Question

A 60-Hz single-phase transformer with capacity of 150 kVA has the following parameters: RP = 0.35 Ω RS = 0.002 Ω Rc = 5.2 kΩ XP = 0.5 Ω XS = 0.008 Ω Xm = 1.1 kΩ The primary transformer voltage is 2.8 kV and the secondary is 230 V. The transformer is connected to a variable load (0 to 300 kW) with a lagging power factor of 0.83 and a load voltage equal to the rated transformer secondary. Determine: (a) the total input impedance of the transformer when the secondary is shorted; and (b) the input current, voltage, power and power factor at full load (150 kW). (c) Plot the voltage regulation versus load, and determine the load that produces 5% regulation.

Answer 1

Answer:

Explanation:

Total Resistance, Rt = (0.35 + 0.002 + 5200 + 0.5 + 0.008 + 1100) = 6300 ohms      

a) Capacity of transformer, Pt = 150KVA = 150,000 W

Input Voltage, Vp = 2.8 KV = 2800 V

Current, Ip = 150000/2800 = 53.57 A

Input impedance, Zp = Vp/Ip = 2800/5357 = 52.27 ohms

b) i)   Input current = 53.57 A

ii) Voltage, V = Ip * Rt = 53.57 X 6300 = 337.5 KV

iii) Power, P = I² * Rt = (53.57)² X 6300 = 18.08 MW

iv) Power factor = 0.83