Estimating Object-to-Ground Impedance

[lookintomyeyes]

Hi All,

I am looking at the preliminary line route for a new transmsision line, and there is a large building that may be subject to induced voltage due to its proximity to the ROW. I want to calculate the expected induced current to determine if it will be within allowable limits.

I've done some reading of Section 7.8 of the EPRI Red Book, which provides a calculation for the induced voltage and currents, but it requires a value for the "60Hz power frequency impedance of the object to ground".

Unfortunately, I'm a mechanical engineer, not an electrical one, so I have no idea how to find any information on the Z value for a metal-clad building that can be described as a "half cylinder on a ground plane".

Any help would be VERY appreciated, thanks!

 

[7anoter4]

I cannot find in EPRI Red Book in ch.7.8 where is your citation from .
By the way CH.7.8 ELECTRIC FIELD INDUCTION IN OBJECTS does not calculate the "flowing" current but the Electric Field level[static induction].

 

[lookintomyeyes]

Hi 7anoter4,

The Section I was reading was Pg 7-34, Section 7.8 "Electric Field Induction in Objects", from the EPRI Red Book 3rd edition. You might have a different version?

V_og = I_sc * Z_og

Where:
V_og = Voltage induced between object and ground
I_sc = current that would flow in a short circuit from the object to ground
Z_og = the power-frequency impedance of of the object to ground, measured at 60Hz (for North America)


My (perhaps poor) understanding is that since I have a large object (very large barn) parallel to an AC transmission line, a voltage will be induced on it. If the building is not already grounded properly, anyone/anything forming a path to ground will be at risk of a shock from the short circuit current that will flow when they connect the building to ground.

Ie, I'm hoping to calculate the expected magnitude of this induced current and associated voltage, to determine if its above the tolerable level.


Hope that clarifies things a bit?

 

[7anoter4]

I have indeed the third edition and I found this equation[ 7.8-2].
The impedance is actually a capacitive reactance
See -in the same ch.7.8.1 -:
"If the object is well insulated from ground, the impedance to ground is due only to this capacitance: Zog = 1/(2*pi*f*Cs)".
Now you have to calculate Cs
From Table 7.2-1 Electric Field for Simple Geometry
For Plate-to-plate Capacitance C=eps/H (F/m2) [That means one has to multiply by m^2 -surface area in order to get total C]
H=average heigth of the roof eps=8.85 pF/m [1/10^12 F/m]
Table 7.8-1 Equivalent Charge-Collecting Area and Spark-Discharge Capacitance of Objects in a Uniform Electric Field
indicates the collecting area of a half-cylinder above the ground as S=4*r*(L+r).
The Vog voltage depends on Transmission Line maximum voltage and the line geometry.
See also:
7.8.3 Accuracy Expected in Calculating Short-Circuit Currents

 

[waross]

The object will form a capacitive voltage divider. Two capacitors in series. One cap is the metal structure, the ground and the separation to the ground. The second capacitor is formed by the structure, the transmission line and the air separating them. The voltage will be the inverse of the values of the capacitors. The voltage in the vicinity of a very high voltage transmission line may be quite high but the current is low. A person making contact with the building may experience a shock similar to a static shock when the capacitor discharges through his body, but after the initial shock the charging current may not be detectable.
Recollections of working in close proximity to a 500kV transmission line.
We found that in most instances we could avoid the discharge shocks by bonding our wrists to the object being worked on with a piece of bare wire.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[lookintomyeyes]

Thanks, I'll check out those two equations!