 # Principles of transformers in parallel connection

introduction

To provide a load greater than the rated value of an existing transformer, two or more transformers can be connected in parallel with the existing transformer. Transformers are connected in parallel when the load on one of the transformers is greater than its capacity.

Reliability is increased with parallel operation compared to a single larger unit.

The cost associated with servicing spare parts is less when two transformers are connected in parallel. It is generally economical to install another transformer in parallel instead of replacing the existing transformer with a single larger unit.

The cost of a spare unit in the case of two parallel transformers (of equal value) is also lower than those of a single large transformer. In addition, it is preferable to have a parallel transformer for reliability reasons.

With this, at least half of the load can be supplied by an out-of-service transformer.

Condition for parallel operation of the transformer

For parallel connection of the transformers, the primary windings of the transformers are connected to the source bus bars and the secondary windings to the load bar machines.

Various conditions to be fulfilled for the correct parallel operation of the transformers: </ p>

1. Same voltage and turn ratio (primary and secondary rated voltage is the same)
2. Same impedance and X / R ratio
3. Identical tap changer position
4. Identical KVA rating
5. Same phase angle shift (vector groups are identical)
6. Same evaluation frequency
7. Same polarity
8. Same phase sequence

Some of these conditions are practical and some are mandatory.

the practical conditions are: same voltage ratio and revolution ratio, same impedance in percentage, same KVA index, same position of the tap-changer

The mandatory conditions The conditions are the same change of phase angle, same polarity, same phase sequence, and same frequency. When the proper conditions are not met, the parallel operation is possible but not optimal.

1. Same voltage ratio and turn ratio (on each outlet)

If the transformers connected in parallel have slightly different voltage ratios, then due to the unevenness of the values ​​generated in the secondary windings, a current flowing will flow in the loop formed by the secondary windings at no load, which can be much larger than the no-load current.

The current will be high enough as the leakage impedance is low. When the secondary windings are loaded, this flowing current will tend to produce an unequal load on the two transformers, and it may not be possible to take a full load of this group of two parallel transformers (one of the transformers can be overloaded).

If two transformers of different voltage ratios are connected in parallel with the same primary supply voltage, there will be a second voltage difference.

Now when the secondaries of these transformers are connected to the same bus, there will be a current flowing between the secondaries and therefore between the primaries as well. Since the internal impedance of the transformer is small, a small voltage difference can generate a sufficiently high circulating current, which will lead to unnecessary overvoltage. 2 R loss.

Primary and secondary ratings should be the same. In other words, transformers should have the same rotation ratio, i.e. the transformation ratio.

2. Same percentage of impedance and X / R ratio

If two transformers are connected in parallel with similar impedances per unit they will mainly share the load in the ratio of their KVA ratings. Here, the load is mostly equal because it is possible to have two transformers with equal impedances per unit, but different X / R ratios. In this case, the line current will be less than the sum of the transformer currents and the combined capacity will be reduced accordingly.

A difference in the ratio of the reactance value The resistance value of the impedance per unit results in a different phase angle of the currents carried by the two transformers in parallel; one transformer will operate with a higher power factor and the other with a lower power factor than the combined output. Therefore, the real power will not be shared proportionally among the processors.

The current shared by two transformers operating in parallel must be proportional to their MVA ratings. </ p>

The current carried by these transformers is inversely proportional to their internal impedance.

From the above two statements, it can be said that the impedance of transformers operating in parallel is inversely proportional to their rated power of MVA. In other words, the percent impedance or the impedance values ​​per unit must be the same for all transformers operating in parallel.

When connecting single-phase transformers in three-phase banks, proper impedance matching becomes even more critical. In addition to following the three rules for parallel operation, it is also a good idea to try and match the X / R ratios of the impedances in series to keep the three-phase output voltages in balance.

When single-phase transformers with the same KVA characteristics are connected in a Y-bank, impedance asymmetries can cause a large load imbalance between the transformers.

Allows you to examine different types of cases among Impedance, Ratio, and KVA.

If single-phase transformers are connected in a YY bank with an isolated neutral, the magnetizing impedance should also be equal on an ohmic basis.

Otherwise, the transformer with the larger magnetizing impedance will have the highest percentage of excitation voltage, which will increase the core losses of that transformer and eventually lead to its saturation.

Case 1: Impedance equality, ratios, and even kVA

The standard method of connecting transformers in parallel is necessary to have the same turn ratios, percentage of impedance, and kVA ratings. Connecting transformers in parallel with the same parameters allow an equal distribution of the load and the absence of circulating currents in the windings of the transformer.

Example Connection of two transformers at 2000 kVA, 5.75% impedance in parallel, each having the same rotation ratio for a load of 4000 kVA.

• Load on transformers-1 = KVA1 = [(KVA1 /% Z) / ((KVA1 /% Z1) + (KVA2 /% Z2)]] X KVAl
• kVA1 = 348 / (348 + 348) x 4000 kVA = 2000 kVA.
• Load on transformers-2 = KVA1 = [(KVA2 /% Z) / ((KVA1 /% Z1) + (KVA2 /% Z2)]] X KVAl
• kVA2 = 348 / (348 + 348) x 4000 kVA = 2000 kVA
• KVA1 = KVA2 = 2000KVA

Case 2: Equal Impedances, Ratios, and Different KVA

This parameter is not common for new installations, sometimes two transformers with different kVA and the same percentage of impedance are connected to a common bus. In this situation, the current division causes each transformer to carry its rated load. There will be no circulating currents because the voltages (rotation ratios) are the same.

Example Connection of 3000 kVA and 1000 kVA transformers in parallel, each with an impedance of 5.75%, each with the same rotation ratios, connected to a common load of 4000 kVA.

• Loading on transformer-1 = kVA1 = 522 / (522 + 174) x 4000 = 3000 kVA
• Loading on transformer-1 = kVA2 = 174 / (522 + 174) x 4000 = 1000 kVA

From the above calculation, it can be seen that differentRated kVA values ​​of transformers connected to a common load, this current division only charging each transformer to its nominal kVA value. The key here is that the percentage of impedance is the same.

Case 3: Unequal impedance but same ratios and kVA

Most of the time, this parameter has been used to improve the production capacity of the plant by connecting in parallel existing transformers which have the same kVA value, but with a different percentage of impedance.

This is common when budget constraints limit the purchase of a new transformer with the same parameters.

We must understand is that the current divisions in inverse proportions with respect to the impedances, and a larger current flows through the smaller impedance. So, the lower impedance percentage transformer may be overloaded when under heavy load, while the other higher impedance transformer is lightly loaded.

Example Two 2000 kVA transformers in parallel, one with an impedance of 5.75% and the other with an impedance of 4%, each with the same rotation ratio, connected to a common load of 3500 kVA.

• Loading on Transformer-1 = kVA1 = 348 / (348 + 500) x 3500 = 1436 kVA
• Loading on Transformer-2 = kVA2 = 500 / (348 + 500) x 3500 = 2064 kVA

It can be seen that because transformer percent impedances do not match, they cannot be loaded into their combined rating in kVA. The load distribution between the transformers is not equal. At a rated load less than the combined kVA, the 4% impedance transformer is overloaded by 3.2%, while the 5.75% transformer is loaded at 72%.

Case 4: Unequal impedance and same KVA ratio

This particular of transformers rarely used in industrial and commercial installations connected to a common bus with different kVA and unequal percentages of impedance. However, it may happen that two single-ended substations can be linked by buses or cables in order to provide better voltage support when starting a large load.

If the percentages of impedance and kVA are different, be sure to load these transformers.

Example Two transformers in parallel, one of 3000 kVA (kVA1) with an impedance of 5.75% and the other of 1000 kVA (kVA2) with an impedance of 4%, each with the same rotation ratio, connected to a common load of 3,500 kVA.

• Loading on Transformer-1 = kVA1 = 522 / (522 + 250) x 3500 = 2366 kVA
• Loading on Transformer-2 = kVA2 = 250 / (522 + 250) x 3500 = 1134 kVA

As the impedance percentage is lower in the 1000 kVA transformer, it is overloaded with a lower rated load than the combination.

Case 5: Ratio of impedance equality and KVA inequality

Small voltage differences cause a lot of currents to flow. It is important to stress that parallel transformers must always be on the same tap connection. The circulating current is completely independent of the load and the division of the load. If the transformers are fully loaded, there will be a considerable amount of overheating due to the circulating currents.

The Takeaway Remember that circulating currents do not flow on the line, they cannot be measured if the monitoring equipment is upstream or downstream of the common connection points.

Example Two 2000 kVA transformers connected in parallel, each with an impedance of 5.75%, the same X / R ratio (8), a transformer 1 with tap set at 2.5% of the nominal value and a transformer 2 with rated outlet What is the percentage of current flowing (% IC)

• % Z1 = 5.75, so% R ‘=% Z1 / √ [(X / R) 2 + 1)] = 5.75 / √ ((8) 2 + 1) = 0.713
• % R1 =% R2 = 0.713
• % X1 =% R x (X / R) =% X1 =% X2 = 0.713 x 8 = 5.7
• Let% e = voltage ratio difference expressed as a percentage of normal and k = kVA1 / kVA2
• Circulating current% IC =% eX100 / √ (% R1 + k% R2) 2 + (% Z1 + k% Z2) 2.
• % CI = 2.5X100 / √ (0.713 + (2000/2000) X0.713) 2 + (5.7 + (2000/2000) X5.7) 2
• % CI = 250 / 11.7 = 21.7

The flowing current is 21.7% of the full load current.

Case 6: Unequal impedance, KVA, and different ratios

This type of setting would be unlikely in your training. If the ratios and impedance are different, the circulating current (due to the uneven ratio) must be combined with the part of the load current of each transformer in order to obtain the actual total current in each unit.

For a unit power factor, 10% circulating current (due to uneven turn ratios) produces only half a percent of the total current. At lower power factors, the flowing current will change drastically.

Example Two transformers connected in parallel, 2000 kVA1 with an impedance of 5.75%, an X / R ratio of 8000 kVA2 with an impedance of 4%, an X / R ratio of 2000 kVA1 with tap adjusted to 2.5% nominal value and at 1000 kVA2 taken at nominal value

• % Z1 = 5.75, so% R ‘=% Z1 / √ [(X / R) 2 + 1)] = 5.75 / √ ((8) 2 + 1) = 0.713
• % X1 =% R x (X / R) = 0.713 x 8 = 5.7
• % Z2 = 4, so% R2 =% Z2 / √ [(X / R) 2 + 1)] = 4 / √ ((5) 2 + 1) = 0.784
• % X2 =% R x (X / R) = 0.784 x 5 = 3.92
• Let% e = voltage ratio difference expressed as a percentage of normal and k = kVA1 / kVA2
• Circulating current% IC =% eX100 / √ (% R1 + k% R2) 2 + (% Z1 + k% Z2) 2.
• % CI = 2.5X100 / √ (0.713 + (2000/2000) X0.713) 2 + (5.7 + (2000/2000) X5.7) 2
• % CI = 250 / 13.73 = 18.21.

The flowing current is 18.21% of the full load current.

3. Same polarity

The polarity of the transformer signifies the instant redirection of the electromotive force induced in secondary. If the instantaneous directions of the secondary electromotive force induced in two transformers are opposite when the same input power is supplied to both transformers, the transformers are said to be of opposite polarity.

Transformers must be correctly connected with regard to their polarity. If they are connected with the wrong polarities, the two CEMs induced in the secondary windings in the parallel act together in the local secondary circuit and produce a short circuit.

The polarity of all transformers working in parallel must be the same, otherwise, there will be huge current currents flowing through the transformer, but no load will be supplied by these transformers.

If the instantaneous directions of the secondary electromotive force induced in two transformers are the same when the same input power is supplied to both transformers, the transformers are said to be of the same polarity.

4. Same phase sequence

The phase sequence of the line voltages of the two transformers must be the same for the parallel operation of three-phase transformers. If the phase sequence is incorrect, on each cycle each phase pair will be short-circuited.

This condition must be strictly followed for the parallel operation of transformers.

5. Same angular phase shift (zero relative phase shift between secondary line voltages)

The windings of the transformer can be connected in a variety of ways which produce different amplitudes and phase shifts of the secondary voltage. All transformer connections can be classified into separate vector groups.

Group 1: Zero phase shift (Yy0, Dd0, Dz0)
Group 2: 180 ° phase shift (Yy6, Dd6, Dz6)
Group 3: -30 ° phase shift (Yd1, Dy1, Yz1)
Group 4: +phaseshift 30 ° (Yd11, Dy11, Yz11)

To have a zero relative phase shift of the secondary voltages, the transformers belonging to the same group can be put in parallel. For example, two transformers with Yd1 and Dy1 connections can be put in parallel.

The transformers of groups 1 and 2 cannot be a parallel queen with the transformers of their own group. However, transformers in groups 3 and 4 can be put in parallel by reversing the phase sequence of one of them. For example, a transformer with a Yd1 1 connection (group 4) can be put in parallel with one with a Dy1 connection (group 3) by reversing the phase sequence of the primary and secondary terminals of the Dy1 transformer.

We can only parallel Dy1 and Dy11 by crossing two incoming phases and the same two outgoing phases on one of the transformers, so if we have a DY11 transformer, we can cross the B & C phases on the primary and secondary to change the +30 degree phase shift into a -30 degree shift that will be parallel to Dy1, assuming all of the above points are met.

6. Same KVA classification

If two or more transformers are connected in parallel, the distribution of the load between them depends on their rating. If all are of the same value, they will share equal charges

Transformers with unequal kVA ratings will share a load practically (but not exactly) in proportion to their ratings, provided the voltage ratios are the same and the impedance percentages (at their own kVA rate) are the same, or almost in these cases, a total of more than 90% of the total of the two evaluations is normally available.

It is recommended that transformers with kVA ratings that differ by more than 2: 1 do not operate in parallel permanently.

Transformers with different kva values ​​can operate in parallel, with a load distribution such that each transformer carries its proportional share of the total load To achieve an accurate load division, it is necessary that the transformers are wound with the same transformation ratio. and that the percentage of the impedance of all transformers is equal when each percentage is expressed on the basis of kva of its respective transformer. It is also necessary that the ratio of resistance to reactant in all transformers is equal.

For satisfactory operation, the circulating current, regardless of the combination of ratios and impedance, should probably not exceed 10% of the rated full load current of the smallest unit. For free training of Engineering and automation courses click here.

7. Identical tap changer and how it works

The only important point to remember is that the tap changer switches should be in the same position for all three transformers and should check and confirm that the secondary voltages are the same.

When the voltage tap needs to be changed, the three taps change switches must be used identically for all transformers. The OL parameters of SF6 must also be identical. If the substation is operating at full load, tripping of one transformer can cause cascade tripping of all three transformers.

In transformers, the output voltage can be controlled either by On-load tap-changer (manual tap-change) or by On – On-load tap-changer – OLTC (automatic change-over).

In the transformer with OLTC is a closed-loop system, with the following components:

1. AVR (Automatic Voltage Regulator) – a programmable electronic device). With this AVR we can set the output voltage of transformers. The output voltage of the transformer is passed to the AVR through the LT panel. The AVR compares the SET voltage and the output voltage and passes error signals, if any, to the OLTC through the RTCC panel for the tap change. This AVR is mounted in the RTCC.

2.RTCC (Remote Tap Changer) – This is a panel consisting of the AVR, display for tap position, voltage and LEDs for up and down relays, selector switches for automatic manual selection… In AUTO mode, the voltage is controlled by the AVR. In manual mode, the operator can increase/decrease the voltage by manually changing the taps using the push button on the RTCC.

3. OLTC is mounted on the transformer – This is an RTCC controlled motor that changes the taps in the transformers

Both transformers should have the same voltage at all taps and when you run transformers in parallel it should run at the same tap position. If we have OLTC with RTCC panel, one RTCC should work as master and another as a follower to keep the same transformer tap positions.

However, current can flow between the two tanks if the impedances of the two transformers are different or if the on-load tap-changer (OLTC) taps are temporarily mismatched due to mechanical delay. Circulating current can cause the protection relays to malfunction.

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