Problem 2. A three phase transformer rated at 5 MVA, 115/13.2 kV has a per phase series impedance of 0.007 + j0.075 per unit (shunt parameters are ignored). The low voltage side is connected to a short distribution line which modeled by a series per phase impedance of 0.02 +j0.1 per unit on the base of 10 MVA. The line supplies a balanced three phase constant impedance load at 4 MVA, 13.2 kV with power factor of 0.85 lagging
• a) Draw an equivalent circuit of the system indicating all impedances in per unit. Choose 10 MVA, 13.2 kV as the base values of load.
• b) Calculate the per unit source voltage that must be supplied to the high voltage side of transformer to supply the load ar rated voltage.
• c) Find the per unit current based on the high voltage side of the transformer.
For a transformer we have -
Given base values as 10 MVA and 13.2 KV
For a transfromer PU
Base current = Base KV/ Base V
IB = 10M/ 1.732 *13.2 K = 437.386 A
Base impedance -
ZB = KVB 2 /MVAB = 17.424 ohms
per unit system equivalent circuit -
Also ZPU will be same on either side of transformer ie, primary and secondary
Given Z = 0.007 + j0.075 then
Now Zpu old = Zact / ZB
= 0.007 + j0.075 / 17.424
= 0.4 + j 4.3 m ohms
now
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