Chapter 4 synchronous machine

CHAPTER 4 SYNCHRONOUS MACHINE

4.1 INTRODUCTION

Synchronous machine is designed to be operating at synchronous speed, nsync.

The rate rotation of the magnetic fields in the synchronous machine is given as:

Where, nm = rate rotation of synchronous machine’s magnetic field, rpmfe = electrical frequency/frequency supply,HzP = number of poles in the machine.

The synchronous machine can be used to operate as:[a] Synchronous generator [b] Synchronous motor

Used principally in large power applications because of their - high operating efficiency, - reliability and - controllable power factor .

Rotates at constant speed in the steady state.

The rotating air gap field and the rotor rotate at the same speed.

Applications:- Used for pumps in generating stations- Electric clock- Timers- Mills- Refineries - Assist in power factor correction and etc

It can draw either leading or lagging reactive current from the ac supply system.

It is a doubly excited machine:

- Rotor poles are excited by a DC current- Stator are connected to the ac supply

The air gap flux is the resultant of the fluxes due to both rotor and stator.

88

Figure 4.1

4.2 SYNCHRONOUS GENERATOR

Construction

Synchronous generator is also known as alternator. This machine consists of two main parts:i. Field winding (rotor winding)

- winding that produce the main magnetic field in the machine. ii Armature winding (stator winding)

- winding where the main voltage is induced.

Main construction:- Rotor- Stator

Stator - has a 3 phase distributed windings(AC supply)– similar to induction machine.- Stator winding is sometimes called the armature winding.

Rotor - has a winding(DC winding) called the field winding.- Field winding is normally fed from an external dc source through slip rings and

brushes.- Rotor can be divided into two groups:

High speed machines with cylindrical (or non salient pole) rotors

89

Low speed machines with salient pole rotors

Figure 4.2

Two types of rotor:

[a] Salient pole rotor – magnetic pole stick out from the surface of the rotor Its rotor poles projecting out from the rotor core. Is use for low-speed hydroelectric generator. Need large number of poles to accumulate in projecting on a rotor (large

diameter but small length). Almost universal adapt. Has non uniform air-gap.

Figure 5.3: Two poles salient pole rotor

[b] Non salient or cylindrical pole rotor – magnetic pole constructed flush with the surface of the rotor.

Has its rotor in cylindrical form with dc field winding embedded in the rotor slots.

90

Has uniform air-gap. Provide greater mechanical strength. Per-unit more accurate dynamic balancing. For use in high speed turbo generator. 2 / most 4 poles machine use. Simple to model & analyze.

Figure 4.4: Two poles round (cylindrical) rotor

Two type of armature winding:[a] Single layer winding.[b] Double layer winding.

Generated voltage

The magnitude of the voltage induced in the given stator phase is given as:

or where φ = flux in the machine

f = frequency of the machineω = speed rotation of the machineK = Constant representing the construction of the machine

From the equation, it can concluded that (i) EA proportional to flux and speed(ii) Flux proportional to field current, If

(iii) Thus EA also proportional to If

91

The internal generated voltage EA Vs field current If plot is shown below:

Figure 4.5: Magnetization curve

Equivalent Circuit of a Synchronous Generator

Per-phase equivalent circuit of synchronous generator

EA = internal generated or generated emf per phaseIA = armature currentVθ = per phase terminal voltageXs = synchronous reactance

Zs

92

IF

EA

VΦ

I

1.0k 1.0m

+-

EG

Ra Xs

load

Phasor Diagram of The Synchronous Generator

The load for synchronous machine may be of three types:(i) Pure resistive load (unity power factor)(ii) Inductive load (lagging power factor)(iii) Capacitive load (leading power factor)

(i) Synchronous generator with pure resistive load (unity power factor) :

1.0k 1.0m

+-

1.0

k

EG

Ra Xs

load

The equation:

(ii) Synchronous generator with inductive load (lagging power factor):

93

V IARA

jXSIAEA

IA

Unity pf

Ia

Vt

Direction Ia out from the generator because generator supply power to the load

I 1.0k 1.0m

+-

1.0

k 1

.0m

EG

Ra Xs

load

The equation:

(iii) Synchronous generator with capacitive load (leading power factor):

The equation:

NOTE:AT NO LOAD CONDITION, IA = 0 A, THUS EA = Vφ∟00 VOLT.Example 1

I

1.0k 1.0m

+- 1

.0k

1.0

u

EG

Ra Xs

load

94

V

IARA

jXSIA

EA

IA

Lagging pf

VIARA

jXSIAEA

IA Leading pf

A 3-phase Y-connected synchronous generator supplies a load of 10MW at power factor 0.85 lagging and the terminal voltage is 11kV. The armature resistance is 0.1 ohm/phase and synchronous reactance of 0.66 ohm/phase. Calculate the armature current, the internal generated voltage and voltage regulation. Draw the phasor diagram.

Solution

Ia

1,0

k 1,0

+-

Eg

load

Power factor = 0.85 (lagging)

θ = cos-1 PF = cos-1 0.85 = 31.790

Pout = 10MW = √3 VLIL cos θ

IL = IA = Pout / (√3 VL cos θ) = 10M / (√3 x 11k x 0.85) = 617.5 A

EA = Vφ∟00 + (IA∟-θ)(RA + jXS) = (11k/√3 )∟00 + (IA∟-31.790) (0.1 + j0.66) = 6624.07∟2.720 V

Power flow diagram and Torque in Synchronous Generator

95

2.72º31.79º

Stray losses

Friction and windage losses

Core losses

I2R losses (copper losses) Pout=3VTILcos

Cop.Loss=3(IA)2RA

Pin=Input torque x gen. speed in r/sec

Pin=Pout + Total LossesIf not given in the question, Pstray = 0

Usually given in the question

Can be calculated when RA is given and IA is knownIf the question don’t mentioned about RA, the copper losses = 0

==

V

IARA

jXSIA

EA

IA

Lagging pf

=6624.07V

0.1Ω

j0.66ΩVt

P copper loss(3Ia2Ra)

P in Pm Pout

P iron loss, Pμ (stray+friction+windage+core+etc)

For delta connection,

Vφ = VL

96

Pin=appm

Synch GeneratorPin = Pout + P iron loss + P copper loss

= Tmwr

IA = IL / √3= Pout / (3 Vφ cosθ)= S / 3 Vφ

For Y connection,

Vφ = VL / √3IA = IL= Pout / (√3 VL cosθ) or Pout / (3 Vφ cosθ)= S / √3 VL or S / 3 Vφ

Real output power can be determined using the following equation:

OR

Since XS >> RA, then RA can be ignored. The new phasor diagram will be:

From the pahsor diagram, it can be seen that IA cosθ can be represented as:

Insert IA cosθ in the real output power equation will give:

At maximum condition, δ = 900, thus the equation will be:

EA

O

jXSIA

V

IA

a b

c

EAsin = XSIAcos

97

Other equation for induced torque for synchronous generator:

4.2.6 Efficiency and Voltage Regulation

The voltage regulation of synchronous generator is given as:

The efficiency of the synchronous generator is given as:

Example 2

A 480V, 60Hz, delta-connected, 4 pole synchronous generator has the OCC shown in figure 1. This generator has a synchronous reactance of 0.1 ohm per phase, and armature resistance of 0.015 ohm per phase. At full-load, the machine supplies 1200 A at 0.8 PF lagging. Under full load conditions, the friction and windage losses are 40kW and the core losses are 30 kWi) What is the speed rotation of the magnetic field in rpm?ii) How much is the field current must be supplied to the generator to make the terminal voltage 480V at no load?iii) If the generator is now connected to the load and the load draws 1200 A at 0.8 PF lagging, how much field current will be required to keep the terminal voltage to 480V? What is the voltage regulation of this generator?iv) How much power is now generator is supplying? How much power is supplied to the generator by the prime mover? What is the generator’s efficiency?

98

v) If the generator is now connected to a load drawing 1200 A at 0.8 PF leading, how much field current will be required to keep VT = 480 V and what is generator’s voltage regulation?

Figure 1

Solution

For delta connection:

Vφ = VL

IA = IL / √3

i) The speed rotation of magnetic field:nm = (120fe )/ P = (120 x 60) / 4 = 1800 rpm

ii) If at no load condition.

At no load, IA = 0 A. Thus,

99

EA = Vφ∟00 Volt. = 480 ∟00 V.

Refer to OCC, when EA = 480 V, If = 4.5 A.

iii) If at load (0.8 power factor lagging). IL = 1200 A.

PF= cos θ = 0.8 (lagging)

θ = cos-1 0.8 = 36.870

IA = IL / √3 = 1200 / √3 = 692.82 A.

Vφ = VL = 480 V

For lagging load, the internal generated voltage is given as:

EA = Vφ∟00 + (IA∟-θ) (RA + jXS) = 480∟00 + (692.82∟-36.870) (0.015 + j0.1) = 529.88 + j49.19 = 532.16 ∟5.300 V

From OCC, when EA = 532.16 V, If = 5.7 A.

The voltage regulation, VR:

VR = [ (EA - Vφ) / Vφ ]x 100% = [(532.16 – 480) / 480] x 100% = 10.83%

iv) The output power:

PF= cos θ = 0.8 (lagging)

Pout = √3 VL IL cosθ = √3 (480) (1200) (0.8) = 798.129kW

The input power:

PIN = Pout + Pstray + Pf&w + Pcore + Pelect

100

Pelect = 3 IA2RA

= 3 (692.82)2(0.015) = 21.6kW

PIN = 798.129k + 0 + 40k + 30k + 21.6k = 889.729kW

The generator’s efficiency:

η = (Pout / PIN ) x 100% = (798.129k / 889.729k) x 100% = 89.7%

v) If at load 0.8 leading.

IL = 1200 A

IA = IL/√3 = 1200 / √3

= 692.82 A.

PF = cos θ = 0.8 (leading)

θ = cos-1 0.8 = 36.870

EA = Vφ∟00 + (IA ∟+θ)(RA + jXS) = 480∟00 + (692.82∟+36.870)(0.015 + j0.1) = 446.74 + j61.66 = 450.98∟7.860

Refer to OCC, when EA = 450.98 V, If = 4A

The generator’s voltage regulation:

VR = [(EA - Vφ) / Vφ] x 100% = [(450.98 – 480) / 480] x 100% = 6.06 %

Measuring Synchronous Generator Model Parameters

101

3 quantities that to be determined, Relationship between field current, If and EA

Synchronous reactance, XS

Armature resistance, RA

2 tests to be conducted, open circuit test – terminal of generator is open-circuited, generator run

at rated speed, If is gradually increased in steps, and terminal voltage is measured. Produced open-circuit characteristic (OCC) graph.

short circuit test – terminal of generator is short-circuited through an ammeter, IA or IL is measured as If is increased. Produced short-circuit characteristic (SCC) graph.

From the short circuit test, the armature current, IA is given as:IA = (EA/ (RA + jXS)

The magnitude of IA is given by:

|IA| = |EA| / √(RA2 + jXs

2)

The internal machine impedance is given as:

ZS = √(RA2 + jXS 2 ) = EA / IA

102

If (A)

VT (V)Air-gap line

Open-circuit characteristic (OCC)

Short-circuit characteristic (SCC)

If (A)

IA (A)

Since Xs >>RA, the equation reduce to:

Xs = Vφoc / IA

= EA/IA

5.2.8 The Synchronous Generator Operating Alone

General conclusions from synchronous generator behavior are– If lagging loads (+Q or inductive load) are added, V and VT decrease

significantly but voltage regulation VR is positive large.– If unity power factor loads (no reactive load) are added to a generator,

there is a slight decrease in V and VT and VR is positive small– If leading loads (-Q or capacitive power loads) are added, V and VT will

rise and VR is negative.

5.3 SYNCHRONOUS MOTOR

’

IAI’A

VV’

EA

E’A

jXSIAjXSI’A

+Q or inductive load added, V and VT VR = large +ve

Lagging PF

’

IA

I’A

V V’

EA

E’A jXSIA

jXSI’A

Leading PF

-Q or capacitive load added, V and VT

VR = -ve

103

’

IA I’A VV’

EA

E’A

Unity PF+P or resistive load added, V and VT VR = small +ve

Synchronous motor converts electrical power to mechanical power.

The equivalent circuit of synchronous motor:

The difference between synchronous generator equivalent circuits and the synchronous generator equivalent circuit is the direction of IA.

The phasor diagram of synchronous motor with various power factor are shown below:

i) Unity Power Factor

ii) Lagging Power Factor

104

VΦ

iii) Leading Power Factor

Power flow for synch motor is reverse from synch motor

P iron loss, Pμ (stray+friction+windage+core+etc)

P in Pm Pout

P copper loss(3Ia

2Ra)

Efficiency and Voltage Regulation

The voltage regulation of synchronous motor is given as:

105

Synch MotorPin = Pout + P iron loss + P copper loss

= Tmwr

The efficiency of the synchronous motor is given as:

Example 3

A three phase, Y connected synchronous generator is rated 120 kVA, 1.5 kV, 50 Hz, 0.75 pf lagging. Its synchronous inductance is 2.0mH and effective resistance is 2.5 Ω. (i) Determine the voltage regulation at this frequency.(ii) Determine the rated voltage and apparent power if the supply frequency is going to

be twice.(iii) Determine the voltage regulation if the frequency is increased to 120% of the

original frequency.

Solution

(i)

+- DC

M1

1,0m 1,0k 120kVA

1.5kV

50HZ

2.5 ohm2.0 mH

Ea

Therefore

106

(ii)

Rated Voltage:

Apparent Power

(iii)

New

at 60Hz

Rated Voltage:

107

Apparent Power

Example 4

A 2300V, 120hp, 50Hz, eight poles, Y-connected synchronous motor has a synchronous inductance of 6.63mH/phase and armature resistance of 1Ω/phase at rated power factor of 0.85 leading. At full load, the efficiency is 90 percent. Find the following quantities for this machine when it is operating at full load.

(i) Draw a phasor diagram to represent back emf, supply voltage and armature current.

(ii) What is its voltage regulation?

(iii) Output power.

(iv) Input power.

(v) Developed mechanical power.

(vi). Draw the power flow diagram.

Solution

1hp = 746W2300V, 120hp=89.52kW, 50Hz, 8 poles, efficiency = 90%Y connected(IΦ=IL, VΦ=VL/√3)

108

Xs = 2πfL=2π(50)(6.63m)=2.08Ω Z = Ra+jXs = 1 + j2.08 Ω (i)

η =

(ii)

(iii) Pout = 120hp = 89.52kW

(iv)

(v) Pmech = Pin-Pcl= 99.47kW - 3Ia2R= 99.47kW - 3(29.38)2(1)= 96.88kW

(vi) P iron loss, Pμ (stray+friction+windage+core+etc)

P in Pm Pout

P copper loss

Steady-state Synchronous Motor Operation

Torque Speed Characteristic Curve

109

1) Speed of the motor will be constant (because it’s locked to the electrical frequency). The result torque speed characteristic is shown in the figure above.

2) The speed is constant from no load torque until max torque, thus SR=0%.

Effect Load Changes on Synchronous Motor

When the load is increased:

1) θ is change from leading to lagging.2) jXSIA is increased, thus IA is increased too.3) Torque angle, δ, is increased4) |EA| is constant

Effect Field Current Changes on Synchronous Generator

nm

ind

pullout

rated

nsync

Maximum torque when =90°

IA1 IA2

IA3

IA4

V

EA1

EA2

EA3

EA4

P1P2

P3 P4

110

When the field current is increased:

1) |EA| increased2) Torque angle, δ, is decreased3) IA first is decreased and then increased.4) θ changed from lagging to leading.5) Real power supply is constant6) VL is constant.

Synchronous V Curve

Starting Synchronous Motor

3 methods to start a synchronous motor:i) Reduce electrical frequency, fe

ii) Use external prime moveriii) Use damper windings or amortisseur windings.

4.4 RELATIONSHIP BETWEEN SYNCHRONOUS MOTOR AND SYNCHRONOUS GENERATOR

P (=constant)

P (=constant)IA1

IA2

IA3

IA4

EA1 EA2 EA3 EA4

V

Lagging power factor

PF = 1.0

Leading power factor

P = P1

P = P2

IA

IF

111

IA

EA

V

IA

EA

V

4.5 INDUSTRIAL APPLICATION

The three phase synchronous motor is used when a prime mover having a constant speed from a no-load condition to full load is required. Such as fans, air compressors and pumps.

Also used to drive mechanical load and also to correct the power factor.

Only used as a correct power factor of an industrial power system, as a bank capacitor used for power factor correction, also called a synchronous capacitor.

Rating up to 10hp are usually started directly across the rated three-phase voltage. Synchronous motor of larger sizes are started through a starting compensator or an automatic starter.

Tutorial 4

112

Consume PConsume P

MotorMotor

EEAA lags lags VV

Supply Supply PP

GeneratorGenerator

EEAA leads leads VV

Consume QConsume Q EEAAcoscos < < VVSupply Supply QQ EEAAcoscos > > VVP \ QP \ Q

IA

EAV

IA

EA V

IA

EA

V

IA

EA

V

1. A 90kVar, 445 V, 60 Hz 3-phase synchronous motor is delta connected. The motor has a synchronous inductance 1.3 mH/phase and armature resistance of 4 ohm/phase. Calculate the back emf and power angle if the motor is operating at 0.85 lagging power factor.

2. A 2400 V, 60 kW, 50 Hz, 6 poles, delta-connected synchronous motor has a synchronous reactance of 4 Ω/phase and armature resistance of 2 Ω/phase. At full load, the efficiency is 92 %. Find the followings for this machine when it is operating at full load at rated power factor 0.85 lagging.

(i) Phasor diagram to represent back emf, supply voltage and armature current.

(ii) Voltage regulation.

(iii) Input power

(iv)Developed mechanical power

(v) Power flow diagram

3. A 3-phase, 50 Hz, Y-connected synchronous generator supplies a load of 10MW. The terminal voltage is 22 kV at power factor 0.85 lagging . The armature resistance is 0.2 ohm/phase and synchronous inductance is 0.3 mH/phase. Calculate the line value of emf generated.

4. A hydraulic turbine turning at 200 rpm is connected to a synchronous generator. If the induced voltage has a frequency of 60 Hz, how many poles does the rotor have.

5. A 2300 V 3-phase, star connected synchronous motor has an armature resistance of 0.2 ohm/phase and a synchronous reactance of 2.2 ohm/phase. The motor is operating on 0.5 power factor leading with a line current of 200A. Determine the value of generated or counter emf per phase. Also draw the phasor diagram. From the phasor diagram, discuss what happen to the counter emf if the power factor is increased to 0.8 leading. No need to recalculate the new emf being generated.

6. A 200kVA, 600 V, 50 Hz three-phase synchronous generator is Y- connected. The generator has a synchronous reactance 0.1 ohm/phase and armature resistance of 2 ohm/phase. Calculate the voltage regulation if the generator is operating at 0.75 leading power factor. Draw the phasor diagram.

7. A 2000V, 500hp, 3-phase Y-connected synchronous motor has a resistance of 0.3 ohm/phase and synchronous reactance of 3 ohm/phase respectively. Determine the induced emf per phase if the motor works on full-load with an efficiency of 92% and p.f = 0.8 leading.

113