# Calculations of MV Cables

July 23, 2018 | Author: Anonymous kdFzbQ4jf | Category: Electrical Substation, Electrical Impedance, Power Engineering, Physics, Physics & Mathematics

#### Short Description

Calculations of MV Cables...

#### Description

Exer Ex erci cise se 6

Upstream network 63kV Short-circuit power 2000 MVA

T2

T1

G1

15 MVA

20 MVA

Usc Us c =1 =10% 0%

Usc Us c =1 =10% 0%

CB1

15 MVA

G

CB2

Usc Su Usc Sub b =1 =15% 5% Usc Tra Trans= ns= 20% 20%

CB3 10KV 10K V bu busb sbar ar

10KV 10 KV bu busb sbar ar

Subs Su bsta tati tion on A

CB4

CB5

T4

T5

CB6

15 MVA

15 15MVA MVA

Usc=10%

Usc=10% Usc=10%

CB18

CB7 In each each subst substation ation 1 single transformer in operation

11 km km Su Subs bstat tation ion Subst Su bstat ation ion B B

CB19

0.5 0.5 km km Subs Su bsta tatio tion nC Subst Su bstat ation ion C

CB10

CB11

T8 5 MVA

3KV

CB8

5 MVA

Usc=8%

CB9

T6

T9

T7

Usc=8%

CB16

10 MVA

10 MVA

Usc=8%

Usc=8%

CB17

3KV CB12

CB13

3KV MV network design & devices selection

Answ An swer er Bo Book ok – Oc Octt 20 2008 08

11

Choice of cables for substations B and C (detailed method cont’d.)

Exercise 6 Substation A

Single-pole AI – PR cables, buried directly in dry, calcareous ground, temperature 20°C, continuous operation,

10KV busbar

CB6

Substation B :

Operational current Ir: 577 A 10 MVA 10 kV 3 2 cables in // imposed Installation mode factor = 1 column 1 Temperature factor = 1 Proximity factor = 0.75 Ground factor = 1

11 km km Substation SubstationB B CB9 T7

CB8 T6

10 MVA Usc=8%

10 MVA Usc=8%

Chosen theoretical currents:

CB13

CB12

577 : 0.75 = 770 A i.e. 385 A per cable column 1: 1852 = 380A 2402 = 440A

3KV 2

cables to substation B: 2 x 240 mm per phase (14% remaining for possible extension)

Substation A

10KV busbar

Substation C Operational current Ir: 288 A Installation mode factor Temperature factor Proximity factor Ground factor

5 MVA 10 kV

= = = =

CB7

3

1 1 1 1

0.5 0.5 km km Substation Substation C C CB11

CB10

Chosen theoretical currents:

T9 5 MVA

T8 5 MVA

288 : 1 = 288 A

Usc=8%

Usc=8%

column 1 952 = 260 A 1202 = 300 A cables to substation C: 1 x 120 mm2 per phase

CB17

CB16 3KV

MV network design & devices selection

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Verification of short-circuit current withstand of cables

Exercise 6

Upstream protection circuit breaker tripping time: 0.4 sec. Permissible temperature rise in cables: 160°(+90°=25 0°) Isc S> k t S>

with k = 94

23689 0.4

= 159.45 mm 2

94

The cable planned for Substation C (120 mm2 ) would not withstand the short-circuit currents. It should have a cross-section of 185 mm2 ●

Calculation of short-circuit currents in substations B and C

Impedance of the 2 cables of substation B 1000 m of 2 x 240 mm2 R=ρ

L 3 1000 = 36.10− • = 0..15Ω S 240

i.e. for the two cables in parallel R = 0.075 Ω - Calculation of reactance x = 0.15  Ω  /km i.e. for the two cables in parallel X = 0.075 Ω - Calculation of impedance:

Z = ●

0. 075 2 + 0 .075 2 = 0.1060 Ω

Impedance of the cable of substation C

500 m of 1 x 185 mm2 − 3 500 L = = 0 . 0973 Ω A single cable R = ρ 36 . 10 . 185 S Calculsations to be done according to verification of results.

X = 0 .15 Ω  / km = MV network design & devices selection

0.15 = 0 .075 Ω 2 Answer Book – Oct 2008

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