Cable

Brandenburg University of Technology Cottbus Department of Power Distribution and High-Voltage Technology

Electrical Distribution Systems I

Dr. Klaus Pfeiffer LG 3 Walther-Pauer-Straße 5 03046 Cottbus Phone: (0355) 69-4035 [email protected] September, 2005

Brandenburg University of Technology Department of Power Distribution and High Voltage Technology Pfeiffer 1

Script M.V. and L.V. cables Contents

1 Types of cables

2 Cable laying

3 Rating

4 Electrical parameters

Exercises

Electrical Distribution Systems - Part I Cable


Medium-voltage- and low-voltage-cables Cables are insulated wires under protective sheath to conduct electrical energy. 1. Types of cables – distinguishing features According to the insulation we distinguish between two types:

• plastic-insulated cables • paper-insulated compound-impregnated cables

Nowadays plastic-insulated cables are almost solely used. Conductor material Copper Aluminium Insulation material Polyvinyl chloride PVC Cross-linked polyethylene XLPE Outer sheath Polyvinyl chloride PVC Cross-linked polyethylene XLPE Furthermore we distinguish between

- Single-core cables - Multicore cables (Multiconductor cables)

Number of conductors for multicore cables Low-voltage level Medium-voltage level

3 (L1, L2, L3) 3 (L1, L2, L3) 4 (L1, L2, L3, PEN) 5 (L1, L2, L3, PE, N)



The number of conductors for cables in low-voltage grids depends on the grid configuration itself. The following grid configurations are therefore examples. TN-S grid S =neutral conductor (N) and protective earth conductor (PE) are separated

TN-C grid C =neutral conductor (N) and protective earth conductor (PE) are combined

Other grid configurations (TT-grid or IT-grid) require several numbers of conductors too.



1.1 Abbreviated type designation The abbreviated type designation consists of the following parts:

- Abbreviated design designation - Number of conductors and cross sectional area in mm2 - Short symbol for structure and shape of conductor - If existing: Nominal cross sectional area of the shielding or the concentric conductor

in mm2 - Nominal voltages U0 / U and insulation voltage Um

Abbreviated design designation

Designation for Identifier

Standard type N

Conductor material Copper --- Aluminium A

Insulation PVC Y XLPE 2X Cross-linked, halogen-free polymer HX

Concentric conductor made by copper wires (only for L.V.-cables) (The concentric conductor consists of a copper wire braiding or of wavelike applied round copper wires. A conductive copper tape connects the wires.)

braiding C wavelike CW

Shielding made by copper wires (only for M.V.-cables) Single-core cables or multicore cables with combined shielding S Multicore cables with single conductor shielding SE

Outer protection covering (outer sheath) PVC Y PE 2Y Aluminium tape glued with PE-covering (FL)2Y Cross-linked, halogen-free polymer HX Not cross-linked, halogen-free polymer H

Cable for U0 /U = 0,6/1 kV without concentric conductor with green-yellow designated conductor -J without green-yellow designated conductor -0



Short symbol for structure and shape of conductor Single-wired round conductor (solid conductor) RE

Normal-twisted round conductor RM

Solid segmental conductor SE

Multiple-wired segmental conductor SM

1.1.1 Examples for abbreviated type designation 1.

Designation for single-core cables in a 3-phase system with one PE-conductor NYY-0 3x(1x95 RM) / (1x95 RM)

2.



3.

4.

5.

Designation for single-core cables in a 3-phase system N2XSY 3x(1x120 RM / 16)



6.

7.

8.



9.

10.

Note: Medium-voltage multicore-cables with aluminium conductors and XLPE-insulation are not manufactured. These cables are designed with paper insulation and lead sheath. Around the conductor insulation a belted insulation is arranged. That’s why these cables are called belted cables (see example NAKBA 6/10 kV).



1.2 Examples

NYY-0 0,6/1 kV 1 Conductor 2 Conductor insulation (PVC) 3 Outer sheath (PVC)

NYY-J 0,6/1 kV 1 Conductor 2 Conductor insulation (PVC) 3 Covering 4 Outer sheath (PVC)

NYCWY 0,6/1 kV 1 Core for mechanical stability 2 Conductor 3 Conductor insulation (PVC) 4 Covering 5 Concentric conductor made

by copper wires (wave shaped)

6 Conductive copper tape 7 Outer sheath (PVC)

N2XSY 6/10 kV 1 Conductor 2 Inner conducting layer 3 Conductor insulation (XLPE) 4 Outer conducting layer 5 Conductive tape 6 Shielding made by copper

wires 7 Conductive copper tape 8 Plastic film 9 Outer sheath (PVC)

N2XSEY 6/10 kV 1 Conductor 2 Inner conducting layer 3 Conductor insulation

(XLPE) 4 Outer conducting layer 5 Conductive tape 6 Shielding made by copper

wires (single conductor shielding)

7 Conductive copper tape 8 Covering 9 Plastic film 10 Outer sheath (PVC)



NAKBA 6/10 kV

1 Conductor 2 Impregnated paper

insulation 3 Tracer 4 Belted insulation 5 Lead sheath 6 Inner protection covering 7 Steel tape armour 8 Polypropylene thread in

compound 2. Cable laying Arrangement of single-core cables

Low-voltage level

Medium-voltage level

Buried laying

- laying directly in the ground - buried laying with covering cap - laying in a tube

The shielding is earthed at the beginning and at the end of the cable



Laying in air

- laying in a cable duct on concrete - laying on a cable ladder (seat area of the cables less than 10% of total area) - laying on a cable tray perforated (30% perforation of total area)

non-perforated - laying on a cable bridge

Cable ladder Accumulation of cable ladders

Perforated cable tray

3. Rating

3.1 Thermal rating

Permissible temperatures for cables with copper and aluminium conductors

Maximum operating

temperature [°C]

Maximum temperature at short-circuit

[°C]

XLPE-insulated cables 90 250

PE-insulated cables 70 150

PVC-insulated cables with

- Cross-sectional area A≤300mm2 70 160

- Cross-sectional area A>300mm2 70 140



3.2 Current carrying capacity The rated current Ir is specified for normal operation conditions. Thus the current carrying capacity Iz under normal operation conditions is

Iz = Ir

Deviant operation conditions will be considered by reduction factors f:

Iz = f1 · f2 · Ir

f1 reduction factor considering deviant temperature f2 reduction factor considering accumulation of cables in one cable route

3.2.1 Buried laying Normal operation conditions

- 1 multicore cable or 3 single-core cables - laying depth t = 0,7 m - ground temperature ϑE = 20°C - capacity factor m = 0,7

- specific thermal resistance of the ground humid ground ρE = 1 K·m/W dry ground ρE = 2,5 K·m/W

- no external temperature influence (e.g. from heating ducts) Deviant operation conditions

- ground temperature ϑE ≠ 20°C - capacity factor m ≠ 0,7 - specific thermal resistance of the ground ρE ≠ 1 K·m/W - several cables in one cable route (accumulation)

Determination of capacity factor m: The area beneath the straight line of value m (dashed line) is equivalent to the area beneath the load curve (solid line).



3.2.2 Laying in air Normal operation conditions

- 1 multicore cable or 3 single-core cables - air temperature ϑE = 30°C (must not be increased by cable power losses) - no external temperature influence (e.g. from solar radiation)

Deviant operation conditions

- air temperature ϑL ≠ 30°C - several cables in one cable route (accumulation)

The reduction factors f are given in standards. They depend on the following parameters:

Buried laying Laying in air

f1 Ground temperature ϑE Capacity factor m Specific thermal resistance of the ground ρE

Air temperature ϑL

f2

Number of cables Arrangement of the cables Capacity factor m Specific thermal resistance of the ground ρE

Number of cables Arrangement of the cables

For buried laying two more conversion factors have to be considered: reduction factor for covering caps f3 = 0,90 reduction factor for laying in a tube f4 = 0,85 For normal operation conditions the load current is assumed to be constant. An intermittent load current yields a higher current carrying capacity. This case will not be considered, when the cables were selected. That’s why for the intermittent operation there are reserves at current carrying capacity.



3.3 Examples for calculation of current carrying capacity Iz 3.3.1 Planning of a cable connection between transformer and switchgear Given: parameters of a station service transformer in a biomass power plant

transformer rated apparent power SrT = 2 MVA transformer rated voltage (L.V.-side) UrT = 420 V

cable laying : - in a cable duct (air temperature ϑL = 50°C) - on a cable ladder

application of single-core cables

Answer:

A2750 kV0,42 3

MVA2 U3

SIrT

rTrT =

⋅=

⋅=

First approach: Selection of single-core cables NYY-0 1x500 in triangular arrangement rated current Ir of cable NYY-0 1x500 Ir = 747 A (p. 253) Deviant operation conditions:

- air temperature ϑL = 50°C f1 = 0,71 (p. 266)

- assumption of 6 cable systems (6 parallel cables per phase) on 2 cable ladders

f2 = 0,93 (p. 268)

Current carrying capacity Iz = 0,71 · 0,93 · 6 · 747 = 2960 A



Proofing of the result The current carrying capacity is greater than the rated transformer current. Iz > IrT ⇒ 2960 A > 2750 A Second approach: Selection of single-core cables NYY-0 1x400 in triangle arrangement rated current Ir of cable NYY-0 1x400 Ir = 646 A (p. 253) Iz = 0,71 · 0,93 · 6 · 646 = 2560 A Iz < I ⇒ 2560 A < 2750 A ⇒ 6 cable systems with NYY-0 1x400 are insufficient. The number of cable systems (parallel cables per phase has to be increased to n = 7). In that case the reduction factor f2 decreases if the worst comes to the worst to f2 = 0,90. (p. 268 for 3 cable ladders) Current carrying capacity Iz = 0,71 · 0,90 · 7 · 646 = 2890 A Proofing of the result The current carrying capacity is greater than the rated transformer current Iz > IrT ⇒ 2890 A > 2750 A Selection of the PEN-conductors The selection and arrangement of the PEN-conductors is to be made in that way, that

10ZZ

1K

0K < (empirical value)

Z0K Zero sequence impedance of the cable connection Z1K Positive sequence impedance of the cable connection

Otherwise the phase-to-ground short-circuit current will be to low to find a setting for the overcurrent protection. This requirement will be meet, if

A0,5APEN ⋅≈ (empirical value)

is selected and the PEN-conductors will be arranged between the cable systems.



Result of the cable selection The following two cable connections meet the technical requirements.

NYY-0 6x3x(1x500) / 3x(1x500) 0,6/1 kV Um = 1,2 kV

NYY-0 7x3x(1x400) / 4x(1x400) 0,6/1 kV Um = 1,2 kV

The decision, which cable system will be applied, is determined from - asset costs - operational costs

3.3.2 Planning of a cable connection for connecting a motor to a switchgear

Given: parameters of the motor (primary air fan in a biomass power plant) PrM = 400 kW

UrM = 690 V cos ϕ = 0,86 η = 0,97

cable laying : - on a perforated cable tray - other cables are already laid on this

cable tray without any distance - air temperature ϑL = 45°C

application of multicore cables

Answer:

rated motor current

A 4010,970,86kV 0,693

kW400 ηcosU3

PIrM

rMrM =

⋅⋅⋅=

⋅⋅⋅=

ϕ

First approach:

Selection of multicore cables NYY-J 3x300/150 rated current Ir of cable NYY-J 3x300/150 Ir = 511 A (p. 253)

Deviant operation conditions: - air temperature ϑL = 45°C f1 = 0,79 (p. 266) - 2 cable trays, on each cable tray 6 cables, laid without any distance f2 = 0,73 (p. 270) Current carrying capacity Iz = 0,79 · 0,73 · 511 = 295 A



Proofing of the result The current carrying capacity is lower than the motor rated current. Iz < IrM ⇒ 295 A < 401 A This selected cable doesn’t meet the requirement. Second approach: Increasing of the cable cross-sectional area Selected cable: NYY-J 3x400/185 Ir = 597 A (p. 253) Iz = 0,79 · 0,73 · 597 = 344 A Proofing of the result Iz < IrM ⇒ 344 A < 401 A The current carrying capacity is still insufficient. Third approach: Multicore cables with cross-sectional area greater than A > 400 mm2 are not produced. That’s why parallel cables have to be used. Selected cable: NYY-J 2x3x240/120 Ir = 2 · 445 A = 890 A (p. 253) Because of laying now 7 cables on one cable tray, the reduction factor f2 decreases to f2 = 0,71 (approximation) (p. 270) Iz = 0,79 · 0,71 · 890 = 499 A Proofing of the result Iz > IrM ⇒ 499 A > 401 A The current carrying capacity is remarkable higher than required. That’s why we can try to selected a cable with a smaller cross-sectional area. Fourth approach Selected cable: NYY-J 2x3x185/95 Ir = 2 · 374 A = 748 A (p. 253) Iz = 0,79 · 0,71 · 748 = 419 A Proofing of the result Iz > IrM ⇒ 419 A > 401 A This selected cable meets the requirements very well.



Result of the cable selection 2 parallel connected multicore cables NYY-J 2x3x185/95 0,6/1 kV Um = 1,2 kV have to be applied. 3.3.3 Planning of a cable connection for connecting wind turbines to the grid Given: 6 wind turbines with Pr = 1,8 MW (electrical power output at cos ρ = 1) the total power output of the wind turbines has to be connected to a

110/20 kV substation (20-kV-side)

cable laying : buried laying directly in the ground ground temperature ϑE = 20 °C

capacity factor m = 1 specific thermal resistance of the ground ρE = 1,5 K·m/W

Answer: current at rated power output

A 123kV 023

MW 8,16U3

P6I r =⋅

⋅=

⋅⋅

=

First approach: Selection of single-core cables NA2XS2Y 3x(1x240), laying side by side Ir = 455 A (p. 256) Deviant operation conditions:

- capacity factor m = 1 - specific thermal resistance of the ground ρE = 1,5 K·m/W

These conditions yield to following reduction factors: f1 = 0,86 (p. 261) f2 = 0,85 (this value requires a distance of 7 cm between the single-core cables) (p. 264) Current carrying capacity Iz = 0,86 · 0,85 · 455 = 332 A



Proofing of the result Iz > I ⇒ 332 A > 312 A This selected cable meets the requirements. Result of the cable selection NA2XS2Y 3x(1x240RM/25) 12/20 kV Um = 24 kV 3.3.4 Planning of a cable connection for connecting a station service transformer to the 20-kV-grid Given: parameters of a station service transformer in a biomass power plant

transformer rated apparent power SrT = 2 MVA transformer rated voltage (H.V.-side) UrT = 20 kV

cable laying : - air temperature ϑL = 40°C - on cable ladder (without other cables)

Answer: current at rated power output

A 58kV 023

MVA 2U3

SIrT

rT =⋅

=⋅

=

First approach: Selection of single-core cables, laying in triangular arrangement N2XS2Y 3x(1x35) Ir = 200 A (p. 257) Note: 35 mm2 is the smallest possible cross-sectional area. Deviant operation conditions: - air temperature ϑL = 40°C f1 = 0,91 (p. 266)

Because of no other cables on the cable ladder (no accumulation)

f2 = 1 (p. 268)



Current carrying capacity Iz = 0,91 · 1 · 200 = 182 A Proofing of the result Iz > I ⇒ 182 A > 58 A The current carrying capacity is remarkable higher than required. Result of the cable selection N2XS2Y 3x(1x35RM/16) 12/20 kV Um = 24 kV Note: These examples only consider the current carrying capacity of the cables. More criteria for a cable selection are

- thermal short-circuit strength - minimum value of short-circuit current at the end of the cable (to be detected by a

protection device) - voltage drop under normal operation (in low-voltage switchgears) - voltage drop at motor start-up (in low-voltage switchgears)

These additional criteria had not been considered in these examples. The details for determination of current carrying capacity are contained in the following German Standards:

- DIN VDE 0276-603: Bemessungsströme für Niederspannungskabel - DIN VDE 0276-620: Bemessungsströme für Mittelspannungskabel - DIN VDE 0276-1000: Umrechnungsfaktoren



4. Electrical parameters 4.1 Π- equivalent circuit of a cable

R ohmic resistance Xb effective reactance Cb effective capacitance G insulation conductance Note: Usually the values for R, Xb, Cb and G are given per length (per km). 4.1.1. Ohmic resistance

R∆RR r ′+′=′ ϑ rRϑ′ D.C. ohmic resistance at permissible operation temperature ϑr

R∆ ′ additional resistance

( )[ ]201RR r2020r −⋅+⋅′=′ ϑαϑ 20R′ D.C. ohmic resistance at 20°C

AlR20 ⋅

=′ρ

ρ specific ohmic resistance copper ρ = 0,0178 Ω mm2/m

aluminium ρ = 0,0278 Ω mm2/m

20α temperature coefficient copper α20 = 0,00393

aluminium α20 = 0,00403



The additional resistance considers

- skin effect - proximity effect - eddy current losses in the shielding or in the concentric conductor - hysteresis losses in metal sheath or armour

The calculation of the additional resistance is very complex. Therefore they will be measured and given by the manufacturer. In a datasheet the D.C. ohmic resistance and the A.C. ohmic resistance (at same temperature and at rated current) is to be found. The difference between D.C. and A.C. ohmic resistance is the additional resistance. 4.1.2 Effective inductance Cables without shielding

Cables with shielding Voltages induced in the shielding yield to a current flow in the shielding. This current generates a magnetic field, which is directly opposed to this caused by the phase currents. That’s why the inductance will be decreased by an additional inductance ∆L’

L∆LL bbS ′−′=′

äq

0b r

aln2

L ⋅=′πµ

räq = 0,779 · r (for solid conductors)

µ0 = 4π · 10-7 H/m

äq

m0b r

aln2

L ⋅=′πµ

median geometric distance

a1,26a22aaaaaaa 333TRSTRSm ⋅=⋅=⋅⋅=⋅⋅=



4.1.3 Effective capacitance

Cables without shielding

The electrical field outwards is indefinite, because it strongly depends on the electrical properties of the environment. The mutual capacitances Cg (between the conductors) and the capacitances between the conductors and ground (CE) cannot be specified. Low-voltage cables (with exception of type NYCWY) don’t have shieldings. For the calculation of phase-to-ground short circuit in TN-low-voltage grids (solidly grounded) cable capacitances are not needed. Cables with shielding Single-core cables

0Cg =

Eb CC =

rrln

εε2πCS

r0b

⋅⋅=

r conductor radius rS radius of the shielding (≈radius of the insulation)

ε0 dielectric constant 90 10

941 −⋅⋅

=π

ε ⎥⎦⎤

⎢⎣⎡mF

εr relative dielectric constant at operational temperature PVC εr = 8 XLPE εr = 2,4



Multicore cables with combined shielding (3 conductor)

gEb C3CC ⋅+=

CE phase to ground capacitance Cg mutual capacitance (phase to phase) 4.2 Effects on the electrical parameters

- The ohmic resistances are primarily determined by the cross-sectional area and the conductor material (and their specific ohmic resistance).

- The effective inductances (and so the effective reactances) will be decreased, if the distance between the conductors is decreasing.

- The effective capacitances will be increased, if the distance between the conductors is decreasing.



4.3 Comparison of cable- and overhead line-parameters

Cable Overhead line

Inductance LbK < LbF (0,08 … 0,16) Ω/km < (0,25 … 0,4) Ω/km

Capacitance CbK > CbF (0,2 … 1,0) µF/km > (6 … 15) · 10-3 µF/km

RXb (0,1 … 1) (1 … 2) single conductor

(2 … 10) conductor bundle 4.4 Dielectric losses These are the losses caused by leakage currents in the insulation.

dielectric loss factor C

wII tan =δ

dielectric loss factor tan δ at operational temperature PVC tan δ = 40 · 10-3 XLPE tan δ = 0,55 · 10-3 Dielectric losses

δ tanUI3UI3P bCbwD ⋅⋅⋅=⋅⋅=

δδ tanQ tanUCωP C2bbD ⋅=⋅⋅⋅=

Conductance of the leakage

2b

D

U

PG =

Iw active current IC capacitive reactive current (charging current)

ω angular frequency Ub operating voltage (phase to phase voltage) QC capacitive charging power IC capacitive charging current



4.5 Capacitive charging current

Faultless operation, symmetrical load

Capacitive charging current YUCωI bC ⋅⋅= UY phase to ground voltage

Capacitive charging power bCYCC UI3UI3Q ⋅⋅=⋅⋅=

2bb

2b

b2

bC UCω3

UCω3UCω3Q ⋅⋅=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅⋅=⋅⋅⋅= Y

Ub operating voltage (phase to phase voltage) 4.6 Capacitive earth-fault current

CcCbCE III +=

*bECb UCωI ⋅⋅= *cECc UCωI ⋅⋅=

YU3UU *c

*b ⋅==

YUCω3I ECE ⋅⋅⋅=

if CE = Cb then ICE = 3 · IC



4.7 Examples for capacitive charging current and capacitive earth-fault current Example 1 Given: 8 wind turbines are connected by a ring main system to a 20-kV-substation the cable connection based on N2XS2Y 3x(1x185RM/25) 12/20 kV has a

total length of 4,8 km

A 4,753kV20 km4,8

kmµF0,273s314 UCωI 1

bC =⋅⋅⋅=⋅⋅= −Y

kVar 164,5kV20 A 4,753UI3UI3Q bCYCC =⋅⋅=⋅⋅=⋅⋅=

cable N2XS2Y has a shielding ⇒ CE = Cb


A14,3 I3I CCE =⋅=

Example 2 Given: the 20-kV-substation is connected via a transformer to a 110-kV-substation

with a 110-kV-cable cable typ: N2XS(FL)2Y 3x(1x800RM/35) 64/110 kV cable length : 12 km

A 6,253kV110 km 21

kmµF0,22s314 UCωI 1

bC =⋅⋅⋅=⋅⋅= −Y

MVar 02,01kV 101A 6,253UI3UI3Q bCYCC =⋅⋅=⋅⋅=⋅⋅=

cable N2XS(FL)2Y has a shielding ⇒ CE = Cb


A157,8 I3I CCE =⋅=

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