44.MET April 2019 Q.7(b)

A-100 kVA transformer has 400 turns on the primary and 80 turns on the secondary. The primary and secondary resistances are 0.3 Ω and 0.01 Ω respectively and the corresponding leakage reactances are 1.1 and 0.035 Ω respectively. The supply voltage is 2200 V. Calculate 
(i) equivalent impedance referred to primary  
(ii) the voltage regulation and the secondary terminal voltage for full load having a power factor of 0.8 leading.

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Given

KVA = 100
N1 = 400
N2 = 80
R1 = 0.3 Ω
R2 = 0.01 Ω
X1 = 1.1 Ω
X2 = 0.035 Ω
V1 = 2200 V

To find 

(i) Equivalent impedance referred to primary  
(ii) The voltage regulation and the secondary terminal voltage for full load having a power factor of 0.8 leading.

Solution

As we know that Voltage transformation ratio,

K = N2 / N1
   = 80 / 400
   = 0.2

Equivalent Resistance referred to primary

R01 = R1 + (R2 / K²)
        = 0.3 + (0.01 / 0.2²)
        = 0.55 Ω

Equivalent Reactance referred to primary

X01 = X1 + (X2 / K²)
       = 1.1 + (0.035 / 0.2²)
       = 1.975 Ω 

i) Equivalent Impedence referred to primary

   Z01 = R01 + j X01
          = 0.55 + j 1.975
          = 2.05 ∠ 74.44°

ii) Equivalent Impedence referred to secondary

 Z02 = K² x Z01
        = 0.2² x (0.55 + j 1.975)
        = 0.022 + j 0.079

So R02 = 0.022 and X02 = 0.079 [Z02 = R02 + j X02]

No load secondary voltage ,

V2 = K x V1
     = 0.2 X 2200
     = 440 V

Full load current,

I2 = KVA / V2
    = 100,000 / 440
    = 227.3 A

F.L Voltage drop across secondary = I2 (R02 cos Ф - X02 sin Φ)
                                                        = 227.3(0.022 x 0.8 - 0.079 x 0.6)
                                                        = - 6.77 V

So Voltage regulation % = (- 6.77 x 100) / 440
                                        = - 1.54

Secondary terminal voltage = 440 - (-6.67)
                                             = 446.7 V