chapter 16: synchronous generators - oakland...

3/2/00 Electromechanical Dynamics 1

Chapter 16: Synchronous Generators


Generator under Load

• The behavior of a synchronous generator depends upon the connected load– two basic load categories

• isolated loads

• infinite bus

– isolated loads with a lagging pf

• current lags the terminal voltage, E

• the voltage drop across the synchronous reactance, EX, leads the current by 90°

• the induced voltage, E0, generated by the flux, Φ, is equal to the phasor sum of E and EX



– isolated loads with a leading pf

• current leads the terminal voltage, E

• the voltage drop across the synchronous reactance, EX, leads the current by 90°

• the induced voltage, E0, generated by the flux, Φ, is equal to the phasor sum of E and EX

– note that E0 always leads E by the angle δ

• for lagging loads E0 is greater than E

• for leading loads E is greater than E0



• Example– a 36 MVA, 20.8 kV, 3-phase

generator• synchronous reactance is 9 ohms

• nominal current is 1000 A

• no-load saturation curve is given

– adjust the excitation so that the terminal voltage is fixed at 21 kV

• calculate the excitation current

• draw the phasor diagrams for the following load conditions– no-load

– resistive load of 36 MW

– capacitive load of 12 MVAr


Regulation Curves

• Voltage regulation is the behavior of the generator’s terminal voltage as the load varies

• Regulation is a function of the load current– the regulation curve is a plot of the terminal voltage, VT, with

respect to load current, I, ranging from no-load to full-load• for a fixed field excitation current

• for a given load power factor

– family of curves are developed for various field excitation currents and for different load power factors

– percent regulation is defined as:

%100%,

,, ×−

=ratedT

ratedTNLT

V

VVVR


Regulation Curves

• Example– consider the regulation curves for a 36 MVA, 21 kV generator

– calculate the percent regulation corresponding to the unity power factor curve


Synchronization of a Generator

• Often two or more generators are connected in parallel to supply a common load in large utility systems– connecting a generator to other generators is called paralleling

– many paralleled generators behaves like an infinite bus• voltage and frequency are constant and can not be easily altered

– before connecting a generator to an electrical grid, it must be synchronized

• the generator frequency is equal to the system frequency

• the generator voltage is equal to the system voltage

• the generator voltage is in phase with the system voltage

• the phase sequence of the generator is the same as that of the system


Synchronization of a Generator

• To synchronize a generator– adjust the speed regulator of the prime mover so that

frequencies are close

– adjust the excitation so that generator voltage and system voltage are equal

– observe the phase angle by means of a synchroscope, which indicates the phase angle between two voltages

• the pointer rotates proportional to the frequency difference

• a zero mark indicates a zero degree phase angle

• the speed regulator is adjusted so that the pointer barely creeps across the dial

– on the zero mark, the line circuit breaker is closed


Connecting to an Infinite Bus

• An infinite bus system is so powerful that it imposes its own– voltage magnitude and frequency

– once an apparatus is connected to an infinite bus, it becomes part of it

– for a synchronized generator, the operator can only vary two machine parameters

• the field excitation current, IX

• the prime-mover’s mechanical torque, T



• Varying the exciting current– impacts the induced voltage E0

– causes a current to flow that is 90 degrees out-of-phase due to the synchronous reactance

– does not affect the flow of active (real) power

– does cause reactive power to flow

SjX

EEI

−= 0



• Varying the mechanical torque– by opening up the control valve of the prime-mover, an

increase torque is developed

– the rotor will accelerate, E0 will increase in value and begin to slip ahead of phasor E, leading by a phase angle δ

– Although both voltages have similar values, the phase angle produces a difference of potential across the synchronous reactance

• a current will flow, but this time almost in phase with E

• real (active) power will flow


Active Power Delivered

• The active power delivered by a synchronous generator is given by

– PE = 3-phase power delivered by the generator

– E = induced generator voltage

– VT = generator terminal voltage

– XS = synchronous reactance, per phase

– δ = phase angle between E and VT

δsinS

TE X

VEP =



0

0.5

1.0

1.5

2.0

P[pu]

0 30 60 180δ [degrees]

90 120 150



• Example– a 36 MVA, 21 kV, 1800 rpm, 3-phase, 60 Hz generator is

connected to the power grid• synchronous reactance of 9 Ω per phase

• line-to-neutral exciting voltage is 12 kV

• line-to-line system voltage is 17.3 kV.

– calculate • the active power delivered when the power angle δ is 30°

• the peak power that the generator can deliver before losing synchronism


Transient Reactance

• A synchronous generator connected to a system is subject to switching events– short-circuits, load energization, etc.

• In many cases, the equivalent circuit doesnot reflect the behavior of the machine– the equivalent circuit is only valid

for steady-state operation

– for sudden, large current changes another reactance is needed

• reactance X' whose value varies as a function of time

– the reactance for a short circuit

• prior to the fault, the reactance equals the synchronous value

• at the instant of fault, the reactance falls to a much lower value, X'd


Transient Reactance

• The reactance X'd is called the transient reactance– can be as low as 15% of the synchronous reactance

– consequently, the initial short-circuit current is much higher than that corresponding to the synchronous reactance


Transient Model

• Example– a 250 MVA, 25 kV, 3-phase generator delivers its rated

output at unity power factor• a synchronous reactance of 1.6 pu

• a transient reactance of 0.23 pu

– a short circuit suddenly occurs on the connecting transmission line, close to the generator

– calculate• the induced voltage, E0, prior to the short circuit

• the initial value of the short-circuit current

• the final value of the short-circuit current if the circuit breaker should fail to open


Power Transfer

• We are often interested in the active power that can be transmitted between source A and source B– using Kirchhoff’s voltage law

– the active power absorbed at source B is

– applying the geometry law of the sines for a triangle

– substitution results in

ABBA IjXEE +=

BABBB IEP θcos=

θθψδ cos90sinsinsinAAAAB EEEIX =

+==

δsinX

EEP AB

B =


Power Transfer

• Example– a transmission line connects two generators

• generator A operates at E = 20 kV∠ 5°

• generator B operates at E = 15 kV∠ 42°

• the transmission line has a reactance of 14 ohms

– calculate the active power that flows over the line• which machine is receiving the power


Machine Efficiency

• The physical size of the synchronous machine has a profound effect upon:– efficiency, power output, relative cost, and temperature rise

– losses in the machine• I2R losses in the stator windings

• Idc2Rf losses in the rotor field winding

• iron core losses and mechanical losses

– keeping all machine parameters and materials the same• an increase in all linear dimensions causes

– voltage increases by the square

– output power increases by the 4th power

– losses increase by the 3rd power


Homework

• Problems 16-22, 16-23, and 16-24

chapter 16: synchronous generators - oakland...

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