Thank you Sir, i got the scanned copy from your link provided I am currently doing my thesis for professional exam let i checked over my hand calculations too. it's great!
sir, you haven't changed the motor voltage when you calculate the motor impedance, it's still 460 divided by 480. it's supposed to be 0,46^2/0,48^2 right?? would it make a different result?
You didn't find an error on conductor 2, you created a new error on your second go. You forgot the term (1/.15xx). Your error was adding the rectangular coordinates. If you look you added a 9 where it didn't belong. Very good presentation, I'm very happy to learn the PU method and short circuit calcs. Haven't got to the end of the video yet
Thanks for pointing that out. I was too close to it and trying not to delay too much for the live stream. The fact you picked it up means you are indeed working the math and learning the method. I appreciate that. This Thursday I want to do the same problem but do the calculation for ground fault current. see ya online!!!!
@@TDUNPLUGGED I was planning on picking up power system analysis and design 5th edition (Glover, Sarma and Overbye) this June. So these videos will give me some perspective prior to reading and solving the chapter problems. Live is tough and given how close your answer was I would've made the same decision.
I recently saw an arc flash label on a 12 KV piece of gear that said 16 cal/cm^2 at 30" with a 37'6" arc flash boundary. Either they calculated by some crazy math that shrapnel might only travel that far, or someone just ran the program and printed the label like an engineer might do! .......................................................................................................................................................................................................................... Oh, and it was a label that said Eaton on it!
Need to avoid rounding to a set number of decimal points, and instead round to significant figures instead. Especially when working with PU, which can be very small numbers that end up being multiplied - that rounding error can grow significant after the move to actual units.
Hey Natalio, thanks for the note. Conductor 2 has 5 in parallel. When you have parallel conductors you divide by that number of conductors because if you took the long road and calculated the equivalent impedance it would work out that you divide by the number of conductors.
@@TDUNPLUGGED start following you. It boost my talent, Am working at fire dept. in the philippines. I've check your data in fault calc. What you've used was aluminum metal and not steel. Anyway, the procedure is the same. Many Thanks...
Thank you Sir, i got the scanned copy from your link provided I am currently doing my thesis for professional exam let i checked over my hand calculations too. it's great!
sir, you haven't changed the motor voltage when you calculate the motor impedance, it's still 460 divided by 480. it's supposed to be 0,46^2/0,48^2 right?? would it make a different result?
1:02:40 - it seems you calculator is set at radians and not degrees when converting polar to rectangular.. could that affect the result?
You didn't find an error on conductor 2, you created a new error on your second go. You forgot the term (1/.15xx).
Your error was adding the rectangular coordinates. If you look you added a 9 where it didn't belong.
Very good presentation, I'm very happy to learn the PU method and short circuit calcs.
Haven't got to the end of the video yet
Thanks for pointing that out. I was too close to it and trying not to delay too much for the live stream. The fact you picked it up means you are indeed working the math and learning the method. I appreciate that. This Thursday I want to do the same problem but do the calculation for ground fault current. see ya online!!!!
@@TDUNPLUGGED I was planning on picking up power system analysis and design 5th edition (Glover, Sarma and Overbye) this June. So these videos will give me some perspective prior to reading and solving the chapter problems.
Live is tough and given how close your answer was I would've made the same decision.
I recently saw an arc flash label on a 12 KV piece of gear that said 16 cal/cm^2 at 30" with a 37'6" arc flash boundary. Either they calculated by some crazy math that shrapnel might only travel that far, or someone just ran the program and printed the label like an engineer might do! ..........................................................................................................................................................................................................................
Oh, and it was a label that said Eaton on it!
Good day. Just checked your OLD. Why Utility Z and C-3 has the same value of 0.0002 =j0.0017 @82.87 ?
Conductor 3 ZPU should be 0.0449 + j0.0299 and not the same as the utility!!!! thanks for finding this Ronnie!!!
i fail to understand why chapter 9 table 10 lists uncoated copper wires, then has a note 1 saying that the values are based on type RHH wires.
Need to avoid rounding to a set number of decimal points, and instead round to significant figures instead. Especially when working with PU, which can be very small numbers that end up being multiplied - that rounding error can grow significant after the move to actual units.
I don't get why we have to divide by 5 for conductor 2.
Hey Natalio, thanks for the note. Conductor 2 has 5 in parallel. When you have parallel conductors you divide by that number of conductors because if you took the long road and calculated the equivalent impedance it would work out that you divide by the number of conductors.
notify me when it's arch fault calc. pls
Probably going to hit this discussion tonight.
@@TDUNPLUGGED start following you. It boost my talent, Am working at fire dept. in the philippines.
I've check your data in fault calc. What you've used was aluminum metal and not steel. Anyway, the procedure is the same. Many Thanks...