Chalmers University of Technology DAT300 THE ELECTRICAL POWER SYSTEM Stefan Lundberg [email protected]Department of Energy and Environment Division of Electric Power Engineering Chalmers University of technology Chalmers University of Technology History 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
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DAT300 THE ELECTRICAL POWER SYSTEM · Elåret 2013 Chalmers University of Technology Electric energy consumption for households in Sweden (investigated 2007) Elåret 2010 Källa:
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� 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
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t (ms)
U I
325V
14.1A Upeak
Ipeakα
β
-5 0 5 10 15 20
IRMS
URMS
T=1/f
f
tIti
tUtu
peak
peak
πω
βω
αω
2
)cos()(
)cos()(
=
−=
−=
Direct Current (DC) / Alternating Current (AC)
Iti
Utu
=
=
)(
)(
2)(
1
0
2
RMS
peakT I
dttiT
I == ∫
RMS = Root-Mean-Square
Only for sinusoidal waveforms
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The two main factors that formed the power system
• Transformer (only works on AC)
• Robust and cheep motor (rotating flux)
Why is AC used?
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f
tIti
tUtu
peak
peak
πω
βω
αω
2
)cos()(
)cos()(
=
−=
−=
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)
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ωC
1j−Ljω
Impedance
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
=
−=−=
=
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+
U1
–
+U3
–
+U2
–
Tre enfassystem
BelastningTre enfas-generatorer Elnät
Ett trefassystem
UR UTUS
I Σ= 0En trefas-generator
+
–
+
–
+
–
Why three phase system?
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Three phase voltage and current
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Line-to-line phasors for the voltages
fL-L 3UU =
Line to Line Phase (to ground)
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dttitutp )()()( =
∫=T
dttituT
P0
)()(1
{ }∫ ++=T
TTSSRR dttitutitutituT
P0
)()()()()()(1
)()()()()()()( titutitutitutp TTSSRR ++=
)cos(2)(
)cos(2)(
ϕω
ω
−=
=
tIti
tUtu
RMS
RMS
Single phase Three phase
[ ]VA* jQPIUS RMSRMS +==
ϕϕ
ϕϕ
sin3sin3
cos3cos3
*3*3
,
,
,
RMSRMSLLRMSRMS
RMSRMSLLRMSRMS
RMSRMSLLRMSRMS
IUIUQ
IUIUP
jQPIUIUS
−
−
−
==
==
+===
Power – Rate of energy flow [W]
αβϕ −=Angle between voltage and current
Apparent power
[ ]VArsinϕRMSRMS IUQ =
Active power[ ]WcosϕRMSRMS IUP =
Reactive power
Instantaneous
power
average
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Power – Rate of energy flow [W]
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Power – Rate of energy flow [W]
3-phase Power [W]
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Reactive power flow – 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
Iline
B
Consider an alternating current Iline flowing in a line
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Reactive power flow – What is reactive power?
� 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∆
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Reactive power flow – What is reactive power?
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.
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Reactive power flow – What is reactive power?
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
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( )L
rspsss
X
EEIEIEP
δsin real
*
===
( )L
rssqsss
X
EEEIEIEQ
)cos( imag
* δ−===
Active/reactive power at sending endEs
LXδ,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 flow
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Voltages at the ends of a transmission line (same phase)
s = 1 (sending end)r = 2 (receiving end)
δ
( )δ
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2212121 jcos
jsin
qp IIX
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 flows = 1 (sending end)r = 2 (receiving end)
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Structure of the Electric Power System
Chalmers University of TechnologyS
ven
ska K
raftnät: 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
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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 the load power?
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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
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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