Question

The 500 KVA, 69kV/11kV, 60 Hz transformer has total resistance Rp of 100 Ω and total leakage reactance of Xp of 600 Ω. Calculate

1. The per unit impedance of the transformer in percent (just magnitude)

2. The voltage regulation of the transformer when it delivers 200 KVA with a lagging power

   factor of 90% while the secondary voltage is fixed at 11 kV.

3. The actual primary voltage V1.

4. The actual line current I1.

Answer 1

Given KVA = 500 KVA

Primary Voltage = V1 = 69 KV

Secondary Voltage = V2 = 11 KV

Rp = 100 Ohms

Xp = 600 Ohms

Impedance Zp = √ Rp² +Xp² = √ 100² +600² = 608.3 Ohms

Primary Current = KVA/kV1 = 500/69 = 7.24 Amperes

Base Impedance = Zb = Primary Voltage / Primary Current = 69000 / 7.24 = 9530 Ohms

Per Unit Impedance = Zpu = Zb/Zp = 608.3/9530 = 0.064 pu

Transformation Ratio = K = V2/V1 = 11000/69000 = 0.159

Now,

Given load = 200 KVA

Secondary Current = I2 = 200/V2 = 200/11 = 18.18 Ampers

Primary current = I1 = I2*K = 18.18*0.159 = 2.89 Amperes

Power factor = Cos(x) =0.9

Sin(x) = √ 1-sin(x)² = √ 1-0.9² = 0.435

Primary Voltage = 69000 Volts

Actual Primary Voltage = V = V1 + (I1*Rp*cos(x) ) + (I1*Xp*sin(x) )

V= 69000 + (2.89*100*09) + (2.89*600*0.435)

V = 70014.4 Volts

Voltage regulation = V-V1 / V1 = (70014.4-69000) / 69000 = 0.0147

Percentage voltage regulation = Voltage Regulation*100 = 0.014 *100 = 1.47 percent.

Actual primary line Current = 2.89 Amperes

Actual Primary voltage = 70014.4 Volts