15/04/2014
EEEN60301 Power System Modelling
Power Transformers
Monday 23rd October 2013
Power Transformers
Prof Peter Crossley
Ferranti Building, C14
School of Electrical and Electronic Engineering
p.crossley@manchester.ac.uk
01613064803(office)
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Transformers in Power Systems
400 kV transformer core and winding,
*from Electrical Engineering Portal
400 kV Power transformer, *from ALSTOM Grid
400 kV transformer end insulation, *from
CIGRE brochure 323
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Outline
• Concept (‘Ideal’ Transformer)
• Equivalent Circuit (‘Real’ Transformer)
– No load condition, load condition
• Determination of Circuit Parameters
– Open circuit test, short circuit test
• Transformer Operation Performance
– Voltage regulation, efficiency
• Transformer Design and Construction
– Turns, three phase, auto transformer
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Basic Electromagnetism
Ampère’s law
 H dl   N  i
Faraday’s law
d
e  N
dt
André-Marie Ampère (1775-1836)
* from Wikipedia
Michael Faraday 1791-1867
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Concept of Transformer
Primary
Secondary
• ‘Transform’ voltage
(and current);
• ‘Transform’ ac
voltage (not dc
voltage);
• Sine wave
‘transforms’ to sine
wave
e  N
Right-hand grip rule
d
dt
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‘Ideal’ Transformer
• Voltage ratio
V1 N1

V2 N 2
E1 N1

E2 N 2
•
•
•
•
No resistance
No leakage flux
No core loss
Permeability μr =∞
e  N
d
dt
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‘Ideal’ Transformer
• Current ratio
• Ampère’s law
 H dl   N  i
• μr ∞
• Magnetic field intensity (H)
H
I1 N 2

I 2 N1
N1 I1  N 2 I 2  0
B
 r 0
0
   r  0
Flux density B    H
Magnetic flux   B  A
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‘Ideal’ Transformer
• Power, Var flow
S  P  jQ  V I
V1 I1  V2 I 2
P1  jQ1  P2  jQ2
V1 N1

V2 N 2
I1 N 2

I 2 N1
P1  P2
Q1  Q2
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Transfer of Impedance
• Assumed an impedance ZL connected at
secondary side, refer this impedance into
the primary side as ZL’
• Derive the equation:
V2  I 2 Z L
V1 N1

V2 N 2
I1 N 2

I 2 N1
Z L ' (
N1 2
) ZL
N2
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‘Real’ Transformer
• Coil resistance: R1, R2
• Coil leakage flux: ΦL1(X1), ΦL2(X2)
• Permeability μ is limited, so magnetising current
is required to establish Φm: Xm
• Core losses, the effect of eddy current and
hysteresis loss: Rm
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‘Real’ Transformer
• No load condition
Equivalent Circuit
V1   E1  I 0 ( R1  jX 1 )
V2  E2
E1
E2

2 E sin  t   N
N1
N2
d
2E

cos t
dt
N
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‘Real’ Transformer
• Load condition
Equivalent Circuit
V1   E1  I1 ( R1  jX 1 )
V2  E2  I 2 ( R2  jX 2 )
E1
E2

N1
N2
I1  I 2
N2
 I0
N1
V2  I 2 Z L
I0 
 E1
Zm
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‘Real’ Transformer
• Load condition
Equivalent Circuit (referred to primary side)
Transformer impedance
Z w  ( R1  R'2 )  j ( X 1  X '2 )
Z w  ( R1  R'2 ) 2  ( X 1  X '2 ) 2
Simplified Equivalent Circuit (referred to primary side)
(Neglect magnetising current and core losses)
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Determination of Circuit Parameters
• Open-circuit test:
– Apply rated voltage V1N
– Keep secondary side open
– Measure Voltage Voc, Current Ioc, Power Poc
Voc2
Rm 
Poc
Xm 
I2=0, I1=0;
Ignore R1,X1 (<< Rm,Xm)
Voc
V
I oc2  ( oc ) 2
Rm
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Determination of Circuit Parameters
• Short-circuit test:
– Apply rated current I1N
– Shot circuit secondary side
– Measure Voltage Vsc, Current Isc, Power Psc
( R1  R'2 ) 
Psc
I sc2
( X 1  X '2 )  (
V2=0, V1 small;
Ignore Rm,Xm
Vsc 2
)  ( R1  R'2 ) 2
I sc
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Transformer Operation Performance
• Voltage Regulation:
Difference between voltage magnitude at no-load
and load conditions expressed as a percentage
of load value.
Re g (%) 
V2noload  V2load
load
2
V
100
Voltage regulation is due to the voltage drop on
transformer impedance at load condition.
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Transformer Operation Performance
• Voltage Regulation:
Simplified Equivalent Circuit
(referred to secondary side)
V  V2noload  V2load  AB
VRe g (%) 
I 2 ( R'1  R2 ) cos  2  I 2 ( X '1  X 2 ) sin  2
100
V2load
VRe g (%) 
I 2 Rw cos  2  I 2 X w sin  2
100
V2load
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Transformer Operation Performance
• Transformer impedance (resistance,
leakage reactance) can also be
expressed as a percentage voltage drop:
Vz (%) 
I FL Z w
100
V2
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Transformer Operation Performance
• Efficiency:
– Efficiency is defined in terms of power transfer:
Efficiency 

Output power
Output power

Input power
Output power  Losses
Output power
Output power  I 2 R loss  Core loss
– Core loss is combination of eddy current and
hysteresis losses, constant at constant voltage
– I2R or copper loss varies with the load
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Transformer Design and Construction
• Volts per turn and flux density
For a given core, the cross-sectional area (A) of
the limb is a constant, the relationship between
volts per turn (E/N) in the winding and the flux
density (B) remains constant at a given
frequency (f).

2E
cos t
N
E
 4.44 f  
N
E
 4.44 f  A  Bm
N
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Transformer Design and Construction
• Exercise:
The maximum flux density within the magnetic
core of a 50 Hz, 400 kV/132 kV transformer is
restricted to 1.55 Tesla as the core has a circular
cross section with a diameter of 1 m. Calculate
the volts per turn for the winding and the number
of turns for HV and LV windings.
(This transformer is connected as Yyn*)
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Transformer Design and Construction
• Solution:
The phase voltage for HV winding is
The phase voltage for LV winding is
E1
3
E2
3

400

132
3
3
 230kV
 76kV
Volt per turn is calculated as
E
 4.44 fBm A  4.44  50 1.55    0.52  270 (V )
N
The HV winding has N1=230000/270=851 turns
The LV winding has N2=76000/270=281.5 turns
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Transformer Design and Construction
• Transformer construction
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Transformer Design and Construction
• Auto Transformer
– single winding per phase, low voltage terminal
is made from a tap part way down the winding
– more economical than two-winding
transformer for voltage ratio up to 3:1
– auto-transformers are usually star connected
and share the same neutral, often undesirable
except in transmission system where solid
earthing at all voltage level
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Transformer Design and Construction
• Auto Transformer
N1: series winding
N2: common winding
I1
+
V1
N1
I2
I2-I1
V1/(N1+N2)=V2/N2
-
N2
+
V2
-
I1N1=(I2-I1)N2
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Transformer Design and Construction
• Three phase winding connection
A
– Star/ Y connection
B
C
Advantages
– more economical for a high voltage winding
– neutral point available for earthing
– permits reduced insulation level of the neutral
– permits taps and tap changer to be located near
neutral
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Transformer Design and Construction
• Three phase winding connection
A
– Delta/ D connection
C
B
Advantages
– more economical for a high current, low voltage
winding
– in combination with a start connected winding, it
reduces the zero-sequence impedance current
in that winding
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Transformer Design and Construction
• Phase relationship
A conventional notation indicating the connections
of the high-voltage and low-voltage windings and
their relative phase displacement expressed as a
combination of letters and clock-hour figure.
First symbol: HV side (minute hand, 12 o’clock)
Second symbol: LV side (hour hand)
Third symbol: phase displacement expressed as
the clock hour number
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Transformer Design and Construction
• Phase relationship
HV
delta
D
LV
star
interconnected star
delta
star
interconnected star
Y
Z
d
y
z
Group number
I
II
III
IV
Phase displacement
0o
180o
-30o
30o
Example:
Group I connection Yy0
Clock hour number
0
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1
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Transformer Design and Construction
• Phase relationship
Group I Yy0
Group IV Dy11
Group IV Yd11
Group IV Yz11
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Summary
1. Transformer Equivalent Circuit
Z2 '  (
N1 2
) Z2
N2
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Summary
2. Determination of Circuit Parameters
Open-circuit test
Voc2
Rm 
Poc
Xm 
Voc
V
I oc2  ( oc ) 2
Rm
Short-circuit test
( R1  R'2 ) 
Psc
I sc2
( X 1  X '2 )  (
Vsc 2
)  ( R1  R'2 ) 2
I sc
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Summary
3. Voltage Regulation
Re g (%) 
V2noload  V2load
V2load
100
4. Efficiency

Output power
Output power  I 2 R loss  Core loss
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Summary
5. Volts per Turn
E
 4.44 f  
N
E
 4.44 f  A  Bm
N
6. Phase Relationship
Group I Yy0
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Reading…
• Power System Analysis
– John J. Grainger, William D. Stevenson.
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