VFD Inrush
Volts Stray Voltage Combined 1.0 Pri N-Ref Sec N-Ref Pri-Sec Cow Contact 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 30 Tue Sep 2014 10:50 10:51 10:52 10:53
The Impact of Capacitors on Stray Voltage
Amps Volts
Capacitance i = C dv/dt 17
18
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
What Happens When A Single Capacitor Bank Fuse Blows
Capacitance X(Ohms) = 29
300 kvar Bank 100 kvar per phase 100 kvar/7.2 kv = 13.9 Amps per phase Equivalent load on primary neutral as 4-100 Amp (at 240 V) load boxes. Equivalent to 400 Amps of single phase farm load at 240 V
Capacitor Bank Switching Transients
The Fencer Circuit
Parts of a fencer installation. Energizer Lead-out wire Fence Earth Return Rod
Electric fence installation and operation improper installation often leads to voltages on waterers and other contact points
Electrical Circuit Electricity needs to complete a circuit in order to flow and do the job it is required to do.
Picture of a simple electrical Circuit. Lead-out Wire fence fencer earth
CAPACITANCE Capacitive Reactance (ohms) = XC = 1 / 2p fc Example: (C = 1 mf) F = frequency in hertz (Hz) C = capacitance in mfarads (mf) @ 60 Hz XC = 2650 ohms @ 1000 Hz XC = 159 ohms @ 10000 Hz XC = 15.9 ohms @ 100000 Hz XC = 1.59 ohms 46
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
Counterpoise Calculations
Vcc = K x Vsn-ref Assuming Vsn is approximately equal to Vpn
Vcc = K x Vpn-ref Vcc can be reduced by reducing Vpn-ref or K
Vpn-ref = It x Rt
Vcc = K x It x Rt
Modify the Circuit
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
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
How Do You Reduce K? Installation of an equipotential plane Ground Ring New Stalls Isolation
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
DISTRIBUTION CONDUCTOR IMPEDANCES Conductor OHMS PER 1000 FT 336.4 KCM ACSR 0.134 1/0 ACSR 0.232 #2 ACSR 0.312 #4 ACSR 0.454 8A CW 0.684 3/12 CW 1.400 6I STEEL 1.920
TYPICAL COUNTERPOISE INSTALLATION ALL RODS - 5/8 inch x 8 foot CU. counterpoise buried 24 inches deep. Nearest rod placed 10 ft. from pole.
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
Example
Load Box Test Rt = 0.35 Ohms K = 43% Rp = 0.44 Ohms Vcc = 0.65 Volts (Peak Load Box) Rf = 1.77 Ohms
Overnight Recording Vp (max) = 3.07 Volts Vcc (max) = 1.23 Volts (3.07 Volts x 0.43 = 1.32 Volts)
Calculations Primary neutral is #2 Al; 0.027 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)
Calculations Continued 1/0 Cu counterpoise = 0.01 Ohms/100 1600 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)
Results!!!!!!!!
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 %
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)
Example 2
Load Box Test Rt = 0.43 Ohms K = 19% Rp = 0.54 Ohms Vcc = 0.66 Volts (Peak Load Box) Rf = 1.89 Ohms
Overnight Recording Vp (max) = 2.77 Volts Vcc (max) = 0.66 Volts (2.77 Volts x 0.19 = 0.53 Volts)
Calculations Primary neutral is 1/0 ACSR; 0.017 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)
Calculations Continued 1/0 Cu counterpoise = 0.01 Ohms/100 2530 feet of 1/0 Cu = 0.253 Ohms 0.253 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)
Results!!!!!!!!
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 %
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)