lecture #2 power engineering - hi
TRANSCRIPT
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
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 2
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...
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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:
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 4
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
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 5
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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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 6
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:
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 7
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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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 8
Load and real/reactive power
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 9
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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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 10
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.
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 11
3 phase power systems
(line to line/phase voltages/currents, 3 phase power, star/delta connections, 3
phase 3 wire/4 wire)
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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.
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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
+
+
+-
--
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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
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 15
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
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson
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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
= = −
= − = − −
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 1818
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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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 19
How are the 3 phases labelled?
•North America: a, b, c•Europe: (old) R,S,T•Europe: (new) L1, L2, L3
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 20
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
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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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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 2222
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
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 23
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:
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 24
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 )
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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:
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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:
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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:
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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
= ⋅ = ⋅
= ⋅
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 29
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
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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!
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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!
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 32
Substation layout
Sour
ce: L
aker
vi &
Hol
mes
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 33
Substation equipment
ENGLISHBusbar
GeneratorTwo-wdg transformer
Power lineCircuit breaker
DisconnectorSurge arrester
Current transformerPotential transformer
ÍslenskaSkinnurRafaliSpennirHáspennulínaAflrofiSkilrofiEldingavariStraummælispennirSpennumælispennir
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 34
Substation layout
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 35Lecture 2
230/69 kV Substation
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 36
Circuit breaker, Disconnect,Current transformer
Circuit breaker
Disconnect DisconnectCurrent CT
Bus bar
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 37Gas insulated 245 kV switchgear in a switchyard in
Burfell
Source: http://www.rafhonnun.is
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 38
Burfell station switchyards
Newer indoor switchyard
Old outdoorswitchyard
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 39
Problems with outdoor switchyards?
A new house for an indoor switchyard at the Burfellpower station
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 40
Búrfell – Gas insulated switchgear
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 41
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
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 42
Substation Brennimelur
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 43
Disconnector switches/ Isolators
• Interrupts small current– Load current
• Visible interruption• Very manual control
Scandinavia USA
Open Closed
Open Closed
Source: Nicklasson
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 44
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
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 45
Current transformer
• Reduces current– Typically 1000/2 A
• Current monitored– Control center– Protection equipment– P, Q transducers
Sour
ce: N
ickl
asso
n
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson 46
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
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson
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47
Surge / lightning arrester
• Overvoltage trap• Alternative to air gap• Short-circuit to ground
Source: Nicklasson
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson
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48
References• E. Lakervi, E.J. Holmes: Electricity
Distribution Network Design Peter Peregrinus 1995, 2nd Ed
• http://www.rafhonnun.is/
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson
ExamplesExample 1
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson
ExamplesExample 1 –solution
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson
Example 2
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson
Example 2 -solution
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson
Example 3 (ands solution)
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson
Example 4
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Lecture #2 Power Engineering - Egill Benedikt Hreinsson
Example 4 - solution