Download Single Phase Transformer© Nafees Ahmed By

Single Phase Transformer©
By
Nafees Ahmed
Asstt. Prof. Department of Electrical Engineering
DIT, University, Dehradun, Uttarakhand
1.
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3.
4.
1
References:
V. Del Toro “Principles of electrical engineering” Prentice Hall International
B. L Theraja “ Electrical Technology” S. Chand
www.google.com
www.wikipedia.org
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 It is a static device which transfers energy form one electrical circuit to another electrical
circuit without change in frequency.
 Very high efficiency (up to 96% or up to 98%, Because there is no rotating part)
Types of transformer:
Depending on types of phase
1. 1-ϕ transformer
2. 3- ϕ transformer
Depending on use
a. Step up transformer: Output voltage is higher then applied voltage
b. Step down transformer: Output voltage is less then applied voltage
Depending of construction
1. Shell type transformer
2. Corel type transformer
HL
VV
L H
V V
HL
VV
L H
V V
LV
LV
HV
HV
LV
LV
HV
HV
LV
LV
Core type
Shell type
Figure 1
Construction and working principle:
Construction:
Laminated Core
Flux
Secondary
Secondary
(HV/LV)
(HV/LV)
Primary
(LV/HV)


2
Figure 2
Laminated core of Si-Steel (Soft Steel)
Soft Steel is used to reduce hysteresis losses
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
Laminated core is to reduced eddy current losses
Principle: Based on low of electromagnetic induction.
Ideal Transformer: Conditions for ideal transformer are
1. Copper losses are zero:
=> Winding resistance = 0
2. Iron (Core) losses are zero: => Hysteresis & eddy current losses =0
3. Leakage flux is zero:
=> Permeability   of iron is infinity
=>
4. Magnetizing current is zero
S
1 l
0
 A
EMF Equation:
(Prove that emf per turn for a transformer is constant)
ϕ
v1
e1
N1
N2
e2
v2
Figure 3: Core type 1-Ph transformer
Let an alternating voltage (v1) is applied to the primary side
So an alternating flux will produce in the core
Let
   m Sint      (1)
Because of this an emf will induce in primary as well as in secondary windings
Primary induced emf
e1   N1
d
dt
d  m Sint 
dt
e1   N1 m Cost
e1   N 1
e1  N1 m Sint   / 2
Or
Where
3
e1  E m1 Sint   / 2      (2)
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E m1  N1 m  Max Value of induced emf
So its RMS value
E1 
Em1 N1m N1m 2f


 4.44 fm N1
2
2
2
E1  4.44 f m N1      (3)
Similarly RMS values of secondary induced emf
E 2  4.44 f m N 2      (4)
For ideal transformer
Induced emf in primary (E1) = Applied Voltage (V1)
Induced emf in secondary (E2) = Terminal Voltage (V2)
From equations (3) & (4)
E1 E 2

 4.44 f m  Constant
N1 N 2
Hence
emf/turn = constant
Voltage and current transformation ratio:
Again from equations (3) & (4)
V2 E 2 N 2


 Constant say K
V1 E1 N 1
K= Voltage (Current) transformation ratio
For ideal transformer
Input VA = Output VA
V1I1 = V2I2

V2 I 1

V1 I 2
Hence
V2 E 2 N 2 I 1



K
V1 E1 N 1 I 2
Transformer on NO load: Consider an ideal transformer with magnetizing current:
 Ideal transformer without iron loss:
 Im & ϕ will be in same phase
 V1 & E1 will be equal and opposite
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V1= -E1
ϕ
Im
E1
V1
N1
N2
E2
ϕ
Im
V2
E1
E2
Figure 4b:
Phasor diagram of ideal transformer
Without Iron loss
Figure 4a
 Ideal transformer with iron loss: + a little copper loss
 No load current (I0) will be very less (2% - 5% of full load current)
 No load power factor angle (ϕ0) will be very high ( 780 to 870)
 Im= Magnetizing or wattles component of no load current (I0)
 Ie= Active or wattfull component of no load current (I0)
V1= -E1
ϕ
I0
Ie=I0Cosϕ0
I0
Φ0
ϕ
Im=I0Sinϕ0
V1
E1
N1
N2
E2
V2
E1
Figure 4c
No load power
E2
Figure 4d:
Phasor diagram of ideal transformer
with Iron loss
P0  V1I oCos0 = Total iron loss
I0
Neglecting copper loss I 02 R
I e  I 0Cos0
I m  I 0 Sin0
V
R0  1
Ie
V
X0  1
Im
I0
Im
Ie
V1
X0
R0
E1
Figure 4e: Equivalent Circuit
Note: No load power factor is very less because ϕ0 is very high. (ϕ0 is high because Ie<<Im)
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Transformer on load:
ϕ
I1 =I0 + I1
’
I2
L
V1 E1
N1
E2
N2
V2
O
A
D
Figure 5a
MMF induced in primary winding =MMF induced in secondary winding
 N1I1'  N 2 I 2
N
 I1'  2 I 2  KI 2
N1
Note:
 I1’ and I2 will be in phase opposition
 Load may be pure resistive, inductive or capacitive resulting in unity, lagging and leading
power factor respectively.
 So the phasor diagram may be on unity, lagging and leading power factor.
I1
V1= -E1
I ’ =KI
1
V1= -E1
I1
’
V1= -E1
I1’ =KI2
I1
I1 =KI2
Φ1
Φ1
Φ0
Φ0
I0 ϕ
2
Φ1
Φ0
ϕ
I0
I0 ϕ
Φ2
I2
Φ2=0
E1
E2=V2
Φ2
I2
E1
E2=V2
E1
E2=V2
I2
Figure 5b:
Phasor diagram at Unity PF
Figure 5c:
Figure 5d:
Phasor diagram at lagging PF
Phasor diagram at leading PF
Resistance and leakage reactance:
In transformer windings do have some resistance
Let
R1 = Resistance of primary winding
R2 = Resistance of secondary winding
The flux which is linked with primary as well as secondary windings is known as common flux.
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There is some flux which does not link with primary or secondary winding know as leakage flux
as shown in figure 6a.
Let
ϕL1 = leakage flux of primary winding (Flux which does not link with secondary winding)
Common Flux
I1 R1
R2 I2
Leakage Flux
L
O
V1
A
V2
D
Figure 6a
ϕL2 = leakage flux of secondary winding (Flux which does not link with primary winding)
L1I1 and L 2I 2
So at no load I1 & I2 both are negligible hence leakage flux is negligible.
Leakage flux produces a self induced back emf in their respective windings. They are
therefore equivalent to small reactance in series with the respective windings. This reactance is
known as leakage reactance.
Let
X1 = Leakage reactance of primary winding
X2 = Leakage reactance of secondary winding
Common Flux
I1 X1
R1
R2
X2 I2
L
V1
N1
E1
N2
O
E2
A
V2
D
Figure 6b
I1 X1
I1’
R1
R2
X2 I2
I0
Im
Ie
V1
R0
X0
L
E1
E2
O
A
V2
D
Figure 6c: Equivalent Circuit
Note:
7
V1≠ E1 & V2 ≠ E2 as we are taking previously.
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Phasor diagram of actual transformer ON Load:
Load may be pure resistive load( Unity power factor), inductive load (Lagging power factor) and
capacitive load (Leading power factor) so as the phasor diagram.
Common Flux
I1 X1
R1
R2
X2 I2
L
V1
N2
N1
E1
E2
O
A
V2
D
Figure 7a
1. Phasor diagram for pure resistive load (at unity power factor):
V1
I 1X 1
I1R1
I1’ =KI2
I1
-E1
Φ1
Φ0
I0 ϕ
Φ2=0
I2
I2R2
V2
E1
I2X2
E2=V2
Figure 7b:
Phasor diagram at Unity PF
Note:
a. Resistive drop I2R2 is parallel to I2 and inductive drop I2X2 is perpendicular
(leading) to I2.
b. Similarly resistive drop I1R1 is parallel to I1 and inductive drop I1X1 is
perpendicular (leading) to I1.
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c. This is also true for rest of following two diagrams (lagging and leading pf).
If we neglect I1R1, I1X1 and I2R2, I2X2 drops, then
Φ1 = Φ2 = Φ1’= 0 or say Φ
E2 
V2  I 2 R2 2  I 2 X 2 2
E2  V2  I 2 R2
E1 
V1 
E2
K
 E1  I1 R1Cos  I1 R1 Sin 2  I1 X 1Cos  I1 R1 Sin 2
Neglecting I1 X 1Cos  I1R1Sin 
2
V1   E1  I1 R1Cos  I1 R1 Sin
  0
V1   E1  I1 R1
2. Phasor diagram for inductive load (at lagging power factor):
V1
I 1X 1
I1R1
-E1
I1’ =KI2
I1
Φ1
Φ0
I0
ϕ
Φ2
I2
I2R2
V2
E1
I2X2
E2=V2
Figure 7c:
Phasor diagram at lagging PF
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If we neglect I1R1, I1X1 and I2R2, I2X2 drops, then
Φ1 = Φ2 = Φ1’= Φ (say)
E2 
V2  I 2 R2Cos2  I 2 X 2 Sin2 2  I 2 X 2Cos2  I 2 R2 Sin2 2
E2  V2  I 2 R2Cos2  I 2 X 2 Sin2
E1 
V1 
neglecting I 2 X 2Cos2  I 2 R2 Sin2
E2
K
 E1  I1 R1Cos  I1 R1 Sin 2  I1 X 1Cos  I1 R1 Sin 2
Neglecting I1 X 1Cos  I1R1Sin 
2
V1   E1  I1 R1Cos  I1 R1 Sin
3. Phasor diagram for capacitive load (at leading power factor):
I1 I1X1
I1’ =KI2
V1
I1R1
-E1
Φ1
Φ0
ϕ
I0
Φ2
I2
E1
V2
I2X2
E2=V2
I2R2
Figure 7d:
Phasor diagram at leading PF
If we neglect I1R1, I1X1 and I2R2, I2X2 drops, then
Φ1 = Φ2 = Φ1’= Φ (say)
E2 
V2  I 2 R2Cos2  I 2 X 2 Sin2 2  I 2 X 2Cos2  I 2 R2 Sin2 2
E2  V2  I 2 R2Cos2  I 2 X 2 Sin2
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neglecting I 2 X 2Cos2  I 2 R2 Sin2
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E1 
V1 
E2
K
 E1  I1R1Cos  I1R1Sin 2  I1 X1Cos  I1R1Sin 2
Neglecting I1 X 1Cos  I1R1Sin 
2
V1   E1  I1R1Cos  I1R1Sin
Note: The above results can be obtained directly if we replace Φ with – Φ in the case of lagging
power factor.
Equivalent circuit of transformer:
Let R1 & X1 are resistance and leakage reactance of primary windings and R2 & X2 are
resistance and leakage reactance of secondary windings respectively.
I1 X1
I1’
R1
X2 I2
R2
I0
Im
Ie
R0
V1
X0
L
O
E2
E1
A
V2
D
Figure 8a: Transformer Equivalent Circuit
We know
V2 E2 N 2 I1


 K
V1 E1 N1 I 2
 V1 
N 
1
V2   1 V2
K
 N2 
I1 X1
2
 N1 


 N  R2
 2
I1’
R1
I0
V1
R0
2
Im
Ie
X0
 N2 


 N I 2
 1
N 
E1   1  E2
 N2 
 N1 


 N  X2 L
 2
O
A
 N1 


 N V2
 2
D
Figure 8b: Transformer Equivalent Circuit
Referred to primary side
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2
X1
I1
R1
 N1 


 N  R2
 2
I1’
I0
Ie
R0
V1
 N2 


 N I 2
 1
2
Im
N 
E1   1  E2
 N2 
X0
 N1 


 N  X2 L
 2
O
A
 N1 


 N V2
 2
D
Figure 8c: Approximately Equivalent Circuit
Referred to primary side after shifting parallel branch
2
I1
X1
 N1 


 N  R2
 2
R1
 N2 


 N I 2
 1
2
 N1 


 N  X2
 2
L
O
V1
A
 N1 


 N V2
 2
D
Figure 8d: Approximately Equivalent Circuit
Referred to primary side after neglecting parallel branch
Equivalent resistance and reactance of transformer:
Let
R1 = Resistance of primary winding
R2 = Resistance of secondary winding
X1 = Leakage reactance of primary winding
X2 = Leakage reactance of secondary winding
I1R1 = Resistive drop in primary winding
I2R2 = Resistive drop in secondary winding
I1X1 = Reactive drop in primary winding
I2X2 = Reactive drop in secondary winding
1. Equivalent resistance and reactance referred to secondary side:
K2X1 K2R1
R2
X2
I2
I1/K
L
KV1
V2
O
A
D
Figure 9a: Approximately Equivalent Circuit
Referred to secondary side
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The above circuit is obtained after considering the following relations
V2 N 2 I1


K
V1 N1 I 2
 V2  KV1
I2 
&
Total resistive drop  I 22 K 2 R1  I 22 R2

 I 22 K 2 R1  R2
 I 22 R02
Where
I1
K

R02 = Equivalent resistance referred to secondary side
2
2
2
Total reactive drop  I 2 K X1  I 2 X 2

 I 22 K 2 X1  X 2
 I 22 X 02
Where

X02 = Equivalent reactance referred to secondary side
X02
R02
I2
L
KV1
V2
O
A
D
Figure 9b: Approximately & simplified Equivalent
Circuit Referred to secondary side
KV1
V2
ϕ
I2
I2X02
I2R02
Figure 9c: Phasor Diagram
(Assuming lagging load)
KV1 
V2  I 2 R02Cos  I 2 X 02Sin 2  I 2 X 02Cos  I 2 R02Sin 2
KV1  V2  I 2 R02Cos  I 2 X 02Sin
If ϕ = 0
13
Neglecting I 2 X 02Cos  I 2 R02Sin
V2  KV1  I 2 R02Cos  I 2 X 02Sin
V2  KV1  I 2 R02
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If ϕ = 900
V2  KV1  I 2 X 02
If ϕ is leading
Replace ϕ with – ϕ in above equation, so
V2  KV1  I 2 R02Cos    I 2 X 02Sin  
V2  KV1  I 2 R02Cos  I 2 X 02Sin
2. Equivalent resistance and reactance referred to primary side:
X1
I1
R1
R2 / K 2
KI 2
X2 / K2
L
O
V1
A
V2 / K
D
Figure 9d: Approximately Equivalent Circuit
Referred to primary side
The above circuit is obtained after considering the following relations
V2 N 2 I1


K
V1 N1 I 2
 V1 
Total resistive drop
1
V2
&
I1  KI 2
K
 I12 R1  I12 R2 / K 2

 I12 R1  R2 / K 2
 I12 R01
Where

R01 = Equivalent resistance referred to primary side
2
2
2
Total reactive drop  I1 X 1  I 2 X 2 / K

Where
14

 I12 X 1  X 2 / K 2
 I12 X 01
X01 = Equivalent reactance referred to primary side
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X01
R01
KI2
L
V1
V2/K
O
A
D
Figure 9e: Approximately & simplified Equivalent
Circuit Referred to primary side
V1
V2/K
ϕ
I1
I1X01
I1R01
Figure 9f: Phasor Diagram
(Assuming lagging load)
V1  V2 / K  I1R01Cos  I1 X 01Sin   I1 X 01Cos  I1R01Sin 
V1  V2 / K  I1R01Cos  I1 X 01Sin
Neglecting I1 X 01Cos  I1R01Sin
2
2
If ϕ = 0
V1 
V2
 I1 R01
K
If ϕ = 900
V1 
V1
 I1 X 01
K
If ϕ is leading
Replace ϕ with – ϕ in above equation, so
V1  V2 / K  I1R01Cos    I1 X 01Sin  
V1  V2 / K  I1R01Cos  I1 X 01Sin
Voltage regulation:
It is change in the voltage of secondary side from no load to full load.
Let
V1 = Full load secondary voltage
E2 = No load secondary voltage
So % voltage regulation
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No load voltage  Full load voltage
100
No load voltage
E V
 2 2 100
E2
Voltage drop in sec winding

100
E2
I R Cos  I 2 R02Sin
 2 02
100    (1)
E2

Where
Cos ϕ = Power factor
R02 = Equivalent resistance referred to secondary side
X02 = Equivalent reactance referred to secondary side
+Ve sign for lagging power factor and –Ve sign for leading power factor.
 Condition for zero voltage regulation:
From equation (1) it is clear that voltage regulation will be zero for leading power
factor only. So condition of zero voltage regulation
I 2 R02Cos  I 2 R02Sin
0
E2
R
 tan   02
X 02
Losses in transformer: The losses are of two type core loss and copper loss
(1) Core or iron losses: It is also known as fixed losses, again of two type
(a) Eddy current losses: Given by
Pe  Ke Bm2 f 2t 2v
Watt
   (1)
Where
Ke = Constant
Bm = Maxi value of flux density
f = frequency
t = thickness of laminations
v = Volume of core
We know
E  4.44 fm N
m
E
 4.44 f
N
Area ( A)
Area ( A)
E
 4.44 fBm N
Area ( A)
Bm 
E/ A
   (2)
4.44 fN
Put the value of Bm in equation (1)
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2
 E/ A  2 2
 f t v
Pe  K e 
 4.44 fN 
2
1

 22
Pe  K e 
 Et v
 4.44 AN 
For a given transformer N, A, t and V are constants so
Pe  E 2
Pe  V 2
 E  V (Voltage )
So eddy current losses are independent of frequency.
(b) Hysteresis losses: Given by
Ph  Bmx fv
Where
Watt
   (3)
η = Constant
Again put the value of Bm from equation (2) to equation (3)
x
 E/ A 
 fv
Ph   
 4.44 fN 
x
1

 x 1 x
Ph   
 E f v
 4.44 AN 
For a given transformer N, A and V are constants so
Ph  E x f 1 x
Ph  V 2 f 1 x
 E  V (Voltage )
So eddy current losses dependent of voltage and frequency both.
(2) Copper Losses: Also know as variable losses because they depend on load current.
R1 = Resistance of primary winding
R2 = Resistance of secondary winding
I1 = Full load current in primary winding
I2 = Full load current in secondary winding
Full load copper losses
PC  I12 R1  I 22 R2
Copper losses at x time of full load
PCx  x 2 PC
17
Where
Prepared By: Nafees Ahmed
x
Any load current
1
; for half load x  etc
Full load current
2
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Tests on transformer
1. Open Circuit (OC) test or No load test:
By OC test we can find out
 Iron losses (Pi)
 No load current (I0)
 Cosϕ0, Ie, Im, R0 & X0
Wattmeter
1-Phase Auto
Auto
Transformer
M
V
CC
COM PC
L
A
V
Rated Voltage
LV
HV
Figure: 10
Iron losses
No load current
Let
Note:
(i)
(ii)
Pi = Reading of wattmeter (P0)
I0 = Reading of Ammeter
V = Reading of voltmeter
P0 = Pi = VI0Cosϕ0
P
 Cos0  i
VI 0
Ie = I0Cosϕ0
Im = I0Sinϕ0
V
V
R0 
& X0 
Im
Ie
Rated voltage is applied at LV side.
This test is generally done on LV side (Why?)
2. Short Circuit (SC) test:
By OC test we can find out
 Copper losses (PC)
 Equivalent resistance or leakage reactance (R01 & X01 OR R02 & X02)
referred to metering side.
Full load Cu losses
PC = Reading of wattmeter (Wsc)
Let
Short circuit current Isc = Reading of Ammeter
Wsc  I sc2 Req
Wsc  I sc Z eq
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R
Z
 R01 or R02 
eq  Z 01 or Z 02 
eq
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X eq  Z eq2  Req2
X
eq
 X 01 or X 02 
Wattmeter
CC
V
V
Shorted
COM PC
L
d
1-Phase Auto
Auto
Transformer
M
A
Rated Current
HV
LV
Figure: 11
Note:
(i)
(ii)
(iii)
Rated Current is applied at HV side.
This test is generally done on HV side (Why?)
Why the position of ammeter and voltmeter is changed as compared to OC
test?
Efficiency of transformer:

O / P power
100
I / P power

O / P power
 100
O / P power  losses

O / P power
 100
O / P power  Iron loss  Cu loss
 Efficiency at full load

V2 I 2Cos2
 100
V2 I 2Cos2  Pi  PC

P2
100
P2  Pi  PC
Where
P2  V2 I 2 Cos2  Rated VA  Cos2
 Efficiency at x time of full load

19
xP2
 100    (1)
xP2  Pi  x 2 PC
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Here
Cos2  Load PF
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 Condition for maximum efficiency
Differentiating equation (1) w. r. t “x” and putting dη/dx=0


xP2  Pi  x 2 PC P2  xP P  2 xPC 
d

 100  0
2
dx
xP2  Pi  x 2 PC




 xP2  Pi  x 2 PC P2  xPP  2 xPC   0
 x 2 PC  Pi
 Cu loss  Iron loss
x
or
Variable loss  Constant loss
Pi
PC
Example 1: A 50 KVA, 4400/220 V transformer has R1 = 3.45 Ω , R2 = 0.009 Ω. The values of
reactance are X1 = 5.2 Ω, X2 = 0.015 Ω. Calculate
a. Equivalent resistance as referred to primary
b. Equivalent resistance as referred to secondary
c. Equivalent reactance as referred to primary
d. Equivalent reactance as referred to secondary
e. Equivalent impedance referred to both side
f. Total copper loss first using individual resistance of two windings and
secondly as referred to each sides.
Solution:
K
N 2 V2
220
1



 0.05
N1 V1 4400 20
a. Equivalent resistance as referred to primary
R01  R1  R2 / K 2
R01  3.45  0.009 / 1 / 20
2
R0 1  7.05 Ω
Ans
b. Equivalent resistance as referred to secondary
R02  K 2 R1  R2
R02  1 / 20 3.45  0.009
2
R0 2  0.0176Ω
20
Ans
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c. Equivalent reactance as referred to primary
X 01  X 1  X 2 / K 2
X 01  5.2  0.015 / 1 / 20
2
X 0 1 11.2 Ω
Ans
d. Equivalent reactance as referred to secondary
X 02  K 2 X 1  X 2
X 02  1 / 20 5.2  0.015
2
X 0 2  0.028Ω
Ans
e. Equivalent impedance referred to both side
Z 0 1  7.05  j11.2Ω Ans
Z 0 2  0.0176 j0.028Ω Ans
f. Total copper loss first using individual resistance of two windings and
secondly as referred to each sides.
I1 
VA Rating 50 103

 11.3636 A
V1
4400
I2 
VA Rating 50  103

 227.2727 A
V2
220
Copper loss  I12 R1  I 22 R2
 11.3636 3.45  227.2727 0.009
 910.38 W Ans
Taking referred values
2
2
W
Copper loss  I12 R01
 11.3636 7.05 W
 910.38 W Ans
Copper loss  I 22 R02
2
 227.2727 0.0176 W
 909.09 W Ans Error due to approxim at
ion
2
21
Prepared By: Nafees Ahmed
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Example 1: Following results were obtained on a 100 KVA, 11000/220 V, single phase
transformer
(i)
OC Test (LV Side)
220 V,
45 A,
2 KW
(ii)
SC Test (HV Side)
500 V,
9.09 A,
3 KW
Determine equivalent circuit parameter of transformer referred to low voltage side and
efficiency at full load unity power factor.
Solution:
Equivalent circuit parameters:
From OC test:
Pi = P0 = 2 KW,
I0 = 45 A,
V2 = 220 V
P0 = Pi = V2I0Cosϕ0
P
 Cos0  i  0.202
V2 I 0
Ie = I0Cosϕ0=9.09 A
Im = I0Sinϕ0=44.07 A
R0 
From SC test:
PC = Wsc = 3 KW,
V2
 24.20 Ω
Ie
Isc = 9.09 A,
Ans & X 0 
V2
5 Ω
Im
Ans
Vsc = 500 V
Wsc  I sc2 R01
 R 0 1  36.31 Ω Ans
Vsc  I sc Z 01
 Z 0 1  55 Ω Ans
2
2
X 01  Z 01
 R01
 X 0 1  41.31 Ω
Efficiency at full load and unity power factor:
Ans
x  RatedVA  Cos2
100
x  RatedVA  Cos2  Pi  x 2 PC
1100 1

100
1100 1  2  12  3

η  95.24% Ans
Note:
100 1000
 9.09 A
11000
Here SC Test is done on full load, so 3 KW is the full load cu loss.
Full load HV side current
22
Prepared By: Nafees Ahmed
I1 
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