The Fig. 1 shows the circuit diagram for conducting the direct loading test on the three phase alternator. The star connected armature is to be connected to a three phase load with the help of triple pole single throw (TPST) switch. The field winding is excited by separate d.c. supply. To control the flux i.e. the current through field winding, a rheostat is inserted in series with the field winding. The prime mover is shown which is driving the alternator at its synchronous speed.
Procedure : The alternator is first driven at its synchronous speed Ns by means of a prime mover.
 |
| Fig. 1 Circuit diagram for direct loading test on alternator |
Now Eph α Φ ..... (From e.m.f. equation)
By giving d.c. supply to the field winding, the field current is adjusted to adjust the flux so that rated voltage is available across the terminals. This can be observed on the voltmeter connected across the lines. The load is then connected by means of a TPST switch. The load is then increased so that ammeter reads rated value of current. This is full load condition of the alternator. Again adjust the voltage to its rated value by means of field excitation using a rheostat connected. The throw off the entire load by opening the TPST switch, without changing the speed and the field excitation. Observe the voltmeter reading. As load is thrown off, there is no armature current and associated drops. So the voltmeter reading in this situation indicates the value of internally induced e.m.f. called no load terminal voltage. Convert both the reading to phase values. The rated voltage on full load is Vph while reading when load is thrown off is Eph. So by using the formula,
the full load regulation of the alternator can be determined. The value of the regulation obtained by this method is accurate as a particular load at required p.f. is actually connected to the alternator to note down the readings.
Note : But for high capacity alternators, that much full load can not be simulated or directly connected to the alternator. Hence method is restricted only for small capacity alternators.
Example : While supplying a full load, running at synchronous speed, the terminal voltage of an alternator is observed to be 1100 V. When the load is thrown off, keeping field excitation and speed constant, the terminal voltage is observed to be 1266 V. Assuming star connected alternator, calculate its regulation on full load.
Solution : On full load, terminal voltage is 1100 V.
So VL = 1100 V
... Vph = VL/√3 = 635.0853 V
When load is thrown off, VL = 1266 V. But on no load,
VL = Eline
... Eline = 1266 V
... Eph = 1266/√3
= 730.925 V
Sponsored links :
Thank u...it is very helpful...
ReplyDeleteVery nice explanation
ReplyDelete