PARALLELING TRANSFORMERS

[calmate]

All,

I have a "old 138kv" switch yard that has 2ea 10MVA transformers in series, the first connects from the utility 138/34.5 (141.5 tap D-Y) then connects to a 34.5/7200 (33.6 tap D-D). they installed another 138kv switchyard in late 1990's and it has a 10MVA 138/6.9 (138 tap D-Y)transformer. Under normal operation, the old switchyard is tied to 4 local generators and the new is a dead end bus. There are times when they tie the "old" to the "new" for switch yard maintenance ONLY and not for continious use ( both transformers in parallel with local generation), the Plant Manager asked me to look into operating both in parallel with local generation all the time
Does anyone have a "reference manual" or "white paper" on calculating circulating currents with respect to different voltage ratios,different tap settings, different impedances etc?

Sean

 

[mpparent]

Do a search on this site and you'll find some discussions on circulating currents.

Mike

 

[stevenal]

Good info at beckwithelectric.com

 

[CKent]

I remember participating in a discussion sometime in Aug 2003 regarding paralleling transformer (see thread238-68183)...I'm doing a study now on technical loss...I have come to the issue of circulating current in parallel XFs...circulatin current (Ic) arises due to difference in the turns ration of the XFs in parallel...Ic adds up to the current in the XF with lower turns ration while it subtracts to the current in the XF with higher turns ration...this effectively changes the capacity of the XFs in parallel...however, in general I'm being confused on how this affects the losses of the combined parallel combination...of course the one with the lower turns ration will have a resulting additional loss due to this circulating current...however, isn't it that the other XF with higher turns ration effectively will have reduce loss because the current it its wdg will be lesser than the actual load...so assuming that the 2 XFs have equal resistance, isn't it that the effective loss of the parallel combination is actually equal as in the condition wherein there is no circulating current (in cases with equal turns ratio)...

In the thread I mentioned above, one of the respondents has this to say in paralleling XFs...
"But an advantage is the reduced losses, especially near full load of one transformer. This applies equally to I^2X and I^2R loss - on a 25MVA transformer, times about one hundred substations, it adds up. Halving the current quarters the loss. But the primary advantage is supply availability - it was this that was the primary driver for us doing it."

How could this be?

 

[waross]

My experience is with much smaller transformers, so my observations may be more of a thought provoking question than a statement of fact.
If transformers of different terminal voltages are connected in parallel then the terminal voltages are forced to be the same. The magnitude of the current will be determined by the impedances of both transformers in series. (Actual impedances, not percentages) If the voltage regulation curves of both transformers are plotted and superimposed, then the point at which the curves cross is the point of zero circulating current due to mis-matched voltages.
At zero load the transformer with the higher open circuit voltage will be supplying energy to the other transformer. As the load increases, the higher voltage transformer will take all the load until the terminal voltage falls to equal the second transformers open circuit voltage.
At that point, the energy transfer to the second transformer will be zero, and any further increase in load will be supported by both transformers and circulating currents due to mismatch will cease..
Depending on the capacity of the supply and other loads on the supply, the energy transfer may result in an increase of voltage on the second supply system, and the magnitude of the circulating current will be reduced accordingly.
In a case of an extreme voltage mis-match, the crossing point of the voltage regulating curves may be outside of normal operating parameters.
In the case of mismatched percent impedances the transformer with the lower percentage impedance will take a disproportionate share of the load.
It will be possible to force the higher percent impedance transformer to take an increased share of the load by setting it to a higher tap, however, at light loading there will be uneven load distribution and there may be circulating currents and energy transfer to the transformer with the lower percent impedance.

The original question:
I calculate your voltage difference at 4.17%.
At no load I would expect a circulating current caused be the voltage difference and limited by the impedances of both transformers and non infinite systems.
Assuming a transformer percent impdance of 7% when the load reaches (4.17%/7%) 67% the circulating currents will cease.
Further load increases will be shared by both transformers, but not necessarily equally.
Again, this post is in the nature of a thought provoking question.
If I have made serious errors, by all means correct me.
respectfully

 

[jghrist]

Do I understand correctly that when operating in parallel:

1. Both the old 138-34.5 kV and the new 138-6.9 kV transformers are connected to the utility 138 kV system.

2. The old 138-34.5 kV transformer is connected to the 34.5-7.2 kV transformer and the generators are connected to the 7.2 kV bus.

3. The new 6.9 kV bus is connected to the old 7.2 kV bus.

Questions:

1. Is the new 6.9 kV winding grounded? Solidly or resistance?

2. Are the generator neutrals grounded?

3. What taps are available on the transformers? If any of them have typical 2 +/-2.5% taps, then you should be able to get the voltages to within 1% at the 6.9/7.2 kV bus.

 

[itsmoked]

waross; In your interesting description I see no reason why the circulating current due to the missmatch would ever drop to zero?!? That voltage missmatch would always be present wouldn't it? Just like a current hogging transistor it will always hog.

 

[amps21]

Would it be ok to say that it is a case of parallelling 2 transformers of different impedences. Considering equal impedences for all the transformers we should have high impedence for the 2 units and low for the one unit. So the division would be 66% - 33 % under normal operation. Is this right? Th.is distribution shall vary more if the X/R ratios are not equal which in this case, i am sure the are not

I agree with Itssmoked, the circulating current is a result of difference (Va-Vb)/(Za+Zb). So irrespective of loads, the currents due to volt difference should be constant.

 

[waross]

itsmoked thank for your reply.
Consider two transformers of equal KVA and equal impedance (10%).
The terminal voltage of the first is 1000 Volts.
The terminal voltage of the second is 950 volts.
Both transformers have an impedance of 10%.
With no external load, the transformer with a terminal voltage of 1000 volts will be delivering energy to the second transformer via a circulating current.

When the First transformer is loaded to 50%, its terminal voltage will drop to 1950 volts and the circulating current will cease.
When the first transformer is fully loaded, the terminal voltage will drop to 900 Volts.
The second transformer will be 56% loaded and its voltage drop due to regulation will be 50 volts. 50/950 = 5.6%
5.6%/10% = 56% loaded.
If you have tap changers trying to maintain two different voltages on paralleled transformers, It's sort of like trying to drive a 4X4 in lock-up with different size tires on the front and back. It is possible but not recommended. The circulating currents will persist or get worse. The tap changers will drive higher voltage transformer to the highest tap and the lower voltage transformer to the lowest tap in a futile attempt to match the voltages. The figures then would be;
1000 volts plus the maximum tap.
950 volts minus the lowest tap.
yours

 

[amps21]

Waross, I agree on your impedence voltage drop treatment of the subject. But in this case i dont think that the 2 units will be loaded the same. There shall not be the same impedence voltage drops in the units. I think that the unit combo of 2 units shall @ any given time shall carry 66% of the load X and the other unit shall carry the rest 33% in addition to its 100% share. Correct me if i am wrong.

Thanks

 

[amps21]

I am sorry : what i want to say is that the lower impedence unit shall supply 66% and the higher impedence unit (pair) shall supply 33%

Thanks

 

[waross]

Apologies for the typo. I wrote 56% instead of 53%
Hi jghrist
I agree with your assessment of the tap changer possibility.

Hi amps21 and itsmoked.
I can hear some of the other regulars “Don’t get him started, Please don’t”
You have to remember that as the load increases the voltage drops. As the voltage drops, the circulating currents are reduced. At no load, one transformer is feeding KVAs into the other transformer and the other system. That is the cause of the circulating current. As the load on the higher voltage transformer increases, the terminal voltage drops. As the terminal voltage drops, the magnitude of the circulating current decreases. At the point that the voltage drop due to loading has reduced the terminal voltage of the higher voltage transformer to equal the open circuit voltage of the second transformer, the circulating current ceases, and further load increases are shared by both transformers. The division of load depends on the magnitude of the load.
In our example transformers;
At the point that the circulating current ceased, one transformer had all the load and the other had not yet started to load up. As the load increased and the voltage dropped further the second transformer almost but not exactly SHARED the INCREASE IN LOAD until the point that one transformer was 100 % loaded and the other was 53% loaded.
At no load on similar sized transformers, the voltage RISE of the second transformer would probably be about 25 Volts and it would be returning KVAs to its supply system.
The voltage drop of the first transformer would be about 25 Volts and it would be supplying KVAs to the second transformer. I say probably because it’s very close, but an exact figure may require the use of the quadratic equation. That sucker can sneak up on you if you are not careful how you state an example.
I think we have two similar concepts mixed up. Both concepts use different variables in the same formula.
One is paralleling transformers with different terminal voltages.
The other is paralleling transformers with different percent impedances. Once we develop confidence in the solution of these two concepts, we can combine them and calculate the circulating currents for transformers with different percent impedances and different terminal voltages.
This is stuff you probably know better than I do. You just have to look at it from a different angle.
The basic formula is the Voltage Regulation Formula.
Transformer Rated Voltage x Percent Impedance Voltage x Percent Loading = Internal Voltage Drop (Rated Voltage Minus Terminal Voltage).
In the following examples all percentages must be expressed as decimal fractions.
Percentage loading = Voltage Drop/ Percent Impedance Voltage/ Rated Voltage

Example
Rated Voltage = 1000 Volts
Percent Impedance Voltage = 0.1 (10%)
Voltage Drop 100 Volts
Percent Loading = 1.0 (100%)
Terminal Voltage = 900 Volts (1000 Volts Minus 100 Volts)

Example

Rated Voltage = 950 Volts
Percent impedance Voltage = 0.1 (10%)
Voltage drop = 50 Volts
Percent Loading = 53%
Terminal Voltage = 900 Volts (950 Volts Minus 50 Volts.

Lets keep the arithmetic easy and stay with 1000 volts and 10% impedance. Let's stay with single phase also.

We can work backwards from the terminal voltage, or the internal voltage drop and calculate the loading.

Let's tabulate the results for both transformers.
We can use Rated Voltage X Percent Loading x Percent Impedance Voltage = Voltage Drop.
Then you can pick a terminal voltage and look up the percentage loading for each transformer.
1000 Volts 0% Loading, Internal Voltage drop 0 Volts.
990 Volts 10% Loading, Internal Voltage drop 10 Volts.
980 Volts 20% Loading, Internal Voltage drop 20 Volts.
970 Volts 30% Loading, Internal Voltage drop 30 Volts.
960 Volts 40% Loading, Internal Voltage drop 40 Volts.
950 Volts 50% Loading, Internal Voltage drop 50 Volts.
940 Volts 60% Loading, Internal Voltage drop 60 Volts.
930 Volts 70% Loading, Internal Voltage drop 70 Volts.
920 Volts 80% Loading, Internal Voltage drop 80 Volts.
910 Volts 90% Loading, Internal Voltage drop 90 Volts.
900 Volts 100% Loading, Internal Voltage drop 100 Volts.

For the second transformer with a lower terminal voltage the table will look like this;
We will use the same Formula:
Rated Voltage X Percent Loading x Percent Impedance Voltage = Voltage Drop.
950.0 Volts 0% Loading, Terminal Voltage drop, 0.0 Volts.
940.5 Volts 10% Loading, Terminal Voltage drop, 9.5 Volts.
931.0 Volts 20% Loading, Terminal Voltage drop, 19.0 Volts.
921.5 Volts 30% Loading, Terminal Voltage drop, 28.5 Volts.
912.0 Volts 40% Loading, Terminal Voltage drop, 38.0 Volts
902.5 Volts 50% Loading, Terminal Voltage drop, 47.5 Volts.
893.0 Volts 60% Loading, Terminal Voltage drop, 57.0 Volts.
883.5 Volts 70% Loading, Terminal Voltage drop, 66.5 Volts.
874.0 Volts 80% Loading, Terminal Voltage drop, 76.0 Volts.
864.5 Volts 90% Loading, Terminal Voltage drop, 85.5 Volts.
855.0 Volts 100% Loading, Terminal Voltage drop, 95.0 Volts.

We can take the Voltage drop and divide by the Percent Impedance and by the Rated Voltage to get the Percent Loading.
You can construct similar tables for any transformers with any percent impedance and any terminal voltage. You can compute overloads and when the sign is reversed and the terminal voltage is above the rated voltage you will be calculating circulating currents. When you get a little confidence you will do the calculations directly without the need to compute tables.
I have carefully stated the loads as percentage loading on the transformers to keep the arithmetic simple. I can look up the Quadratic Equation and use it if I have to, but we will all be happier if I state the examples in such a way as to avoid it.

We have been assuming that the transformers in these examples were basically equal except for the different Voltage ratings.
Suppose that the 1000 Volt transformer is rated at 100 KVA
Suppose that the 950 Volt transformer is rated at 500 KVA
Suppose that the 950 Volt transformer is loading up and it is proposed to parallel the 100 KVA transformer to share the load.
What will happen?
The 100 KVA transformer reaches 100% load when the terminal voltage drops to 900 Volts.
The 950 Volt transformer is 53% loaded at 900 Volts at the terminals.
The total load that can be supported by the two transformers together without overloading either is 100 KVA plus 265 KVA (53% of 500 VA) = a total of 365 KVA.
The addition of the second transformer actually decreased the capacity .

This combination of ratings is for example only and is not likely to occur in practice.
What is likely to occur is a situation were two distribution transformers of dissimilar rating and/or size are to be paralleled, usually in an emergency.

 

[waross]

I was having so much fun I forgot the question.
In your problem calmates, the following suggestion may give you a starting point.
1> start with a typical current or percentage load and calculate the combined Impedance voltage of the two transformers in cascade.
Calculate a terminal voltage.
Use this terminal voltage to see if the other transformer is still backfeeding or has started to take up some load.
With two transformers in cascade, the voltage drop under load may be much more than the single transformer. The actual voltage diference under load and the resulting circulating currents may be much less than the 4% difference at no load would indicate.
I would expect circulating currents at no load, but you may not have circulating currents at part or full load and you may actually have fairly good load sharing at full load.
Under load you can probably optimize loading by tweaking the generator voltages a part of a percent. Your system may quite well have been designed with different voltages to compensate for the different impedance voltages and live with circulating currents at light load to facilitate load sharing at heavy load.

 

[amps21]

Waross, Thanks for your explaination. Question:

What is the effect of the X/R ratios of the transformer being same or different? For a given load power factor, the individual unit load currents shall not be in phase with the total load current if the X/R ratios of the units is not the same. There will be a relative angular displacement in between currents of the 2 units to vectorially give the load current. The quadrature component thus will be nothing but circulating current. This Circulating current shall stay. This current can be easily calculated using R&X values of the transformer.

Please correct me if I am wrong. I do not see the mention of this in your elaboration.

Thanks

 

[itsmoked]

Thanks Waross...I B-E-L-I-E-V-E.

 

[fsmyth]

amps21 brings up a good point - what ARE the differentials
in this scenerio? How does a good/poor PF affect the
circulating currents and the backfed portion?
<als>

 

[waross]

Hi amps21, itsmoked, and fsmyth.
This answer will be very short. I am not sure. My scenario is based on smaller distribution transformers with similar characteristics.
Perhaps someone else can give us an explanation.

 

[waross]

Load sharing between transformers of different Percent impedance voltages.
Transformers with equal Impedance voltages will share the load in the ratio of their KVA ratings.
You can arrive at imaginary ratings for the purpose of calculating load sharing by re-rating the KVA of a transformer to a value that results in the same percent impedance rating.
Compare the computed load to the actual transformer rating to make sure that it is not overloading.
Example;
2 transformers. Both 100 KVA
#1.- 2.8% Impedance voltage.
#2.- 3.2% Impedance voltage.
The transformer with the lower percent impedance voltage will load up first so we will re-rate the 3.2% transformer. 100 KVA x 2.8 / 3.2 = 87.5 KVA
When there is 100% load on transformer #1 there will be 87.5% load on transformer #2
50% load will be; 50% x 100 KVA = 50 KVA on transformer #1 and 50% x 87.5 KVA = 43.75 KVA on transformer #2