Environmental Engineering Reference
In-Depth Information
Worked Example 4.1
A synchronous generator of 10 ohms synchronous reactance is supplying 5 MW and 2 MVAR
to an 11 kV network. Calculate the generator internal voltage.
Rearranging Equations (4.8a) and (4.8b) gives
XP
V
V
sin
δ =
A
B
and
XQ
V
V
cos
δ =−
V
A
B
B
Therefore
XP
V
tan
δ =
2
Q
B
In this case P =
5, Q =
2, V B = 11, X = 10.
Hence
δ
= 19.52 ° , and V A = 13.6 kV
4.2.7 Four - Quadrant Operation
Consider the case where a lossless synchronous machine is run up to synchronous speed by
an external prime mover. The fi eld current is then gradually increased until the terminal
voltage of the machine (the same as the internal generated voltage, as no current is taken) is
made to be equal to the voltage of the local bus to which the machine is to be connected.
Precise adjustments to the speed of the prime mover are made so that through some external
instrumentation it is possible to detect an instant at which the internal voltage V A exactly
matches V B in magnitude and phase. The synchronous machine can now be safely connected
to the bus. This process is known as synchronisation and must be carried out each time a
synchronous machine is to be connected to the mains.
Now arrange for the prime mover to apply an accelerating torque. This will result in a
positive load angle
and according to Equation (4.8a) a negative active power, i.e. active
power injected into the power system. As expected, the machine is generating. With a braking
torque on the shaft, i.e. with the machine motoring, the load angle is negative and active
power is supplied to the machine from the power system.
Returning to the idling state, consider now what happens if the fi eld current is increased
so that V A is made larger than V B , but no external torque is applied so that
δ
and the active
power are zero. Equation (4.8b) shows that the reactive power is negative; i.e. the synchro-
nous machine injects reactive power into the system, and acts as a generator of reactive power.
In this state the machine is said to be overexcited .
Conversely, if the excitation current is decreased so that V A
δ
V B , the reactive power
fl ow is reversed, the machine is a consumer of reactive power and it is said to be
underexcited .
The synchronous machine is capable of operating in each of the four quadrants of the
quadrant diagram shown in Figure A18 in the Appendix.
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