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1 VFD Inrush

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7 Volts Stray Voltage Combined 1.0 Pri N-Ref Sec N-Ref Pri-Sec Cow Contact Tue Sep :50 10:51 10:52 10:53

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14 The Impact of Capacitors on Stray Voltage

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16 Amps Volts

17 Capacitance i = C dv/dt 17

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19 The current leads the voltage on a capacitor. In other words, the charge has to build up on the plates first (current first) before the voltage across the plates builds up. 19

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24 What Happens When A Single Capacitor Bank Fuse Blows

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29 Capacitance X(Ohms) = 29

30 300 kvar Bank 100 kvar per phase 100 kvar/7.2 kv = 13.9 Amps per phase Equivalent load on primary neutral as Amp (at 240 V) load boxes. Equivalent to 400 Amps of single phase farm load at 240 V

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32 Capacitor Bank Switching Transients

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36 The Fencer Circuit

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38 Parts of a fencer installation. Energizer Lead-out wire Fence Earth Return Rod

39 Electric fence installation and operation improper installation often leads to voltages on waterers and other contact points

40 Electrical Circuit Electricity needs to complete a circuit in order to flow and do the job it is required to do.

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45 Picture of a simple electrical Circuit. Lead-out Wire fence fencer earth

46 CAPACITANCE Capacitive Reactance (ohms) = XC = 1 / 2p fc Example: (C = 1 mf) F = frequency in hertz (Hz) C = capacitance in mfarads 60 Hz XC = Hz XC = Hz XC = Hz XC = 1.59 ohms 46

47 The current through (or into) a capacitor is zero if the voltage across it is not changing with time. A capacitor is thus an open circuit to dc. The higher the frequency, the more the capacitor looks like a short circuit. The fencer impulse, being a very fast transient, looks like a high frequency signal to the capacitance of the fence to ground. 47

48 Counterpoise Calculations

49 Vcc = K x Vsn-ref Assuming Vsn is approximately equal to Vpn

50 Vcc = K x Vpn-ref Vcc can be reduced by reducing Vpn-ref or K

51 Vpn-ref = It x Rt

52 Vcc = K x It x Rt

53 Modify the Circuit

54 Because V CC is a function of K, Itotal, and R T, if V CC is too high it can be reduced by: Reducing R T Reducing Itotal (and thus I PRI NEUTRAL and I SEC NEUTRAL ) Reducing K

55 How do Reduce In? Voltage conversion to a higher distribution voltage, i.e. 7200V to 14400V - P = VI, so for constant power V then I If the feeder is 3 phase, balancing the feeder in effect reduces I PN without changing I Reduce on farm load

56 How Do You Reduce K? Installation of an equipotential plane Ground Ring New Stalls Isolation

57 How Do You Reduce R T? R T is the parallel combination of R F and R PN, i.e. R T = (R F x R PN )/(R F + R PN ) Reduce R F, i.e. Improve on-farm grounding Reduce R PN -Install a larger neutral wire -Repair defective connections -Add grounding including deep grounding -Install counterpoise

58 DISTRIBUTION CONDUCTOR IMPEDANCES Conductor OHMS PER 1000 FT KCM ACSR /0 ACSR #2 ACSR #4 ACSR A CW /12 CW I STEEL 1.920

59 TYPICAL COUNTERPOISE INSTALLATION ALL RODS - 5/8 inch x 8 foot CU. counterpoise buried 24 inches deep. Nearest rod placed 10 ft. from pole.

60 Counterpoise Estimate Calculation Rp is resistance of ground system to reference. From the peak Vp from the overnight recording, calculate the farm equivalent load current that alone would raise Vp to that level (Vp/Rt=Ieq) Based upon existing primary neutral wire resistance, calculate length of neutral to equal measured Rp. Run counterpoise (1/0 Cu?) beneath power line that distance. Expected resultant Rp is the parallel combination of counterpoise with primary neutral for parallel distance. Expected Vp is the expected Rp x Ieq. Expected Vcc = Vp x K

61 Example

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63 Load Box Test Rt = 0.35 Ohms K = 43% Rp = 0.44 Ohms Vcc = 0.65 Volts (Peak Load Box) Rf = 1.77 Ohms

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65 Overnight Recording Vp (max) = 3.07 Volts Vcc (max) = 1.23 Volts (3.07 Volts x 0.43 = 1.32 Volts)

66 Calculations Primary neutral is #2 Al; Ohms/100 Rp = 0.44 Ohms 0.44 Ohms/0.027 x 100 = 1630 feet Vp(max) / Rt = 3.07 V/0.35 Ohms = 8.77 Amps (equivalent farm load to drive Vp to that level)

67 Calculations Continued 1/0 Cu counterpoise = 0.01 Ohms/ feet of 1/0 Cu = 0.16 Ohms 0.16 Ohms ǁ 0.35 Ohms = 0.11 Ohms 0.11 Ohms x 8.77 Amps = 0.96 (Vp predicted) 0.96 x K (43%) = 0.41 Volts (Vcc predicted)

68 Results!!!!!!!!

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70 Load Box Test Post Counterpoise Pre-Counterpoise Rt = 0.18 Ohms Rt = 0.35 Ohms Rf = 1.71 Ohms Rf = 1.77 Ohms Rp = 0.36 Ohms Rp = 0.44 Ohms K = 26% K = 43 %

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72 Overnight Recording Post Counterpoise Pre-Counterpoise Vp(max) = 1.4 Volts Vp(max) = 3.07 Volts Vcc(max) = 0.37 Volts Vcc(max) = 1.23 Volts (1.4 V x 0.26 = 0.37 V) (3.07 x 0.43 = 1.32 Volts)

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75 Example 2

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77 Load Box Test Rt = 0.43 Ohms K = 19% Rp = 0.54 Ohms Vcc = 0.66 Volts (Peak Load Box) Rf = 1.89 Ohms

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79 Overnight Recording Vp (max) = 2.77 Volts Vcc (max) = 0.66 Volts (2.77 Volts x 0.19 = 0.53 Volts)

80 Calculations Primary neutral is 1/0 ACSR; Ohms/100 Rp = 0.43 Ohms 0.43 Ohms/0.017 x 100 = 2530 feet Vp(max) / Rt = 6.44 Amps (equivalent farm load to drive Vp to that level)

81 Calculations Continued 1/0 Cu counterpoise = 0.01 Ohms/ feet of 1/0 Cu = Ohms Ohms ǁ 0.43 Ohms = 0.16 Ohms 0.16 Ohms x 6.44 Amps = 1.03 (Vp predicted) 1.03 x K (19%) = 0.19 Volts (Vcc predicted)

82 Results!!!!!!!!

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84 Load Box Test Post Counterpoise Pre-Counterpoise Rt = 0.24 Ohms Rt = 0.43 Ohms Rf = 1.48 Ohms Rf = 1.89 Ohms Rp = 0.25 Ohms Rp = 0.54 Ohms K = 15% K = 19 %

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86 Overnight Recording Post Counterpoise Pre-Counterpoise Vp(max) = 0.75 Volts Vp(max) = 2.77 Volts Vcc(max) = 0.12 Volts Vcc(max) = 0.66 Volts (0.75 V x 0.14 = 0.11 V) (2.77 x 0.19 = 0.53 Volts)

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Any wave shape can be reproduced by the sum of sine waves of the appropriate magnitude and frequency.

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