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Class 11 Regulation and Efficiency of Transformer.ppt MADAM

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Class 11 Regulation and Efficiency of Transformer.ppt MADAM Powered By Docstoc
					                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
loss under no load condition.
• 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:
               1. Full-load copper loss
               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 &
1/4th rated load.
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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