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Hi all,
Currently i'm doing earthing calculation based on BS 7430. our plant has limited area, only 18m x 22m. soil resistivity is 244 ohm-meter.
according user requirement, resistance shall be max 4 ohm.
after calculate with the hollow square, 9 rods, i just got 5,565 ohms.
please advise how to get resistance below 4 ohms?
your response is highly appreciated.
thank you
Hi All,
do you ever try to combine hollow square and line configuration?
please share if you have experience on this.
thank you
Hi All,
do you ever try to combine hollow square and line configuration?
please share if you have experience on this.
thank you
I did a simulation with your parameters received 5,67Ohms, for 8pcs of 3m copper bound steel rods, in hollow square configuration, after adding another cross-connection copper wire (all wire with 10mm dia., around 150mm2).
I tried double square configuration, keep squares 3m distance one from another (square in square), add more rods, total 15. Still, resistance too high : 5.29Ohms.
You need to improve earth with the active electrodes (with salt), etc, which is beyond our calc, and should be done by supplier of such electrodes.
Cheers
I did a simulation with your parameters received 5,67Ohms, for 8pcs of 3m copper bound steel rods, in hollow square configuration, after adding another cross-connection copper wire (all wire with 10mm dia., around 150mm2).
I tried double square configuration, keep squares 3m distance one from another (square in square), add more rods, total 15. Still, resistance too high : 5.29Ohms.
You need to improve earth with the active electrodes (with salt), etc, which is beyond our calc, and should be done by supplier of such electrodes.
Cheers
Thank for your response.
please see below picture for my configuration.
[ATTACH=CONFIG]4235[/ATTACH]
my calculation :
R=Ï/2Ï€L [log_e [8L/d]-1]
Where
R : the resistance of rod or pipe, in ohms (Ω)
L : the length of the rod or pipe, in meters (m) = 15 m
d : the diameter of the rod or pipe, in meters (m) = 0.0603 m
Ï : the soil resistivity in ohm meters (Ω-m) = 244 ohm-m
Substituting the values.
R=244/(2Ï€×15) [log_e [(8×15)/0.0603]-1]
R=2.5889 [7.5959-1] Ω
R=17.076 Ω
According to BS 7430, Resistance of vertical parallel:
R_n=R[(1+λa)/n] => a=Ï/2Ï€Rs
Where
Rn : the resistance of multiple electrodes
R : the resistance of one electrode in isolation, in ohms (Ω) = 17.076 Ω
s : the distance between rods, in meters (m) = 8.25 m
Ï : is the resistivity of the soil, in ohm meters (Ω-m) = 244 Ω-m
λ : A factor for electrodes
: an hollow square = 7.65 (refer to BS 7430 table 2)
: parallel in a line = 1.66 (refer to BS 7430 table 4)
n : no. of electrodes arrange
The resistance of rod electrodes in hollow square R1:
a=244/(2×Ï€×17.076×8.25)
a=0.252686
R1=17.076[(1+7.65×0.275657)/9]
R1=5.898 ohm
please whether my calculation is correct or not.
actually in BS7430:2011 page 93, give an example for combination between hollow square and line configuration to reduce the resistance. but i'm still not sure since some value of variables are wrong.
any suggestion?
thank you
Thank for your response.
please see below picture for my configuration.
[ATTACH=CONFIG]4235[/ATTACH]
my calculation :
R=ρ/2πL [log_e [8L/d]-1]
Where
R : the resistance of rod or pipe, in ohms (Ω)
L : the length of the rod or pipe, in meters (m) = 15 m
d : the diameter of the rod or pipe, in meters (m) = 0.0603 m
ρ : the soil resistivity in ohm meters (Ω-m) = 244 ohm-m
Substituting the values.
R=244/(2π×15) [log_e [(8×15)/0.0603]-1]
R=2.5889 [7.5959-1] Ω
R=17.076 Ω
According to BS 7430, Resistance of vertical parallel:
R_n=R[(1+λa)/n] => a=ρ/2πRs
Where
Rn : the resistance of multiple electrodes
R : the resistance of one electrode in isolation, in ohms (Ω) = 17.076 Ω
s : the distance between rods, in meters (m) = 8.25 m
ρ : is the resistivity of the soil, in ohm meters (Ω-m) = 244 Ω-m
λ : A factor for electrodes
: an hollow square = 7.65 (refer to BS 7430 table 2)
: parallel in a line = 1.66 (refer to BS 7430 table 4)
n : no. of electrodes arrange
The resistance of rod electrodes in hollow square R1:
a=244/(2×π×17.076×8.25)
a=0.252686
R1=17.076[(1+7.65×0.275657)/9]
R1=5.898 ohm
please whether my calculation is correct or not.
actually in BS7430:2011 page 93, give an example for combination between hollow square and line configuration to reduce the resistance. but i'm still not sure since some value of variables are wrong.
any suggestion?
thank you