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.