lecture #2 power engineering - hi

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 1

Complex power

Consider the phasors V and I for voltage and current. These are represented by the complex numbers:

*S V I= ⋅This is the definition of “complex power”!!

We can now define, as a complex number, the quantity:

0j

j

V V e

I I e φ−

=

=

V

IIr

Ix

φ

Ix

Ir

V V e

I I e

j

j

=

= −

0

φ

V V e

I I e

j

j

=

= −

0

φ

23-Sep-11


Complex, real and reactive power

cos sin

jS V I e

S V I j V I

φ

φ φ

=

= +

0j

j

V V e

I I e φ−

=

=From...

We get:

S P jQ= +and finally:

*S V I= ⋅..and...

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 3Complex power, impedance and admittance

VI VYZ

= =

With the concept complex power we can expand our model of circuit impedance and admittance in an AC circuit:

2 2

2 2*

*

V IS S

Z YS Z I S Y V

= =

= =

*S V I=From: ..and...

…we get:

2 2 2( )S R jX I P R I Q X I= + = =

…in particular:

23-Sep-11


Phasors (vectors) and complex power

Re

Im

Projection of a vectoron the Re-axis

Note: The projection of a revolving current and/or voltage complex phasor (also called vector) on the Re axis represents the instantaneous values of the current and/or voltage

This does not apply to complex power vectors

Voltage or current phasor

23-Sep-11


Reactive power in a resistance

Z=R

I

V

+

-

V VIZ R

= =

2** VV VS V I

R R⋅

= ⋅ = =

VI

Circuit: Phasors:

φ= 0°

2

0

VP

RQ

=

=

A resistance neither generates or consumesreactive power

Therefore:

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The inductor as a consumer of reactive power

Z=jωL

I

V

+

-

V V VI jZ j L Lω ω

= = = −

2** VV VS V I j j

L Lω ω⋅

= ⋅ = =

V

I

Circuit: Phasors:

φ= 90°

2

0

VQ

LP

ω=

=

Q is positive ⇒Inductance consumesreactive power

Therefore:

23-Sep-11


The capacitor as a generator of reactive power

I

V

+

-

VI j CVZ

ω= =

* *

2

( )S V I V j C V

j C V

ω

ω

= ⋅ = ⋅ − ⋅

= −

V

ICircuit: Phasors:

φ= -90°

2

0Q C VP

ω= −

=

Q is negative ⇒Capacitance generatesreactive power

1Zj Cω

=

Negative consumption = generation

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Load and real/reactive power

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Reactive Power

• Reactive power is supplied by– generators– capacitors– transmission lines– loads

• Reactive power is consumed by– loads– transmission lines and transformers (very high losses

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Reactive Power

• Reactive power doesn’t travel well - must be supplied locally.

• Reactive power must also satisfy Kirchhoff’s law - total reactive power into a bus MUST be zero.

23-Sep-11


3 phase power systems

(line to line/phase voltages/currents, 3 phase power, star/delta connections, 3

phase 3 wire/4 wire)

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 12Advantages of three phase. Why 3 phases systems?

• Smooth flow of power (instantaneous power is a constant). Constant torque (reduced vibrations)

• The power delivery capacity is tripled (increased by 200%!) by increasing the number of conductors from 2 to 3 (increase by 50%)

• Reduced cost (same power less wire or more power same wire)

• Greater "power per kg" in motors, generators, and transformers.

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 133 separate identical and simple 1 phase systems

6 conductors!

Generation Transmission Load

Z

Z

Z

IaVa

Vb

Vc

IbIc

+

+

+-

--

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 14Symmetrical voltages will lead to symmetrical currents

V af

V cf

V bf

Z

Z

Z

IaVafVbf

Vcf

IbIc

++

+

-

-

-

I a

I c

I b

Voltage phasors Current phasors

0a b cI I I I= + + =

With a zero total current, the 3 return conductors are not needed for a symmetrical power system

Identical impedances!

If the system is symmetric, the total current = 0

The angle between voltage and current

23-Sep-11


A 3 phase 3 wire system

Z

Z

Z

IaVa

Vb

Vc

IbIc

+

+

+-

--

Both neutrals may or may not be grounded

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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 16A 3 phase 4 wire system (with ground wire)

Z

Z

Z

IaVa

Vb

Vc

IbIc

+

+

+-

--

Both neutrals may or may not be grounded

Ground wire

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson

23-Sep-11

17A 2 phase 3 wire system (with ground wire)

Z

Z

iav a

v b ib

+

+-

- Ground wire

( ) 2 cos( ) ( ) 2 cos( )

( ) 2 cos( ) ( ) 2 cos( )2 2

a a

b b

v t V t i t I t

v t V t i t I t

ω ω φπ πω ω φ

= = −

= − = − −

23-Sep-11


A 1 phase system with 2 wires

ZL/2

ZL/2

ia

–

-i a

+

The system is balanced against the earth

1 1( ) 2 cos( )v t V tω=

1 2

2V

–

+

z r j Lω= +

z r j Lω= +

25 km

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How are the 3 phases labelled?

•North America: a, b, c•Europe: (old) R,S,T•Europe: (new) L1, L2, L3

23-Sep-11


Symmetrical 3 phase systems

• 3 phase voltages and currents are defined as being symmetric if….– All three phasors are of equal length– A phase difference of 120° is between phases

• A 3 phase system is defined as being symmetric or balanced, if...– All voltages and currents are symmetric– Impedances in all 3 phases are identical

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 21Symmetrical phase and line to line voltages

V af

V bf

V cf

N

Phase voltage: V f

Line to line voltage = voltagebetween phases : V L

1. All three phase voltages are of equal length

3. The phase difference is in both cases 1/3 of 360º or 120º

2. All three line to linevoltages are of equal length

“neutral”

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Symmetry in a 3 phase system

1. All three phase voltages are of equal length

3. The phase difference is in both cases 1/3 of 360º or 120º

2. All three line to line voltages are of equal length

23-Sep-11


Line to line voltage - phase voltage• Line to line voltage (VL): Voltage

between phases• Phase voltage (Vf also called Vp ):

Voltage from phase to neutral

( )3L

f pV

V V= =

3x

xA triangle with 120° top angle:

23-Sep-11


Voltage in 3 phase systems• When we talk about a system

voltage in a 3 phase power system (such as 220 kV or 400 V ), we always mean the RMS value of the voltage between phases (or the line to line voltage, VL )

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 253 different representation of symmetrical 3 phase

quantities :

Re

Im

a- Phase

a-Phase

b- Phase

b-Phase

c- Phase

c-Phase

time

Voltage or current

( ) 2 sin( )

( ) 2 sin ( 120 )

( ) 2 sin ( 120 )

a

b

c

v t V t

v t V t

v t V t

ω

ω

ω

=

= − °

= + °

( ) 2 sin( )

( ) 2 sin( 120 )

( ) 2 sin( 120 )

a

b

c

i t I t

i t I t

i t I t

ω φ

ω φ

ω φ

= −

= − °−

= + °−

Formulas:

Vectors:

Wave-forms:

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 26Instantaneous power in a 3 phase system

( ) 2 sin( )

( ) 2 sin ( 120 )

( ) 2 sin ( 120 )

a

b

c

v t V t

v t V t

v t V t

ω

ω

ω

=

= − °

= + °

3 phase voltage:

( ) 2 sin( )

( ) 2 sin( 120 )

( ) 2 sin( 120 )

a

b

c

i t I t

i t I t

i t I t

ω φ

ω φ

ω φ

= −

= − °−

= + °−

3-phase current:

3 ( ) ( ) ( ) ( ) ( ) ( ) ( )phase a a b b c cp t v t i t v t i t v t i t= + +

From the above formulas, we get the total instantaneous power:

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 27Instantaneous power in a 3 phase system(2)

3 ( ) ( ) ( ) ( ) ( ) ( ) ( )phase a a b b c cp t v t i t v t i t v t i t= + +

[

]

2 sin( )sin( )

sin( 120 )sin( 120 )sin( 120 )sin( 120 )

I V t t

t tt t

ω ω φ

ω ω φω ω φ

= −

+ − ° − °−

+ + ° + °−

By inserting the formulas, we get:

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 28Instantaneous total power in a 3 phase system(3)

[ ]1sin sin cos( ) cos( )2

x y x y x y= − − +

Use the following trigonometric identities to simplify:

cos( ) cos( 120 ) cos( 120 ) 0x x x+ − ° + + ° =And we get finally the following formula

3 1

1

( ) 3 cos 3

cosphase phase

phase

p t I V P

P I V

φ

φ

= ⋅ = ⋅

= ⋅

23-Sep-11


Total 3 phase power• Therefore, the total instantaneous power in all 3 phases is

constant - or - 3 times the real power in each phase• The power oscillates in each phase (although the sum of

power in the phases is constant)• No reactive power appears in the formula!!• Reactive power is, however, very much present in each

individual phase

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 30A mechanical analogy with a 3 phase hydraulic generator

The total power delivery in a three phase system is smooth!

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 31A mechanical analogy with a 1 phase hydraulic generator

The power delivery in a one phase system is bumpy!

23-Sep-11


Substation layout

Sour

ce: L

aker

vi &

Hol

mes

23-Sep-11


Substation equipment

ENGLISHBusbar

GeneratorTwo-wdg transformer

Power lineCircuit breaker

DisconnectorSurge arrester

Current transformerPotential transformer

ÍslenskaSkinnurRafaliSpennirHáspennulínaAflrofiSkilrofiEldingavariStraummælispennirSpennumælispennir

23-Sep-11


Substation layout

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 35Lecture 2

230/69 kV Substation

23-Sep-11


Circuit breaker, Disconnect,Current transformer

Circuit breaker

Disconnect DisconnectCurrent CT

Bus bar

23-Sep-11

Lecture #2 Power Engineering - Egill Benedikt Hreinsson 37Gas insulated 245 kV switchgear in a switchyard in

Burfell

Source: http://www.rafhonnun.is

23-Sep-11


Burfell station switchyards

Newer indoor switchyard

Old outdoorswitchyard

23-Sep-11


Problems with outdoor switchyards?

A new house for an indoor switchyard at the Burfellpower station

23-Sep-11


Búrfell – Gas insulated switchgear

23-Sep-11


Indoor switchyards

Switchyards in a power station are based on conducting elements in gas insulated chambers. The gas is SF6, which has especially good insulation

SF6 molecule

23-Sep-11


Substation Brennimelur

23-Sep-11


Disconnector switches/ Isolators

• Interrupts small current– Load current

• Visible interruption• Very manual control

Scandinavia USA

Open Closed

Open Closed

Source: Nicklasson

23-Sep-11


Circuit breakers

• Interrupts large current– Several kA– Short-circuit current– Hidden contacts

• Control– Protection systems– Manual remote

Iceland USA

Open Closed

Sour

ce: N

ickl

asso

n

23-Sep-11


Current transformer

• Reduces current– Typically 1000/2 A

• Current monitored– Control center– Protection equipment– P, Q transducers

Sour

ce: N

ickl

asso

n

23-Sep-11


Voltage/potential transformer

• Reduces voltage– Typically x kV/110 V

• Voltage monitored– Control center– Protection equipment– P, Q transducers

• C voltage divider

Sour

ce: N

ickl

asso

n

23-Sep-11


23-Sep-11

47

Surge / lightning arrester

• Overvoltage trap• Alternative to air gap• Short-circuit to ground

Source: Nicklasson

23-Sep-11


23-Sep-11

48

References• E. Lakervi, E.J. Holmes: Electricity

Distribution Network Design Peter Peregrinus 1995, 2nd Ed

• http://www.rafhonnun.is/

23-Sep-11


ExamplesExample 1

23-Sep-11


ExamplesExample 1 –solution

23-Sep-11


Example 2

23-Sep-11


Example 2 -solution

23-Sep-11


Example 3 (ands solution)

23-Sep-11


Example 4

23-Sep-11


Example 4 - solution

lecture #2 power engineering - hi

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