Question

Question3 3000 kVA / 0.09 p.u. Alternators 2 40004VA 0.12 p.u 6kV Busbars 4000 LVA 0.05 p.. Transmission Lines Lines 4000 kV 0.05 p.ย. Step-Up Transformers 5000 kVA 5000 kVA 0.06 р.u. 0.06 p.u. 11 kV Basbars -150 kVA 0.05 pu. Secondary feeder line Step-Down 1500 kVA 0.03 p.u 415 V Busbars Fault Point Figure Q3 Figure Q3 shows a power system comprising two three phase alternators supplying 6 kV busbars. I wo paralilel transmission lines fed from these busbars supply 11kv busbars, each line containing a step-up transformer Secondary feeders from the 11 kV busbars supply a number of ransurors to the load points of the system at 415 V via step down transformers one of the distributors has developed a 3-phase fault at a point clos e to the output of its step down transformer. Question 3 (continued) Working to a base of 6MVA: a) Calculate the combined per unit (p.u.) reactance of (4 marks) the alternators. b) Calculate the p.u. reactances of each transformer, transmission line and secondary feeder. (4 marks) c) Sketch a single line schematic diagram to show how the reactances of the system components are effectively connected together. (8 marks) d) Calculate the total p.u. reactance of the system up to the fault point. (4 marks) (2 marks) e) Calculate the short-circuit MVA of the system. Calculate the short-circuit line current of the fault. (3 marks)

Answer 1

a)on Base 3000 3000 Kvh Combined u. Reactande - (0:18)I1 (o-18) D:0 fo 5000KVA SOO

0 075 0:19 0.12 Fult (1φ) 0' Callu Hind tntaLh reactan ce b. 0383S 0

Pu 01 65 19 443 KA