the synchronous machine (sm) in the power system (2) · 2011-09-30 · 30-sep-11 lecture 06 power...

30-Sep-11

1Lecture 06 Power Engineering - Egill Benedikt Hreinsson

The synchronous machine (SM) in the power system (2)

(Where does the electricity come from)?

30-Sep-11


Lecture overview

• Synchronous machines with more than 2 magnetic poles

• The relation between the number of poles and rotation speed

• The magnetic field in the SM• Circuit model for a SM and power when we have a

salient pole rotor• Introduction to the dynamics of machines• Photos of practical SM’s

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3 phase, 2 pole synchronous machine

Stator

A round rotor Rotor

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A 3 phase, 2 pole generator

The stator windings for phase “a” with direction “into the figure”

The stator windings for phase “-a” with direction “out of the figure”

The stator

The rotorRotor windings (or field windings) for Direct current with direction “into the figure”

Rotor windings (or field windings) for Direct current with direction “out of the figure”

Magnetic poles, “N” and “S”

A salient pole rotor

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N

N

S

S

-a

-a

b

-c

a

-b

c

c

a

-c

b

-b

A 3 phase, 4 pole generator

The stator windings for phase “a” with direction “into the figure”

The stator windings for phase “a” with direction “out of the figure”

Stator

Rotor

Rotor windings (or field windings) for Direct current with direction “into the figure”

Rotor windings (or field windings) for Direct current with direction “out of the figure”

Magnetic poles, “N”and “S”

A salient pole rotor

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A 3 phase, 6 pole machine

We have and increased number of poles both on the rotor and stator

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Speed of rotation and number of poles

2 2 60 120e mp p n p nf f ⋅

= ⋅ = ⋅ =

ωe = “Electrical radians”ωm = “Mechanical radians”f e= Frequency of the AC voltage (50 or 60 Hz)f m= Revolutions per secondp = Number of poles (2,4,6...)n = Revolutions per minute (rpm)

150046001030020

30002

230.826

n (f = 50 Hz)p

fe = 50 Hz

60006000 p n np

= ⋅ → =

2m ep

ω ω=

An example for the number of poles:

Legend:

f e= 60 Hz78007800 p n n

p= ⋅ → =

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The Magnetic Field in the SM

No load conditionδ = 0 A phase angle α between the rotor

and stator current (stator field) This angle is also called δ

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The Magnetic Field in the SM (2)

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Reactive Power from a Synchronous Machine

V

EjXI

I

φ

δ

Assume the machine is connected to an “infinite bus” with constant voltage V.

Locus for vector E withconstant real power and

variable exitationand reactive power

Locus for vector Iwith constant real power and variable

exitationThe length of this line is proportional to

and hence the reactive power generation and exitation

cosE Vδ −

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11Lecture 06 Power Engineering - Egill Benedikt HreinssonPhasor diagram with a constant magnetization

2a sjI XtV

t t

s s

V VjjX X−

=

2 2f f

s s

E Ej

jX X= −

1 1f f

s s

E Ej

jX X= −

1aI

2aI

1δ 2δ 1a sjI X

1fE

2fE1δ

2δ

2φ1φ

The locus for a vector of the constant excitation voltage

A locus for the vector of armature current in the case of constant excitation voltage

f t s a

f ft ta

s s s s

E V jX I

E EV VI jjX jX jX X

= +

= − = +

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Salient pole rotord - and q - axis are differentd - axis is directged along the pole direction.q –axis is perpendicular to the pole directions in electrical degrees

•Geometry•Flux•Inductance•Currents and voltages

d-axis

q-axis

d-axis

q-axis

q d d q qE V jX I jX I= + +d djX I

q qjX I

Vφ

qEqV

q-ás

d-ás

δ

I

qI

dV

dI

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P and Q for salient pole rotor

sin cos

cos sinsin cos

q q de

d q

qe

d q

E V VP V V

X X

E V VP V VX X

δ δ

δ δδ δ

−= ⋅ + ⋅

−= ⋅ + ⋅

We substitute:

q qd

d

E VI

X−

=

* *( ) ( )e e d q d qP jQ V I V jV I jI+ = ⋅ = + ⋅ +

sindV V δ= cosqV V δ=

dq

q

VI

X=

e d d q qP V I V I= +

sin cose d qP I V I Vδ δ= +

d djX I

q qjX I

Vφ

qEqV q-axisd-axis

δ

I

qI

dV

dI

( ) ( )e e d q d qP jQ V jV I jI+ = + ⋅ −

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2

field reluctance

2 22

1 1sin sin 22

sin coscos

qe

d q d

qe

d q d

E V VP P PX X X

E VQ V

X X X

δ δ

δ δδ

⎛ ⎞= + − = +⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞

= − +⎜ ⎟⎜ ⎟⎝ ⎠ Note when:

δ

Pe

2 2

sin sin cos sin cosqe

d d q

E V V VPX X X

δ δ δ δ δ= − ⋅ + ⋅

sin 2sin cos2

δδ δ =We use:

reluctancePfieldP

d qX X=

Power

P and Q for salient pole rotor

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The swing equationm

m edW P Pdt

= −

Pe is the power delivered from one or more machines into the power systemJ is the moment of inertia (one or more machines)Pm is the shaft mechanical power from the turbine pis the number of poles

The energy balance in rotation, or the swing equation is as follows: 2 2

20 0

0 0

12

m em m

m e

fW J W Wf

ωωω

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2e efω π=Kinetic energy in rotation as a function of frequency:

0 02e efω π=2m ep

ω ω=2m mfω π= 0 02m mfω π=

Wm is the kinetic rotation energy of one or more machineswe , fe is the electrical angle and frequency for the electric oscillation wm , fm is rotation angle and frequency of the rotor mechanical rotation. (The index “0”, means that the quantity is a constant at the nominal frequency, 50 or 60 Hz)

The relation between rotation angle and the electrical angle of the stator current:

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The swing equation (2)2 2

20 0

0 0

12

m em m

m e

fW J W Wf

ωωω

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠We differentiate this equation.....

We substitute into the swing equation:

.....and then we get: 02

0

2m m m mm

m

dW d W dJdt dt dt

ω ω ωωω

= =

02

0

2 m m e

m m m

W d P Pdtω

ω ω ω= −

02

0

2 mm e

m

W d T Tdtω

ω= −

mm e

dW P Pdt

= −

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The inertia constant, H

2

0

12 m

m

rating rating

JWHS S

ω= =

The unit for H is: Ws/VA = s

We define the following quantity:

We examine now the former version of the swing equation....

02

0

2 m mm e

m

W d P Pdt

ω ωω

= −

and make the following approximation

02

0 0 0 0 0 0

1 1 11m m m m

m m m m m m

ω ω ω ωω ω ω ω ω ω

⎛ ⎞+ Δ Δ= ⋅ = +⎜ ⎟

⎝ ⎠

where the deviation form the nominal frequency, Δω is small relative to ω

The kinetic energy of the rotating massRated power for the generators in this mass

H =

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The swing equation (3)We then get the following equation:

we divide by the rating |Srating | and get:

0

0

2 mm e

m

W d P Pdtω

ω= −

0

0

2 m m e

m rating rating rating

W d P PdtS S Sω

ω= −

( ) ( )0

2 mm n e n

m

dH P Pdtω

ω= −

By defining the power from the turbine (or system) as a fraction of the rated power, we get:

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Two equilibrium points

• Synchronizing torque from δ deviation

•dPe /dδ > 0 for δ < 90° - stable equilibrium

• dPe /dδ < 0 for δ > 90° - unstable equilibrium

δ

Pe

Pm

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H on different MVA bases

• Machine base– Steam turbines 4 - 9 s– Gas turbines 3 - 4 s– Hydro turbines 2 - 4 s– Synchronous compensator 1-1.5 s

• Common base– H ~ generator size– Infinite bus has infinite H

Narrow range!

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Hydroeelectric rotor

The multiple poles are shown clearly on the rotor

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Inside the Búrfell power house

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Rotor in a hydro station

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A rotor from Laxá hydro station, North Iceland

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A stator from Laxá hydro station, North Iceland

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The stator in a hydroelectric generator

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267 MVA, 83.3 rpm stator

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References• P.C. Sen: "Principles of Electric Machines and Power

Electronics"; 2. edition: John Wiley & Sons, 1997• O.I. Elgerd: “Electric Energy Systems Theory”, McGraw-Hill,

1983• A.E. Fitzgerald, C. Kingsley Jr., S.D. Umans: Electric

Machinery, McGraw-Hill 2003• P. Kundur: Power System Stability and Control, McGraw-Hill

1994• D.P.Kothari, I.J. Nagrath: Modern Power System Analysis,

McGraw-Hill 2003

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Example 7

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Example 8

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Example 9

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Example 7 - solution

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Example 8 - solution

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Example 9 solution

the synchronous machine (sm) in the power system (2) · 2011-09-30 · 30-sep-11 lecture 06 power...

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