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TRANSFORMER

C
ip
Np
vp
P
A transformer consists of a core made of laminated iron
separated by insulators and a coil of Np turns wound
around the core. This coil is supplied with an a.c voltage
supply vp which then produces a current ip. Due to this
current , a flux  is produced which is given by an
equation
= Npip/S
…….(1)
where S is the reluctance
Since current varied with time ,  also varied with time.
A back electromagnetic force (e.m.f) will be produced
which is given by the equation.
d
vp  N p
dt
……(2)
Substitute  = Npip/S into the above equation , then
vp 
Np2


d
(ip )
dt
……(3)
If ip is sinusoidal, the flux produced also sinusoidal, i.e
 = m sin 2ft
therefore
……(4)
d m sin 2ft 
vp  N p
dt
vp = NP2fmcos 2ft = NP2fmsin (2ft + /2)
The peak value = Vpm = NP2fm
……(5)
……(6)
and vp is leading the flux by /2.
The rms value Vp 
Vpm
2
 0.707  N P 2πf m  4.44 N P f m
……(7)

C
secondary
primary
is
ip
N
vp
p
P
N
Ss
vs
Load
When another coil is wound on the other side of
the core with no of turns Ns , then the fux will
induce the e.m.f vS as given by
d
vs  Ns
dt
……(8)
From (2) and (8) we get
vs Ns

vp N p
Vs N s

Vp N p
Or in rms value
…….(9)
…….(10)
With load , is will flow in the load, mmf at load will
equal to the mmf at input, then (in rms value)
NpIp = NsIs
rearrange
Ip
Ns

Is N p
…….(11)
…….(12)
For ideal transformer, the energy transferred will be the
same as input. Thus power at primary is same power at
secondary.
Pp = Ps
or
IpVp = IsVs
Primary
NP
VP
Secondary
: NS
VS
Symbol for ideal transformer
A 250 kVA,11000V/400V, 50Hz single –phase transformer has
80 turns on the secondary. Calculate
(a) The appropriate values of the primary and secondary
currents;
(b) The approximate number of primary turns;
(c) the maximum value of the flux.
(a) Full-load primary current
P 250 103
Ip 

 22.7 A
Vp
11000
Full-load secondary current
P 250 10-3
Is 

 625 A
Vs
400
(b) Number of primary turns
recall
NP 
Vs N s

Vp N p
Ns
80
 VP 
11000  2200
Vs
400
(c) Maximum flux
recall
E  4.44 N f  m
Es
400
m 

 22.5mWb
4.44 N s f 4.44  80  50
IO
EP
VP
NP
NS
VS
Ideal transformer with no load
Io is the no load current when the secondary is open
circuit. This current consists of Iom that is required to
produce the flux in the core (it is in phase) and Io1 is to
compensate the hysteresis and eddy current losses.
EP
VP= emf of supply to the primary coil
VP
IO
and 90o leads the flux.
IOI
o
EP=emf induced in the primary coil and
same phase as VP.
IOm
VS=emf induced in the secondary coil
VS
and 90o lags the flux.
Phasor diagram for
Iom=magnetizing current to produce flux
no load transfomer
and it is in phase with flux.
Io1=current to compensate the losses
due to hysteresis and eddy current.
2
2
I

I

I
Io=the no load current and given by
o
om
o1
I o1
Power factor cos o 
Io

Transformer converts the energy to high electrical
voltage and transmits in the high voltage line. At the
load, the high voltage energy is converted to low
voltage. In this way, it will compensate the losses during
transferring of the voltage energy.
Transformer 1
Low
voltage
generator
High voltage line
Transformer 2
Low voltage
load
A single-phase transformer has 480 turns on the primary and
90 turns on the secondary. The mean length of the flux path
in core is 1.8m and the joints are equivalent to the airgap of
0.1mm. The value of the magnetic field strength for 1.1 T in
the core is 400A/m, the corresponding core loss is 1.7W/kg at
50Hz and the density of the core is 7800kg/m3.
If the maximum value of the flux is to be 1.1T when a p.d of
2200V at 50Hz is applied to the primary, calculate:
(a) the cross-sectional area of the core;
(b) the secondary voltage on no load;
(c) the primary current and power factor on no load
E  4.44 N f  m
(a) recall
Ep
2200
m 

 0.0206Wb
4.44N p f 4.44  480  50
recall
  B A
 m 0.026
A

 0.0187m 2
B
1.1
(b)
recall
Practically 10% more
allow for insulator
Vs N s

Vp N p
Ns
90
NP 
 VP 
 2200  413V
Vs
480
(c) magnetomotive force (mmf) for the core is
HC C  400 1.8  720 A
mmf for the airgap is H a  a  B   a  1.1 7  0.0001  87.5 A
o
4 10
Total mmf is
recall

 720  87.5  807.5 A
H  N I
Maximum magnetizing current
Rms value
I om 
H 807.5

 1.682 A
N
480
Iom  0.707 I om  0.707 1.682  1.19 A
Volume of core    A  1.8  0.0187  0.0337m3
mass of core
 Vol.  density  0.0337  7800  263kg
Core loss= loss rate x mass
 2631.7  447W
Core-loss component of current
No load current
Power factor
P
447
I o1 

 0.203 A
V p 2200
2
I o  I om
 I o21  1.192  0.2032  1.21A
cos  
I 01 0.203

 0.168 lagging
Io
1.21
E1
I1
I2
V1
V2
L(2)
V1 , E 1
Loaded transformer
L(2)= load with power factor of cos 2
V1 = emf at supply
E1=induced voltage at primary
V2=emf at load
E2=induced voltage at secondary
I1= primary current
I2=secondary current
Phasor diagram
- I2‘ I
1
1
O
2
I2
V2 , E 2
Io

A single-phase transformer has 1000 turns on the primary and
200 turns on the secondary. The no load current is 3A at a
power factor 0.2 lagging when secondary current is 280A at a
power factor of 0.8 lagging. Calculate the primary current and
the power factor. Assume the voltage drop in the windings to
be negligible.
Recall Equation 12
therefore
Ip
Is

IP 
Ns
Np
NS
200
 IS 
 280  56 A
NP
1000
cos2'  0.8
coso  0.2
 sin 2'  0.6
V1 , E 1
- I 2‘
 sin o  0.98
O
I1 cos 1  I 2' cos 2'  I o cos o
 56  0.8  3 0.2  45.4 A
2
I2
 56  0.6  3 0.98  36.54 A
I1 
1
1
Solve for horizontal and vertical components
I1 sin 1  I 2' sin 2'  I o sin o
I
V2 , E 2
45.42  36.542  58.3A
o
36.54


38
50'
1
tan 1 
 0.805
45.4
Power factor cos1  cos 38o50'  0.78 lagging
Io

I1 R1
E1
Path of leakage
I1 ’
L1
RC
Lm
L2
R2
V2 ’
Equivalent circuit of transformer
Flux leakage is due to secondary current that produce flux
which encounter the primary flux. Some of the flux will link to
its own windings and produce induction. This is represented by
inductance L1. Similarly with the flux in secondary and
represented by L2.
V2
There are four main losses
•Dissipated power by wire resistance of the windings (I2R)
•Power due to hysteresis
I1 R1 L1
I1 ’
L2
R2
•Power due to eddy current
•Power via flux leakages.
V1
RC
Lm
E1
E2
Equivalent circuit of transformer
R1= wire resistance of primary windings
L1=inductance due to leakage flux in primary windings
RC=resistance represent power loss due to in hysteresis and eddy
current
Lm= inductance due to magnetizing current Iom
L2=inductance due to leakage flux in secondary windings
R2=wire resistance of secondary windings
V2
V1
-I2’
E1
I1
1
I1 Z
1
I1R1
I0
I1X1
I2R2
I2 X2
2
I2
I2Z2
V2
E2
Phasor diagram for a transformer on load
We can replace R2 by inserting R2’ in the primary thus the
equivalent resistance is
I 2R  I 2R
1
2
2'
2
2
2
Giving us
 I2 
 V1 
R2 '  R2    R2  
 I1 
 V2 
Similarly
 N1 
 V1 
  X 2  
X 2'  X 2 
 N2 
 V2 
2
2
 V1 
Re  R1  R2 '  R1  R2  
 V2 
2
 V1 
X e  X 1  X 2'  X 1  X 2  
 V2 
then
where
2
Ze  Re2  X e2
Re  Z e cos e
(b)
Ze
I1
V1
I2
E1=V2’
E2=V2
X e  Z e sin e
and
Xe
tan e 
Re
Transformer simplified circuit
To
load
Phasor diagram of simplified equivalent circuit of transformer
V1
o-2
E1=V2'

I1Z e
I1Ze
I1

e
I1Xe
I 1Re
I1Xe


e

I1Re

Magnified Ze portion
I2
E2,V2
Complete
Voltage regulation 
recall
no - load voltage  full - load voltage
no - load voltage
Vs N s

Vp N p
Secondary voltage on no-load
 N2 

V2  V1 
 N1 
V2 is a secondary terminal voltage on full load
 N2 
  V2
V1 
N1 

Voltage regulation 
 N2 

V1 
 N1 
Substitute we have
 N1 

V1  V2 
N2 

Or Voltage regulation 
V1
per unit
 N1 

V1  V2 
N2 


100
V1
per cent
From phasor diagram can be proved that
I1 Z e cose  2 
Per  unit voltage regulation 
V1
Or
I1 Re cos 2  X e sin 2 
Per  unit voltage regulation 
V1
A 100kVA transformer has 400 turns on the primary and 800
turns on the secondary. The primary and secondary resistances
are 0.3W and 0.01W respectively, and the corresponding
leakage reactances are 1.1W and 0.035W respectively. The
supply voltage is 2200V. Calculate:
(a)The equivalent impedance referred to the primary circuit;
(b)The voltage regulation and the secondary terminal voltage for
full load having a power factor of (i) 0.8 lagging and (ii) 0.8
leading.
(c)The percentage resistance and leakage reactance drops of the
transformer
(a)
2
2
2
2
 V1 
 400 
Re  R1  R2    0.3  0.01
  0.55W
 80 
 V2 
 V1 
 400 


X e  X 1  X 2    1.1  0.035
  1.975W
 80 
 V2 
Z e  Re2  X e2 
(b) (i)
0.552  1.9752
 2.05W
P 100 103
Full  load primary current  
 45.45 A
V
2200
I1 Re cos 2  X e sin 2 
Per  unit voltage regulation 
V1
45.450.55  0.8  1.975  0.6

 0.0336 per unit
2200
 3.36 per cent
Sec. terminal voltage on no-load  VP 
The decreasing of full-load voltage is
NS
80
 2200 
 440V
NP
400
 440  0.0336  14.8V
Therefore the secondary full-load voltage
 440 14.8  425.2V
(b) (ii) power factor 0.8 leading
Voltage regulation 
45.450.55  0.8  1.975  0.6
 0.0154 per unit
2200
 1.54 per cent
The increasing of full-load voltage is
Therefore the secondary full-load voltage
 440  0.0154  6.78V
 440  6.78  446.78V
Or
 full - load
  equivalent resistance

  
primary current   referred to primary
Resistance drop per unit  
primary vo ltage



 full - load
  equivalent resistance

  
secondary current   referred to secondary
Resistance drop per unit  
secondary voltage on no - load



 Resistance drop

I1R e
45.45  0.55
 0.0114

V1
2200
 0.0114%
Per unit
Alternative
Recall Equation 12
Full load secondary current
Equivalent resistance
referred to secondary
Ip
Is

Ns
Np
NP
400
IS 
IP 
 45.45  227.2 A
NS
80
 N2
Re  R2  R1 
 N1
Secondary voltage on no-load
 VP 
2

 80 
  0.01  0.3
  0.022W
 400 

2
NS
80
 2200 
 440V
NP
400
 full - load
  equivalent resistance

  
secondary current   referred to secondary
Resistance drop per unit  
secondary voltage on no - load
 0.0114 per unit  1.14%


  227.2  0.022
440
equivalent leakage resistance referred to primary
2
V 
 400 
X e  X 1  X 2  1   1.1  0.035
  1.975W
 80 
 V2 
2
P 100 103
Full  load primary current  
 45.45 A
V
2200
 full - load
  equivalent leakage resistance

  
primary
current
  referred to primary
Leakage ractance drop per unit  
primary vo ltage
 0.0408 per unit  4.08%


  45.45  1.975
2200