Chalmers University of Technology DAT300 DAT300 THE HE ELECTRICAL LECTRICAL POWER OWER SYSTEM YSTEM 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 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 Chalmers University of Technology History of the power systems in Sweden History of the power systems in Sweden First 3-phase transmission system installed in Sweden between Hellsjön and Grängesberg 1893 voltage 9650 V, 70 Hz, 70 kW First 400 kV system Harsprånget Hallsberg 1952 Series compensation introduced 1954 Chalmers University of Technology Fundamentals of Electric Power Fundamentals 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 Chalmers University of Technology f t I t i t U t u peak peak π ω β ω α ω 2 ) cos( ) ( ) cos( ) ( = - = - = Direct Current (DC) / Alternating Current (AC) Direct Current (DC) / Alternating Current (AC) I t i U t u = = ) ( ) ( 2 ) ( 1 0 2 RMS peak T I dt t i T I = = ∫ RMS = Root-Mean-Square Only for sinusoidal waveforms Chalmers University of Technology 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?
7
Embed
DAT300 TTHHEE EELECTRICAL PPOWER SSYSTEM€¦ · Elåret 2010. Chalmers University of Technology Electric energy consumption for households in Sweden (investigated 2007) Elåret 2010
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
� 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
Chalmers University of Technology
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
Chalmers University of Technology
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?
Chalmers University of Technology
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)
Chalmers University of Technology
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
=
−=−=
=
Chalmers University of Technology
+
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?
Chalmers University of Technology
Three phase voltage and currentThree phase voltage and current
Chalmers University of Technology
Phasors for the voltagesPhasors for the voltages
fL-L 3UU =
Line to Line Phase (to ground)
Chalmers University of Technology
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
Chalmers University of Technology
Power Power –– Rate of energy flow [W]Rate of energy flow [W]
Chalmers University of Technology
Power Power –– Rate of energy flow [W]Rate of energy flow [W]
33--phase Power [W]phase Power [W]
Chalmers University of Technology
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
Chalmers University of Technology
Reactive power flow Reactive power flow –– What is reactive power?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∆
Chalmers University of Technology
Reactive power flow Reactive power flow –– What is reactive power?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.
Chalmers University of Technology
Reactive power flow Reactive power flow –– What is reactive power?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
Chalmers University of Technology
( )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
Chalmers University of Technology
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
Chalmers University of Technology
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: