A three phase 6000 V Alternator gave the following open circuit characteristic at normal speed. ted and full load currant foluing

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Problem 5: A three phase 6000 V Alternator gave the following open circuit characteristic at normal speed. With the alternator
short circuited and full load current folwing, the field current is 16 A. Neglecting the armature resistance and using
synchronous impedance method, Determine the voltage regulation of the alternator supplying the full load of 200 KVA, at 0.8
power factor lagging.
Field current in amper
Terminal Voltage in volt
14
4000
18
5000
23
6000
30
7000
43
8000
gooo
THENS
TERMINAL VOLTS
8000
7000
6000
5000
200
4000k
18 22 26 30 34 36 42
FIELD AMPERES, I
OPEN CIRCUIT CHARACTERSTICS
46 50 54
Transcribed Image Text:Problem 5: A three phase 6000 V Alternator gave the following open circuit characteristic at normal speed. With the alternator short circuited and full load current folwing, the field current is 16 A. Neglecting the armature resistance and using synchronous impedance method, Determine the voltage regulation of the alternator supplying the full load of 200 KVA, at 0.8 power factor lagging. Field current in amper Terminal Voltage in volt 14 4000 18 5000 23 6000 30 7000 43 8000 gooo THENS TERMINAL VOLTS 8000 7000 6000 5000 200 4000k 18 22 26 30 34 36 42 FIELD AMPERES, I OPEN CIRCUIT CHARACTERSTICS 46 50 54
Expert Solution
Step 1

Given data:

3-ϕ , 6000 v alternator

If=16 A

The open circuit voltage is 4700 v

The phase O.C. voltage=47003=2714 V

Full load current =200 ×10003×6000×0.8=24 A

Short circuit current corresponding to the filed current of a 16 amperes in the 24 amperes synchronous impedance.

Zs=O.C. voltageS.C. current    =271424=113 Ω

 

 

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