(Source: Kundur’s book) - Department of Electrical...
8
17 (Source: Kundur’s book)
Transcript of (Source: Kundur’s book) - Department of Electrical...
17
(Source: Kundur’s book)
18
Complex Power Flow Through Transmission Lines
S R
S R
V VA BI C D I
0R RV V= S SV V d=
BB B Zq ¢= =12AZ YA A q¢ ¢
= = +S RRV AVIB
SR
R S
VV D BI C A I
S RS
AV VIB
)*
(3 3 RR RS V I * * 2 2
*
| || | | || |3 3 ( ) 3 ( )| | | |
R S R S R RB B A
V V A V V V A VB B B
( ( ) ( )2
) || || |( ) ( )
| ||| |
| |S LR L L
B AL L
BL RA V V VBB
( ) (2
( ))
)3
2
(
**
*
| || |( ) ( )
|| || |
| || |3 3
|S S S L L R LS L L
S SL
S BR
A B
A VA V VVVSV V
BB BI
det 1AD BCD A
2
( 1) / ( 1) /C AD B
A B
A0, B90o
• For a lossless line, B=jX’, A=0, B=90o, and A=cosl
( ) ((3 )
)(3 )
| || |sinS L L R L L
R S
V VX
PP
2( ) ( ) ( )
(3 )
| | | || |cos cosR L L S L L R L L
R
V V VQ
X X
2
( ) ( ) ( )(3 )
| | | || |cos cosS L L S L L R L L
S
V V VQ
X X
( )R BC R
( ) S BC R
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Sending & Receiving End Power Circle Diagram
• When do two circles intersect?
2( )| || |
( )| |S L L
S B A
A VC
B
2( )| || |
( )| |R L L
R B A
A VC
B
( ) ( )| || || |
S L L R L LV VR
B
| | | | 2S RC C R ( ) ( )2 2( ) ( )
2 | || || | (| | | | )| | | |
S L L R L LS L L R L L
V VA V VB B
( ) ( )2 2
( ) ( )
2 | || || |
| | | |S L L R L L
S L L R L L
V VA
V V
| | | cosh | | cosh( ) | | cosh( ) | 1A j zy
1 (=1 iff |VS|=|VR|)
A special case having two circles intersecting:A lossless line with |VS|=|VR| and |A|=|cosl|1
A necessary condition:
P
QSending end
circle
Receiving end circle
B-A
(PS, QS)
(PR, QR)
CS
CR
R
R
+B
B
A
A
(3 ) ( )R R BS C R (3 ) ( ) S S BS C R
20
21
Power Transmission Capacity• Thermal loading limit:
– A conductor is stretched if its temperature increases due to real power loss, which will increase the sag between transmission towers
– With the current-carrying capacity (Ithermal) of the conductor provided by the manufacturer, the thermal loading limit is
• Steady-state stability limit (ignoring losses)
– Theoretical limit P3,max with max=90o
– Practical line loadability with max=30o to 45o
3thermal rated theramalS V I
| | sin
sin( )
cX B Z
L LCC
max
( ) ( ) ( ) ( )3 , max max
2( ) ( ) max
| || | | || |sin sin
sin| | | | sin ( )( )( )
sin
S L L R L L S L L R L L
c
S L L R L L rated
rated rated C
V V V VP
X ZV V VV V Z
maxsin| || |
sin(2 / )Spu RpuV V SIL
maxsin| || |sinSpu RpuV V SIL
1 5000km
if =60Hzf LCf
(They are all real numbers)
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Example 5.6
3
| || |sin
sinSpu RpuV V SIL
P
Solution: (a)
3 max
| || |sin 90 1167MW =700/sin(36.87 )
sinSpu RpuV V SIL
P
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Sending & Receiving End Power Circle Diagram
R=1167 (MVA) = P3(max)
CS=0+j1196 (MVA)
CR= -j969 (MVA)
Assume <30o
Practical line loadability =583.5MW
-1500 -1000 -500 0 500 1000 1500-2500
-2000
-1500
-1000
-500
0
500
1000
1500
2000
2500
P (MW)
Q (M
var)
CS
CR
R30o
30o
-700 -600 -500 -400 -300 -200 -100 0 100 200 300 400 500 600 700-300
-200
-100
0
100
200
300
400
P (MW)
Q (M
var)
1167
583.5
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Line Loadability Curves• Assume VRVS=400kV, Ithermal=3000A, SIL=499.83MW and max=30o
– SThermal =2078MW– Line loadability vs. Line length:
50 100 150 200 250 3000
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
Line Length (km)
Load
ing
Lim
it (M
VA)
SIL=499.83
Thermal Limit=2078
Theoretical limit
Practical line loadability
( ) ( )3 max
| || |S L L R L LV VP
X
o 3 max3 max,300.5P P
0 500 1000 1500 2000 2500 3000 3500 4000-5000
0
5000
Line Length (km)
Load
ing
Lim
it (M
VA
)
0 500 1000 1500 2000 2500 3000 3500 4000-5000
0
5000
Line Length (km)
Load
ing
Lim
it (M
VA
)
