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About RalphChristie

  • Birthday 29/12/1975

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  1. I suppose your system is solidly earthed. Somewhere between the panel box and your source one of your wires is off, either the neutral or the earth, most probably the earth.
  2. GGOSS Hehehe.....sorry for that one. Anyway, thanks for the name, I was looking for information on some of their relays and could just remember it was "micro"-something. Regards Ralph
  3. Microelecttrica http://www.microelettrica.com
  4. See if this pdf-file can be of help: http://www.squared.com/us/products/nema.ns...FILE/M-379F.pdf
  5. BigMax Most probably way to late, but I saw this post only today... For anyone interrested in short-circuit effects on busbars check: http://www.cda.org.uk/megab2/elecapps/pub22/sec6.htm http://www.cda.org.uk/megab2/elecapps/pub22/sec8.htm Regards
  6. Bob: Residual earth fault protection: Uses three CTs and the earth fault relay is connected in the residual point of the three CTs. Since a residual current only exists when a fault current flows to earth, you can obtain a relative low setting. We use settings in the order of 5-10% FLA, but it will depends on the type of CTs, the burden of the relay on the CTs, leakage currents etc. See it many times with two or three O/C elements in the circuit, and the E/F relay (residual) connected between the star point of the relay group and the CTs. Need to coordinate with similar relays downstream. Sensitive Earth Fault Protection Normally used on systems where the earth fault current is limited to a very low value (Neutral earthed through resistors, Peterson coil, reactance, etc) Scheme consists out of a ring CT (Core Balance CT) around the three phase conductors, and the relay connected to the secondary side of the CT. A very low setting is possible, even sometimes lower then 1% FLA. Need to coordinate with similar relays downstream. Restricted Earth Fault Protection Is a type of zone of protection, or a differential scheme. You get two types, low impedance and high impedance. Low impedance is normally current operated and high impedance voltage operated. High impedance REF is the most commonly used. Relay connected in parallel with three phase CTs and one neutral CT. It will look just for faults inside the transformer, and need not to coordinate with relays downstream (can operate very fast). With this kind of protection you can protect over 90% transformer windings. Need to use special class CTs, previously called class X CTs. I am not really familiar with low impedance REF, but have seen it incorporated into some the new microprocessor relays. Standby earth fault protection Relay connected to the neutral CT. Normally a kind of back-up protection with a relative long delay or many times a long-time inverse curve. Need to coordinate with downstream E/F relays. Hope this helps Regards Ralph
  7. Just an extra comment: Overcurrent Protection is not the same as Overload Protection. There is a general misconception that an IDMT O/C relay is there and to be set to cover overloading of a transformer. If the thermal characteristic of a transformer is plotted against the normal inverse curve, (which is on log-log graph scale) it will be seen as a straight line crossing the IDMT characteristic at some point. For small overloads the relay will trip before the transformer heats up to its limit. For sudden heavy overloads the transformer will cook before the relay trips. The NI IDMT-relay is therefore NOT suitable for overload duty - it is for fault protection. Some degree of overload protection may be obtained with these curves. For overload monitoring of a transformer use: 1.) a relay with a suitable thermal characteristic. 2.) an EI characteristic in the IDMT range, the time being approx inverse proportional to the square of the current. 3.) Winding temp protection.
  8. chaterpillar: You have to have coordination between your relays downstream, the transformer relay, and the relays upstream. You do not want your main breaker to trip for a fault at a downstream motor's terminals. Therefore you have to do a coordination study to ensure that your downstream relays will trip before your upstream relays. That is called protective coordination. Thus, you have to set the transformer's secondary feeder on such a way that it does not trip before a downstream relay. If the downstream relay do not clear the fault for any reason, then the upstream relay has to operate. Now for the relay curves: You can use either a direct time or a IDMTL-curve (Inverse definite minimum time lag) For a direct time: your relay will operate in the set time if the set current is exceeded. For the IDMT-curves: The bigger the current into a relay, the faster will the relay operate. There are a lot of curves in use, but the most common are: IEC-world: standard/normal inverse very inverse extremely inverse long-time inverse short-time inverse ANSI/IEEE-world Moderately inverse inverse very inverse extremely inverse short-time inverse etc. The IEC and ANSI-curves are not the same, but uses the same principle. These curves have two setpoints: Current and Time Note that it is not a specific time, but a time-multiplier. The triptime of a relay would be dependant on the current factor (fault-current/set-current) and the time multiplier. Download the NPAG-Book at Areva's site (free download) and read especially through chapter 9 (Overcurrent and earth fault protection): http://www.areva-td.com/servlet/ContentSer...089880316992&am p;rid=&lid=en&pid=1017999014820&tab=Chapters&id=1056536208254 (copy and paste whole address) A few comments on some of your settings: For a infinite bus, your max. three phase short circuit current will be almost 7kA. Assuming you are using a standard/normal inverse curve from the IEC-world: Your O/C setting of 330A is almost 8MVA. That is where your relay will begin to pick-up. The time multiplier is set on 0.3 (not 300msec - it is a factor, not a time) For a fault of 7kA your relay will trip theoretically in 0.67sec For a fault of 1kA your relay will trip theoretically in 1.87sec Your O/C instantaneous element will never operate, because you will never see such a high current. For your earth-faults (remember with a solid earth your earth-fault currents can be very high) For any earth-fault above 40A your relay will operate in 0.1second due to the fact that you use a definite time setting Hope this will help you Regards Ralph
  9. chaterpillar: This is a very difficult question to answer. All your settings would depend on how many levels of protection you have downstream, and how much time you are allowed upstream. It would also be dependant on the kind of relays you are using (electromechanical, electronic, etc) and what kind of protective schemes you are using on the transformer (REF, differential, etc) Normally your residual O/C and E/F curves would be Normal Inverse Curves. You E/F settings would depend on your transformer earthing arrangement. (solidly, high/low resistance, reactance, ungrounded, etc.) Without your total system lay-out it would be impossible to give any setting recommendations. Regards Ralph
  10. I am not sure what you are asking, but the X/R-ratio determines the peak asymmetrical fault current in a power system. See also: http://www.ecmweb.com/mag/electric_beware_...mplistic_fault/ Regards Ralph
  11. Tree winding transformer is a transformer with three windings per phase, thus like in a single trsf with a High Voltage, Medium Voltage and Low Voltage winding. Most common reason for the addition of a third winding is for the provision of a delta-connected tertiary winding. (connection in delta to provide a low-impedance path for third-harmonic currents) Other reasons: To limit the fault level on the LV system by subdividing the infeed - thus double secondary transformers. Interconnection of several power systems operating at different supply voltages.
  12. Just a few points: Regarding the short-circuit current of 30kA (330MVA) This is teoretical the highest current (seen from the secondary side of the trsf) your transformer will provide during a 3ph short on the secondary bushings with an infinite bus. Under real conditions your source impedance will also have an effect on the short circuit current, and the actual current will be lower. How much lower depends on your source. Short Circuit Currents produce very high mechanical stresses in the equipment through which they flow, these stresses being proportional to the square of the currents. Thus, the longer the fault duration, the bigger the damage. You have to remove the faulted part as fast as possible. (A trsf of this size is not manufactured in a week, it can take up to a year's time.) Normally there is a max time in which equipment can handle SCC, say 30kA for 3sec. (just an example) Equipment manufactured a few years ago were much more robust than equipment manufactured today, and normally they will withstand this current longer. Remember, everytime this current flows, you are reducing the life of the transformer. The time this current will flow depends also on your protective devices. Some kind of devices, like differential relays do not have to coordinate with downstream devices, and can take the faulted part out of the system in a few cycles. Other devices, like IDMT O/C and E/F relays have to coordinate with downstream and upstream devices, and can take longer to remove the faulted part. For more on protective relays you can check the IEEE Red Book. hope this advice is going to help you Regards Ralph
  13. No. You can: 1. Obtain the data on the test-report of the transformer, or 2. Contact the manufacturer and ask for the data, or 3. Test the transformer yourself to obtain the data, or 4. As a last resort assume the X/R ratio. Regards Ralph
  14. The trsf X/R ratio is not on the nameplate, but can be determined from the transformer test sheet or transformer losses, if you've got it. %R = ((Total watt loss - No-load watt loss) x 100) / transformer rating in volt-amperes X/R = [(%Z^2 - %R^2)/(%R^2)^0.5 As a last resort you cam assume your X/R ratio. Range of X/R ratio's (from IEEE Std C37.010-1979) shows for a 30MVA rating something between 16 and 50 (average of 33) Yes, you are correct. (MVA SC = sqrt3 * SCA * V) At 30MVA base and 6.3kV base MVA SC would be 330MVA
  15. Sorry, should read: O - Mineral oil or synthetic insulating liquid with a fire point equal or lower than 300°C
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