Relation between unbalanced loading and transformer lossby b.shahvir Tags: loading, loss, relation, transformer, unbalanced 

#1
Dec1509, 12:08 AM

P: 190

Hi all,
Is there a definite relation between unbalanced loading of a 3 phase transformer and increase in transformer loss? If so, how? Does unbalanced loading affect variable (copper) loss or fixed (core) loss? Pls. explain in detail. If I carry out load balancing between the o/p phases of the Xmer, will it reduce losses? Pls. help! Thanks & Kind regards, Shahvir 



#2
Dec1509, 03:05 AM

Sci Advisor
P: 4,003

The windings of a 3 phase transformer are usually 3 independent single phase windings. They look like one transformer, but magnetically they are independent. Other countries may be different.
Suppose each of these can handle 33.3 KVA so the whole transformer can handle 100 KVA. If you try to pass 40 KVA through one of the transformers, it will be overloaded and the transformer overload protection will work to protect that transformer by shutting down that transformer and the other transformers as well. Now, the other transformers may only be handling 25 KVA each making a total of 90 KVA (25 + 25 + 40 = 90 ). So, a 100 KVA transformer shuts down with only 90 KVA passing through it. That is one reason why it is important to balance the load on a transformer. 



#3
Dec1509, 03:30 AM

P: 190

There might be benefits of balanced loading from an operational point of view, but my topic is related to power (energy) loss in a 3 ph Xmer due to unbalanced loading.




#4
Dec1509, 07:22 AM

P: 997

Relation between unbalanced loading and transformer loss
I'd say that as long as all xfmr units stay within their rated limits, then no, the losses will not increase. Say we gang 3 1phase units together for 3phase operation. Under full load, and balanced, each xfmr has 1.0 pu volts (per unit), and 1.0 pu amps. The core losses are related to the 1.0 pu V, and the copper losses are related to the 1.0 pu A.
If the load is *unbalanced*, are you referring to 1 phase at 100% load, and the others at less than 100%? Or, do you mean that 1 phase is greater than 100% loaded? It matters. If 1 phase is at 100% load, with the other 2 at less, than the core losses are the same, as each xfmr sees 1.0 pu V, but the lighter loaded xfmr phases see less than 1.0 pu A. Hence these losses are lower. Of course if 1, or 2 xfmr phases are loaded beyond 100%, the losses do indeed increase. Did I help? Claude 



#5
Dec1509, 08:40 AM

Sci Advisor
P: 4,003

I can think of a way that the copper losses would be greater if the load was unbalanced.
Because of the I squared factor, it is better to keep the load balanced. Suppose the windings are identical and have a net resistance of 1 ohm. Suppose the total load is 200 amps from two of the windings. If both windings deliver 100 amps, the total power loss in each winding is 100 * 100 * 1 or 10000 watts. So in the two windings, the loss is 20000 watts. Now, suppose one winding delivers 70 amps and the other delivers 130 amps (still 200 total.) The first winding is dissipating 70 * 70 * 1 = 4900 watts The second is dissipating 130 * 130 * 1 = 16900 watts Total power = 4900 + 16900 or 21800 watts So, that is an extra 1800 watts (21800  20000) that would be dissipated by not having a balanced load. 



#6
Dec1509, 09:44 AM

P: 190

Thanx all for help. The above reply is more so palatable & easy to comprehend. One last thing... can unbalanced loading ever affect fixed (core) losses? Regards, Shahvir 



#7
Dec1709, 06:20 AM

P: 219

Unbalanced system loads can cause significant zerosequence magnetic flux to occur for some threephase connections. Unless a magnetic return path for this flux is provided, the flux returns outside the core and can cause eddycurrent heating in other transformer conductive components, such as the tank. The existence of zerosequence flux either within or outside the core depends on both core configuration and winding connections. Threephase transformercore assemblies do not usually provide fullcapacity return legs for zerosequence flux. Thus, if sufficient zerosequence flux occurs, it will be forced to return outside the core. Providing of magnetic return path for zerosequence magnetic flux is possible but in this case we front to additional core loss too, also providing of zero sequence circuit for zerosequence flux preventing is possible (delta tertiary winding) but in this case we front of additional ohm losses in new circuit.
 Creative thinking is enjoyable, Then think about your surrounding things and other thought products. http://electricalriddles.com 



#8
Dec1809, 08:15 AM

P: 997

Of course, if sensitive (signal level) circuitry is in close proximity to the core, then the zero sequence flux can indeed result in interference. When I worked in xfmr design for a defense & aerospace company, we used 3phase core type construction (3legged E core). These cores are magnetically coupled between the 3 phases, so a delta winding is not really needed. But because of potential emi concerns, we always used a delta, at least on 1 winding. This provides a path for zero sequence currents, preventing zero sequence flux from interfering with neighboring circuits. Claude 



#9
Dec1809, 10:27 AM

P: 190

The above explanation gives a clear picture that although the total Amps are same in both the cases (i.e. 200A), the power loss in both the cases are different, i.e. power loss in the 2nd case is more ( 'I' squared factor )! This in itself is a unique phenomenon and warrants elaborate discussions and/or examples... in the sense that can this case be the only cause of power loss due to unbalanced loading? Regards, Shahvir 



#10
Dec1809, 10:03 PM

Sci Advisor
P: 4,003

To explain this a bit more...
Suppose both transformer windings were 1000 volts open circuit. If the loads on both of them were 100 amps, the voltage drops across each winding resistance would be (100 amps times 1 ohm) or 100 volts. So the voltage getting to both loads would be 900 volts (1000  100 ) and the power with 200 amps flowing would be (900 volts * 200 amps) or 180000 watts If the load on one was 70 amps, the voltage drops across the winding resistance would be 70 amps times 1 ohm or 70 volts. So the voltage getting to the load would be 1000 70 or 930 volts. So, the power in that load would be 930 volts * 70 amps or 65100 watts. If the load on the other winding was 130 amps, the voltage drops across the winding resistance would be (130 amps times 1 ohm) or 130 volts. So the voltage getting to the load would be 1000 130 or 870 volts. So, the power in that load would be 870 volts * 130 amps or 113100 watts. The total power from both windings is now 65100 + 113100 or 178200 watts. Note that this is (180000  178200) or 1800 watts less than the balanced case. So, the extra power dissipated in the winding is exactly the power that didn't get to the loads. 



#11
Dec1909, 01:07 AM

P: 219

For similar discussion you can refer to Transformer Riddle No.5 from http://electricalriddles.com 



#12
Dec2009, 03:51 AM

P: 190

Output power = Input power  Fixed (iron) loss in core In this case, the input side Wattmeter will measure input power + fixed power loss, while the output side Wattmeter will measure only output power consumed by load...the difference of both being fixed losses! Also, with the same arrangement, if one was to measure variable (copper) losses, which in turn, depends on Xmer loading, then; Output power = Input power  Variable (copper) loss in Xmer winding In this case, Input power = Output power (as per equivalent circuit arrangement of Xmer). Caveat: If I were to use a kWh meter on the input side, core losses would consume 'extra units' (power loss) as the core loss is represented by a dissipating resistance in parallel to the supply voltage and load. However, copper loss is in series with the supply voltage and load and hence cannot be recorded by kWh meter as consuming extra units. Rather it results in loss of terminal voltage which would be subtractive power not transferred to load. Hence, variable power loss (copper loss) being in series with the load, cannot be measured by the input side Wattmeter or kWh meter! Is this presumption correct? Pls. comment. 



#13
Dec2009, 06:46 AM

Sci Advisor
P: 4,003

The quote you gave was made in the context of the effect of an unbalanced load.
Transformers, especially large ones are not tested on full load. The method you gave would just give the total loss of the transformer. If there was a problem, how would you know if it was copper loss or iron loss causing it? What is done is to do a short circuit test to test for copper loss and an open circuit test to test for iron loss. Briefly, with just a current meter as a load, you increase the primary voltage until the maximum load current flows in the meter and measure the power. It will only take a small primary voltage to do this, so the Iron losses will be small. Then you measure the power into the primary with no secondary current but full primary voltage. So, there should be minimum copper loss. There are many references to this if you would like to try it on Google. One of the first that came up was this: http://www.ece.vt.edu/ece3354/labs/xformermodel.pdf when I typed in: "transformer short circuit test procedure" 



#14
Dec2009, 10:44 AM

P: 190

Thanks. However, how does one define power loss in electrical machines? From the recurrent explanations, the term power loss would have dual meanings;
1) Power loss due to losses in Xmer core. This power consumed by Xmer iron is recorded by input side power meters as it is in parallel to the input voltage supply and load. This would be extra power lost to Xmer core as heat. 2) Power loss due to winding resistances. This power is consumed by winding resistance and causes loss in terminal voltage. This loss is in series with supply voltage and load and hence is the power which never reached the load. Since this power (or voltage) never reached the load, this subtractive power is also termed as a 'loss'. Hence, out of the above 2 cases, which one gives an exact definition of 'power loss' in electrical machines? 



#15
Dec2009, 06:08 PM

Sci Advisor
P: 4,003

The power out of a transformer is equal to the power in, minus losses.
However, that is not the whole story. If you were testing a cheap transformer it may have very little inductance in the primary and this could cause it to draw a lot of current from the mains. This is reactive current which is not in phase with the voltage, but has to be generated and yet is usually not paid for. Transformer behaviour is very complex and needs vector diagrams to describe in full. Since this has already been done in text books, you could look there if you are still interested. 


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