II. Transformers
Transformer
Transformer comprises two or more windings coupled by a common
magnetic circuit (M.C.).
If the primary side is connected to an AC voltage source v1(t), an AC
flux  (t) will be produced in the M.C.
Primary side
Secondary side
1
Ideal transformer voltage relationship
e1 ( t )  N 1
d ( t )
dt
e2 (t )  N 2
d ( t )
dt
e1 N 1

e2 N 2
Ideal transformer symbol
v1 N 1

v2 N 2
Another representation
n
N1
N2
2
There are two common types of transformer construction:
(a) core-type
(b) shell-type
1. Analysis of the transformer
with no-load
Secondary side
open-circuited:
(no sec. current)
3
Modelling of Magnetic Core
rc: core loss resistance
(hysteresis & eddy-current loss)
xm: magnetizing reactance
Ie : exciting current
gc: core loss conductance
gc - jbm: magnetizing admittance
1
gc 
rc
1
 jbm 
jx m
bm : magnetizing susceptance
Phasor diagram
Im lags E1 by 90º
Ie  Ic  Im
Ic in phase with E1
Modelling of Leakage Flux
Let us express the voltage drop due to leakage flux in the primary winding below
v1 
d  1
dt
v  1  L 1

d  1 die
di
 L 1 e
dt
die dt
di e
dt
L 1 :
where
  1  N 1  1
leakage inductance of primary winding
primary leakage reactance:
x1  ω L 1
ω  2f
Modelling of Copper Loss
r1 :
resistance of primary winding
4
Equivalent circuit model of primary side

primary leakage
impedance



exciting branch
impedance
Exciting current is only a few percent of rated primary current of the transformer
2. Ideal Transformer Operation
• No leakage fluxes
• Negligible winding internal resistances
• B-H characteristic of the magnetic material is singlevalued, and linear
– No hysteresis loss
• Magnetic core has a very high r, i.e. Core reluctance
is negligible.
• No copper, no core losses (Efficiency  = 100%)
• Interwinding capacitatances are negligle at power
frequencies (50Hz, 60Hz)
5
Basic Relations
1. From Faraday’s Law: e1 = N1 d /dt, and
e2 = N2 d /dt
So,
e1 N 1

e2 N 2
2.
Since winding resistances & leakage fluxes are
negligible:
v1 N 1

v2 N 2
3.
F1= F2
i1 N 2

i2 N 1
4. Ideal transformer symbol
6
4. No power loss
Conservation of power:
v1 i1  v 2 i2
5. Under Load
Ideal transformer under load
v1 N1

v2 N2
i1 N 2

i 2 N1
v 1i 1  v 2 i 2
v 2  i 2Z2

I 1  I 2
Eqv. crt. referred to primary
Secondary impedance
referred to primary side:
2
N 
Z 2   1  Z 2  n 2 Z 2
 N2 
7
Terminology
V1 , I 1 , Z 1 : actual primary quantities
V2 , I 2 , Z 2 : actual secondary quantities
V1, I 1 , Z 1 : primary quantities referred to secondary side
V2 , I 2 , Z 2 : secondary quantities referred to primary side
V1 
N2
V1 ,
N1
N
V2  1 V2 ,
N2
I 1 
N1
I1 ,
N2
N
I 2  2 I 2 ,
N1
2
N 
Z 1   2  Z 1
 N1 
2
N 
Z 2   1  Z 2
 N2 
3. Equivalent circuit representation of a
practical transformer
Transformer under load
8
Equivalent Circuit representation

 




secondary leakage
impedance
primary leakage
impedance

exciting branch



ideal transformer
r1: Primary winding internal resistance ()
r2: Secondary winding internal resistance ()
x1: Primary winding leakage reactance ()
x2: Secondary winding leakage reactance ()
rc: Core-loss resistance ()
xm: Magnetizing reactance ()
Equivalent circuit referred to primary side
r2: Secondary winding internal resistance
referred to primary side
x2: Secondary winding leakage reactance
referred to primary side
I2: Secondary winding current
referred to primary side
V2: Secondary winding voltage
referred to primary side
ZL: Load impedance referred to primary side
r2  n 2 r2
x 2  n x 2
I
I 2  2
n
V2  nV2
2
n
N1
N2
E1  E 2  nE 2
Z L  n 2 Z L
9
Equivalent circuit referred to secondary side
r1: Primary winding internal resistance
referred to secondary side
r1 
x1: Primary winding leakage reactance
referred to secondary side
x1 
I1: Primary winding current
referred to secondary side
rc 
V1: Primary winding voltage
referred to secondary side
rc : Core-loss resistance referred to sec. side
x m 
xm : Magnetizing reactance referred to sec. side
r1
n
2
n
x1
n2
rc
n
2
xm
n2
N1
N2
I 1  nI 1
V1 
V1
n
E 2  E1 
E1
n
Equivalent circuit referred to secondary side
gc : Core-loss conductance referred to sec. side
n
bm : Magnetizing admittance referred to sec. side
g c 
1
rc
g c  n 2 gc
 
 jbm
N1
N2
1
jx m
  n 2 bm
bm
10
Simplification
VTh, ZTh
Let us apply Thévenin equivalent circuit to the primary side
rc || jx m 
VTh  V1
r1 
rc || jx m
 V1
jx1   rc || jx m 
Z Th  r1  jx1  || rc || jx m 
where
1
g c  jbm
rc || jx m  r1  jx1
VTh  V1
 r1  jx1
Z Th  r1  jx1
Simplification
E1  VTh  Z Th I 2
 VTh
 V1
where Z Th  r1  jx1
E1  V1
Error made in approximations is at most 1% – 2% compared to actual values.
11
Approximate Equivalent Circuit
referred to primary side
Further simplification gives us the figure below
req  r1  r2
x eq  x1  x 2
Approximate Equivalent Circuit
referred to secondary side
Further simplification gives us the figure below
  r1  r2
req
  x1  x 2
x eq
12
Approximate Equivalent Circuits for Large Transformers
(referred to primary side)
(of a few 100 kVAs)
rc || jx m is very large
(in the MVA range)
x eq  req
(4 - 10 times)
AC Power
 : angle between voltage V and current I
:
P = Vrms I rms cos
[W, Watts]
Reactive Power
:
(Imaginary Power)
Q = Vrms I rms sin 
[VAR, Volt Ampere Reactive]
Complex Power
:
S = P  jQ
Apparent Power
:
S = Vrms I rms [VA, Volt Ampere]
Power Factor
:
cos 
Real Power
P
S
13
Transformer Power Flow
Pin = V1 I 1 cos 1
Pout = Pload = V2 I 2 cos 2
Pin = Pcu1  Pcore  Pcu 2  Pload
Pcu : Copper loss
Pcore : Core loss
Pcu1  I 12 r1  I 1 2 r1
Pcore  E12 gc  E1 2 gc
Pcore  V12 g c
Pcu 2  I 22 r2  I 2 2 r2
4. Short-circuit and Open-circuit Tests
• Measure voltage (V), current (I) and power (P) in
order to determine the equivalent circuit
parameters of the transformer:
– For leakage impedance parameters
• with secondary short circuited
– For exciting branch parameters
• with secondary open circuited
14
i. Short-circuit Test
Vsc , I sc , Psc
A reduced voltage Vsc of 2% - 10 % of rated voltage is applied to allow rated primary current.
where
1
 req  jx eq
g c  jbm
Short circuit equivalent circuit
Short-circuit Test
req 
Psc
z eq 
2
I sc
V sc
I sc
where
x eq 
2
z eq  req2
req
req  r1  r2
where r1  r2
r1  r2 
x eq  x1  x 2
where x1  x 2
x1  x 2 
at 50 Hz
Form factor
r1 AC  1.1 r1 DC
r2 AC  1.1 r2 DC
z eq  req  j x eq
2
x eq
2
Measured DC resistance
15
ii. Open-circuit Test
Voc , I oc , Poc
Rated voltage is applied to the transformer under no-load and exciting current flows,
which is a few percent of rated current.
I 2  0
where
and
I e  I1
Open circuit equivalent circuit
Open-circuit Test
gc 
Poc
Yc 
Voc2
I oc
Voc
where Yc  g c  j bm
bm 
2
Yc  g c2
where g c 
1
rc
and
bm 
1
xm
16
5. Voltage regulation (VR%)
• The change in secondary terminal voltage
(load voltage) from no-load to full-load
– expressed as a percentage (%) of the rated value
– ideally VR% = 0.
VR% 
V1  V2
 100%
V1(rated)
or
VR% 
V1  V2
 100%
V2(rated)
Simplification by approximation


  j x eq
 I2
V1  V2  req
 cos   I 2 x eq
 sin 
V1  V2  I 2 req
VR% 
 cos   I 2 x eq
 sin 
I 2 req
V2
 100%
17
Zero regulation, i.e. VR% = 0
• For zero regulation, phase angle ( ) of the load is given by
VR% 


I2
 sin   100%
r  cos   x eq
V2 eq
 cos   x eq
 sin   0
req
 
 req

 
 x eq
   tan 1 
Note that  < 0, so
Load must be capacitive for zero regulation
NOTE
• For an inductive load, ZL = RL + jXL
– Always V1 > V2, i.e. VR% > 0
• For a capactive load, ZL = RL – jXL
 sin   req
 cos  )
– Usually V1  V2, i.e. VR%  0 ( where  xeq
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6. Efficiency (%)
• The ratio of the output power given to the
load and the input power taken from the
electrical supply
– expressed as a percentage (%)




Pout
100%
Pin

Pout
100%
Pout  Plosses
Pout
100%
Pout  Plosses
Pout
Plosses  Pcu  Pcore
Pout
100%
 Pcu1  Pcu 2  Pcore
V2 I 2 cos 
V2 I 2 cos   I 12 r1  I 22 r2  V12 g c

Pcu  Pcu1  Pcu 2
100%

Pcu  I 22 req
V2 I 2 cos 
100%
  V12 g c
V2 I 2 cos   I 22 req
or

V2 I 2 cos 
100%
V2 I 2 cos   I 2 2 req  V12 g c
19
Maximum efficiency
d
Let us find the value of I2 which maximizes the efficiency dI  0
2
where
d
0
dI 2



V2 I 2 cos 
  Pcore
V2 I 2 cos   I 22 req
 
100%

  Pcore  V2 cos   2 I 2 req
 V2 I 2 cos   0
V2 cos  V2 I 2 cos   I 22 req


  Pcore  V2 cos   2 I 2 req
 I2  0
V2 I 2 cos   I 22 req
 0
Pcore  I 22 req

Pcore  I 22 req
Pcore  Pcu
i.e.,
  I 2 2 req
V12 g c  I 22 req
Thus, maximum efficiency is achieved if core loss equals to the copper loss.
Examples
1. A 12kVA, 220/440V, 50 Hz single phase transformer has the
following test data:
•
•
No-load test: 220V, 2A, 165W (measured at primary side)
Short-circuit test: 12V, 15A, 60W (measured at secondary side)
a) Calculate the equivalent circuit parameters referred to primary
side
b) Calculate the primary terminal voltage on full-load at a power
factor of: 0.8 pf lagging.
20
2. Given a 250kVA, 4160:480V, 60 Hz transformer, the following
parameters are obtained by tests
• r1 = 0.09  and x1=1.7 
• r2 = 1.2  10-3  and x2 = 2.26  10-2 
Neglecting core losses,
a) Calculate the primary voltage and voltage regulation for rated
load at 76% pf lagging
b) Repeat a) for a load of 76% pf leading
c) Calculate the transformer efficiency for parts a) and b) with a core
loss Pcore = 547W.
3. The parameters of the exact equivalent circuit of a 150kVA,
2400/240V transformer are
• r1 = r2 = 0.2 
• x1 = x2 = 0.45 
• rc = 10 k and xm = 1.55 k
Using both the exact and approximate equivalent circuit of the
transformer, determine
a) Voltage regulation
b) Efficiency
for rated load at 0.8pf lagging
21
4. At 10kVA, 8000:230V transformer has a leakage impedance
referred to primary of 90+ j400. Exciting branch parameters are
• rc = 500 k and xm = 60 k
a) If primary voltage V1 = 7967V and actual load impedance
ZL = 4.2+3.15, find the secondary voltage of the transformer
b) If the load is disconnected, and a capacitor of –j6 is connected
in its place, what will be the load voltage?
22