Previous in Forum: AC/ DC Power Supply   Next in Forum: ACS700 ABB Drive
Close
Close
Close
5 comments
Rate Comments: Nested
Anonymous Poster

Earth Pit Calculation

03/25/2009 11:33 PM

Hai,

We are an OEM,and Im doing a big project, for that along with the Electrical and automation design we need to give the EARTH PIT Design and calculation. So can anyone help me to find out what are the parameters i need to consider while calculating this. How to do the design.

Thanks

Reply
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.
Guru
United Kingdom - Member - Indeterminate Engineering Fields - Control Engineering - New Member

Join Date: Jan 2007
Location: In the bothy, 7 chains down the line from Dodman's Lane level crossing, in the nation formerly known as Great Britain. Kettle's on.
Posts: 32175
Good Answers: 839
#1

Re: Earth Pit Calculation

03/26/2009 4:02 AM

What national code is applicable? Where is the earth pit based?

__________________
"Did you get my e-mail?" - "The biggest problem in communication is the illusion that it has taken place" - George Bernard Shaw, 1856
Reply
Participant

Join Date: Jan 2009
Location: INDIA
Posts: 1
#3
In reply to #1

Re: Earth Pit Calculation

03/26/2009 6:49 AM

Location is India only.

Reply
Commentator

Join Date: Mar 2009
Location: delhi india
Posts: 79
Good Answers: 2
#2

Re: Earth Pit Calculation

03/26/2009 4:06 AM

Dear Sir

you have to first check the surface where you are planning to make earth pit by making cross pit in that ground .then you have to check which type of earthing you have to done either CU earthing or GI earthing or chemical earting acc to your soil situvation .then after this you have to design pit so that electrical equpiment earthing seprete and elctronics equpiment seprete.after this you have to calculate indivisble earthing and after this you have to calculate grid earhing

Reply
Participant

Join Date: Mar 2009
Posts: 1
#4

Re: Earth Pit Calculation

03/27/2009 11:57 PM

Please when you will get the total information,then kindly convey to my mail. I would also like to know the same question.

Thanks & Regards

A.mukherjee

asokmukherjee1@yahoo.co.in

Reply Off Topic (Score 5)
Anonymous Poster
#5

Re: Earth Pit Calculation

07/21/2010 5:00 AM

Eg:- Supply voltage – 11 kV, fault level at 11 kV side at substation – 350 MVA, lengthof 11 kV feeder from substation to factory – 3 km, 11 kV conductor size – 95 Sq. mm,spacing of conductor – 1 m, resistance of line – 0.5 Ω/ km. Rating of transformer atfactory – 900 kVA, 11 kV/433 V, % impedance – 6 Ω (2 Nos in parallel). Soil

resistivity, ρ = 200 Ω-m

Take base values: 100 MVA, 11 kV
Base MVA x 100
Short Circuit MVA
= 100 x 100
350
% Source impedance =
= 28.57 %
95 = 5.5 mm
π
11 kV cond. radius =
Spacing between cond. = 1000mm
-7
e
-7
e
2 log 1000 +0.5 x 10 x 3000
5.5
Inductance of line for eqvt. spacing = (2 log d + 0.5) x 10 H/m
r
= ⎛ ⎛ ⎞ ⎞
⎜ ⎜ ⎟ ⎟ ⎝ ⎝ ⎠ ⎠
= 0.0033 H
XL = 2π F L = 1.03 Ω
R = 0.5 x 3 = 1.5 Ω
2 2
L L Z = x + R = 1.82 Ω
2
2
1.82 x 100 x 100 = 150.4%
11
% line impedance = line impedance (Ω) x Base MVA x 100
Base kV
=
⎛ ⎞
⎜ ⎟
⎝ ⎠
Total % impedance up to factory = 150.4 + 28.5 = 178.97%
S/C MVA at 11 kV side at factory = (100 x 100)/ 178.97 = 55.87%
Considering the future expansion (say new substation) in the source side , S/C
MVA is taken as 250 MVA.
[Note: unless otherwise specified, a minimum fault level at 11 kV shall be
taken as 250 MVA]
250 x 103 Fault level at 11kV side = = 13.122 kA
3 x 11
% impedance corresponding to 250 MVA fault level = 100 x 100 = 40%
250
14
6 x 100 = 666.67%
0.9
% impedance of transformer at new base MVA ( ie,100 MVA) =
666.67
= 333.33%
2
since two transformers are in parallel the effective impedance =
Total % impedance up to 433 V bus = 40 + 333.33 = 373.33%
S/C MVA at 433 kV bus
100 x 100 = 26.78 MVA
373.33
26.78 x 103 = 35.71 kA
3 x 433
Corresponding fault current =
Earthing design:
Current density of copper – 118 A/mm2 (for 3 sec)
3
Size of conuctor at 11 kV side = 13.122 x 10 = 111.2 mm2
118
Nearest standard size = 25 x 6 mm cu strip
3
Size of conductor at MV side = 35.71 x 10 = 302.32 mm2
118
Nearest standard size = 63 x 6 mm cu strip
3
7.57 x 10 = 7570 = 309 A/ m2
ρ t 200 x 3
Permissible current density at electrode =
Plate electrode of 1.2 x 1.2 x 0.012m is used for earthing
Total area of both sides of plate electrode = 1.2 x 1.2 x 2 = 2.88 m2
3
Area required to dissipate fault at 11 kV side = 13.122 x 10 = 42.46 m2
309
Number of plate required = 42.46 = 14.74
2.88
Therefore 15 plate electrodes are to be provided.

Reply
Reply to Forum Thread 5 comments
Copy to Clipboard

Users who posted comments:

Anonymous Poster (1); asokmukherjee1@yahoo.co.i (1); dineshpandit80 (1); PWSlack (1); rajeev.nair (1)

Previous in Forum: AC/ DC Power Supply   Next in Forum: ACS700 ABB Drive

Advertisement