The "Broken Delta Winding" and what we really measure!
ฝัง
- เผยแพร่เมื่อ 15 ก.ย. 2018
- What voltage do we measure at the broken delta winding?
There is only one correct answer. In the following video we derive the voltage and answer this seemingly trivial question.
Further informations: PROCONPLANET.COM
Thank you. This really helped me understand the broken delta voltage detector.
Very good. Thanks 🙏🙏🙏🙏🙏
One more hint from us: The three small blue vectors have to be rotated by 180 °. It is added from "arrow origin" to "tip". Since we are interested in the video only for the amounts of the vectors, it has for the derivation of our size no further influence. Best regards
Thanks u sir
love it !!, explain how polarization works in distance relays like negetive sequence , or ground polarization and which methods to use when
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Excellent presentation and explanation, thank you. To obtain 100V across the VT broken delta winding during a full on earth fault on the primary system what is the necessary voltage ratio of the VT?
Thank you for your feedback. The voltage ratio of the delta winding is Ur/root(3) / 100/3.
@@electricalengineeringacade5770 Thank you for the reply, which confirms my understanding and is exactly in agreement with what I was taught (too many years ago) should be the broken delta VT ratio specified for an isolated or high impedance earthed system. One suggestion to improve the presentation it is to state the VT ratios on the diagram that appears about 33 seconds into the presentation, even though such information may be seemingly clear and go without saying. Many times on very high impedance earthed systems I have seen the broken delta VT is 100/root(3) volts per phase. With this ratio the voltage across the broken delta during a full on earth fault becomes 173 V i.e. 3x the Vo voltage. All will still be well provided the voltage element of the connected relay is adequately rated (usually is) and that the protection engineer has calculated the correct voltage setting (occasionally not).
Excellent presentation. I have a following question. We have 25/6kV transformer and we are providing 59N at primary side addition to 51N. 25kV side ground current is limited to 600A. when both 51N and 59N detects earth fault trip will be issued to the breaker. I set 51N as 60A (10% of maximum ground fault current) and I set 30V for 59N. My doubt is, is 60A enough to generate 30V in broken delta side? do you have formula to explain this in detail. Thanks in advance!
Hey, that's not as easy as it may sound. The star treatment will determine wether the zero system voltage will be reliable or not. You can't make it up to the setting of the earth fault current. You should get a calculation tool (like Relay SimTest of Omicron) and calculate the entire case in detail. best regards and thank you for your post!
@@electricalengineeringacade5770
Team Thanks for your response. Yes we did analysis via software and found the setting current is adequate. I thought there will be some calculation. Thanks a lot.
I have a question about specifying a three-phase Y-broken delta grounding transformer that you might be able to help.
As the primary voltage rises from a root(3) factor during the phase to ground fault and considering that the system has to tolerate this fault for 10 seconds (adopting a system for 10 seconds) then the rated voltage on the primary side ( Y side) of the transformer must:
a) be equal to the root (3) x the supply system rated voltage
or
b) be equal to the rated voltage of the system because the grounding transformer already supports this overvoltage during those 10 seconds.
Thank you
a
Thanks
Hi everyone. Great video. i have some doubts regarding a system with the following characteristics:
Ground Resistance Transformer star-open delta (with resistor). Primary: 6,6kV / Secondary: unknown voltage.
Main network voltage is 11kV, we have a 11kV/6,6kV transformer to supply a switchboard 6,6kV that will supply 2 massive cranes. We also have a 6,6kV generator to supply the cranes in case of power failure. The idea is to avoid the installation having to swap from the network neutral to the generator neutral.
What should be the power rating for the GRT (Ground Resistance Transformer) ?
What should be the voltage ratio of the GRT ?
What should be the value of resistance in ohms for the resistance in the secondary side with the open triangle ?
What should be the desired short circuit current in the primary side neutral (of the GRT) ?
What should be the current ratio of the associated current transformer ?
Hope you can help. Regards to all
The value of the resistance is not that easy to determine. It's of course connected to the aim and the function of the resistance. There can be many user cases. The value of resistance is important for the needed power of the GRT. In summary we are talking about complex calculations, which cant' be done in a nutshell .... sorry bro
Hi, does the VT Primary Star Point have to be earthed/grounded in order for NVD to work?
It's about getting a fixed point of earth potential.
Great video , what if a high impedance or partial ground fault would occur on U-PG1 ? would the voltage rise to sqrt(3) U-PG2 and U-PG3 vectors include the potential difference from U-PG1 to earth as it is not 0 ? what would be the Ue or Uo value be in this case ? as an example, U-PG1 was reduced to 50% of its un-faulted value.
Following this question. Would like to know the answer to the question above.
Thank you for your feedback. No of course not. To achieve this conditions you need a real ground fault. Otherwise the described effects are smaller, depending of the size of the fault resistance ... warm wishes, Your EEA-Team
Hi great video! What is the phase voltage of a single winding , is it 100/3? This is the part that is confusing me! thanks
The single winding has 33,3 V in normal operation and 57,7 Volt during earth fault conditions. The angles of the fault-free phases differ about 30° each as well. Please look at the phasors, it's the best way to understand it. Thank you and kind regards!
@@electricalengineeringacade5770 ..Hi, thanks for that, but to be honest your video is a bit confusing because you never mention in it that the steady state secondary Ph-N voltage is 33.3 Volts, this is why your video is "different" from what is seen elsewhere....although at the end of the day we know that the result is the same, thanks again
The primary of the potential transformer is Star Solidly Earthed and secondary is Open Delta. My question: Since the Primary is solidly earthed, could we consider that this potential transformer is like an Earthing Transformer? Thank you
Can you explain your question closer and more detailed? Warm wishes!
What if you just lose one phase due to fuse filure. IMO then there is no shift in neutral. But we will stil meassure voltage from da dn, and it should be -Voltage of broken phase?
That's the huge difference in using the broken delta windings vs. the star windings in order to measure/calculate the zero sequence voltages. But in this case you also don't have to worry, modern relays come with fuse failure monitoring and are able to distinguish between real faults or secondary problems... warm wishes, your EEA Team
In normal condition what will be the voltage at open delta terminals.. i.e (da, dn).. ??
If in normal condition means in faulty-free operation condition the voltage at open delta is zero. In practice it never will be really zero, due to unbalanced phases you will always measure a small voltage for instance about 5 Volt.
@@alexandermuth7447... Thanks... Please uploaded more videos
why does it say everywhere that Uen = 3*UO?
It's because it's confusing for the most people. The relation to the star winding is the best you can do. But you are right: Be careful, even some vendors do it wrong ... kind regards!
I have a question. The phase/ground amplitude must be 33,3 V at the beginning. When a ground fault occurs, the vectors becomes 57,7 V. After that sum the vectors by 57,7 with 120 degrees and you have 100 volt. This means that U0 is 33,3 V, not 57,7 V. I'm I right??
Absolutely correct. U0 is 33,3, which means that Uen is actually 3*U0! This video is not correct.
@@tarekyared4404 Thx my friend
No U0 ist 57,7 Volt.
By the way: In case of an ground fault the angle shifts from 120 degrees to 60 degrees ;-) Warm wishes!
Ok o99 IP
I am sorry, but this video is not correct. The output of a broken-delta is 3U0. I have verified this result via circuit simulation and with White Papers from Schweitzer Engineering Laboratories.
For example, how can you say it is not 3 times the zero sequence voltage? Please consider trivial circuit analysis with proper polarity for a delta connection. Via trivial circuit analysis, the delta winding mesh voltage is Va + Vb + Vc. This trivial result is clearly 3*V0, by DEFINITION of V0. You do not need vector analysis. Please correct the video.
Thank you for your feedback! It is correct. I understand your argument, but this is the trap that a lot of engineers fall in, even SEL in this case. There can only be one secondary zero sequence voltage. Calculated on the star winding you get 57,7 V. This is U0. When U0 = 57,7 than we have to declare 100 Volt is the 1,7321 fold of U0. You have to see this in context. If you declare the delta windings to god, than you are right. But then you have two different secondery zero sequence voltages (57,7 V and 33,3 V) one for the star- and one for the delta winding. This is odd. Dont't do that. Otherwise you could say there are two different secondary phase to earth voltages, which also wouldn't make sense. I know, its hard to understand ;-) Warm wishes!