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I find it great, straight forward to the logical explanation of getting the answer. We thank u for this effort and keeping this unselfish knowledge sharing you made sir...
Excellent explanation! Just a thing: for Transformer 2, when calculating Zpu (New), the rate Vbo and Vbn is mistaken. It should be 800/100 instead of 1000/800.
Hi, thank you for your feedback, there was typo in the statement, I have replied in a similar comment below. The correct ration is 1000/800, because Vbo = 1000V and Vbn = 800V. The calculation is correct. Thank you !
Hi Joseph, thank you for your input, there was a typo in the tutorial but the final answer is correct, if you check in the comments below I have already answered a similar observation that was raised.
Hi John, thanks for your input, that was a clear typo, at the time of the recording I did not have a good input mechanism, I was doing a lot of copy paste with formula to avoid writing them. So, in this case Vbo = 1000V & Vbn = 800V as shown in the calculation. Given transformer values are always considered as old, 1000V is a primary old value for T2 and 800V is coming from T1 and this is now considered new value to T2 primary. There's a transfer function equation normally used for that, I will be uploading more tutorials on this topic. I hope that clarifies things.
@@CMTEQ I just got confused because my professor's lecture and the book, I think, have exactly the same thing as yours. Anyway, thank you! Keep on creating!
Hi Davies, I do not have that tutorial yet, hopefully I will get to it sooner. If you have any question on the topic (conductor spacing, inductance and capacitance between them) don't hesitate to ask.
Hi there, thank you for your comment, that's how the PF is converted into degrees angle, you however don't need to include 180°/π , as π=180 in degree so the fraction will simplify to 1. So, θ=arccosine(08) = 36.87°
Hi @Deltech, check my previous reply above. If you have a scientific calculator you can just enter the following: θ=arccosine(08) = 36.87° to get your power factor in Degrees.
thanks sir but am bit confused how do you use 800 as a base voltage on the line one while is seems like it supposed to be 1000v as a voltage on a primary side of a second transform according to the books that how the did it.
Hi, thanks for the engagement, I believe you talking about the calculation of Zpu(Transformer2) @ 7:30, someone pointed out the same question, it was a typo between Vbase(old) and Vbase(new) for transformer 2 when recording the video, but if you look at the answer and the formula the values are correct and the answer is correct, I just swapped them when explaining. Let me know if that clear the cloud.
Hi Andile, that was a typo, at the time of the recording I did not have a good input mechanism, I was doing a lot of copy paste with formula to avoid writing them. So, in this case Vbo = 1000V & Vbn = 800V as shown in the calculation. Given transformer values are always considered as old, 1000V is a primary old value for T2 and 800V is coming from T1 and this is now considered new value to T2 primary. There's a transfer function equation normally used for that, I will be uploading more tutorials on this topic. I hope that clarifies things.
Yes that is correct, this impedance is derived from the formula S=V*I =>(S is apparent power). We know that impedance causes a voltage drop which is given by the formula V=I*Z => I = V/Z. Replacing I in the S formula gives S=V*(V/Z) = V^2/Z. Then Z = V^2/S. This formula is important because its takes account of the Load impedance which will tell us what the value of the power factor is, whether its a Lagging, Leading or Unit Power factor. That power factor is needed to find the Impedance angle when calculating Zb. The Load apparent power is S=P/cosθ, as u can see when Pf is 1, S=P, meaning there's no loss, power is being delivered to the system. You may do some reading on the power triangle. I can also maybe make a Tutorial on power triangle.
Thank you for your comment, please consider to subscribe as there will be more tutorial on fault calculations. Watch this tutorial it will help you understand better. Part 1: th-cam.com/video/tGtu4fyg1fM/w-d-xo.html Part 2: th-cam.com/video/9YDAJj-ECus/w-d-xo.html
@@CMTEQ thank you dear. I have designed a DC power connected to Ac grid through inverter and LCL filter in PSCAD. Now I want to convert the whole system to per unit system. How to deal with LCL filter.? Thank you in anticipation
Very good question, the purpose of this exercise is precisely to ensure protection of your equipment and personal in real life and also ensure continuity of supply. All you have to do is to read your equipment name plate and draw a per unit diagram & find the possible fault Current and any location.
Hi Fati, thank you for the comment there's a typo on the Values of Vb(old) and Vb(new), the values are swapped between them, but in the formula they are used correctly, the answer for Zpu(T2) is correct. The base voltage on the load side is also correct, as Zone2 Vbase = 800 this give then Vbase(Load)=400V. Let me know if this clarify your concern.
Salut Fati, Merci pour votre commentaire, il y une erreur de typo entre les valeurs de Vb(new) et Vb(old), les valeurs ont ete echangees entre eux, la reponse de Zpu(T2) est correcte. La tension de base du cote de la charge est aussi correct a 400 Volts, par ce que la tension de la Zone2 est Vb = 800V, cela done Vb(Load)=400V. Dites moi si cela clarifie votre souci. Considerez aussi vous subscrire merci.
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Typo @5:55 The correct values are: Vbase(old) = 1000V and Vbase(new) = 800V
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I realize I'm kinda off topic but does anybody know of a good site to watch newly released tv shows online?
You Google search that, you can also subscribe to this channel for NEW Tutorial every week since you looking for new SHOW.
I find it great, straight forward to the logical explanation of getting the answer. We thank u for this effort and keeping this unselfish knowledge sharing you made sir...
You're very welcome, I'm passionate about sharing this knowledge.
Excellent explanation! Just a thing: for Transformer 2, when calculating Zpu (New), the rate Vbo and Vbn is mistaken. It should be 800/100 instead of 1000/800.
Hi, thank you for your feedback, there was typo in the statement, I have replied in a similar comment below. The correct ration is 1000/800, because Vbo = 1000V and Vbn = 800V.
The calculation is correct.
Thank you !
V base new (in zone 2 )= Vbase old X Transformer ratio
= 250 x (800/250)= 800
So at T2
Vbase new= Vbase old x Transformer ratio
= 800(500/1000)= 400
Hi Joseph, thank you for your input, there was a typo in the tutorial but the final answer is correct, if you check in the comments below I have already answered a similar observation that was raised.
For the Zpu of transformer 2, why was 1000V set as the Vbo value in the formula when it was declared that Vbo is 800V in the given?
Hi John, thanks for your input, that was a clear typo, at the time of the recording I did not have a good input mechanism, I was doing a lot of copy paste with formula to avoid writing them.
So, in this case Vbo = 1000V & Vbn = 800V as shown in the calculation.
Given transformer values are always considered as old, 1000V is a primary old value for T2 and 800V is coming from T1 and this is now considered new value to T2 primary.
There's a transfer function equation normally used for that, I will be uploading more tutorials on this topic.
I hope that clarifies things.
@@CMTEQ I just got confused because my professor's lecture and the book, I think, have exactly the same thing as yours. Anyway, thank you! Keep on creating!
I was confused there
I hope you got your answers
youre a great teacher its been well understood!!!!
Thank you, I'm glad it was helpful !
Do you have a video on Three phase Composite conductor.?
Hi Davies, I do not have that tutorial yet, hopefully I will get to it sooner. If you have any question on the topic (conductor spacing, inductance and capacitance between them) don't hesitate to ask.
For those wondering: the p.f. Conversion to degrees is angle = arccos(p.f.) * 180°/π
Hi there, thank you for your comment, that's how the PF is converted into degrees angle, you however don't need to include 180°/π , as π=180 in degree so the fraction will simplify to 1.
So, θ=arccosine(08) = 36.87°
@@CMTEQ can you tell me how you convert the power factor into angle
Hi @Deltech, check my previous reply above.
If you have a scientific calculator you can just enter the following: θ=arccosine(08) = 36.87° to get your power factor in Degrees.
thanks sir but am bit confused how do you use 800 as a base voltage on the line one while is seems like it supposed to be 1000v as a voltage on a primary side of a second transform according to the books that how the did it.
Hi, thanks for the engagement, I believe you talking about the calculation of Zpu(Transformer2) @ 7:30, someone pointed out the same question, it was a typo between Vbase(old) and Vbase(new) for transformer 2 when recording the video, but if you look at the answer and the formula the values are correct and the answer is correct, I just swapped them when explaining.
Let me know if that clear the cloud.
Thank you so much. Very detailed explanation.
Its a pleasure !
Sir there is a something i don't get on Transformer 2 Vbo is 1000V or is 800V
Hi Andile, that was a typo, at the time of the recording I did not have a good input mechanism, I was doing a lot of copy paste with formula to avoid writing them.
So, in this case Vbo = 1000V & Vbn = 800V as shown in the calculation.
Given transformer values are always considered as old, 1000V is a primary old value for T2 and 800V is coming from T1 and this is now considered new value to T2 primary.
There's a transfer function equation normally used for that, I will be uploading more tutorials on this topic.
I hope that clarifies things.
@@CMTEQ thank you sir
So, the base impedance on the load side is calculated using the load apparent power?
What is the load apparent power?
Yes that is correct, this impedance is derived from the formula S=V*I =>(S is apparent power).
We know that impedance causes a voltage drop which is given by the formula V=I*Z => I = V/Z.
Replacing I in the S formula gives S=V*(V/Z) = V^2/Z.
Then Z = V^2/S.
This formula is important because its takes account of the Load impedance which will tell us what the value of the power factor is, whether its a Lagging, Leading or Unit Power factor. That power factor is needed to find the Impedance angle when calculating Zb.
The Load apparent power is S=P/cosθ, as u can see when Pf is 1, S=P, meaning there's no loss, power is being delivered to the system.
You may do some reading on the power triangle.
I can also maybe make a Tutorial on power triangle.
@@CMTEQ Thank you for clarification
accelent, well explained
Thank you 🙂
Excellent explanation sir. Thanks
You are welcome
sir ye angel ka kuch samajh nii aaya please help
Thank you for your comment, please consider to subscribe as there will be more tutorial on fault calculations.
Watch this tutorial it will help you understand better.
Part 1: th-cam.com/video/tGtu4fyg1fM/w-d-xo.html
Part 2: th-cam.com/video/9YDAJj-ECus/w-d-xo.html
Great job. I learned alot
Glad you enjoyed it! Please stay tuned for more upcoming tutorial on fault current calculation
@@CMTEQ thank you dear. I have designed a DC power connected to Ac grid through inverter and LCL filter in PSCAD. Now I want to convert the whole system to per unit system. How to deal with LCL filter.?
Thank you in anticipation
Great stuff, that's what electrical engineers do, keep up making things.
Thank you, Sir🫡
You are welcome. Stay tuned for more.
Thank you.
You're welcome!
Thank you
Tanks
It's my pleasure 🙏
Thank you
You are welcome
you get the answer.. So how to applied in the real situation?.
Very good question, the purpose of this exercise is precisely to ensure protection of your equipment and personal in real life and also ensure continuity of supply.
All you have to do is to read your equipment name plate and draw a per unit diagram & find the possible fault Current and any location.
You have a mistake when you calcul T2
Vbase =(500\1000)*800=400
Hi Fati, thank you for the comment there's a typo on the Values of Vb(old) and Vb(new), the values are swapped between them, but in the formula they are used correctly, the answer for Zpu(T2) is correct. The base voltage on the load side is also correct, as Zone2 Vbase = 800 this give then Vbase(Load)=400V. Let me know if this clarify your concern.
@@CMTEQ plise
You can repeat your answer on arabic or french
Salut Fati, Merci pour votre commentaire, il y une erreur de typo entre les valeurs de Vb(new) et Vb(old), les valeurs ont ete echangees entre eux, la reponse de Zpu(T2) est correcte. La tension de base du cote de la charge est aussi correct a 400 Volts, par ce que la tension de la Zone2 est Vb = 800V, cela done Vb(Load)=400V. Dites moi si cela clarifie votre souci. Considerez aussi vous subscrire merci.
@@CMTEQ meci beaucoup
@@CMTEQ tkanks
And you can put a vidéo aboute GMD and GMR
There’s a mistake on Vbase in zone 3
Hi Joseph, it can help if you can elaborate on where the mistake is on Zone 3
🙏🙏🙏 well explained 👏
Thank you for the feedback, I'm glad it was helpful.
Thanks
You are welcome