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8/13/2019 Lightning Protection Full

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DIRECT STROKE LIGHTNING PROTECTION

For

KULAURA & SHERPUR 132/33KV GRID SUBSTATION

Desig! s"##$%! ere'io! 'es'ig & o((issioig

o) '*o e* 132/33 KV +ir is"$+'e, s*i'-ge+r

.AIS s"0s'+'io +' K"$+"r+ +, S-er#"r o'"re% 0+sis

C$ie'4 PO5ER GRID CO6PAN7 OF BANGLADESH

LTD

N+(e D+'e Sig+'"re

Desige,0%

Rashad Sarwar 22-10-2013

C-ee,

0%

Md. Rafiqul

Islam

22-10-2013


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DSLP CALCULATION: KULAURA & SHERPUR 132/33KV GRID SUBSTATION

1. Introd!t"on:

Lightning stroke can cause fatality structural damage, and could lead to malfunction ofthe electric equipment. The lightning stroke will vary by characteristics from area toarea. The lightning itself is an emission or discharge of electricity from cloud to ground,from ground to cloud and from cloud to cloud. When the lightning strikes the ground, itchooses a path with low resistance. According to the I standard !!"#$!!% &thestroke occurs in two steps, the first is ioni'ation of the air surrounding the centre and the

development of stepped leaders, which propagate charge from the cloud into the air(.The second step is return stroke, according to the same standard, &the return stroke isthe e)tremely bright streamer that propagates upward from the earth to the cloudfollowing the same path as the main channel of the downward stepped leader(.

*irect stroke lightning protection for +ulaura -herpur $/0+1 grid -ub#station isprovided by shield wire. ach lightning conductor consists shield wire, the earthing gridand the conductor which connects the shield wire with the earthing grid meant to carry

the lightning current away safely to the ground.


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-hield wire 78verhead ground wire0-ky wire95

A wire suspended above the phase conductors positioned with the intention of havinglightning strike it instead of the phase conductors.

-triking *istance5

The length of the final 6ump of the stepped leader as its potential e)ceeds thebreakdown resistance of this last gap: found to be related to the amplitude of the first

return stroke.

-urge Impedance5

The ratio of voltage and current of a wave that travels on a conductor.

3. R##r#n!# do!)#nt$:

I std !!"#$!!%5 I ;uide for *irect Lightning -troke -hielding of -ubstations.

*. C'(!('t"on:

Area wise calculations have been shown one by one and all results have beentabulated in the summary


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DSLP CALCULATION +OR 132KV SIDE

Cond!tor D't':

-2 @

*iameter 7/)r9 = ./"B m

-ub conductor -pacing 7l9 = .CB m 7Assumed9 7.B, I !!"#$!!%9

= ./D$"D m

Where,

r = radius of sub#conductors in m


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= $B +10m

4y solving the equation for 2>, we have 2> = .DCB m

0= 2H 2c= .$$% m 7.D,I !!"#$!!%9

Where,

h = average height of the conductor in m

2>0 = >orona radius of the bundle conductor

2 = quivalent radius of the bundle conductor

@ence s = %D.D%! ohms

The allowable stroke current is obtained by the equation

I- = /./ 4IL0s

Where 4IL = 2ated Lighting impulse withstand voltage


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Prot#!t"on ,- $"n(# $"#(d 0"r#:

@ere, K8 is the origin of the 2olling -phere

K-W denotes the location of the shield wire

?ow,

Allowed strike distance 7-9 = $!.!/ m

@eight of shield wire 7@9 = $% m


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3inimum hori'ontal separation of shield wire and ob6ect to be protected at height7A9

>= 72#T9 = .!B!/m

Co),"n#d %rot#!t"on ,- t0o $"#(d 0"r#



?ow,


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L = J7-/G /9 = $.""!B m

3a)imum allowable hori'ontal separation of the shield wires ensuring protection ofob6ect at height 7A9

M = /L = /$.DD"! m

DSLP CALCULATION +OR TRANS+ORER AREA!on$"d#r"n )'"))$"#(d 0"r#4

Cond!tor D't':

-2 @



= ./D$"D m

Where,



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= $B +10m

4y solving the equation for 2>, we have 2> = .DCB m

0= 2H 2c= .$$% m 7.D,I !!"#$!!%9

Where,






I- = /./ 4IL0s



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Prot#!t"on ,- $"n(# $"#(d 0"r#:



?ow,




11/24


>= 72#T9 = .!B!/m

Co),"n#d %rot#!t"on ,- t0o $"#(d 0"r#



?ow,


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L = J7-/G /9 = $.""!B m


M = /L = /$.DD"! m

DSLP CALCULATION +OR TRANS+ORER AREA't t# 'ntr- (o!'t"on n#'r'to T/+ HV B$"n4

Cond!tor D't':

-2 @



= ./D$"D m

Where,


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= Limiting corona gradient

= $B +10m4y solving the equation for 2>, we have

2> = .DCB m

0= 2H 2c= .$$% m 7.D,I !!"#$!!%9

Where,






I- = /./ 4IL0s


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@ence - = $!.!/ m

Prot#!t"on ,- $"n(# $"#(d 0"r#:



?ow,




15/24


>= 72#T9 = .!B!/m

Co),"n#d %rot#!t"on ,- t0o $"#(d 0"r#



?ow,

Allowable strike distance 7-9 = $!.!/ m


16/24


M = /L = /$.DD"! m

DSLP CALCULATION +OR TRANS+ORER AREACon$"d#r"n )"n"))$"#(d 0"r# #"t4

Cond!tor D't':

-2 @


-ub conductor -pacing 7l9 = .CB m 7nergypac -tandard9

@eight of the ob6ect 7h9 = $ m

In case of a twin conductor bundle, the equivalent radius is given by

2 = 7r)l9$0/ 7eqn. >.B, I !!"#$!!%9

= ./D$"D m

Where,



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= $B +10m

4y solving the equation for 2>, we have 2> = .D!BB m

0= 2H 2c= .%C!DC m 7.D,I !!"#$!!%9

Where,




@ence s = BC.CC%D% ohms


I- = /./ 4IL0s



18/24

Prot#!t"on ,- $"n(# $"#(d 0"r#:



?ow,

Allowed strike distance 7-9 = $!.C"%$ m


19/24

T= J7-/#7-#A9/9 = $%.C//C$ m

3a)imum hori'ontal separation of shield wire and ob6ect to be protected at height7A9

>= 72#T9 = $.D%%CC$ m

Co),"n#d %rot#!t"on ,- t0o $"#(d 0"r#



Allowable strike distance 7-9 = $!.C"%$ m


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M = /L = //./"CB m

DSLP CALCULATION +OR 33KV SIDE

Cond!tor D't':

-2 @



= ./D$"D m

Where, r = radius of subconductors in m

l = spacing between ad6acent conductor in m


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4y solving the equation for 2>, we have

2> = .D m

0= 2H 2c= .%C/$D m 7.D,I !!"#$!!%9

Where,


2>0

= >orona radius of the bundle conductor2 = quivalent radius of the bundle conductor

@ence s = %!.%$ ohms


I-= /./ 4IL0s


s = surge impedence of the conductor carrying the surge


22/24

Prot#!t"on ,- $"n(# $"#(d 0"r#:



?ow,

Allowed strike distance 7-9 = $/."!%!C m

@eight of shield wire 7@9 = $m

@eight of ob6ect to be protected 7A9 = D.! m

levation difference between the ob6ect to be protected and origin of the rollingsphere


23/24

>= 72#T9 = .%DDD"m


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CALCULATION SUAR5

2%: P a g e

E N E R G Y P ! E N G I N E E R I N G " # $

T-e o'e' o) '-is (+'eri+$ (+% 0e s"0;e' 'o o#%rig-' #ro'e'io ",er '-e B+g$+,es- o#%rig-' #ro'e'io +'!2999 Co#%ig o)

'-is ,o"(e' or gi

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