Resistor Manufacturer

[james001]

I have ungrounded system and want to install the 59N relay to detect the ground fault. When I did my calc, I concluded that I need 6.7 Ohm, 6500W resistor and connect it on the secondary side of the open delta PT, 120Vac

Do you know where I can get this type of resistor? I tried Post Glover and they said they have to custom make that and it takes 2.5 weeks. I need something that size but on the shelf type, it's urgent..

Please help

Thx

jtt

 

[scottf]

james001-

Do you mean that you will be connecting the resistor across 3 - PTs with the secondaries connected in open delta, or across 1 - PT on an open delta system? If it's across 1 PT, are you sure your PT will handle 6500 W??

 

[james001]

Scott,

it is connected across 3PT's, there will be 3 PT connected wye primary-open delta secondary with resistor connected in between.

jtt

 

[dpc]

You might contact PowerOhm or Avtron. If you are going to be tripping to clear a detected ground fault, the power rating of the resistor does not need to be on a continuous basis. So you can probably get by with a resistor smaller than a 6.5 kW continuous rating.

 

[busbar]

The transformer configuration most often needed for this duty is referred to as grounded-wye/broken-delta. Be sure not to exceed the thermal-burden rating of the VTs, and if the accuracy burden is exceeded, they won’t be usable for other potential-measurement tasks.

GE used to use plain-vanilla banks of finned strip heaters like Chromalox or HotWatt

http://www.hotwatt.com/stripmfgf.htm

in series/parallel combinations. Another possibility is groups of aluminum-shelled stud-terminated Arcol resistors similar to

http://www.mouser.com//index.cfm?handler=fra_pdfset&pdffile=255

For either product, consider wiring using high-temperature crimp terminals and 250°C TGGT motor-lead wire.

 

[scottf]

busbar-

During steady-state operation, the resistor across the open delta will not impose a burden on the VTs. The VTs will only see a burden when there is an unbalance in one of the VTs secondary voltage.

This open delta configuration will obviously have to be done with a seperate secondary winding than the one being used for metering/normal protections.

 

[jbartos]

Suggestion: A suitable load bank with proper kW rating will do and leave you option to adjust Resistance around 6.7 Ohms. It may also have its cooling. The heater strips will vary their resistances with increasing temperature. It may be a good idea to compare characteristics related to the resistor temperature. Visit

http://www.nooutage.com/loadbanks.htm

etc. for more info

 

[busbar]

scottf — I agree that the resistor only dissipates heat [loading the PTs] when a neutral-shift condition takes place. {An example of this arrangement is

http://www.selinc.com/techpprs/6143.pdf

Figure 24,

http://www.selinc.com/datasheets/351s.pdf

Figure 5, or

http://pm.geindustrial.com/FAQ/Documents/SGC/GET-6497A.pdf

Page 26:

For many years, it has been and still is recommended that line-to-line rated PTs connected line-to-ground be used on high resistance grounded generators. This practice minimizes the possibility of ferro-non-linear resonance when the generator and transformer are connected as a unit. It should be noted it is possible to have a PT ferro-resonance problem with the unit generator transformer arrangement if the generator is disconnected and the PTs are left connected to the delta winding of the GSU transformer, which is then used to serve station auxiliary load. With the PTs connected to an ungrounded system, the possibility of ferro-non-linear resonance is almost a certainty. To suppress ferro-resonance for this operating condition, resistance loading should be applied across each phase of the secondary winding. Resistance loading equal to PT thermal rating may be required.

If the detection of ground faults is desired during this mode of operation… Connect a secondary winding in broken delta and use an IAV51 K across the broken delta.

My point is that PTs used for ground detection would not likely be satisfactory for, say, a watthour meter.

james — 6500W and 6.7Ω seems to correspond to about 31 volts. For 120V PTs with a solid ø-g fault, closer to 208V would be seen across the broken-delta connection. If you are in a serious pinch, what about using a string of incandescent lamps or about 9KW of “240V” baseboard heaters [sans thermostats]?

 

[jbartos]

Suggestion: Reference:
1. J. Lewis Blackburn "Protective Relaying Principles and Applications," Second Edition, Marcel Dekker, Inc., 1998,
Table 7.1 on page 199
Recommends, for example, for:
4.16kV Nominal Voltage System, VT ratio 4200:120V, Resistor R=125Ohms, 350Watts
which gives V**2=125 x 350=43750 and V=209V
The original posting indicates:
R=6.7Ohm and 6500Watts, or
V**2=6.7 x 6500=43550 and V=208.7V
So what is really new? Is there a resistor problem or a calculation problem or an empirical problem?

 

[busbar]

jbartos -- thank you for the corrected calcualtions.