Recent content by Fuse4

I agree with protoslash, with the time of 150 ms, verify that opening mechanism is equal in both recloser and has the good opening speed conditions to avoid miscordination (low bateries produce slow opening conditions). In primary tests include the time of the relay and the mechanism and is a...

For earth fault both neutrals of the transformers YNyn should be conected to ground

Dear all. I am reviewing this topic since time ago. Here the comments: 1. Obtain the Equivalent Circuit of the Ungrounded System: Is a series VLine, Zfault, C circuit VLine/V3 = Ifault *(Zfault + Xco/3) (Complex quantities) 2. For Bolted Fault (Zf = 0) The total capacitive charging current is...

For ungrounded systems with only 1 feeder is true that the ZCT (window transformer) do not see earth fault current because the current of the faulted phase is the sum of the other phases (opossite direction) and always the ZCT see zero amper. (IEEE papaer "Considerations for ground fault...

For ground faults, the delta system has the same behaviour that wye system with isolated neutral (the low voltage side).

Dear all I am triying write the title of an graphic and of the X and Y coordinates, if someone can give advice. Thanls.

I agree with HamburgerHelper, perhaps, you can use the negative sequence overcurrent to the high transformer side protection. Check appendix II Faults on the secondary of a delta - wye transformer in paper "Negative-Sequence Overcurrent Element. Application and Coordination in Distribution...

Dear Bloozntooz7868 In this Excellent book "Industrial Power System Grounding Design Handbook" by J.R. Dunki-Jacobs • F.J. Shields • Conrad St. Pierre, you can find about your question. Look the cover of the book in http://www.groundingdesignbook.com/ anf Fig. 12.2...

Jackreacher, was a mistake, thanks. You are with true. Here the values: Define: Vk" = 11 kV < 0° Sk" = SQRT(3)*Vk"*Ik" = 1.73*(11kV<0°)*(13,027kA<-86,24°)= 248,20MVA<-86,24° Zk" = Vk"*Vk"/Sk" = (Rk"+jXk")Ω= 0.488Ω<86.24°=(0.032+j0.486)Ω Rk" = 0.032 Ω Xk" = 0.886 Ω X/R = 15.216 %X =...

Define: Vk" = 11 kV < 0° Sk" = SQRT(3)*Vk"*Ik" = 1.73*(11kV<0°)*(13,027kA<-86,24°)= 248,20MVA<-86,24° Zk" = Vk"*Vk"/Sk" = (Rk"+jXk")Ω= 0.844Ω<86.24°=(0.055+j0.843)Ω Rk" = 0.055 Ω Xk" = 0.843 Ω X/R = 15.327 %X = (0.843 Ω / 0.844Ω)*100% = 99.882%

Dear tctctraining You can apply for free webinars about protection systems too. Here some that I know "2014 Protective Relay School webinar series". I will thank if someone know others. http://www.abb.com/industries/ap/db0003db004333/557dde83db2055c7c1257acb00690105.aspx Regards,

Sorry Here the atttachment. http://files.engineering.com/getfile.aspx?folder=61a8d628-d36b-4e52-acb3-6e43edd9c759&file=Inrush_Transformer_220_kV.docx

Dear all I have a question about inrush power transformer waveform. When we connected to the system the new bank power transformer in 220 kV without loas we have the following record in the relay, we can see the frecuency es different in two times. I will thanks your comments.

Dear Engineers I am looking about the "Selection and ordering data" of the Power Line Carrier Siemens brand equipments (Power Link - Transmission Power, ethernet port, remote management, teleprotection, others). I have two models and I do not know the details. I will thanks your answer...

The resistance depends of the transformers brand. In same time ago I saw a unit with a variable resistor. Here an example: Cahier technique n° 190 Ferroresonance schneider-electric.com...