Synchronous Generators Overview - Clark Science …jcardell/Courses/EGR325/slides/C...Synchronous...

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Smith College, EGR 325 April 3, 2018

Synchronous Generators Overview   Generating 3-phase power

  ‘Field’ winding on rotor to energize with DC current

  Equivalent circuit and phasor diagrams

  Connecting to a power grid   The power system as an “infinite bus”

  Maximum power transfer   Power system dynamics   Maintaining synchronism

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Small Synchronous Machine

Small Synchronous Machine

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Synchronous Generator Operation Discuss at tables, to explain how a synchronous generator works.

  Start with knowledge of inputs and outputs   Input: Mechanical power, electrical DC voltage   Output: 3 phase electrical power

  Discuss where/how each input is connected

  Output – from where?

  Electromagnetics à explain

  What is the ‘synchronous’ speed?

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Conceptual Objective for Today   Determine a simple representation of the maximum power that a generator can deliver (we already know this?) a)  Generator Pmax limit? b)  Thermal (melting point) limit of line? c)   Dynamics of what the system can support in order to

maintain synchronism   How much power can the generator be expected to supply before it essentially ‘stalls’   A function of the generator and the power system

N

S

Ia

Ib

Ic

φf

ωs

Armature windings – on the stator, induced current Field windings – on the rotor, energized by a DC input voltage (to get current)

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Rotating rotor induces currents in

stator windings, of

what waveform?

N

S

Ia

Ib

Ic

φf

ωs

Time

Vaa’ Vbb’ Vcc’ X

X c

b ́a

X c ́

b

N

S

φf ωs

3 Stator Windings (“pole pairs”)

a’

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View of a Single Phase φ f

ω s E f

N

S

I f

V f tdd

~E ff

φ

Ø  If is energized by an external circuit, creates the ‘moving charges’ Ø Ef is directly proportional to the excitation current If (rotor windings) Ø  The frequency of Ef is proportional to the synchronous speed ωs Ø Ef exists because the rotor is energized and is rotating…

Generator Equivalent Circuit

Ø  Ef represents the rotor so we do not need to draw the rotor in the circuit model Ø  Subscript f = ‘field’ with If supplied from external DC source

I a

X s

E f V t

Vt is the power system, an “Infinite Bus”

Ef is function of If Magnitude and phase of Ia are dependant variables

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Generator Equivalent Circuit

satf XIVE +=

V t

I a X s

I a θ

E f δ I a

X s

E f V t

(Phasor for IaXs is 90° shifted from the phasor for Ia)

Two Angles (differences): • θ = θ__ – θ__ = power factor angle • δ = δ__ – δ__ = “power angle”

Power Calculations: S = VI*

θsinIVQ att 3=

θcosIVP at3=

Ia

Xs

Ef Vt

Ia

Vt

Ef

Ia Xs

θ

δ

Vt and Ef are phase quantities (see ‘side trip’ at end of slides)

f t

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* Real Power Transfer Calculation *

δθ sinEcosXI fsa =

s

fa X

sinEcosI

δθ =

δsinXEV3

Ps

ft=

θcosIV3P at=

θ

Ia

Vt

Ef

Ia Xs

θ

δ

Observe from the phasor diagram:

New method to obtain previous equation

Power Characteristics of Generator

δsinXEV3

Ps

ft=

P

δ

Pmax

δl 90o

s

ft

XEV

P3

max =

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Generator Connected to Power System

Pm G

Xs

Infinite bus

Vt Infinite Bus: •  = The entire power system •  Constant Voltage •  Constant Frequency

Connection Through a Transmission Line…

Pm Xl GXs Vt Vo Ia

I a Xs

Ef Xl

Vt Vo

δsinXEV3

Ps

ft=

P

δ

Pmax

δl 90o

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Reactive Power Calculation

tfsa VEXI −= δθ cossin

θsin3 att IVQ =

Ia

Xs

Ef Vt

Ia

Vt

Ef

Ia Xs

θ

δ

Vt and Ef are phase quantities

f t

€

Qt = 3Vt Ia sin θ =3VtXs

E f cosδ − Vt( )€

Qt = 3Vt Ia sin θ

€

Ia Xs sin θ = Ef cosδ − Vt

If Ef cos δ > Vt ; Qt is positive and Current is lagging

If Ef cos δ < Vt ; Qt is negative and Current is leading

If Ef cos δ = Vt ; Qt is zero and Current is in phase

Reactive Power: Can be Negative

1)

2) 3)

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Example   A (3-phase) synchronous generator is connected to

an infinite bus.   The terminal voltage of the generator is 5 kV   The equivalent field voltage is 4.8 kV.   The synchronous reactance of the generator is 10 Ω.

  Compute the maximum power the generator can deliver before it will be pulled out of synchronism

Solution

Ia

Xs

Ef Vt

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Solution

P = 3Vt E f

Xs

sinδ P

P

δ

Pmax

δl 90o

Pmax = 3Vt E f

Xs

=3*5 (4.8)10

=7.2MW

Phase values Line-to-line

P

P

δ

Pmax

δl 90o

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Maximum Power Delivery

P =Vt E f

Xs

sinδI a

X s

E f V t

P

P

δ

Pmax

δl 90o

Power Angle

satf XIVE +=

V t

I a X s

I a θ

E f δ

Two Important Angles: • θ = θ__ – θ__ = power factor angle • δ = δ__ – δ__ = “power angle”

I a

X s

E f V t

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Mid-way Recap of Concepts   From circuits, EGR 220

  Power factor, power factor angle, θ

  From previous slides   Power angle, δ   Maximum power transfer depends upon ___?

  Infinite bus representing the rest of the power system   The actual total mass of all rotating rotors

  Real power, P vs. reactive power, Q   Possible values? < 0? > 0?   Interpretation of above (< or >0)

Synchronous Generators 1)  Understand how mechanical power is converted

into electrical power

2)  Understand maximum power transfer   How the limit on power delivered to the grid is

determined.

3)  Dynamics – the affect of load increase and decrease on the generator speed (and the system frequency)

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Real Power – Frequency   P and f dynamics are coupled

  Demand > Supply: frequency will decrease (more energy drained from system than produced, acts like brakes on the turbines)

  Supply > Demand: frequency will increase (more energy in the power system than consumed, acts like an accelerator so turbines spin faster)

  Generation-based frequency regulation   Generator inertia   Generator governors

Real Power – Frequency

An increase/decrease in load causes the generator rotation to ____________? (increase/decrease)

The angular frequency of the generator is the frequency (or multiple) of the electricity generated.

Mechanical Turbine Electrical Power Load/Demand

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Renewables and Power Balance

  The net load variability with wind and solar variability directly affects the grid frequency, and can harm load motors

Mechanical Turbine Electrical Power Load/Demand

Power Angle à Springs Analogy   First example: Springs

  Twisting a stiff spring vs. a weak spring and notice the relative angular position of both ends

  Restoring force returns it to its resting position   If you twist too far, it cannot return à “loss of synchronism”

  Second example: Hand generators   Relative angular position of shaft   Before and after a disturbance

  A deceleration or an acceleration   An imbalance of PM and PE

  Power angle   Angular position of the generator rotors,   (Relative to a rotating synchronous position)   Angular position of rotor (mechanical angle) = phase angle of voltage phasor (electrical

angle) = power angle

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Power Angle   Generators rotate with angular velocity ωm

  This is the angular velocity of the ‘rotor’

δm = rotor angular position with respect to a synchronously rotating reference

δm is the power angle

  This is also the phase angle of the voltage phasor

  The mechanical angle is the electrical angle   Coupling of the electro-mechanical system

Power Delivered Across a Line   This mechanical angle is the electrical phase Ø  Coupling of the electro-mechanical system

  What is the role of this “power angle?”   We know Z (X) is a fixed parameter.   Goal of good system operations is to keep|Vi|, voltage

magnitude, nearly constant

  This means that δ is what we change in order to change real power flow   How does an operator change δ?

P =

V1 V2X

sinδ

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Connection Through a Transmission Line…

Pm Xl GXs Vt Vo Ia

P =VtV0Xl

sinδ

P

P

δ

Pmax

δl 90o

Graphically: PM & PE vs. δ   Assume a step change to PM changing mechanical

power from pm0 to pm1

GSO reading

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Interpreting Dynamics   In steady-state pe = pm = pm0 andδ = δ0

  pe0 = pm0 = pmax sin(δ0)

  A step change in pm from pm0 to pm1 occurs at time t = 0.

  Due to rotor inertia, the rotor position cannot change instantaneously δ(0+) =δ(0-) = δ0

  This means that electrical power output remains unchanged

pe(0+) = pe(0-)

Interpreting Dynamics   But pm(0+) = pm1

Mechanical power (energy) has changed

  This means that pm(0+) > pe(0+)   i.e., supply > demand

  So, there is a positive, momentary, acceleration of the rotor

  The rotor accelerates and δ increases   Recall δ is the angular difference between the rotor

positions at either end of the lines (as well as the difference in their voltage phase angles)

  Until pe = pm1 at point δ= δ1

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To Increase Power Delivery

P =VtV0Xl

sinδ

Pm Xl GXs Vt Vo Ia

Assume a sudden change to PM changing mechanical power from Pm0 to Pm1

Pm < Pe = area below curve

Pm > Pe = area above curve

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Dynamics   Steady-state, point a

  Pm0 = Pe0 = Pmax sin(δ0)

  Suddenly, Pm increases!   Perhaps a steam valve was opened   This causes the rotor speed to increase

  Accelerating power, Pa = Pm1 – Pe   This causes the rotor speed and rotor angle to increase, momentarily

  With an increase in δ, the power delivery also increases

P =VtV0Xl

sinδ

HW Question 1   A synchronous generator is connected to an

infinite bus through a transmission line. The infinite bus voltage is 15kV and the equivalent field voltage of the machine is 14kV. The transmission line inductive reactance is 4Ω, and the synchronous reactance of the machine is 5Ω.   Compute the (power) transfer capability of the system.   If a 2Ω capacitor is connected in series with the

transmission line, compute the new capacity of the system.

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HW Question 2   A 100 MVA synchronous generator is connected to a 25kV

infinite bus through two parallel transmission lines.

  The synchronous reactance of the generator is 2.5Ω, and the inductive reactance of each transmission line is 2Ω.

  The generator delivers 100 MVA to the infinite bus at 0.8 power factor lagging.

  Suppose a lightning strike causes one of the transmission lines to open. Assume that the mechanical power and excitation of the generator are unchanged.

  Can the generator still deliver the same amount of power to the infinite bus?

A disturbance can also be loss of a Tx line – changing ‘Xeq’ – changing Pmax – changing the curve itself

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Renewables and Power Balance

  The net load variability with wind and solar variability directly affects the grid frequency, and can harm load motors

Mechanical Turbine Electrical Power Load/Demand

Restarting a System   New Jersey and New York remained in electrical

blackout longer than expected.

  Once all the transmission lines are reconnected, what do power system engineers need to do to be able to restart the system?   They cannot simply start all the generators up separately

and say they are done – why not?

  What can the power transfer equation tell us?

  What can knowledge of reactive power, Q, tell us?

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Summary   Synchronous generators

  P and Q   Expressions   Graphs (esp. Pmax) Phasor diagrams

  Connecting to the power grid   “Infinite bus”   Power delivered to load

  Dynamics – energy balance and effect on system frequency

Summary   Power delivery into a power system

  Role of the “power angle”

  Power system dynamics   Spring analogy   Coupling of mechanical and electrical elements via the

power angle   Loss of a transmission line   Adding series compensation   Variability of wind and solar power – maintaining

system energy balance   Restarting a system after a blackout

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X

X c

b ́a

X c ́

b a ́

SIDE TRIP: How is Three-Phase Connected?

X

X c

b ́a

X c ́

b a ́

Y-Connection (Wye)

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X

X c

b ́a

X c ́

b a ́

Delta (Δ) Connection: Source

v ca v bc

v ab

v cn v bn v an

Phase vs. Line-Line Voltages

Phase Voltages Line Voltages

(generator)

(ground, or neutral)

Task: Draw the phasor diagrams for the phase voltages, and for the line-line voltages.

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Phase voltage; Wye Connected

n

€

V an = V a = V∠0°V

v an

v bn

v cn

Reference

v cn v bn v an

n

c

b

a

a

c

b

€

V cn = V c = V∠120°V

€

V bn = V b = V∠−120°V

€

V bc = V bn −V cn = V∠−120°( )− V∠120°( ) = 3 V ∠− 90°

v ca v bc

v ab

n

c

b

a

Line-to-line voltage; Delta connection

€

V ab = V an −V bn = V∠0°( )− V∠−120°( ) = 3 V ∠30°

€

V ca = V cn −V an = V∠120°( )− V∠0°( ) = 3 V ∠150°

Reference

v ab

v ca

v bc

030

v bn

b

n v an

a

v cn

c

Unless stated otherwise: All voltages are line-to-line quantities

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Connections: “Delta” or “Wye”

Source Load Transmission Line

a

c b

Z

Z Z

I a

I b

I c

V ca

a

b c

n V ab

I a

V bc I c

I b

Iab +

-

Line current

Phase current

Ica

Ibc

Phase current

Objective of “Side Trip”

 To be familiar with: Wye and delta connections  “Phase” voltage versus line-line voltage  To understand the factor of √3

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Using Equations and Data   Phase to ground, vs. line to line voltages:

  Be careful with phase vs. line-line values   Line-line given in problem   Values often for a single phase

  Note that the current in the transmission line is a function of the power angle, δ.   Find δ   Find Ia = (Ef – V0)/X   Find Vt = V0 + IaXl

  The line-line Vt = √3 Vt-ph