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
