Ac gen2

Synchronous Machines(AC Generators)

Muhammad Abdul Majid

Construction of synchronous machinesmachines

FundamentalsFundamentals

• AC (or DC) Machines are of two typesAC (or DC) Machines are of two types– Generators

Motors– Motors

• Every AC (or DC) motor or generator has two tparts:

– rotating part (rotor)

– and a stationary part (stator)

The induced voltage in a single coil on la two‐pole stator

The statorThe stator

The induced voltage in a 3‐phase set of lcoils

Induced voltage: Exampleg p

Example 6.1: The peak flux density of the rotor magnetic field in a simple 2‐pole 3‐phase generator is 0 2 T; the mechanical speed of rotation is 3600 rpm; the stator diameter isgenerator is 0.2 T; the mechanical speed of rotation is 3600 rpm; the stator diameter is 0.5 m; the length of its coil is 0.3 m and each coil consists of 15 turns of wire. The machine is Y‐connected.

a) What are the 3‐phase voltages of the generator as a function of time?a) What are the 3 phase voltages of the generator as a function of time?b) What is the rms phase voltage of the generator?c) What is the rms terminal voltage of the generator?

Construction of synchronous machines’ rotormachines rotor

The rotor of a synchronous machine is a large electromagnet. The magnetic poles can be either salient (sticking out of rotor surface) or non‐salient construction.

Non‐salient‐pole rotor: usually two‐ and four‐pole rotors. Salient‐pole rotor: four and more poles.

R t d l i t d t d dd t lRotors are made laminated to reduce eddy current losses.

Construction of synchronous machines’ rotormachines rotor

A synchronous rotor with 8 salient poles

Salient pole with field Salient pole without fi ld i di

pwindings field windings –

observe laminations

Rotor Excitation

h d l hTwo common approaches are used to supply a DC current to thefield circuits on the rotating rotor:

Supply the DC power from an external DC source to the rotor bymeans of slip rings and brushes.

Supply the DC power from a special DC power source mounteddirectly on the shaft of the machine.

Rotor Excitation

Slip rings are metal rings completely encircling the shaft of aSlip rings are metal rings completely encircling the shaft of amachine but insulated from it.

One end of a DC rotor winding is connected to each of the twoslip rings on the machine’s shaft.

Graphite‐like carbon brushes connected to DC terminals ride oneach slip ring supplying DC voltage to field windings regardlessthe position or speed of the rotorthe position or speed of the rotor.

Slip Rings and Brushesp g

Slip rings

Brush

Slip Rings and Brushesp g

Brushless Exciter Schematic1

A brushless exciter: a lowA brushless exciter: a low 3‐phase current is rectified and used to supply the field circuit of the exciter (located on the stator). The output of the exciter’s armature circuit (on the rotor) is rectified and used as the field current of the main machinemachine.

Brushless Exciter Schematic2

To make the excitation of a generatorof a generator completely independent of any external power source, a small pilot exciter is often added to the circuit. The pilot exciter is an AC generator withis an AC generator with a permanent magnet mounted on the rotor shaft and a 3‐phase winding on the stator producing the power for the field circuit of the exciter.the exciter.


A rotor of large synchronous machine with a brushless exciterwith a brushless exciter mounted on the same shaft.

Many synchronous generators havinggenerators having brushless exciters also include slip rings and brushes to provide

f themergency source of the field DC current.


A largeA large synchronous machine with th itthe exciterand salient poles.

Rotation speed of synchronous generatorgenerator

By the definition, synchronous generators produce electricity whose frequency i h i d i h h h i l i l dis synchronized with the mechanical rotational speed.

120m

en Pf =120ef

Where fe is the electrical frequency, Hz;nm is mechanical speed of magnetic field (rotor speed for synchronous m p g ( p ymachine), rpm;P is the number of poles.

Steam turbines are most efficient when rotating at high speed; therefore, to generate 60 Hz, they are usually rotating at 3600 rpm and turn 2‐pole generators.Water turbines are most efficient when rotating at low speeds (200‐300 rpm); therefore, they usually turn generators with many poles.

Internal generated voltage of a synchronous generatorsynchronous generator

Since flux in theSince flux in the machine depends on the field current through it, the internal generated voltage is a function of the rotor field current.current.

Magnetization curve (open‐circuit characteristic) of a synchronous machine

Equivalent circuit of a synchronous generatorgenerator

The internally generated voltage in a single phase of a synchronousThe internally generated voltage in a single phase of a synchronous machine EA is not usually the voltage appearing at its terminals. It equals to the output voltage Vφ only when there is no armature

t i th hi Th th t th t lt Ecurrent in the machine. The reasons that the armature voltage EAis not equal to the output voltage Vφ are:

1 Distortion of the air gap magnetic field caused by the current1. Distortion of the air‐gap magnetic field caused by the current flowing in the stator (armature reaction);

2. Self‐inductance of the armature coils;3. Resistance of the armature coils;4. Effect of salient‐pole rotor shapes.

Armature ReactionArmature Reaction


Assuming that the generator is connected to a lagging load, the load current IA will create a stator magnetic field BS, which will produce the armature reaction voltage Estat. Therefore, the phase voltage will be

V E E= +A statV E Eφ = +

The net magnetic flux will be

net R SB B B= +Rotor field Stator field

Note that the directions of the net magnetic flux and the phase voltage are the same.


A synchronous generator can be Y‐ or ∆‐connected:

3V V f Y V V f ∆3T TV V for Y V V forφ φ= − = − ∆


Note: the discussion above assumed a balanced load on the generator!

Since – for balanced loads – the three phases of a synchronous generator are identical except for phase angles, per‐phase equivalent circuits are often used.

Phasor diagram of a synchronous generatorgenerator

A phasor diagram of a synchronous generator with a unity power factor (resistive load)

Lagging power factor (inductive load): a larger thanLagging power factor (inductive load): a larger than for leading PF internal generated voltage EA is needed to form the same phase voltage.

Leading power factor (capacitive load).

For a given field current and magnitude of loadFor a given field current and magnitude of load current, the terminal voltage is lower for lagging loads and higher for leading loads.

Power and torque in synchronous generatorsgenerators

Open Circuit Curvep

Open Circuit Curve

Example

• A 2300‐V 1000‐kVA 0.8‐PF‐lagging 60‐Hz two‐300 000 0.8 agg g 60 t opole Y‐connected synchronous generator has a

• synchronous reactance of 1.1 Ω and an armature yresistance of 0.15 Ω. At 60 Hz, its friction and windage

• losses are 24 kW, and its core losses are 18 kW. The field circuit has a dc voltage of 200 V, and the

i IF i 10 A Th i t f th fi ld• maximum IF is 10 A. The resistance of the field circuit is adjustable over the range from 20 to 200 ΩΩ

Example

• The OCC is shown in Figure on next slide• (a) How much field current is required to make VT equal to 2300 V when the generator is running at no

• load?• load?• (b) What is the internal generated voltage of this machine at rated conditions?

• (c) How much field current is required to make VT equal to 2300 V when the generator is running at rated

• conditions?conditions?• (d) How much power and torque must the generator’s prime mover be capable of supplying?

Related Example

• Assume that the field current of the generator in above Problem h b dj t d t l f 4 5 Ahas been adjusted to a value of 4.5 A.

• (a) What will the terminal voltage of this generator be if it is connected to a ∆‐connected load with an impedance of 20 30 °Ω?

• (b) Sketch the phasor diagram of this generator.• (c) What is the efficiency of the generator at these conditions?• (d) Now assume that another identical ∆‐connected load is to be

paralleled with the first one. What• happens to the phasor diagram for the generator?• (e) What is the new terminal voltage after the load has been

dd d?added?• (f) What must be done to restore the terminal voltage to its original

value?

Measuring parameters of synchronous generator modelsynchronous generator model

An approximate method to determine the synchronous reactanceAn approximate method to determine the synchronous reactance XS at a given field current:

1. Get the internal generated voltage EA from the OCC at that field Acurrent.

2. Get the short‐circuit current IA,SC at that field current from the SCCSCC.

3. Find approximate XS = EA /(IA,SC)


The approximate value of synchronous reactance varies with the degreeThe approximate value of synchronous reactance varies with the degree of saturation of the OCC.The value is more accurate in unsaturated portion only.

Measuring parameters of synchronous generator model: Exsynchronous generator model: Ex

Example 7.1: A 200 kVA, 480 V, 50 Hz, Y‐connected synchronous generator with a rated field current of 5 A was tested and the following data were obtained:current of 5 A was tested and the following data were obtained:

1. VT,OC = 540 V at the rated IF.2. IL,SC = 300 A at the rated IF.3 When a DC voltage of 10 V was applied to two of the terminals a current of 25 A was3. When a DC voltage of 10 V was applied to two of the terminals, a current of 25 A was

measured.

Find the generator’s model at the rated conditions (i.e., the armature resistance and the approximate synchronous reactance)approximate synchronous reactance).


Example 7.1: A 200 kVA, 480 V, 50 Hz, Y‐connected synchronous generator with a rated field current of 5 A was tested and the following data were obtained:current of 5 A was tested and the following data were obtained:

1. VT,OC = 540 V at the rated IF.2. IL,SC = 300 A at the rated IF.3 When a DC voltage of 10 V was applied to two of the terminals a current of 25 A was3. When a DC voltage of 10 V was applied to two of the terminals, a current of 25 A was

measured.

Find the generator’s model at the rated conditions (i.e., the armature resistance and the approximate synchronous reactance)approximate synchronous reactance).

Since the generator is Y‐connected, a DC voltage was applied between its two phases. Therefore:

2 DCA

DC

VRI

=

10 0.22 2 25

DCA

DC

VRI

= = = Ω⋅


The internal generated voltage at the rated field current is

540V,

540 311.83 3T

A OCVE V Vφ= = = =

The synchronous reactance at the rated field current is preciselyThe synchronous reactance at the rated field current is precisely

2 22 2 2 2

2 2

311.8 0.2 1.02300

AS S A A

A SC

EX Z R RI

= − = − = − = Ω,A SC

We observe that if XS was estimated via the approximate formula, the result would be:

311.8AE

,

311.8 1.04300

AS

A SC

EXI

≈ = = Ω

Which is close to the previous result. The i i i h ll h herror ignoring RA is much smaller than the

error due to core saturation.The equivalent circuit

The Synchronous generator operating aloneoperating alone

Effects of load changes

(7.34.1)


The behavior of a synchronous generator varies greatly under load y g g ydepending on the power factor of the load and on whether the generator is working alone or in parallel with other synchronous generatorsgenerators.

Although most of the synchronous generators in the world operate as parts of large power systems, we start our discussion assuming that the synchronous generator works alone.

Unless otherwise stated, the speed of the generator is assumed constant.


Effect of increasing inductive load

How to compensate the increase in LoadLoad

Increase the field current, that increases flux and in turn EA causing the VT to remainsame.


load increase effect on generators with

Leading PF

Unity PF

The Synchronous generator capability curvescapability curves

The Synchronous generator operating alone: Exampleoperating alone: Example

Example : A 500 V, 60 Hz, Y‐connected six‐pole synchronous generator has a per‐phase synchronous reactance of 1 0 Ω Its full load armature current is 60 A at 0 8 PF lagging Itssynchronous reactance of 1.0 Ω. Its full‐load armature current is 60 A at 0.8 PF lagging. Its friction and windage losses are 1.5 kW and core losses are 1.0 kW at 60 Hz at full load. Assume that the armature resistance can be ignored. The field current has been adjusted such that the no‐load terminal voltage is 500 V.

a. What is the speed of rotation of this generator?b What is the terminal voltage of the generator ifb. What is the terminal voltage of the generator if

1. It is loaded with the rated current at 0.8 PF lagging;2. It is loaded with the rated current at 1.0 PF;3. It is loaded with the rated current at 0.8 PF leading.

c. What is the efficiency of this generator (ignoring the unknown electrical losses) when it is operating at the rated current and 0.8 PF lagging?

d. How much shaft torque must be applied by the prime mover at the full load?how large is the induced countertorque?how large is the induced countertorque?

e. What is the voltage regulation of this generator at 0.8 PF lagging? at 1.0 PF? at 0.8 PF leading?

The Synchronous generator operating alone: Exampleoperating alone: Example

Since the generator is Y‐connected, its phase voltage is

At no load, IA = 0 and EA = 288.6 V and it is constant since the field current was initially

VT(no load) = 500V = EA (line, no load)Vφ = VT/(3)1/2 =288.6 = EA (phase, no load)

adjusted that way.

a. The speed of rotation of the synchronous generator is

n = 120f /P = 120(60)/6 =1200 rpmnm = 120fe/P = 120(60)/6 =1200 rpm1200 rpm =125.7 rad/s

Example continued0 8 lagging PF0.8 lagging PF

jXSIA = (j 1) x( 60 ‐36.87) = 60 53.13from phasor diagram E2A = (Vφ + XSIASin θ)2 +(XSIA Cosθ)2

Example continued0 8 lagging PF0.8 lagging PF

(288.6)2 = (Vφ + (1.0)(60.0)(Sin 36.87))2 +(60( 1.0) Cos 36.87)2( ) ( φ ( )( )( )) ( ( ) )

V φ = 248V

Y connected thereforeY connected thereforeVT = (3)1/2 Vφ = 430 V

Example continuedUNITY PFUNITY PF

E2A = (Vφ)2 +(XSIA)2

(288 6)2 = (Vφ)2 +((1 0)60)2(288.6) = (Vφ) +((1.0)60)Vφ =282.VY connected thereforeVT = (3)1/2 Vφ = 488 V

Example continued0 8 leading PF0.8 leading PF

E2A = (V ‐ XSIASin θ)2 +(XSIA Cosθ)2E A (Vφ XSIASin θ) +(XSIA Cosθ)

V φ =320 v

Terminal characteristics of synchronous generatorssynchronous generators

All generators are driven by a prime mover, such as a steam, gas, water, wind turbines, diesel engines, etc. Regardless the power source, most of prime movers tend to slow down with increasing the load. This decrease in speed is usually nonlinear but governor mechanisms of some type may be included to linearize this dependence.

The speed drop (SD) of a prime mover is defined as:

100%nl fl

fl

n nSD

n−

= ⋅

Most prime movers have a speed drop from 2% to 4%. Most governors have a mechanism to adjust the turbine’s no‐load speed (set‐point adjustment).


A typical speed vs. power plot

A typical frequency vs.

shaft speed is linked to the electrical frequency as

power plot


fe = nmP/120

the power output from the generator is related to its frequency:

P = SP (fnl – fsys)

Operating frequency of the systemSlope of curve, W/Hz


A similar relationship can be derived for the reactive power Q and terminal voltage VT. When adding a lagging load to a synchronous generator its terminal voltage decreasesWhen adding a lagging load to a synchronous generator, its terminal voltage decreases. When adding a leading load to a synchronous generator, its terminal voltage increases.


When a generator is operating alone supplying the load:1. The real and reactive powers are the amounts demanded by the

load.2. The governor of the prime mover controls the operating frequency

of the system.of the system.3. The field current controls the terminal voltage of the power system.

Terminal characteristics of synchronous generators: Examplesynchronous generators: Example

Example 7.3: A generator with no‐load frequency of 61 0 Hz and a slope s of 1 MW/Hz is connected to61.0 Hz and a slope sp of 1 MW/Hz is connected to Load 1 consuming 1 MW of real power at 0.8 PF lagging. Load 2 (that is to be connected to the generator) consumes a real power of 0.8 MW at 0.707 PF lagging.

a. Find the operating frequency of the system before the switch is closed.b Find the operating frequency of the system after the switch is closedb. Find the operating frequency of the system after the switch is closed.c. What action could an operator take to restore the system frequency to 60 Hz after

both loads are connected to the generator?


Example 7.3: A generator with no‐load frequency of 61 0 Hz and a slope s of 1 MW/Hz is connected to61.0 Hz and a slope sp of 1 MW/Hz is connected to Load 1 consuming 1 MW of real power at 0.8 PF lagging. Load 2 (that is to be connected to the generator) consumes a real power of 0.8 MW at 0.707 PF lagging.

a. Find the operating frequency of the system before the switch is closed.b Find the operating frequency of the system after the switch is closedb. Find the operating frequency of the system after the switch is closed.c. What action could an operator take to restore the system frequency to 60 Hz after

both loads are connected to the generator?

The power produced by the generator isThe power produced by the generator is

( )p nl sysP s f f= −P

Therefore:sys nl

p

Pf fs

= −


a. The frequency of the system with one load is

161 601sys nl

p

Pf f Hzs

= − = − =

b. The frequency of the system with two loads is

1.861 59.21sys nl

Pf f Hz= − = − = 1sys nl

p

f fs

c. To restore the system to the proper operating frequency, the operator should increase the governor no‐load set point by 0 8 Hz to 61 8 Hz This will restore the systemthe governor no‐load set point by 0.8 Hz, to 61.8 Hz. This will restore the system frequency of 60 Hz.

Parallel operation of synchronous generatorsgenerators

Advantages of parallel operation of generators:

1. Several generators can supply a bigger load;f f2. A failure of a single generator does not result in a total power

loss to the load increasing reliability of the power system;3. Individual generators may be removed from the power system g y p y

for maintenance without shutting down the load;4. A single generator not operating at near full load might be quite

inefficient While having several generators in parallel it isinefficient. While having several generators in parallel, it is possible to turn off some of them when operating the rest at near full‐load condition.

Conditions required for paralleling

A diagram shows that Generator 2 g(oncoming generator) will be connected in parallel when the switch S1 is closed.However, closing the switch at an arbitrary moment can severely damage bothmoment can severely damage both generators!


If lt t tl th i b th li (i iIf voltages are not exactly the same in both lines (i.e. in aand a’, b and b’ etc.), a very large current will flow when the switch is closed Therefore to avoid this voltagesthe switch is closed. Therefore, to avoid this, voltages coming from both generators must be exactly the same.


Th f th f ll i diti t b tTherefore, the following conditions must be met:

1. The rms line voltages of the two generators must be equal.2 The two generators must have the same phase sequence2. The two generators must have the same phase sequence.3. The phase angles of two a phases must be equal.4. The frequency of the oncoming generator must be slightly higher

h h f f h ithan the frequency of the running system.


If the phase sequences are different, then even if one pair of voltages (phases a) are in phase, the other two pairs will be 1200 out of phase creating huge currents in th hthese phases.Solution: swap any two phases of the incoming macnine


1. If the frequencies of the generators are different, a large power transient may occur until the generators stabilize at a common frequencymay occur until the generators stabilize at a common frequency.

2. The frequencies of two machines must be very close to each other but not exactly equal.

3. If frequencies differ by a small amount, the phase angles of the oncoming3. If frequencies differ by a small amount, the phase angles of the oncoming generator will change slowly with respect to the phase angles of the running system.

4. If the angles between the voltages can be observed, it is possible to close the g g pswitch S1 when the machines are in phase.

General procedure for paralleling generatorsgenerators

When connecting the generator G2 to the running system, the following steps should be taken:taken:

1. Adjust the field current of the oncoming generator to make its terminal voltage equal to the line voltage of the system (use a voltmeter).

2 Compare the phase sequences of the oncoming generator and the running system2. Compare the phase sequences of the oncoming generator and the running system. This can be done by different ways:1) Connect a small induction motor to the terminals of the oncoming generator and

then to the terminals of the running system. If the motor rotates in the same direction, the phase sequence is the same;

2) Connect three light bulbs across the open terminals of the switch. As the phase changes between the two generators light bulbs getbetween the two generators, light bulbs get brighter (large phase difference) or dimmer (small phase difference). If all three bulbs get bright and dark together, both generators have the same phase sequences.

General procedure for paralleling generatorsgenerators

If phase sequences are different, two of the conductors on the oncoming generator must be reversedgenerator must be reversed.

3. The frequency of the oncoming generator is adjusted to be slightly higher than the system’s frequency.

4 Turn on the switch connecting G to the system when phase angles are equal4. Turn on the switch connecting G2 to the system when phase angles are equal.

The simplest way to determine the moment when two generators are in phase is by observing the same three light bulbs. When all three lights go out, the voltage across them is zero and therefore machines are in phaseis zero and, therefore, machines are in phase.

A more accurate way is to use a synchroscope – a meter measuring the difference in phase angles between two aphases However a synchroscope does not check the phasephases. However, a synchroscope does not check the phase sequence since it only measures the phase difference in one phase.

The whole process is usually automated…


A typical speed vs. power plot

A typical frequency vs.


power plot


fe = nmP/120

the power output from the generator is related to its frequency:

P = SP (fnl – fsys)

Operating frequency of the systemSlope of curve, W/Hz


Th f th f ll i diti t b tTherefore, the following conditions must be met:

1. The rms line voltages of the two generators must be equal.2 The two generators must have the same phase sequence2. The two generators must have the same phase sequence.3. The phase angles of two a phases must be equal.4. The frequency of the oncoming generator must be slightly higher

h h f f h ithan the frequency of the running system.

Operation of generators in parallel with large power systemswith large power systems

Often, when a synchronous generator is added to a power system, that system is so large that one additional generator does not cause observable changes to the system. A concept of an infinite bus is used to characterize such power systems.An infinite bus is a power system that is so large that its voltage and frequency do not vary regardless of how much real and reactive power is drawn from or supplied to it. Theregardless of how much real and reactive power is drawn from or supplied to it. The power‐frequency and reactive power‐voltage characteristics are:


Consider adding a generator to an infiniteConsider adding a generator to an infinite bus supplying a load.

The frequency and terminal voltage of all machines must be the same Thereforemachines must be the same. Therefore, their power‐frequency and reactive power‐voltage characteristics can be plotted with a common vertical axis.

Such plots are called sometimes as house diagrams.


If the no‐load frequency of the oncoming generator is slightly higher than the system’s frequency, the generator will be “floating” on the line supplying a small amount of real power and little or no reactive power.power and little or no reactive power.

If the no‐load frequency of the oncoming generator is slightly lower than the system’s frequency, the generator will supply a negative power to the system: the generatornegative power to the system: the generator actually consumes energy acting as a motor!Many generators have circuitry automatically disconnecting them from the line when they start consuming energy.


If the frequency of the generator is increased after it is connected to theincreased after it is connected to the infinite bus, the system frequency cannot change and the power supplied by the generator increases.

Notice that when EA stays constant (field current and speed are the same) EAsinδcurrent and speed are the same), EAsinδ(which is proportional to the output power if VT is constant) increases.


Summarizing, when the generator is operating in parallel to an infinite bus:

1. The frequency and terminal voltage of the generator are controlled by the system to which it is connected.

2 The governor set points of the generator control the real power2. The governor set points of the generator control the real power supplied by the generator to the system.

3. The generator’s field current controls the reactive power supplied by the generator to the system.

Generators in parallel with other generators of the same sizegenerators of the same size

When a generator is working alone, its real and reactive power are fixed and determined by the loadby the load.When a generator is connected to an infinite bus, its frequency and the terminal voltage are constant and determined by a bus.

When two generators of the same size are connected to the same load, the sum of the real and reactive powers supplied by the two generators must equal the real andtwo generators must equal the real and reactive powers demanded by the load:


Since the frequency of G2 must be slightly higher than the system’s frequency the powerhigher than the system s frequency, the power‐frequency diagram right after G2 is connected to the system is shown.

If the frequency of G2 is next increased, its power‐frequency diagram shifts upwards. Since the total power supplied to the loadSince the total power supplied to the load is constant, G2 starts supplying more power and G1 starts supplying less power and the system’s frequency increases.


Therefore, when two generators are operating together, an increase in frequency ( t i t) f th(governor set point) on one of them:

1. Increases the system frequency.2. Increases the real power supplied by that generator, while reducing the real

power supplied by the other one.

When two generators are operating together, an increase in the field current on one of them:

1. Increases the system terminal voltage.2. Increases the reactive power supplied by2. Increases the reactive power supplied by

that generator, while reducing the reactive power supplied by the other.

If the frequency‐power curves of both generators are known, the powers supplied by each generator and the resulting system frequency can be determined.

Generators in parallel with other generators of the same size: Exgenerators of the same size: Ex

Example 4: Two generators are set to supply the same load. Generator 1 has a no‐load frequency ofsame load. Generator 1 has a no load frequency of 61.5 Hz and a slope sp1 of 1 MW/Hz. Generator 2 has a no‐load frequency of 61.0 Hz and a slope sp2of 1 MW/Hz. The two generators are supplying a

l l d f 2 5 MW 0 8 PF l ireal load of 2.5 MW at 0.8 PF lagging.

a. Find the system frequency and power supplied by each generator.b. Assuming that an additional 1 MW load is attached to the power system, find the new

system frequency and powers supplied by each generator.c. With the additional load attached (total load of 3.5 MW), find the system frequency and

the generator powers if the no‐load frequency of G is increased by 0 5 Hzthe generator powers, if the no load frequency of G2 is increased by 0.5 Hz.


The total power supplied by the generators equals to the power consumed by the load:

1 2loadP P P= +

a. The system frequency can be found from:

( ) ( )( ) ( )1 2 1 ,1 2 ,2load p nl sys p nl sysP P P s f f s f f= + = − + −

1 1 2 2 1 61.5 1 61.0 2.5 60 0p nl p nl loads f s f Pf H

+ − ⋅ + ⋅ − as 1 ,1 2 ,2

1 2

60.01 1

p nl p nl loadsys

p p

f ff Hz

s s= = =

+ +

The powers supplied by each generator are:

( ) ( )( ) ( )

1 1 ,1 1 61.5 60 1.5

1 61 0 60 1

p nl sysP s f f MW

P f f MW

= − = ⋅ − =

( ) ( )2 2 ,2 1 61.0 60 1p nl sysP s f f MW= − = ⋅ − =


b. For the new load of 3.5 MW, the system frequency is

f f P1 ,1 2 ,2

1 2

1 61.5 1 61.0 3.5 59.51 1

p nl p nl loadsys

p p

s f s f Pf Hz

s s+ − ⋅ + ⋅ −

= = = + +

( ) ( )1 61 5 59 5 2 0P f f MW ( ) ( )( ) ( )

1 1 ,1

2 2 ,2

1 61.5 59.5 2.0

1 61.0 59.5 1.5

p nl sys

p nl sys

P s f f MW

P s f f MW

= − = ⋅ − =

= − = ⋅ − = The powers are:

c. If the no‐load frequency of G2 increases, the system frequency is

1 ,1 2 ,2 1 61.5 1 61.5 3.5 59 75p nl p nl loads f s f Pf Hz

+ − ⋅ + ⋅ −= = =

1 2

59.751 1sys

p p

f Hzs s

+ +

( )The powers are:

( ) ( )1 2 1 ,1 1 61.5 59.75 1.75p nl sysP P s f f MW= = − = ⋅ − =

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