# Class 11 Regulation and Efficiency of Transformer.ppt MADAM

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```					                Transformer tests
•   The performance of a transformer can be calculated on the basis of its
equivalent circuit which contains the 4 main parameters:
1. Equivalent resistance R01(or R02)
2. Equivalent leakage reactance X01 (or X02)
3. Core loss resistance R0
4. Magnetizing reactance X0.

•   These parameters are determined from the following tests:
a) Open circuit test
b) Short circuit test
Open circuit test
I0 W    A

V1    V                                                               V2=E2

Low Voltage winding              High Voltage winding

Open Circuit Test

• As the primary no-load current I0 is small, Copper loss is negligibly small in
primary & nil in secondary. Hence, wattmeter reading represents the core
• Core loss = W = input power on no-load = V1I0cosΦ0
=> cosΦ0 = W/(V1I0)
Hence, Iw = I0 cos Φ0 & Iμ=I0sin Φ0
Also, X0 = V1/ Iμ & R0 = V1/ Iw
Short Circuit test
•       This test is conducted to determine:
2. Equivalent resistance & reactance referred to metering side.

W        A

LV
supply    V

High Voltage winding     Low Voltage winding

•    A low voltage (5 -10% of normal primary voltage) at correct frequency is applied
to the primary winding & is continuously increased till full load currents flow both
in primary & secondary. Since applied voltage is small, flux linking with core is
very small & hence, iron loss can be neglected & reading of wattmeter gives
total copper losses at full load.
•    If Vsc is the voltage required to circulate rated load currents, then
Vsc
Z 01         ;W  I sc R01  X 01            Z 01  R01
2                           2         2

I sc
Regulation of a transformer

• Voltage regulation is defined as: “the change in secondary
voltage when rated load at a specified power is removed”.

• Let, E2 = sec. terminal voltage at no load
V2 = sec. terminal voltage at full load

E 2  V2
Then,    % regulation =                100
V2
Approximate Equivalent circuit with primary impedances
transferred to secondary

I1/k          R02 = R2 +k2R1 X02= X2+ k2X1   I2

I0/k

kV1=E2     k2R0          k2X0                              ZL
V2

I2
Regulation expressed in terms of secondary values

E2=kV1
I2 Z02             θ2
I2 X02 cosθ2
δ            I2 R02 cosθ2
0                                  I2 R02 sinθ2
θ2          V2      θ2
I2 X02
I2                 I2 R02
I2 X02 sinθ2
E 2  V2  I 2 R 02 cosθ 2  I 2 X 02 sinθ 2   I 2 X 02 cosθ 2  I 2 R 02 sinθ 2 
2                                                           2                         2

 V2  I 2 R 02 cos2 θ 2  I 2 X 02 sin 2 θ 2  2V2 I 2 R 02 cosθ 2
2           2       2                   2   2

 2I 2 R 02 X 02 sinθ 2 cosθ 2  2V2 I 2 X 02 sinθ 2  I 2 X 02 cos2 θ 2  I 2 R 02 sin 2 θ 2
2                                               2   2                     2   2

 2I 2 R 02 X 02 sinθ 2 cosθ 2
2

 V2  I 2 R 02  I 2 X 02  2V2 I 2 R 02 cosθ 2  2V2 I 2 X 02 sinθ 2
2           2       2       2       2

E 2  V2  I 2 R 02  I 2 X 02  2V2 I 2 R 02 cosθ 2  2V2 I 2 X 02 sinθ 2
2           2       2       2       2
Efficiency of a transformer
•    Losses in a transformer –
1.   Core or Iron loss = Hystersis loss + Eddy Current
loss ( from OC test)
2.   Copper loss =      I 2 R1  I 2 R2  I 2 R 01  I 2 R 02
1      2     1       2

(from SC test)

Output        Output
Efficiency        
Input Output  Losses
Output

Output  Iron loss  Cu loss
Condition for maximum efficiency
• Cu loss = I12 R01         Iron loss = Wh  We  Wi
considering primary side
Primary input = V1 I1 cos 1

V1 I1 cos1  losses

V1 I1 cos1
V1 I1 cos1  I1 R01  Wi
2

V1 I1 cos1
I1 R01         Wi
 1              
V1 cos 1 V1 I1 cos 1
Differentiating both sides w.r.t I1, we get
dη         R 01        Wi
 0        
dI1      V1cosφ1 V1 I 21cosφ1

dη
For η to be maximum,                 dI1
0

R 01        Wi
0  0          
V1cosφ1 V1 I 2 1 cosφ1
R 01        Wi
        
V1cosφ1 V1 I 2 1 cosφ1
 I 2 1 R 01  Wi
Cu loss  Iron loss

The current corresponding to maximum efficiency
is:     I 
1
W     i
R01
I1       Wi
I2       
k        R02
Efficiency of Transformer at any load
x  Full Load kVA  p.f
η                                                                      100%
x  Full Load kVA  p.f  Iron Loss  x 2  Full Load Copper Loss

x  Ratio of Actual to Full Load kVA
Ex: A 2300/208 V, 500kVA, 60Hz transformer has the
following data:

Test           V       A          W              Side used
O.C           208      85       1800                L.V
S.C           95      217.4     8200                H.V

Calculate efficiency of the transformer at 0.8 pf lagging &
Solution

x  Full Load kVA  p.f
η                                                                    100%
x  Full Load kVA  p.f  Iron Loss  x  Full Load Copper Loss
2

1
 500  0.8
          4
1                1
(  500  0.8)  ( ) 2  8.2  1.8
4                4
 97.74%

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 views: 1814 posted: 4/15/2010 language: English pages: 13