of the factory) off.

The pf being locally compensated do not come into existence on those times. And the load on the system being a fraction of the full load, it does not move us into the penalty area either _{(we are on the process of putting a central APFC too for this though).}

Assuming the transformer isolaetd at secondary breaker - 0.164 pf

Only loads switched off at the distribution boards - 0.58

You have a problem in the cables or some leakage. Hopefully your RCD/EFCB is in working condition.

The power factor of .58 is when the factory is in "idle" mode." At that point, there is a load of about 80 kW through the transformers. This 80 kW represents some security lights, some ventilation fans in the factory, some computers that cannot be turned off, and some climate control in the office area.

When the transformers have *no load*, the power factor is .164 and the power draw is 7 kW.

So as guesed

This type of pf (0.5-0.6) we have seen in our installation when the lighting loads are on but the main loads (the machines of the factory) off.

Hence I will assume the iron losses at 7KW provided you have the sustem as per assumptions.

This will be matched by the FL Cu losses, so total losses at FL is likely to be 14KW.

Depending on the transformer rating that may be good or bad.

Thank you very much for the response.

The transformer ratings are 1500 KVA, 4800 volt primary and 480 volt secondary. There are two identical transformers. The current factory owner is using 350 kW maximum, normally the factory uses below 300 kW, which is a small fraction of the transformers capacity.

Your question regarding the current is very good. The meter has a CT of 80, so the values need to be multiplied. At no load, the values would be as follows:

Phase A current= 3.68 A voltage = 4840 V

Phase B readings on meter were zero

Phase C current = 6.48 A voltage = 4840 V

Power Factor = 0.164

You are suggesting that the 7 kW of losses that I calculate at no load represent the core losses and they can be assumed constant over the load range of the transformers. The copper losses at rated load (3000 KVA for the sum of the 2 transformers) would also be about 7 kW. Are the Cu losses reasonably linear across the load range? If we are using 15% of rated load (say 450 KVA at a power factor of .8 is 360 kW), would we have 15% of 7 kW as our Cu losses? That would be an additional 1 kW.

I am looking for all the help I can get to understand this situation. I am writing an energy audit report for non-technical energy management class; this is not a problem for an electrical engineering class. Most reports include some basic math on changing light fixtures to more efficient T8 tubes or adding some insulation. I decided to tackle the transformer problem because it seemed interesting to me.

LTU student

The assumption that the copper losses are as big as the core losses is an assumption.

Nothing more.

I have had offers for the same spec transformers where one manufacturer had quadruple the copper losses of the others, but he could deliver in half the time.

Needless to say, he didn't get the order.

I'd advise to look up the manufacturer datasheet, but i suspect that has been lost long ago.

Another way of looking, quick, dirty, not all too precise, would be to asume the copperlosses and the core losses both as an internal resistance.

Now assuming you can get a correct, sufficiently precise Watt measurement on both sides, ie. either directly, or by computation

you could just measure power in, and power out.

the difference between those two measurements will give you the total losses at that specific (primary) voltage and current.

Copper losses are NOT linear with current.

Remeber Ohms law? (U = I.R)

Power is P = U.I

combined U.I= I.I.R or P= I^2 .R

As for cost cutting, did you factor in the down time of the factory for when the transfomer needs service, or fails?

There might he a reason why there are two transformers set on opposite sides of the building......

hie is this exp done in ur lab or out side!!

if the lt is with no load and u take it zero !! then the value will b negative and not valid !!!

I believe ,Transformer losses are found only when there is load !!

this is acc to my knowledge correct me if i am wrong .

Transformer losses are :

1. Core or magnetism losses, you can found when transformer in open circuit as like your explanation

2. Copper losses, you can found when transformer in short circuit, normally LV side must be SC and supply from HV side, measure the HV side by ampmeter when LV side reached full current acc to name plate nominal current

Sum of both measurement = losses in the transformer

You cant only measure on OC, you must measured also on SCC

can i calculate core loss based on the dimentions of the core and thenumber of turns and the voltage and current phasors?