ELEC 1104

Lecture 6:Lecture 6:

O h d li dOverhead lines and underground cablesunderground cables

Power System Layouty y

Transmission lines

Distribution lines

Transmission and Distribution Lines

Transmission lines are hung from steelTransmission lines are hung from steel towers through insulator strings, and they may be single circuit or double circuit lines.ay be s g e c cu t o doub e c cu t es.There are usually one or two earthed wires at the top of tower for lightning protectionat the top of tower for lightning protection.Distribution lines are usually supported on i l t i t d tinsulator pins mount on wooden or concrete poles.

Sag and Spang p

Earth wire

Sag

Earth wire

ConductortowerMinimum

clearance

S

clearance

Span

Earth Wires

Overhead lines are Earth wireusually protected from lightning by installing one or two overhead earth wires positioned to give suitable shielding ShieldLine give suitable shielding over the line conductors. These earth wires are

Shieldangle

conductor

These earth wires are electrically connected to the earthed towers.

Insulators

Overhead lines

Bare conductors stranded of several wiresBare conductors stranded of several wires for greater flexibility and mechanical strength.st e gt .Standard number of strands are in either one of the series:one of the series:(a) 1, 7, 19 ……..(b) 3, 12, 27 …….

Overhead lines

Conductors

Material Hard drawn C i

Hard drawn Al i i

Heat treated Al i iCopper wire Aluminium

wireAluminium

alloySpecific gravity 8.89 2.7 2.70

Breaking stress (tons/sq in)

23-30 10-12 19.2

Conductivity at 20oC 97.4 61 53.5(% of ICAS)Resistivity at 20oC(Ω-m x 10-8)

1.774 2.826 3.22

Resistivity of International Annealed Copper Standard

Coefficient of linear expansion (oC-1)

0.000017 0.000023 0.000023

(IACS) at 20oC = 1.7241 x 10-8 ohm-meter.

Copper vs Aluminiumpp

Aluminium is lighter but its conductivity is lower.g yFor equal conductivity, aluminium conductor has 1.64 times the cross section of copper, but its weight is only about half of that of the copper conductor.Aluminium has low tensile strength and high coefficient of expansion.Cost of aluminium is lower and more stable.

ASCR

Aluminium conductors are often reinforced by ysteel for greater mechanical strength and are known as ACSR (Aluminium Conductor, Steel Reinforced). In ACSR the central strands of the conductor are

d f l i d l f h h hmade of galvanized steel for strength whereas the peripheral strands are made of aluminium for electrical conductivityelectrical conductivity.

6 aluminium1 steel

6 aluminium7 steel1 steel 7 steel

Bundle conductors

Bundle conductors composed of two, threeBundle conductors composed of two, three or four stranded conductors are used for very high voltages.ve y g vo tages.Lower voltage gradient at conductor surfaceB tt h t di i ti d h b ttBetter heat dissipation and hence better current rating.

A bundle of 2 A bundle of 3 A bundle of 4

Corona

A corona is a luminous partial electrical discharge p gdue to ionization of the air surrounding a conductor. The breakdown stress, i.e. the critical field intensity, of air would depend on the atmospheric

di iconditions.For a given voltage, the maximum field intensity

t th d t f d doccurs at the conductor surface and decreases as the conductor radius is increased.

Corona

Corona

There is a certain definite loss associated with corona.The ionization current associated with corona flows in pulses only during the voltage peaks and is therefore rich in harmonics.Ozone is produced in corona and would cause deterioration to any organic materials nearby.Audible noise is produced in corona and hence is a source of noise pollution.

Electrical Parameters

These are distributed parameters by nature:These are distributed parameters by nature:

» Series resistance r Ωm» Series inductance l H/m» Shunt capacitance c F/m» Shunt conductance g S/m

For overhead lines shunt conductance representsFor overhead lines, shunt conductance represents leakage through insulators or corona loss and is usually ignored.y g

Transmission Line Model

Transmission line as a two-portp

VS = sending end voltageVS sending end voltageIS = sending end currentV i i d ltVR = receiving end voltageIR = receiving end current

IS IR

LineVS VR

Transmission parametersp

VS = AVR + BIRVS AVR BIRIS = CVR + DIR

IS IR

A, B, C, DVS VR

Line Model

Nominal π representation (Medium line)Nominal π representation (Medium line)

Z = R + jXI IIS IRI

YC/2 YC/2VS VR

Line Model

I = IR + VRY/2VS = VR + ZI

= (1+ZY/2)VR + ZIR( ) R RIS = I + VSY/2

= Y(1+ZY/4)V + (1+ZY/2)I= Y(1+ZY/4)VR + (1+ZY/2)IRHence

A = D = (1+ZY/2), B = Z,C = Y(1+ZY/4)( )

Line Model

Nominal T representation (Medium line)Nominal T representation (Medium line)

(R + jX)/2I I

(R + jX)/2VIS IRV

YCVS VR

Line Model

V = VR + IR Z/2V VR IR Z/2I = YV = YVR + IR YZ/2I I + I YV + (1+YZ/2)IIS = IR + I = YVR + (1+YZ/2)IRVS = V + ISZ/2

= (1+YZ/2)VR + Z(1+YZ/4)IRHence

A = D = (1+ZY/2),B Y(1+ZY/4) C YB = Y(1+ZY/4), C = Y.

Line Model

Series impedance (Short line)Series impedance (Short line)» VS = VR + ZIR; IS = IR

R + jXIS IR

VS VR

Examplep

Given a 3-phase, 132 kV line 350 km longGiven a 3 phase, 132 kV line 350 km long with parameters

r = 0 108 ohm/km; l = 1 37 mH/km;r = 0.108 ohm/km; l = 1.37 mH/km;g = 0 siemens/km; c = 0.0085 μF/km.

Load: 50 MVA at 0.8 power factor lagging.

To determine sending-end voltage, current and power factor.p

Examplep

Z = (0.108+j2π×50×1.37×10-3) × 350( j )=155.27∠75.91o Ω

Y = (j2π×50×0.0085×10-6) × 350Y (j2π 50 0.0085 10 ) 350 = 934.6 ×10-6∠90o Siemens

VR = 132/√3 = 76.21 kV (phase)I 50 103/√3 132 218 7 AIR = 50×103/√3×132 = 218.7 Aθ = cos-1 0.8 = 36.87o (lagging)

Examplep

Using short line representationUsing short line representation

IS = IR = 218.7∠-36.87o AVS = VR + ZIR

= 76.21+155.27∠75.91o × 0.2187∠-36.87o kV= 76.21+33.96 ∠39.04o = 104.8 ∠11.78o kV

Input power factor = cos (11.78o + 36.87o ) = 0.66 (lagging)

Examplep

Using nominal-π representationUsing nominal π representationA = D =1+YZ/2

= 1 + 0 0726∠165 91o 1 + 0.0726∠165.91= 0.9297+j0.0176 = 0.9298 ∠1.08o

B = Z = 155 27∠75 91o ΩB = Z = 155.27∠75.91oΩC = Y(1+YZ/4)

= j 934 6 ×10-6(1 + 0 0263∠165 91o)= j 934.6 ×10-6(1 + 0.0263∠165.91o)= 910.8 ×10-6 ∠90.38o

Examplep

VS = AVR + BIR= 0.9298 ∠1.08o × 76.21 ∠0o +

155.27∠75.91o × 0.2187∠-36.87o kV 99 95 ∠13 15o kV= 99.95 ∠13.15o kV

IS = CVR + DIR910 8 10 6 ∠90 38 76210 ∠0= 910.8 ×10-6 ∠90.38o × 76210 ∠0o +

0.9298 ∠1.08o × 218.7∠-36.87o

171 78∠ 16 75o A= 171.78∠-16.75o AInput power factor

(13 1 16 ) 0 86 (l i )= cos (13.15o + 16.75o ) = 0.867 (lagging)

Underground Cablesg

Cables contain one or more conductorsCables contain one or more conductorswithin a protective sheath. The conductors are separated from eachThe conductors are separated from each other and from the sheath by solid insulating materialinsulating material. The protective sheath is an impervious

i i l ti d i ll fcovering over insulation and is usually of lead. Its main function is to prevent the ingress of moist re to the ins lationingress of moisture to the insulation.

Underground Cablesg

They may be single-core cables with oneThey may be single core cables with one cable per phase or three-core cables with one common lead sheath.o e co o ead s eat .In single-core cables the stranded conductor is always of round cross-sectionis always of round cross-section. In multi-core cables so-called sector shaped t d l d t b tt tili thstrands are also used to better utilize the

space within the sheath.

Cable Insulations

Common insulating materials used in cablesCommon insulating materials used in cables are:Oil-impregnated paperOil-impregnated paperVulcanised rubbersynthetic polymeric dielectrics such as» polyethylene (PE), » propylene (PP), » polyvinyl chloride (PVC)

Solid CablesSolid Cables

Single Core Three coreSingle Core Three core

Lead sheathsfillersHessian

servings

Paper

g

insulation

Stranded conductors Belt

insulationFabraic

tapes

Solid Cables

Single Conductor, paper-insulated powerSingle Conductor, paper insulated power cable.

Solid Cables

Three-conductor, belted, compact-sector,Three conductor, belted, compact sector, paper-insulated cable.

Solid Cables

Three-conductor, shielded (H-type),Three conductor, shielded (H type), compact-sector , paper-insulated cable.

Solid Cables

Three-conductor solid-type cable withThree conductor solid type cable with protective steel armour.

Cable Parameters

Cables have the same distributed electricalCables have the same distributed electrical parameters as the overhead lines but» Capacitance is much higher due to closer» Capacitance is much higher due to closer

proximity of the conductors.» Shunt loss is no longer negligible. the shunt S u oss s o o ge eg g b e. e s u

loss in the dielectric include– leakage– dielectric hysteresis

Dielectric loss angleg

The dielectric loss is usually measured byThe dielectric loss is usually measured by the dielectric power factor

dielectric p f = cos φ Idielectric p.f. = cos φThe dielectric loss angle is

δ 90 f l φδ = 90 – power factor angle φSince δ is small

V

φδ

δ ≈ sin δ = cos φ = dielectric p.f. V

Cable Ratingsg

The current rating of a cable is limited byThe current rating of a cable is limited by the maximum permissible temperature of its insulations.su at o s.Depending on the expected loading, we have the following ratings:have the following ratings:» Continuous rating» Short time rating» Short time rating» Cyclic rating

Cable Ratingg

The steady loading that results in a finalThe steady loading that results in a final temperature equal to the maximum permissible value is known as the continuous rating. va ue s ow as t e co t uous at g.

tempcurrent

T

Tmax

Continuous rating

T

I

time

I

time

Cable Ratingg

If the loading is applied for a short durationIf the loading is applied for a short duration only, say 1 hour, then the loading without the maximum temperature being exceeded is a u te pe atu e be g e ceeded sknown as the (1 hour) short-time rating.

temp Tmaxcurrent

Short-time rating

Tmax

Irating

T

I

time

Cable Ratingg

For a given cyclic pattern, the maximum loadFor a given cyclic pattern, the maximum load that can be supplied without the maximum temperature being exceeded is known as the te pe atu e be g e ceeded s ow as t ecyclic rating.

temp Tmaxcurrent

cyclic ratingT

Tmax

cyclic ratingI

time

Thermal equationq

Heat Balance EquationHeat Balance EquationHeat generated = heat dissipated

+ h t b b d+ heat absorbed.Heat generated depends on power loss P in the

bl (I2R l d di l t i l )cable (I2R loss and dielectric loss)Heat dissipated depends on the surface area,

method of cooling and temperature differencemethod of cooling and temperature difference.Heat absorbed results in temperature increase

depending on the specific heatdepending on the specific heat.

Thermal equationq

LetLetP = power loss in cableλ = emissivity (watt/m2/oC)λ emissivity (watt/m / C)A = surface area for heat dissipation (m2)θ = temperature rise above ambient (oC)θ = temperature rise above ambient (oC)M = mass (kg)C = specific heat (joule/kg/ oC)Cp = specific heat (joule/kg/ oC)

Thermal equationq

Then with temperature rise dθ in the timeThen with temperature rise dθ in the time period dt,

Heat generated = P dtHeat generated = P dtHeat dissipated = λAθ dtHeat absorbed = MCp dθ

HenceHenceMCp dθ + λAθ dt = P dt

Thermal equationq

This can be written in the formτ(dθ/dt) + θ = θ∞

wherewhereτ = MCp/λA is the heating time constant.θ = P/λA is the steady state temperature riseθ∞ = P/λA is the steady state temperature rise.

The solution isθ(t) = θ∞ - (θ∞ - θ0)e -t/ τ

where θ0 is the initial temperature rise above ambientwhere θ0 is the initial temperature rise above ambient at t = 0.

Examplep

Heat run test of a transformer from coldHeat run test of a transformer from cold» Temperature rise after 1 hr – 15o C» Temperature rise after 2 hr – 27o C» Temperature rise after 2 hr 27 C

D t iDetermine» Final temperature rise if run continuously» Heating time constant of transformer

Examplep

From thermal heating equationFrom thermal heating equationθ∞(1 – e-1/τ) = 15 (1)θ (1 2/τ) 27 (2)θ∞(1 – e-2/τ) = 27 (2)

Dividing (2) by (1)(1 + e-1/τ) = 27/15 = 1.8e-1/τ = 1.8 – 1 = 0.8

∴ θ∞ = 15/(1 – e-1/τ) = 75 1/l (0 8) 4 48τ = -1/ln(0.8) = 4.48

Overhead line vs Underground cablesg

CostCost» Underground cables cost, on average, 8~15

times more than overhead lines.

Operation» Charging current for underground cables is

much higher than that of overhead lines and can l t f th t i ituse up a lot of the current carrying capacity.

The situation gets worse as the voltage increasesincreases.

Overhead line vs Underground cablesg

ReliabilityReliability» Overhead lines have more outages than

underground cables per unit length, but the g p g ,outages are usually shorter in duration.

Fl ibilitFlexibility» Overhead lines can be upgraded to higher

lt if U d d blvoltages if necessary. Underground cables cannot be easily upgraded.

Overhead line vs Underground cablesg

SafetySafety» Underground cables are more safe and are

always used in densely populated areas for this y y p preason alone.

Environmentv o e» Overhead Line Towers (aesthetic problem)» EM Field under overhead lines (effect on» EM Field under overhead lines (effect on

human beings)» Corona (radio interference, noise pollution etc)» Corona (radio interference, noise pollution etc)