DAT300 TTHHEE EELECTRICAL PPOWER SSYSTEM€¦ · Chalmers University of Technology DAT300 TTHHEE...

Chalmers University of Technology

DAT300DAT300

TTHEHE EELECTRICALLECTRICAL PPOWEROWER SSYSTEMYSTEM

Stefan [email protected]

Department of Energy and Environment

Division of Electric Power Engineering

Chalmers University of technology


History of the power systemsHistory of the power systems

AC transmission was first demonstrated at an exhibition in Frankfurt am Main 1891

170 kW transferred 175 km from Lauffen hydropower station to the exhibition area at 13000-14700 V


History of the power systems in SwedenHistory of the power systems in Sweden

First 3-phase transmission systeminstalled in Sweden between Hellsjönand Grängesberg 1893voltage 9650 V, 70 Hz, 70 kW

First 400 kV system HarsprångetHallsberg 1952

Series compensation introduced1954


Fundamentals of Electric PowerFundamentals of Electric Power

� Energy- Ability to perform work, [J], [Ws], [kWh] (1 kWh = 3.6 MJ)

� Voltage- Measured between two points [V], [kV]- Equivalent to pressure in a water pipe

� Current- Measure of rate of flow of charge through a conductor [A], [kA]- Equivalent to the rate of flow of water through a pipe.- Must have a closed circuit to have a current


f

tIti

tUtu

peak

peak

πω

βω

αω

2

)cos()(

)cos()(

=

−=

−=

Direct Current (DC) / Alternating Current (AC)Direct Current (DC) / Alternating Current (AC)

Iti

Utu

=

=

)(

)(

2)(

1

0

2

RMS

peak

T Idtti

TI == ∫

RMS = Root-Mean-Square

Only for sinusoidal waveforms


The two main factors that formed the power system

• Transformer (only works on AC)

• Robust and cheep motor (rotating flux)

Why is AC used?Why is AC used?


f

tIti

tUtu

peak

peak

πω

βω

αω

2

)cos()(

)cos()(

=

−=

−=

Alternating Current (AC)Alternating Current (AC)

β

α

∠=

∠=

RMS

RMS

II

UU

{ }

{ }

=

⇒=

=

⇒=

−

−

−

−

tj

I

j

RMS

tj

RMS

tj

U

j

RMS

tj

RMS

eeIti

eIti

eeUtu

eUtu

ωβ

βω

ωα

αω

43421

43421

)(

)(

)(

)(

Re2)(

Re2)(

Re2)(

Re2)(

Express the sinusoidal voltage and current as complex rotating phasors and use RMS values for the amplitude

Since all phasors are rotating with the same

speed, we select one as the

reference and observe all others relative to this one.

This gives that the rotation disappears and the voltage

and currents can be expressed as complex

number (constant)


ωC

1j−Ljω

ImpedanceImpedance

RR IRU

tRitu

=

= )()( RR

LX

IjXILjU

dt

tdiLtu

L

LLLL

ω

ω

=

==

=)(

)( LL

CX

IjXIC

jU

dt

tduCti

C

LCCC

ω

ω

1

1

)()( C

C

=

−=−=

=


+

U1

–

+U3

–

+U2

–

Tre enfassystem

BelastningTre enfas-generatorer Elnät

Ett trefassystem

UR UTUS

I Σ= 0En trefas-generator

+

–

+

–

+

–

Why three phase system?Why three phase system?


Three phase voltage and currentThree phase voltage and current


Phasors for the voltagesPhasors for the voltages

fL-L 3UU =

Line to Line Phase (to ground)


dttitutp )()()( =

∫=T

dttituT

P0

)()(1

{ }∫ ++=T

TTSSRRdttitutitutitu

TP

0

)()()()()()(1

)()()()()()()( titutitutitutpTTSSRR

++=

)cos(2)(

)cos(2)(

ϕω

ω

−=

=

tIti

tUtu

RMS

RMS

Single phase Three phase

[ ]VA* jQPIUS +==

ϕϕ

ϕϕ

sin3sin3

cos3cos3

*3*3

,

,

RMSRMSLLRMSRMS

RMSRMSLLRMSRMS

LL

IUIUQ

IUIUP

jQPIUIUS

−

−

−

==

==

+===

Power Power –– Rate of energy flow [W]Rate of energy flow [W]

βαϕ −=Angle between voltage and current

Apparent power

[ ]VArsinϕRMSRMS IUQ =

Active power[ ]WcosϕRMSRMS IUP =

Reactive power

Instantaneous

power

average





33--phase Power [W]phase Power [W]


Reactive power flow Reactive power flow –– What is reactive power?What is reactive power?

� The current causes a magnetic fieldaround the conductor

� The field strength is highest close tothe conductor surface

� The field energy density is proportional to the square of the field strenght

� The field is built up and eleminatedwith the double of the networkfrequency in each phase

Consider an alternating current Iline flowing in a line



� The distance between the phases is about 10 m.

� It is not possible to transfer the energy directly between the neighboring phases.

� The energy must be transported to some place where the conductors are connected (a generator or a transformer).

Consider a line segment of, for example, 100 km in a long transmission line

QW∆

QW∆



How much energy is involved?

Consider a line segment of 100 km and a current of 1 kA (rms value); the

energy at the current peak is

kJiLW lineQ 100)21000(1.02

1ˆ2

1 22 ===

7 m

It is the same energy needed to lift a 1500

kg car up to 7 meters.

This is done each 10 ms, in each phase.



Due to the presence of the reactive power, the system cannot be used up

to its thermal limit

Need for reactive power compensation for better utilization of the system


( )L

rspsss

X

EEIEIEP

δsin real

*

===

( )L

rssqsss

X

EEEIEIEQ

)cos( imag

* δ−===

Active/reactive power at sending endEs

LX

δ,rr EE →

I

0,ss EE →

QP,

( )L

srrr

X

EEIEP

δsin real

*

==

( )L

srrrr

X

EEEIEQ

)cos( imag

* δ−−==

Active/reactive power at receiving end Er

Power flowPower flow


Voltages at the ends of a transmission line (same phase)Voltages at the ends of a transmission line (same phase)

s = 1 (sending end)r = 2 (receiving end)

δ

( )δ


2212121 jcos

jsin

qpII

X

EE

X

E

jX

EEI −=

−+=

−=

δδ

22222*

22 j)j( QPIIEIES qp +=+==

Complex power to E2: X

EEIEP p

δsin12222 ==

Re

Im

δδ sincos 111 jEE +=E

X

EEIq

δcos122

−−=

δsin1E

I

δ 22 E=E

δcos12 EE −X

EI p

δsin12 =

Active/reactive power toE2:

Reactive power consumption of the transmission line:

( )X

EEEEE

XQQQ L

2

2122

2121 cos2

1=−+=−=∆ δ

Active power from E1 to E2 :

X

EEPPP

δsin1221 ===

X

EEEIEQ q

)cos( 122222

δ−−==

Power flowPower flows = 1 (sending end)r = 2 (receiving end)


Structure of the Electric Power SystemStructure of the Electric Power System


Svensk

a Kraftnät: w

ww

.svk.se

• Transmission 400, 220 kV

• Regionalnät 130 kV

• Distributionsnät 70, 40, 30, 20,10 kV

• Kunder 400 V (Industri 10-130 kV)

Source: Svenska Kraftnät

Power balancing


Source: Svenska Kraftnät

Power balancingsI

loadR

LXrω

turbineTgenT

J

grid

genturbinegrid

p

p

grid

r

genload

genrgen

turbinerturbine

genturbiner

f

PP

dt

df

nJ

n

f

PP

TP

TP

TTdt

dJ

−=⇒

=

≈

=

=

−=

2

24

2

π

πω

ω

ω

ω

What happens if the turbine power does not match What happens if the turbine power does not match the load power?the load power?


The generator transforms the rotational energy into electric energy

Control gate

Reservoir(energy storage)

The kinetic energy of the water is transformed into rotation of the generator shaft (rotational kinetic energy)

Step-up

transformer

Hydro Power Station

Screen


Profile over the electric energy consumption in

Sweden for a typical summer day, winter day and the

highest consumption day 22th of December 2010

Källa: Svenska Kraftnät och Svensk EnergiElåret 2010

On the 23rd of

February 2011

Sweden used

26 000 MW

between 08-09


240 6 12 18

Baskraft: Kärnkraft, fossil förbränning, (vattenkraft)

Topplast -Gasturbin, vatten, m.m.

Pmax

Pmin

Lastkurva

Production planing


Integrated power during 1 year

24 000 kWh

Göteborg

Latitude 57.7 º

200 m2 of solar cells

Statistical cloudiness

Sun tracking

0 1000 2000 3000 4000 5000 6000 7000 8000 90000

5

10

15

20

25Solgenerator vs Tidpunkt

Tidpunkt (timme)

Eff

ekt

(kW

)

Solar Plant

Time [Hour]

Po

wer

[kW

]

Efficiency:

MPP 0.95

Power electronics 0.95

Solar cells 0.15


Total input energy to Sweden 1973–2010

Källa: SCB

Elåret 2010


Sweden electric energy production


Electric energy production and ussage

in Sweden for 2008–2010, TWh/week

Källa: Svensk Energi

Elåret 2010


Electric energy flow to and from Sweden 2008, GWh

Källa: Svenska Kraftnät och Svensk Energi


Normalized electric production mix for

the Nordic countries

Källa: Svensk Energi

Elåret 2010


Electric energy production in Sweden:

145,0 TWh år 2007 (v 65,5; k 64,3)

146,0 TWh år 2008 (v 68,6; k 61,3; v 2,0)

133,7 TWh år 2009 (v 65,3; k 50,0; v 2,5)

144,9 TWh år 2010 (v 66,8; k 55,6; v 3,5)

146,9 TWh år 2011 (v 66,0; k 58,0; v 6,1)

v=Hydro power

k=nuclear power

v=Wind power


Installed peak power in Sweden 2011.12.31, MWel

Hydro power 16 197

Wind power 2 899

Nuclear power 9 363

Other thermal power 7 988

∑ 36 447


Electric energy consumption in Sweden

divided on different consumers 1970–2010

Källa: SCB

Elåret 2010


Electric energy consumption for

households in Sweden (investigated 2007)

Källa: EnergimyndighetenElåret 2010

The consumption is higher in winter time in the Nordic countries, but in warm countries it is opposite

Typical household


The EndThe End

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DAT300 TTHHEE EELECTRICAL PPOWER SSYSTEM€¦ · Chalmers University of Technology DAT300 TTHHEE...

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