Good morning engineer, I would like to ask you a question about this exercise, I am asked how to calculate the three phase short circuit currents in bus 2, with fault impedances of 3 Ω, this would be posed with the same formula of Isc= Ip.u. x I line (Iline=Sbnew/sqrt(3) Vb(zone)), I fail to understand what would be referred to those 3Ω fault impedances. it is the same exercise only that my professor adds that concept. for example, how do you get a short circuit current and a fault current is it the same thing? thanks a lot !
Hi, yes Fault current and Short circuit current are the same thing, they both describe a condition where the current exceed the rated or nominal value. The 3Ω fault impedance is the actual impedance of the line or element during the fault, so in this case this is Z(actual). In this problem Z actual is j50Ω, and in your case its j3Ω. You can ask your lecturer for further clarification.
Good day, I noticed when solving for the Xpu(new) of generator G1 the Kvb old and Kvb new was not considered, unlike the formula you stated, could there be a reason for this?
Hi Mathias, are you enquiring about Zpu(new) = 0.4jpu? If yes, the values of Vb(old) and Vb(new) where actually used in the formally but they did not affect the answer because they canceled out. Vb(old) = 13.8kV and Vb(new) = 13.8kV for G1 and T1 calculation. I hope that clarifies your concern.
Hi Mathias, actually Vb(old) and Vb(new) are the same for G1, please watch my earlier video where I only focused on per unit, this video is more about fault calculation. Sb(new) = 1000MVA, yes it is true the problem statement states that Vb(new)=20kv which actually only apply to G3 zone. You can not impose a base voltage for the entire system, each zone have its own voltage as you will have Step up and step down transformers along your transmission line. So, on G1 zone Vb(old) is 13.8kV and Vb(new) =13.8kV you cannot inject another voltage there unless you change the Generator design and its output voltage.
Your videos are really helpful. Can you please do a video on fault current analysis when the system is loaded. I learned that all what you solving is an unloaded system.
Hi, thank you for your input, if you are refering to load flow analysis for power system planning, I have those tutorial in the pipeline, stay tuned. If you want me to assist you with some specific, feel free to contact me privately, or Join my channel membership that way I can release a video for your need.
Hi, thank you for spotting the typo. The parameters for G2 should be : 750MVA, 18kV, Xpu:0.18jpu. Generator 2 voltage must match the low voltage side of the transformer to maintain the same base voltage on that section, so Vbase = 18kV from generator not 13.8kV. Let me know if there's amything else you need clarification.
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Good morning engineer, I would like to ask you a question about this exercise, I am asked how to calculate the three phase short circuit currents in bus 2, with fault impedances of 3 Ω, this would be posed with the same formula of Isc= Ip.u. x I line (Iline=Sbnew/sqrt(3) Vb(zone)), I fail to understand what would be referred to those 3Ω fault impedances. it is the same exercise only that my professor adds that concept. for example, how do you get a short circuit current and a fault current is it the same thing? thanks a lot !
Hi, yes Fault current and Short circuit current are the same thing, they both describe a condition where the current exceed the rated or nominal value.
The 3Ω fault impedance is the actual impedance of the line or element during the fault, so in this case this is Z(actual). In this problem Z actual is j50Ω, and in your case its j3Ω.
You can ask your lecturer for further clarification.
Good day, I noticed when solving for the Xpu(new) of generator G1 the Kvb old and Kvb new was not considered, unlike the formula you stated, could there be a reason for this?
Hi Mathias, are you enquiring about Zpu(new) = 0.4jpu?
If yes, the values of Vb(old) and Vb(new) where actually used in the formally but they did not affect the answer because they canceled out.
Vb(old) = 13.8kV and Vb(new) = 13.8kV for G1 and T1 calculation.
I hope that clarifies your concern.
@@CMTEQ thanks for the timely response, it doesn't really clarify because the Vb (old ) is 13.8 and Vb (new) is 20 as stated, so they don't cancel out
Hi Mathias, actually Vb(old) and Vb(new) are the same for G1, please watch my earlier video where I only focused on per unit, this video is more about fault calculation.
Sb(new) = 1000MVA, yes it is true the problem statement states that Vb(new)=20kv which actually only apply to G3 zone.
You can not impose a base voltage for the entire system, each zone have its own voltage as you will have Step up and step down transformers along your transmission line.
So, on G1 zone Vb(old) is 13.8kV and Vb(new) =13.8kV you cannot inject another voltage there unless you change the Generator design and its output voltage.
@@CMTEQ thanks for the clarification, I'll watch more of your videos for better understanding.
Very nice video. Thx for your help.
Thank you, I'm glad it was helpful. Stay tuned for more power systems tutorials!
you are brilliant salut
Thank you so much 😀
@@CMTEQ I have a test on those topic and your videos have been helping a lot
Your videos are really helpful. Can you please do a video on fault current analysis when the system is loaded. I learned that all what you solving is an unloaded system.
Hi, thank you for your input, if you are refering to load flow analysis for power system planning, I have those tutorial in the pipeline, stay tuned.
If you want me to assist you with some specific, feel free to contact me privately, or Join my channel membership that way I can release a video for your need.
Awesome video engineer 👍👋
Great stuff 👌, I'm glad I was useful
Good Video, but I have a doubt: G2 is 13.8 kV and T2 is 18/500, if Vbase = 18, then Xg2 would be 0.141, I do not understand why you got 0.24?
Hi, thank you for spotting the typo.
The parameters for G2 should be : 750MVA, 18kV, Xpu:0.18jpu.
Generator 2 voltage must match the low voltage side of the transformer to maintain the same base voltage on that section, so Vbase = 18kV from generator not 13.8kV.
Let me know if there's amything else you need clarification.
@@CMTEQ thank you
You are welcome 😊