ieee-tpc tutorial line losses 26 july 2010 final2

8/3/2019 IEEE-TPC Tutorial Line Losses 26 July 2010 Final2

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Dale Douglass, PDC

Doug Proctor, Proctor Engineering

July 26, 2010

IEEE Special Session Tutorial


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Executive Summary Electrical losses (or carbon emissions) for

transmission lines are a small part of the totalsystem losses (emissions).

Losses in both phase conductors and shield wiresare proportional to their ac resistance increase

with the square of line current flow.

The ratio of phase conductor to shield wire lossesis on the order of 100.


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System LossesAccording to Wikipedia, electrical losses in the US

power system amount to about 7% of transmission.

Most of these losses are in distribution rather thantransmission lines but the losses in transmission arenot negligible.

Transmission losses occur in the phase conductors and

, to a much smaller extent, in the grounded Shield wiresystem.


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115-kV Structure Configuration


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500 kV Structure Configuration


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Transmitted Powercos3 = IVP

V = Line-line voltage (volts)

I = Line current (amperes)

Cos() = Power Factor (unitless)


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Electrical Losses for 3-Phase

Conductors in an Overhead Line

LITcRPLoss =2)(3

R(Tc) = Resistance of phase bundle at operating

temperature (Ohms/mile)

I = Line current (amperes)

Ploss=Power Loss (watts)

L = Line length (miles)


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Cost of Phase Conductor Losses2$ 3 ( ) ( ) $ / Losses R Tc I L t hrs watt hr =

At a cost of $0.10/kw-hr, the cost of

losses in carrying 35 MW over a 10

mile-long 115kV line with Drake

phase conductors for one year is

$284K and the value of the

transmitted energy is $52.6M


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Impact of Emergency Operation

Consider the 10 mile 115-kV line with a normalcurrent of 300a & emergency current of 1000afor 24hr per year.

23 0.12 300 8736 Normal Losses = 23 0.15 1000 24 Emergency Losses =

Impact of emergency operation on energy losses is

small - 2830 MW-hrs to 2938 MW-hrs


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Reducing AC ResistanceWhat are the Variables?

Crossectional Area of aluminum (primary)

Conductivity of conducting wires (primary) Conductor Temperature (secondary)

Skin Effect (minor up to 1.5 inches)

Proximity Effect (non-existent)

Magnetic Core Interaction (


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IEEE TP&C

Summer Meeting

Minneapolis, MN


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115-kV Structure Configuration


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Two Step Process

Approximate method

First - determine voltages induced on the shield wires by the phases.

Capacitive induction is not considered

Mutual induction between shield wires is not considered The phase voltages and currents are assumed to be balanced

Second determine the currents in the shield wires

The shield wire current is confined to circulate through thegrounding system and earth return

The effect of the shield wire currents on the phase currents isassumed to be minimal and is not considered

The mutual induction between shield wires is now included


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Step 1 - Induced VoltagesVoltage induced by mutual coupling

cycbybayay

cxcbxbaxa

ZIZIZIE

ZIZIZIEx

++=

++=

=

+=

ax

e

ax

axax

D

DfX

jXfZ

10log004652.0

00159.0


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Equivalent Depth of Earth

Return The mutual impedance between the shield wire and

the phase conductors includes the earth return pathassociated with the shield wire circulating current

The equivalent depth of the earth return is required todetermine the mutual reactance:

Where r r 10W m3feD

2160=


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Self Impedance Parameters795 Drake

Resistance Rc= 0.1172 /mi

Reactance Xc= j0.399 /mi

7#8 Alumoweld

Resistance Rsw= 2.44 /miReactance Xsw= j0.749 /mi


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Sample Calculations - Geometry

The geometry from our example:

Dax=Dcy=10.6 ftDbx=Dby=11.7 ft

Dcx=Day=21.8 ft


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Sample Calculations X & Z

The calculated reactancesXax=Xcy=0.537 ohms/mi

Xbx=Xby=0.525 ohms/mi

Xcx=Xay=0.449 ohms/miThe calculated impedances

Zax=Zcy=0.095+j0.537 ohms/mi

Zbx=Zby=0.095+j0.525 ohms/miZcx=Zay=0.095+j0.449 ohms/mi


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The Voltages Induced AreEx Ia Zax Ib Zbx+ Ic Zc+:=

Ey Ia Zay Ib Zby+ Ic Zcy+:=

V/mi

V/mi==

=+=

7.827.246.241.3

3.377.240.157.19

jE

jE

y

x


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Step 2 Induced Currents

Current induced by voltages on the shield

wire has a path to ground

An expression is derived considering only the

voltages and currents in the shield wires.


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Shield Wire Voltage Equation The voltages induced in the shield wire can also beexpressed:

From step 1:yyyxyxy

xyyxxxx

ZIZIE

ZIZIE

+=

+=

cycbybayay

cxcbxbaxa

ZIZIZIE

ZIZIZIEx

++=

++=


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Solving for Shield Wire Current

xy

xxxy

cxbxax

xy

xxcybyay

ay

yy

xy

xx

cybyay

yy

xy

cxbxax

ax

Z

ZZ

aZZaZZ

ZaZZaZ

II

Z

ZZ

aZZaZZ

ZaZZaZ

II

2

22

2

22

)()(

)()(

++++

=

++++

=

Express phase currents

in terms of Ia

Rearrange terms

Self Impedance

Expression for Currents)(

)(

2

2

2

cybyayay

cxbxaxax

ac

ab

aZZaZIE

aZZaZIE

aII

IaI

++=

++=

=

=

)(

)(

2

2

cybyayayyyxyx

cxbxaxaxyyxxx

aZZaZIZIZI

aZZaZIZIZI

++=+

++=+


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Induced Shield Wire Currents The magnitude of the SW current in the 115kV line:

3.8

9.8

=

=

y

x

I

I

The magnitude of the SW current in the 500kV line:

amps

amps

2931

==

y

x

I

I amps

amps


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Losses in the 115kV ExampleThe resistance of 7#8 Alumoweld is:

2.44 ohms/mile

yyxx RIRI22Losses +=

= 0.366 kW/mile


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Losses in the 500 kV Example The resistance of 7#6 Alumoweld is:

1.479 ohms/mile

yyxx RIRI22Losses +=

= 2.67 kW/mile


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Comparison of 115 & 500 kV

Phase and Shield Wire Losses115 kV 1-Drake, 7#8AW 500 kV 3-Bunting, 7#6AW

60 MVA power flow

300 amps

R = 0.128 Ohm/mi

35 kW/mi (0.06%/mi)

2 ea 7#8 shield wires

Induced current = 10 amps

370 W/mi (0.0006%/mi)

1000 MVA load

1000 amps,

R = 0.088/3 Ohm/mi

1 MW/mi (0.1%/mi)

2 ea 7#6 AW

Induced current = 30 amps

2.7 kW/mi (0.0003%/mi)


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Conclusions Electrical losses in transmission lines are a small

part of the total system losses (< 0.1% per mile). Losses in phase conductors are much larger than

shield wires (100:1) Occasional system emergencies will have little

impact on losses. Losses are proportional to ac resistance of

conductors - largely determined by conductivity

and area. Use lots of aluminum to reduce losses (paid

advertisement). Lines are Green. Be proud.

ieee-tpc tutorial line losses 26 july 2010 final2

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