Makes total sense now! The step by step walkthrough is great. Showing the calculator also helps remove the final confusion when trying to figure out how you got the answer on the calculator when the math on paper made sense.
I calculated total MVA contribution by faster programming as ((1/1908)+ (1/978)+(1/4938.))^(-1) = 571.73MVA. Same 1.5kA total I-fault value. Great MVA problems to practice! With the same arrangement they sometimes will ask to calculate fault at the generator's bus which will result in using Tx high voltage instead of 220kV.
Short circuit analysis is one of the strongest topics on the PE exam. I'm glad that i found Zach's course. I can say: Hands down !!! i tried different sources before with no results. Zach really finds the way to make you understand the fundamentals behind each topic.. Thank you!
I've always used the per unit method to solve for fault current because I was never taught the MVA method. Zach does a great job showing how simple and useful the MVA method is. After watching this video, I'll be using the MVA method over the per unit method whenever I can.
These short videos are really well designed. They are just about the correct size and keeps me engaged. The MVA method will be the method of choice for me. Thank you for a job well done Zach!
I never know about the MVA method until I took this video. I always used the per unit method. Now that I know how to solve using MVA method, it is a lot easier than the per unit method. Thanks Zach !
Great job Zach. Fault analysis sounds like such a complicated topic, but it really boils down to doing it step by step. I really like how you show the calculator too. It's good to see some shortcuts!
the fault current video is a good one. It really went into detail of each type of calc and situation from parallel generators to series transformers really helping to understand fault current calcs. the video in the actual course adds in the PU calc comparison which helps understand both methods
I have been struggling to understand why reciprocal of power for series connections and simply adding power together for parallel connections... After looking at the solution for question #540 in the NCEES practice exam, I finally realized that we just need to calculate the impedance of each component, and add them up you would any impedance. Once you have the total impedance, simply use P=U^2/Z to get the power. I tried it for this example and it works like this: 1. Use 50MVA as base for all components. Use 13.8KV as base on the generator side of the transformer, and use 220KV as base on the load side of the transformer. Calculate the new impedance for all components using the new base to get: Zg1=50/55x5.6%=0.059 Zg2=0.054 Z transformer=50/45x4.6%=0.0511 Z load=(4+9j)/(220000)^2x50x10^6=0.0102 2. Calculate the total impedance: G1 and G2 are in parallel, so the total impedance for G1 and G2 is Zg1xZg2/(Zg1+Zg2)=0.0262 This impedance is in series with transformer and 4+9j, so the total impedance is 0.0262+0.0511+0.0102=0.0875 3. P=U^2/Z. Since voltage has a per unit value of 1, P=1/0.0875=11.4286pu. Base is 50MVA, so P=11.4286x50MVA=571.4286MVA. By calculate this way, it helps me to realize how the impedance of transformer actually restricts the fault current the generators can contribute. The method in the video is easy to follow, but from a theoretical standpoint, it creates the illusion that the transformer actually contributes to the fault current. I think many people think the latter way after reading the comment in these threads regarding question 540: engineerboards.com/threads/ncees-540.17353/ engineerboards.com/threads/problem-540-ncees.11835/ engineerboards.com/threads/ncees-power-practice-exam-540.34966/
Hi JS, thank you for your detailed comment. You are correct in that transformers absolutely restrict the fault current that generators can contribute. Transformers as usually referred to "current limiting devices" during fault conditions for this exact reason. A fun and quick exercise to perform on paper to help understand this is to work a fault current problem using the MVA method with one generator and one transformer using hypothetical values while increasing the MVA rating of the generator. If you work the problem enough times, you'll notice that it does not matter how large the generator is, at most only the fault duty of the transformer will be able to "pass through" from the primary side to the secondary side of the transformer.
wonderful examples on how to apply the MVA method. Also, very useful to see the calculator in this video. I had never used the fractions button or the %. now, i never don't use them!
Ah yes, the fraction and percent button. Huge time savers, especially when you only have an average of 6 minutes per problem on the exam. Glad you enjoyed it, Anthony.
Thanks for sharing this video. At the end, you mention that the same problem was going to be solved via the pu method. Could you please share the link tor that video? Thanks!
glad you enjoyed it. I'm not sure if that particular video is on our RUclips channel it most likely follows this video in our online course at www.electricalpereview.com
please I question The safe Fault Level for Planning of 11 kV systems is 95% of the design rating of older equipment which is rated at 250 MVA. Comment on whether or not this will be achieved with the new generator added to the existing network. If not, discuss measures that might be taken to reduce the fault level and propose a solution that will rectify the problem.
There are a lot of moving parts to the per unit system that takes time to really understand before being able to apply it easily. The MVA method is much, much, simpler and most of all fool proof!
Good day. I want to ask, why is that only MVA is considered rather than having the MW and the MVAR for the cable? there will be discrepancy if only MVA was considered unless you include the phase angle. thank you in advance.
Why is the line in parallel with the upstream? You turned the parallel xfmrs into a single equivalent and then went in series with the xfmr. But from there, why is the equivalent then in parallel with the bus/load fault location?
When using the MVA method, the fault duty in volt-amps of parallel devices sum up, while the fault duty of series devices add using the reciprocal of reciprocal sums method (I use the short hand of "//" in the video). The line is in series.
Awesome video. Seems like MVA method is significantly faster than per unit analysis method. Are there any disadvantages of the using the MVA rather than per unit analysis method?
No disadvantage what so ever as long as you understand both methods. If a question only gives you per unit values, and asks for a per unit answer however, then it's best to stay within the per unit system.
Hi Alexander, the MVA method (what this video teaches) is for balanced symmetrical faults. The type of faults in your comments are unsymetrical faults that are solved using symmetrical components instead of the MVA method. Here is a video on our RUclips channel solving a line to ground fault using symmetrical components: ruclips.net/video/kgkyUXRNPGo/видео.html&t And here is the supporting article for this video with additional written explanation and supporting diagrams and calculations: www.electricalpereview.com/symmetrical-components-single-line-ground-fault-electrical-pe-exam/ Hope this helps
Hi Joel, the reason we are not converting all impedances to the same base is because we are using the MVA method to solve for the short circuit current. If we were using the per unit method instead, then we would need to convert all impedances to the same base first.
Hi Rahul, you'll need to know the transformer percent impedance. Divide the transformer's full load amps by the percent impedance to calculate maximum fault current: |Isc| = FLA/%Z
4 + j9 ohms is the line impedance between the transformer and the faulted bus, it is a value that is given in this specific problem without having to calculate it.
What kind of fault would this be? What if they specified Line to ground, double line to ground etc at that fault point. How would you differentiate between them with the MVA method?
This would be for a worse case three-phase bolted fault. For unsymmetrical faults, instead of the MVA method (which is quick and simple), you have to approach it using symmetrical components for the specific fault type. Here is an example video I have on our RUclips Channel on how to solve a line to ground fault using symmetrical components: ruclips.net/video/kgkyUXRNPGo/видео.html
Makes total sense now! The step by step walkthrough is great. Showing the calculator also helps remove the final confusion when trying to figure out how you got the answer on the calculator when the math on paper made sense.
I calculated total MVA contribution by faster programming as ((1/1908)+ (1/978)+(1/4938.))^(-1) = 571.73MVA. Same 1.5kA total I-fault value. Great MVA problems to practice! With the same arrangement they sometimes will ask to calculate fault at the generator's bus which will result in using Tx high voltage instead of 220kV.
Short circuit analysis is one of the strongest topics on the PE exam. I'm glad that i found Zach's course. I can say: Hands down !!! i tried different sources before with no results. Zach really finds the way to make you understand the fundamentals behind each topic.. Thank you!
I've always used the per unit method to solve for fault current because I was never taught the MVA method. Zach does a great job showing how simple and useful the MVA method is. After watching this video, I'll be using the MVA method over the per unit method whenever I can.
These short videos are really well designed. They are just about the correct size and keeps me engaged. The MVA method will be the method of choice for me. Thank you for a job well done Zach!
I never know about the MVA method until I took this video. I always used the per unit method. Now that I know how to solve using MVA method, it is a lot easier than the per unit method. Thanks Zach !
Great Video, I love how Zach takes huge topics and breaks them down into small understandable concepts and procedures
Thank you!
Great job Zach. Fault analysis sounds like such a complicated topic, but it really boils down to doing it step by step. I really like how you show the calculator too. It's good to see some shortcuts!
This video really simplifies fault current analysis.
GREAT Explanation on MVA Method. Finally understood this approach after watching this video. Thanks Zach!!
Another topic that is much clearer to me. Thanks Zach.
the fault current video is a good one. It really went into detail of each type of calc and situation from parallel generators to series transformers really helping to understand fault current calcs. the video in the actual course adds in the PU calc comparison which helps understand both methods
I wish I had this in undergrad power courses. Saves time!!
The explanation of MVA method with an example was super useful. Thanks
You are most welcome Mini, glad to be of service.
Great video on fault analysis. Love the MVA method and your explanations! They make a tough topic easy to grasp.
Thank you! This is great. I like how you go step by step in such a way to make this very clear. I prefer this to the pu method.
I have been struggling to understand why reciprocal of power for series connections and simply adding power together for parallel connections... After looking at the solution for question #540 in the NCEES practice exam, I finally realized that we just need to calculate the impedance of each component, and add them up you would any impedance. Once you have the total impedance, simply use P=U^2/Z to get the power. I tried it for this example and it works like this:
1. Use 50MVA as base for all components. Use 13.8KV as base on the generator side of the transformer, and use 220KV as base on the load side of the transformer. Calculate the new impedance for all components using the new base to get:
Zg1=50/55x5.6%=0.059
Zg2=0.054
Z transformer=50/45x4.6%=0.0511
Z load=(4+9j)/(220000)^2x50x10^6=0.0102
2. Calculate the total impedance:
G1 and G2 are in parallel, so the total impedance for G1 and G2 is Zg1xZg2/(Zg1+Zg2)=0.0262
This impedance is in series with transformer and 4+9j, so the total impedance is 0.0262+0.0511+0.0102=0.0875
3. P=U^2/Z. Since voltage has a per unit value of 1, P=1/0.0875=11.4286pu. Base is 50MVA, so P=11.4286x50MVA=571.4286MVA.
By calculate this way, it helps me to realize how the impedance of transformer actually restricts the fault current the generators can contribute. The method in the video is easy to follow, but from a theoretical standpoint, it creates the illusion that the transformer actually contributes to the fault current. I think many people think the latter way after reading the comment in these threads regarding question 540:
engineerboards.com/threads/ncees-540.17353/
engineerboards.com/threads/problem-540-ncees.11835/
engineerboards.com/threads/ncees-power-practice-exam-540.34966/
Hi JS, thank you for your detailed comment.
You are correct in that transformers absolutely restrict the fault current that generators can contribute. Transformers as usually referred to "current limiting devices" during fault conditions for this exact reason. A fun and quick exercise to perform on paper to help understand this is to work a fault current problem using the MVA method with one generator and one transformer using hypothetical values while increasing the MVA rating of the generator. If you work the problem enough times, you'll notice that it does not matter how large the generator is, at most only the fault duty of the transformer will be able to "pass through" from the primary side to the secondary side of the transformer.
YOU ABSOLUTE LIFE SAVIOUR!
Glad you enjoyed it Jakub
wonderful examples on how to apply the MVA method. Also, very useful to see the calculator in this video. I had never used the fractions button or the %. now, i never don't use them!
Ah yes, the fraction and percent button. Huge time savers, especially when you only have an average of 6 minutes per problem on the exam. Glad you enjoyed it, Anthony.
Thank you. I have calculated as you recommend. That’s great. I have more clear.
Glad it was helpful.
Great Video Zach! MVA method is way easier!!!
Thanks for sharing this video. At the end, you mention that the same problem was going to be solved via the pu method. Could you please share the link tor that video? Thanks!
glad you enjoyed it. I'm not sure if that particular video is on our RUclips channel it most likely follows this video in our online course at www.electricalpereview.com
please I question The safe Fault Level for Planning of 11 kV systems is 95% of the design rating of older equipment which is rated at 250 MVA. Comment on whether or not this will be achieved with the new generator added to the existing network. If not, discuss measures that might be taken to reduce the fault level and propose a solution that will rectify the problem.
How would this be solved if the transmission line were 3 phase, and the phases had different impedances?
For some reason this method makes way more sense to me than the per unit method. I get almost every per unit question wrong
There are a lot of moving parts to the per unit system that takes time to really understand before being able to apply it easily. The MVA method is much, much, simpler and most of all fool proof!
Good day. I want to ask, why is that only MVA is considered rather than having the MW and the MVAR for the cable? there will be discrepancy if only MVA was considered unless you include the phase angle. thank you in advance.
Great video mate
Glad you enjoyed it.
Is it a 3-phase or a 1-phase fault?
I have a better understanding of the MVA method!
Why is the line in parallel with the upstream? You turned the parallel xfmrs into a single equivalent and then went in series with the xfmr. But from there, why is the equivalent then in parallel with the bus/load fault location?
When using the MVA method, the fault duty in volt-amps of parallel devices sum up, while the fault duty of series devices add using the reciprocal of reciprocal sums method (I use the short hand of "//" in the video). The line is in series.
Awesome video. Seems like MVA method is significantly faster than per unit analysis method. Are there any disadvantages of the using the MVA rather than per unit analysis method?
No disadvantage what so ever as long as you understand both methods. If a question only gives you per unit values, and asks for a per unit answer however, then it's best to stay within the per unit system.
Thank you very useful
You are welcome, glad you enjoyed.
Need an example for a line to ground fault, line to line fault, double line to ground fault.
Hi Alexander, the MVA method (what this video teaches) is for balanced symmetrical faults. The type of faults in your comments are unsymetrical faults that are solved using symmetrical components instead of the MVA method.
Here is a video on our RUclips channel solving a line to ground fault using symmetrical components:
ruclips.net/video/kgkyUXRNPGo/видео.html&t
And here is the supporting article for this video with additional written explanation and supporting diagrams and calculations:
www.electricalpereview.com/symmetrical-components-single-line-ground-fault-electrical-pe-exam/
Hope this helps
hey i am just learning about power system. Why did you not start by converting all of the Z's to the same base ?
Hi Joel, the reason we are not converting all impedances to the same base is because we are using the MVA method to solve for the short circuit current. If we were using the per unit method instead, then we would need to convert all impedances to the same base first.
I couldn't get the answer .4914
Sir.. Max. Flt level.. 7.663mva..
Vltge..415..
Hw to find mxm fault curreny.. Could u pls help.. 🙏🌸🌸
Hi Rahul, you'll need to know the transformer percent impedance. Divide the transformer's full load amps by the percent impedance to calculate maximum fault current:
|Isc| = FLA/%Z
Hi
How do you calculate. 4+j9.....
4 + j9 ohms is the line impedance between the transformer and the faulted bus, it is a value that is given in this specific problem without having to calculate it.
@@electricalpereview could u advise .how to get the answer 4914
@@muhdjamil4517 V^2/z=220kv^2/sqrt(4^2+9^2)
What kind of fault would this be? What if they specified Line to ground, double line to ground etc at that fault point. How would you differentiate between them with the MVA method?
This would be for a worse case three-phase bolted fault. For unsymmetrical faults, instead of the MVA method (which is quick and simple), you have to approach it using symmetrical components for the specific fault type.
Here is an example video I have on our RUclips Channel on how to solve a line to ground fault using symmetrical components: ruclips.net/video/kgkyUXRNPGo/видео.html