1/11/2016
Lesson 12: Parallel Transformers
and Autotransformers
ET 332b Ac Motors, Generators Lesson 12_et332b.pptx
and Power Systems
Lesson 12_et332b.pptx
Learning Objectives
After this presentation you will be able to:
Explain what causes circulating currents in parallel and
compute its value.
Compute the load division between parallel
transformers.
Explain how autotransformers operate
Make calculation using ideal autotransformer model
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Lesson 12_et332b.pptx
Parallel Operation of Transformers
Currents circulate
between A and B based
on the voltage
difference and
transformer
impedance even with
EA≠EB no load
EA EB
Ic
ZA ZB
Where: EA = operating voltage of
When voltage ratios are not equal, transformer A
currents circulate between the EB = operating voltage of
windings of each transformer without a transformer B
load connected. Circulating currents ZA = series impedance of A
reduce the load capacity of transformer ZB = series impedance of B
Lesson 12_et332b.pptx
Capacity Loss Due to Circulating
Currents
Find effects using superposition Transformer A current
IA+Ic ITA I A Ic
Transformer B current
ILoad ITB I B Ic
IB-Ic Ic driven by EA – EB
Adding circulating
current to transformer
A increases total
current in winding. Not
seen in load current.
Can cause overload
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Circulating Current Example
Example 12-1: Two 100 kVA single phase transformer operated in
parallel.
Nameplate data:
Transformer V-ratio %R %X
A 2300-460 1.36 3.50
B 2300-450 1.40 3.32
Find Ic magnitude and Ic as percent of transformer secondary ratings
Lesson 12_et332b.pptx
Example 12-1 Solution (1)
Use per unit method – Vbase = secondary voltage
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Example 12-1 Solution (2)
Use formula
Convert per unit to percent
Now convert this to amps using a base
Ans current
29.55% of Transformer A’s
capacity is consumed by Ic.
Lesson 12_et332b.pptx
Load Division Between Parallel Transformers
When turns ratios are equal, the load current divides following the
winding impedance of the transformers. More current flows through
the lowest impedance.
All Transformer Z's and Load Z referred
ZA
IA
to the same side of transformer or all per
ZB unit (%) quantities
IB
1 1 1 1
Iin
Zk YA , YB ... Yk ...Yn
Ik ZA ZB Zk Zn
VT
Zn Use current divider rule
In
Yp YA YB ...Yk ...Yn
Load
Y
Circuit model I k Iin k Findsththe current in
Y the k transformer
p
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Parallel Transformer Example
Example 12-2: A 100 kVA transformer is to be paralleled with a
200 kVA transformer. Each transformer has rated voltages of 4160 -
240 V. Their percent impedances based on the ratings
of each are:
Z% = 1.64+3.16j % 100 kVA
Z% = 1.10+4.03j % 200 kVA
Find: a) rated high side current of each transformer
b) % of total bank current drawn by each transformer
c) maximum bank load that can be handled without
overloading either transformer
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Lesson 12_et332b.pptx
Example 12-2 Solution (1)
a) Rated current of both transformers
Transformer A: 100 kVA
Transformer B: 200 kVA
b) Percent current drawn by each transformer
Convert %Z to actual ohms. Need base impedances
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Example 12-2 Solution (2)
Convert p.u. to ohms
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Lesson 12_et332b.pptx
Example 12-2 Solution (3)
Find the admittance
Total admittance.
Now use current divide rule to find flows through
each transformer.
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Lesson 12_et332b.pptx
Example 12-2 Solution (4)
Find IA and IB in terms of Iin
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Lesson 12_et332b.pptx
Example 12-2 Solution (5)
Transformer A carries 37.17% of the
total load
Transformer B carries 63.35% of the
total load
c) Find the maximum load of the parallel transformers without an overload
Let IA=IratedA=24.04 A and compute Iin using relationships above. Then
find flow through other transformer
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Example 12-2 Solution (6)
TX B not
overloaded
IratedB=48.08
Let IB=IratedB=48.08 A. Find Iin and
then compute the flow in
transformer A
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Lesson 12_et332b.pptx
Example 12-2 Solution (7)
Max load, Iin=64.68 A
Find bank power
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Autotransformers
Autotransformers use a single taped coil to change voltage levels
and current levels – They provide no electrical isolation
NLS = number of turns
"embraced" by low side
NHS = number of turns on high
side
Polarity of induced voltages
determined by direction of
current and winding wraps.
If NLS = 20 and NHS = 80 Step-down action
N HS VHS 80 VHS 120
a a 4 VHS 120 V so VLS 30 V
N LS VLS 20 a 4
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Lesson 12_et332b.pptx
Autotransformers: Step-Down Operation
Some load is transferred via conduction from one side to the other
and some is transferred by transformer action
Like two winding transformers
SHS SLS
VHS I HS VLS I LS
Itr = the current from
transformer action
Transformed
current
Low side current must
increase to maintain
Autotransformer connected in step- power balance so:
down mode. Note direction of Itr I LS I HS I tr
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Autotransformer Current Ratio and
Step-Up Operation
I HS 1 N
Current ratio of autotransformer Where a HS
I LS a N LS
Autotransformer In Step-up Mode
Coils in these diagrams are
series aiding
(induced voltages add)
Note: direction of Itr reversed to maintain power balance
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Lesson 12_et332b.pptx
Autotransformers from Two-Winding
Transformers
Autotransformer action can be obtained by proper connection of
two winding transformer coils
For step-down mode
N1 N HS N1 N 2
VHS
N LS N 2
N HS N1 N 2 VHS
Load N2 a
N LS N2 VLS
Where: N1 = number of turns in primary
(HV)
N2 = number of turns in
secondary (LV)
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Autotransformers from Two-Winding
Transformers
Step-Down Connections
Find VLS with VHS=120 V, N1=500 and N2 =100 VHS
a Where VHS 120 V
VLS
N HS N1 N 2 VHS 500 100
a a 6 VHS 120 V
N LS N2 VLS 100 VLS 20 V
a 6
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Lesson 12_et332b.pptx
Autotransformer Example
Example 12-3: 400 turn autotransformer operating at a
25% tap supplies a 4.8 kVA load at 0.85 lagging P.F. VHS =
2400 V
Find:
a) load current b) incoming line current c) Itr d) apparent
power transformed and conducted
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Example 12-3 Solution (1)
Find turns ratio
Find secondary voltage
a) Find ILS
Ans
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Lesson 12_et332b.pptx
Example 12-3 Solution (2)
b) Find high-side current
Current must decrease to
maintain power balance
c) Find transformed current
Ans
d) Find transformed and conducted apparent powers
Ans
Ans
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ET 332b Ac Motors, Generators and Power Systems
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