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VOLTAGE DROPS IN POWER SYSTEM
Simple DC line (LV, MV)
Double-wire circuit. Assumption: constant cross-section and resistivity. El. traction, electrochemistry, light sources, long-distance transmission, power electronics.
Single loads supplied from one side
a) addition method It adds voltage drops along the power line sections. (Voltage drops are always in both conductors in the section.)
kth section
)A,m,m,m;V(IllS
2UUU 2k1k1kkk1kk1k
Current in kth section
n
kyyk1k II
Maximum voltage drop
n
1k
n
kyy1kk
n
1kk1kn Ill
S2UU
b) superposition method It adds voltage drops for individual discrete loads:
n
1kkkn Il
S2U
kkIl …current moments to the feeder
Relative voltage drop:
V,V;U
U
n
Note. Losses must be calculated only by means of the addition method!
)A,m,m,m;W(IllS
2P 22k)1k(1kkk1k
n
1kk1kPP
Single loads supplied from both sides – the same feeders voltages
Ring grid, higher reliability of supply. Two one-feeder lines after a fault. Calculation of current distribution and voltage drops.
Consider IB as a negative load:
B
n
1kkkBAAB lI
S2Il
S20UUU
Hence (moment theorem)
l
IlI
n
1kkk
B
Analogous (current moments to other feeder)
l
IllI
n
1kkk
A
Of course
n
1yyBA III
Current distribution identifies the place with the biggest voltage drop = the place with feeder division → split-up into two one-feeder lines.
Single loads supplied from both sides – different feeders voltages
Two different sources, meshed grid.
Superposition: 1) Current distribution with the same voltages. 2) Different voltages and zero loads → balancing current
lS
2
UUI BAv
3) Sum of the solutions 1+2
Further calculation is the same.
Or directly:
B
n
1kkkBA lI
S2Il
S2UU
lS
2
UU
lS
2
IlS
2I BA
n
1kkk
B
Direct current transmission HV
Long distance transmission or local stations (AC-DC-AC). Reasons: transmission stability, short-circuit conditions, parameters compensation, power losses, economic aspects, power systems connection.
Possibility of smaller insulation distances and higher transmission capacity than for AC systems. Control by DC voltage. Principle
DC line
Rectifier Inverter
Inverter Rectifier
Configurations:
a) Ground as return conductor. Ground resistance does not depend on line length but on ground connection. Suitable for unoccupied regions (corrosion, EMC).
b) Two wires line. Compared to 3ph less material, lighter towers. c) Two convertors in series. Low balancing ground current caused by
unbalance. During a fault it transforms to a), half power.
450 kV DC Canada
Complex power in AC grids
3 phase )W(cosUI3cosIU3P f )VAr(sinUI3sinIU3Q f
)VA(QPUI3IU3S 22f
Complex (1 phase) j
fffff eSsinjcosIUjQPS Sign according a convention.
Inductive load iu j
fj
ff IeI,eUU Complex conjugated current
jf
jf
*ff IeUIeUIUS iu
CAPIND
jQPIUS ff*
ff
3 phase power lines LV, MV
Series parameters are applied, for LV X→0.
3 phase power line MV, 1 load at the end
Symmetrical load → 1 phase diagram, operational parameters. Complex voltage drop
CAPIND
jIIjXRIZU jčlf
CAPIND
RIXIjXIRIU jčjčf magnitude phase
Phasor diagram (input Uf2, I, φ2) (angle υ usually small, up to 3)
Imagin. part neglecting and modifications
ff
jfčff U3
XQRPU3
IU3XIU3RU
Percentage voltage drop
22ff
f
UXQRP
U3XQRP
UU
3 phase active power losses
222
2l
*l
*f
XI3jRI3IjXR3
IZ3IIZ3IU3S
)A,;W(IIR3RI3P 2j
2č
2
! Even the reactive current causes active power losses! → reactive power compensation
3 phase MV power line supplied from one side
Constant series impedance
)km/(jXRZ 11l1
Voltage drop at the end (needn’t be the highest one, it depends on load character)
n
1kkklfAn IlZU
1
After imaginary part neglecting
CAPIND
IlXIlRUn
1kjkk1
n
1kčkk1fAn
CAPIND
U3
QlXPlRU
f
n
1kkk1
n
1kkk1
fAn
Voltage drop up to the point X (superposition)
n
1XkkAXl
X
1kkklfAX IlZIlZU
11
3 phase MV power line supplied from both sides
Calculation as for DC line (feeder is a negative load, zero voltage drop).
Bl
n
1kkklAB IlZIlZ0U
11
Moment theorems
l
IlI
n
1kkk
B
l
IllI
n
1kkk
A
n
1yyBA III
(In principle it is the current divider for each load.)
Active and reactive current sign change could be in different nodes → maximum voltage drop should be checked in all grid points.
Meshed grids MV
NO!
YES!
Node voltage method
Grid with n nodes. Set series branch parameters kmZ , load currents (nodal currents) kI , min. 1 node voltage fkU (between the node and the ground).
Calculation with series admittances
kmkm
1kmkm jXR
1ZY
Node k
n
km1m
0kkmk 0III
0kfk0k YUI Branches k, m kmfmfkkm YUUI After modifications:
0kfk
n
km1m
kmfmfkk YUYUUI
n
km1m
kmfm0k
n
km1m
kmfkk YUYYUI
Admittance matrix parameters definition: Nodal self-admittance (diagonal element)
0k
n
km1m
km)k,k( YYY
Between nodes admittance (non-diagonal element)
kmforYYY km)k,m()m,k(
(for non-connected nodes 0Y )m,k( )
Hence
n
1mfm)m,k(k UYI
Matrix form fUYI
Set voltages at nodes 1 to k (x), currents at nodes k+1 to n (y)
fy
fx
yy
T
xy
xyxx
y
x
UU
YY
YYII
Hence
fyyyfx
T
xyy
fyxyfxxxx
UYUYI
UYUYI
Calculate fyx U,I
fx
T
xy
1
yyy
1
yyfy UYYIYU
If some nodes are connected to the ground (through an admittance), then the admittance matrix is regular → to set all nodal current is enough.
IYU1
f
Note 1: Similar for DC grid. UGI
Note 2: For power engineering – powers are set, currents are calculated from the powers.
*
U3SI
Results are not precise if nominal voltages are used → iteration methods.
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