Grounding Isolated Transmission Lines FTL-LIN-GDN-001-M, Version 1.0 CCContro EXERCISES 1C Identification of Electrical Hazards 1. __C__ When a conductor is in a coil in an AC circuit, it creates a. an inductor 2. __D__ Which one of the following best completes this sentence? The current induced through capacitive induction from an adjacent energized line not carrying any load current is proportional to the d. A and B: voltage of energized line and the distance between the energized and isolated lines; length of the isolated line exposed to the electric field 3. __A__ How do you mitigate potential hazards from static induction? a. apply a bleed ground 4. What is the primary source of backfeed on a power system? d. onsite customer generation 5. __B__ How do you protect yourself in case system neutrals on distribution circuits may not be at ground potential? b. bond the system neutral into the work zone 6. How do you reduce the likelihood of accidental energization when working on isolated transmission lines that are crossing under or over another power line? c. inspect the crossing line to ensure its integrity, and then set up an equipotential zone 7. _FALSE_ Capacitive Induction can induce voltages on isolated lines. The voltage is never more than 10% of the energized line. 8. _TRUE_ When electromagnetic induction is present; it is proportional to the amount of current flowing in the adjacent energized line. 9. _FALSE_ The magnitude of magnetically inducted current in an isolated line is proportional to voltage of the adjacent energized line. 10. An isolated line in proximity to an energized line not carrying any load current will have a voltage induced on it through ___capacitive coupling___. Exercises 1C 51
Grounding Isolated Transmission Lines FTL-LIN-GDN-001-M, Version 1.0 CCContro APPENDIX B: CIRCUIT CALCULATIONS This section includes the calculations used in the scenarios in “Impact of Grounding in Parallel Circuits” in the Module 1 Manual. Circuit 1 – No Portable Protective Grounds Installed RC Conductor resistance 0.1 Ω RW Worker resistance 1000 Ω RP Pole resistance (below 2200 Ω workers foot to ground) RG Ground resistance 20 Ω The total resistance is: RTotal = RC + RW + RP + RG = 0.1 + 1000 + 2200 + 20 = 3220.1 Ohms The total current is: ITotal = V / RTotal = 80,000 / 3220.1 = 24.84 Amps Circuit voltages developed: VC = ITotal x RC = 24.84 x 0.1 = 2.484 V Voltage across the conductor VW= ITotal x RW Voltage across the worker = 24.84 x 1000 = 24,840 V Voltage across the pole VP = ITotal x RP = 24.84 x 2200 = 54,648 V Voltage across ground (GPR) VG= ITotal x RG = 24.84 x 20 = 497 V 52 Appendix B: Circuit Calculations
Grounding Isolated Transmission Lines FTL-LIN-GDN-001-M, Version 1.0 CCContro Circuit 2 – One Portable Protective Ground Installed RC Conductor resistance 0.1 Ω RW Worker resistance 1000 Ω R P Pole resistance (below workers foot to ground) 2200 Ω RJ1 Portable protective ground – around worker .005 Ω RG Ground resistance 20 Ω The total resistance is: RTotal = RC + ((RW x RJ1)/(RW + RJ1)) + RP + RG = 0.1 + ((1000 x 0.005)/(1000 + 0.005)) + 2200 + 20 = 0.1 + 0.005 + 2200 + 20 = 2220.105 ohms The total current is: ITotal = V / RTotal = 80,000 / 2220.105 = 36.03 Amps Circuit voltages developed: VC = ITotal x RC = 36.03 x 0.1 = 3.603 V Voltage across conductor Voltage across pole VP = ITotal x RP Ground potential rise = 36.03 x 2200 = 79,266 V Voltage across worker VG = ITotal x RG Current through worker = 36.03 x 20 = 721 V Current through portable jumper 1 VW = ITotal x ((R W x RJ1)/(R W + RJ1)) = 36.03 x ((1000 x .005)/(1000 + .005)) = 36.03 x 0.005 = 0.1802 V IW = VW / RW = .1802 /1000 = 0.1802 mA IJ1 = VJ1 / RJ1 = 0.1802 / 0.005 = 36.04 Amps Circuit 3 – Two Portable Protective Grounds Installed RC Conductor resistance 0.1 Ω RW Worker resistance 1000 Ω R P Pole resistance (below workers foot to ground) 2200 Ω RJ1 Portable protective ground – around worker .005 Ω RJ2 Portable protective ground – from pole band to .005 Ω ground rod Rrod Ground rod resistance 2Ω RG Ground resistance 20 Ω Appendix B: Circuit Calculations 53
Grounding Isolated Transmission Lines FTL-LIN-GDN-001-M, Version 1.0 CCContro Resistance of jumper 1 and worker resistance in parallel: R1 = ((RW x RJ1)/( RW + RJ1)) = ((1000 x 0.005)/(1000 + 0.005)) = 0.005 Ohms Resistance of jumper 2, ground rod, ground resistance in parallel with pole and ground resistance: R2 = (((RJ2 + Rrod + RG2) x (RP + RG1)) / ((RJ2 + Rrod + RG2) + (RP + RG1))) = (((0.005 + 2 + 20) x (2200 + 20)) / ((0.005 + 2+ 20) + (2200 + 20))) = ((48,851) / (2242.005)) = 21.789 Ohms Total resistance: RTotal = RC + R1+ R2 = 0.1 + 0.005 + 21.789 = 21.894 Ohms Total current: ITotal = V / RTotal = 80,000 / 21.894 = 3,653.96 Amps Circuit voltages developed: V R1 = ITotal x R1 = 3653.96 x 0.005 = 18.27 V Voltage across worker Current through worker IW = V R1 / RW Voltage of RJ2, Rrod,RG = 18.27 / 1000 = 18.27 mA Current at pole GPR at pole V R2 = ITotal x RG Current at ground rod = 3653.96 x 21.789 = 79,616 V GPR at ground rod IRP = VR2 / Rp = 79,616/ 2220 = 35.86 Amps VRG1 = IRP x RG = 35.86 x 20 = 717.2 V Irod = V R2 / (RG2 + Rrod + RJ2) = 79,616/ 22.005 = 3618.09 Amps Vrod = Irod x (Rrod + RG2) = 3618.09 x (2 +20) = 79,598 V 54 Appendix B: Circuit Calculations
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