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Lesson 12: Parallel Transformers
and Autotransformers ET 332b Ac Motors, Generators and Power Systems
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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
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Parallel Operation of Transformers
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When voltage ratios are not equal, currents circulate between the windings of each transformer without a load connected. Circulating currents reduce the load capacity of transformer
EA≠EB
Currents circulate between A and B based on the voltage difference and transformer impedance even with no load
Where: EA = operating voltage of transformer A
EB = operating voltage of transformer B
ZA = series impedance of A ZB = series impedance of B
BA
BAc
ZZ
EEI
Capacity Loss Due to Circulating
Currents
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Find effects using superposition Transformer A current
IA+Ic cATA III
IB-Ic
cBTB III
Transformer B current
Ic driven by EA – EB
Adding circulating current to transformer A increases total current in winding. Not seen in load current. Can cause overload
ILoad
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Circulating Current Example
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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
Example 12-1 Solution (1)
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Use per unit method – Vbase = secondary voltage
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Example 12-1 Solution (2)
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Use formula
Convert per unit to percent
29.55% of Transformer A’s capacity is consumed by Ic.
Ans
Now convert this to amps using a base current
Load Division Between Parallel Transformers
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When turns ratios are equal, the load current divides following the winding impedance of the transformers. More current flows through the lowest impedance.
VT
ZA
ZB
Zk
Zn
Load
Iin
IA
IB
Ik
In
All Transformer Z's and Load Z referred to the same side of transformer or all per unit (%) quantities
Circuit model
n
n
k
k
B
B
A
AZ
1Y...
Z
1Y ...
Z
1Y ,
Z
1Y
Use current divider rule
nkBAp Y...Y...YYY
p
kink
Y
YII Finds the current in
the kth transformer
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Parallel Transformer Example
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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
Example 12-2 Solution (1)
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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)
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Convert p.u. to ohms
Example 12-2 Solution (3)
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Find the admittance
Total admittance. Now use current divide rule to find flows through each transformer.
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Example 12-2 Solution (4)
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Find IA and IB in terms of Iin
Example 12-2 Solution (5)
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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)
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TX B not overloaded
IratedB=48.08
Let IB=IratedB=48.08 A. Find Iin and then compute the flow in transformer A
Example 12-2 Solution (7)
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Max load, Iin=64.68 A
Find bank power
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Autotransformers
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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
LS
HS
LS
HS
V
V
N
Na 4
20
80a V 30
4
120
a
VV so V 120V HS
LSHS
Step-down action
Autotransformers: Step-Down Operation Lesson 12_et332b.pptx
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Some load is transferred via conduction from one side to the other and some is transferred by transformer action
Autotransformer connected in step-down mode. Note direction of Itr
Like two winding transformers
LSLSHSHS
LSHS
IVIV
SS
Itr = the current from transformer action Low side current must increase to maintain power balance so:
Transformed current
trHSLS III
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Autotransformer Current Ratio and
Step-Up Operation
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Current ratio of autotransformer LS
HS
LS
HS
N
Na Where
a
1
I
I
Autotransformer In Step-up Mode
Note: direction of Itr reversed to maintain power balance
Coils in these diagrams are series aiding (induced voltages add)
Autotransformers from Two-Winding
Transformers
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Autotransformer action can be obtained by proper connection of two winding transformer coils
N2
N1
VHS
Load
For step-down mode
2LS
21HS
NN
NNN
LS
HS
2
21
LS
HS
V
V
N
NN
N
Na
Where: N1 = number of turns in primary (HV)
N2 = number of turns in secondary (LV)
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Autotransformers from Two-Winding
Transformers
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Step-Down Connections
Find VLS with VHS=120 V, N1=500 and N2 =100
LS
HS
2
21
LS
HS
V
V
N
NN
N
Na
6
100
100500a
V 120V WhereV
Va HS
LS
HS
V 206
V 120
a
VV HS
LS
Autotransformer Example
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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)
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Find turns ratio
Find secondary voltage
a) Find ILS
Ans
Example 12-3 Solution (2)
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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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