الخميس، 14 يوليو 2011

Equivalent Impedance

Equivalent Impedance 
       The transformer primary has resistance R1 and reactance X1. While the transformer secondary has resistance R2 and reacatnce X2. Thus we can say that the total impedance of primary winding is Z1 which is,
                         Z1 = R1 + j X1 Ω                                         ..........(1)
And the total impedance of the secondary winding is which is ,
                          Z2 = R2 + j X2 Ω                                            ...........(2)
       This is shown in the Fig. 1.
Fig. 1 Individual impedance

      The individual magnitudes of and are,
                            Z1 = √(R1+ X12)                                       ...........(3)
and                        Z2 = √(R2+ X22)                                        .........(4)
      Similar to resistance and reactance, the impedance also can be referred to any one side.
Let                       Z1e = total equivalent impedance referred to primary
then                       Z1e = R1e + j X1e
                              Z1e = Z1 + Z2' = Z1 + Z2/K2                            ............(5)
Similarly                  Z2e = total equivalent impedance referred to secondary
then                         Z2e =  R2e + j X2e
                                 Z2e = Z2 + Z1' = Z2 + K2 Z1                         ............(6)
       The magnitude of Z1e and Z2e are,
                                   Z1e = √(R1e+ X1e2)                                       ........(7)
and                             Z2e  = √(R2e+ X2e2)          ...........(8)
      It can be denoted that,
                                   Z2e = K2 Z1e              and  Z1e  = Z2e /K2
       The concept of equivalent impedance is shown in the Fig. 2.
Fig 2  Equivalent impedance

Example 1 :A 15 KVA, 2200/110 V transformer has R1 = 1.75Ω,  R2 = 0.0045 Ω the leakage reactance are X1 = 2.6 Ω and X2 = 0.0075 Ω Calculate,
 a) equivalent resistance referred to primary
b) equivalent resistance referred to secondary
c) equivalent reactance referred to primary
d)  equivalent reactance referred tosecondary
e)  equivalent impedance referred to primary
f)  equivalent impedance referred to secondary
g) total copper loss
Solution : The given values are, R1 = 1.75 Ω, R2 = 0.0045Ω, X1 = 2.6 Ω, X2 = 0.0075 Ω
       K = 110/2200 = 1/20 = 0.05
a)   R1e = R1 + R2' = R1 + R2 /K = 1.75 + 0.0045/0.052 = 3.55 Ω
b) R2e = R2 + R1' = R2 + K R1 =
           = 0.0045 + (0.05)2 x 1.75 = 0.00887 Ω
c) X1e = X1 + X2' = X1 + X2/K2 = 2.6 + 0.0075/(0.05)2 = 5.6 Ω
d) X2e   = X2 + X1'= X2 + K X1
              = 0.0075 + (0.05)2 x 2.6 = 0.014 Ω
e) Z1e = R1e + j X1e= 3.55 + j 5.6 Ω
Z1e = √(3.55+ 5.62) = 6.6304 Ω
f)  Z2e =  R2e + j X2e = 0.00887 + j 0.014 Ω
Z2e = √(0.00887+ 0.014 2) = 0.01657 Ω
g) To find the load copper loss, calculate full load current.
(I1) F.L. = (KVA x 1000)/V1 = (25 x 1000)/2200 = 11.3636 A
 total copper loss = ((I1)F.L.)2 R1e =  (11.3636)2 x 355 = 458.4194 W
This can be checked as,
(I2) F.L.= (KVA x 1000)/V2 = (25 x 1000/110 = 227.272 A
total copper loss  =  I12 R1 + I22 R2
                           = (11.3636)2 x 1.75 + (227.373)2 x 0.0045
                           = 225.98 + 232.4365 = 458.419 W

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