230kv cable sizing 1 - copy

7/27/2019 230kv Cable Sizing 1 - Copy

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i. Maximum Permissible Short Circuit Current For Conductor

ST = Area of Conductor = 630 mm2

T = duration of Short circuit = 1.0 sec

K =Constant depending on the material = 226 A.s 1/2/mm2

c = reciprocal of temperature coefficient of resistance = 234.5

f = final temperature = 250c

i = operating temperature of the conductor = 90c

I ADIABATIC = K. ST / t x [ ln ((f+c) / (i+ c))]

= 226 x 630/ (1) x [ln ( ( 250 + 234.5) / (90 +234.5 ) ) ] = 90.14 kA for 1 sec

ii. Maximum Permissible Adiabatic Short Circuit Current For Lead Sheath

tLEAD = thikness of lead = 8.70 mm ( Nominal )

ALEAD = 2599 mm2-Material constants

LEAD = reciprocal of temperature coefficient of resistance = 230 K

K = constant depending on the material = 41.0 A.s 1/2/mm2

-Short Circuit conditions

t = Short circuit duration = 1.0seci = operating temperature of lead sheath = 85 cf = final temperature of lead sheath = 85 c

-Maximum Permissible Adiabatic short circuit current

I ADIABATIC = K. ST / t x [ ln ((f+c) / (i+ c))]

=41 x 2599 / (1) x [ ln ((210 + 230) / (85+230))] = 61.61 kA for 1Sec

iii . Maximum Permissible Non-Adiabatic Short Circuit Current For Lead Sheath

-Non-Adiabatic Short Circuit Current For Lead Sheath

i = volumetric specific heat of media either side of the sheath 2.4 x 106 & 2.4 x 106 (J/K .m3).

i = thermal resistivity of media either side of the sheath 2.5K.m/W & 3.5K.m/Wi = volumetric specific heat of the sheath = 1.45E + 06

= tLEAD = 8.70mm

F = Constant = 0.7


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M = [(2 / 2) + [(3 / 3)] / (2 1 x 10-3) F = 5.016E-02

= heat loss factor= 1+ 0.61 Mt 0.069 ( Mt )2 +0.0043 ( Mt )3 = 1.030

iV . Maximum Permissible Non-Adiabatic Short Circuit Current

Isc = I ADIABATIC x

= 61.61 x 1.030 = 63.48 kA for 1 second

Loss factor for sheath and screen

[ cross bonded screen, flat formation 300 mm spacing]

1 = power loss in the sheath or screen, consists of losses :

1 = 1+ 1

where:

1 =circulating currents

1 = eddy currentsscreen = maximum operating temperature of the screen =85 c

Lead sheath

Alead = area of lead= 1647 mm2DMlead = mean diameter of lead = Dt3 + tpb =97.7 +5.1 = 102.8 mm

= resistivity of lead at maximum operating temperature

= lead / Rlead

Eddy currents

d = mean diameter of screen = 102.8 mm

Rs = resist.lnce of screen at its maximum operating temperature

ts = screen thickness == 5.1 mm

Ds = diameter over screen = 107.9 mm

= 2 x f = 2 x 60 =107.9

m = ( / R s ) x 10-7 =

0 = 6 ( m2 / (1+ m2) (d / 2s)2

1 = (0.86 / m3.08)(d/2s) 1.4m + 0.7

2 = 0


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s = resistivity of screen material at maximum operating temperature

1 =( 4 / (107 s))1/2

gs = 1 +( ts / Ds)1.74 (1 Ds10-3-1.6)

1 = Rs /R [ gs 0 (1+1+2) + (1 ts)4 / (12 x 1012) ]

For cross bonded system

1 =

Total loss factor for sheath and screen

For cross bonded system

1 = 1+ 1

1 =

Loss factor for steel armour

2 = 0

Thermal resistance of the constituent parts of the cable: T 1,T2, T3

Thermal resistance between conductor and sheath T 1

T1 = ( T /2 ) ln (1+ 2 t1/dc )

T = thermal resistivity of layer (km/W)

t 1 = thickness of the layer (mm)d 1 = diameter under the layer (mm)

thermal resistance of the semiconductive tape

Tt1 = ( T /2 ) ln (1+ 2 t1/dc )


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thermal resistance of the conductor screen

Tcs = ( T /2 ) ln (1+ 2 t1/dc )

thermal resistance of the insulation

Tinsulation = ( T /2 ) ln (1+ 2 t1/dc )

thermal resistance of the insulation screenTis = ( T /2 ) ln (1+ 2 t1/dc )

thermal resistance of the semi conductive tape

Tt2 = ( T /2 ) ln (1+ 2 t1/dc )

Thermal resistance between sheath and armour T2

T2 = 0.00 c /W

Thermal resistance of outer sheath T3

T1 = ( T /2 ) ln (1+ 2 t1/dc )

Three Circuits in Flat Formation Laid Directly in Ground with Backfill

L = depth from the surface of the ground to the group cent er= 1450 mm

d1

d2

= distance between circuitsA & B = 1200 mm

= distance between circuits B & C = 1200

p bf = thermal resistivity of backfill = 1.2 C.m/W

p r =thermal resistivity of the soil= 2.0 C.m/W

External thermal reslistance T 4

T'4 = externalthermal resistance

T'\ = correction factor

-External thermal resistance of the cable


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ba' ba be' be bd' bdbe'bebfbf bl'

bi

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= distance from cable b to the Image of cable a = 2915 mm

= distance from cable b to a = 300 mm

= distance from cable b to the Image of cable c = 2915 mm

= distance from cable b to c = 300 mm

distance from cable b to the Image of cable d = 3036 mm

= distance from cable b to d = 900 mm

= distance from cable b to the Image of cable e = 3138 mm

= distance from cable b to e = 1200 mm

= distance from cable b to the Image of cable f = 3265 mm

= distance from cable b to f = 1500 mm

= distance from cable b to the Image of cable I = 3036 mm

= distance from cable b to I = 900 mm

70 kV) up to 500 kV (Um=550 kV) - Test methods and requirements

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230kv cable sizing 1 - copy

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