Volt Drop Calculation
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- Опубліковано 22 сер 2024
- In this video I use Table D3 in the Canadian Electrical Code (2018 Version).
This vide is intended as an instructional aid for academic purposes. Always consult with your local supply authority for further clarification of rules. Thanks for watching
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Thanks so much man. I’m all done with school now, but still didn’t have a thorough understanding of volt drop.
I DO NOW! Keep me coming, way better than some of the rabbit holes I find on UA-cam
Im happy it was helpful. Thank you for the encouragement!
Yeah hes awesome
Hey dude. Thanks a ton for posting this video. It really helped me a lot because I struggle with D3
I'm happy it was able to help you! Thanks for watching!
Excellent explanation. I found the classroom explanations incomprehensible, yours however, was clear and concise. Don't stop.
Thanks for the support!
Terrific explanation of the many rules that factor into the end result. Just awesome!!!
Thank you for the encouraging words!
Thanks for this video. Great information and very well explained!!
I absolutly love your videos! They are definitly my go to suggestion for students who want video tutorials.
Quick question, for a motor voltage drop calculation, do you use your FLA or the FLA x 1.25 to obtain your distance from Table D3 and in determining the distance correction factor (load ampacity/conductors ampacity at the rated insulation temperature). The CEC handbook uses FLA to determine the distance in m from Table D3 and uses the FLA x 1.25 to calculate the percent of allowable ampacity. Just curious if there is a mistake or if there is an explanation for this? I couldnt find an explanation or example anywhere! (None at least based on the CEC.)
Thanks for the support! I would use the FLA x 1.25 at D3, similarly to how I would use the load ÷ .8 for a continuous load before going to D3. For non continuous motors, FLA x T27% before using D3. That's my initial gut interpretation, but I may change it after looking into it a bit more lol. I'll let you know what I dig up!
@schulerruler awesome thank you! And then would you use that same value (fla x 1.25) to calculate your percent of allowable ampacity to use in determining your distance correction factor in note 3 of the table D3?
allowable ampacity of 1 AWG at 75 degree should be 130A, not 150A. why you select the 150A from Table 2?
A 1/0 would be able to go significantly farther carrying the same load as the no. 1awg. The intent of the video is demonstrating the allowable distance a 1/0 can carry the given load. Under no voltage drop, yes a no. 1awg could carry the load, but once we start to exceed the 3% voltage drop, we must start increasing the size to accomodate.
Better than my teacher thank you
Thank you. Informative!
Great video. 15.8 x 2 x 3 x 1.04 = 98.592, but close enough.
I think according to Table 19, the temperature should be 90°C in this question.
Table D3 note 3 confirms this
Hi very nicely explained. which software are you using for making this white board videos and on which OS you are using it. I mean which App you are using to write and draw
Hey you have lots of great videos, but just so everyone knows we should not be using the 75 degree column here, note 3 under Table D3 says to use conductor insulation temperature, so we would use 90
If you are determining the distance correction factor you compare the load to the allowable ampacity of the conductor based on table 2 values with lowest temperature in mind which in this case is 75° C.
When we select the distance correction factor however it is the percentage loading (rounded up) and then the insulation temperature to select which in this case is 90° C.
@@schulerruler This seems to be a point of confusion for a lot of people, sorry if I wasn't making great sense. There is strong evidence that we should not be using the 75 degree column, there is note 3, but also if you look at the actual example in the code book, they use #12 wires ampacity under the 90 degree column, not the 75 degree. They divide 16A by 30A, not by 25A
The example is using 90 degrees because in this question the only temperature reference is the insulation temperature of 90 degrees. If your conductor is terminated at 75 degrees, which my example is, then you would not select the conductor ampacity from the 90 degree column in table 2.
Please remember that the rules trump the appendix notes. We still must obide by 4-006 when determining conductor ampacity in a circuit.
@@schulerruler The example says "from the centre of distribution to the load", these conductors have to be terminated under something in order to even have a voltage to measure.
I did a practice question from the CSA exam today and it used the 90 degree column for getting the ampacity to load ratio.
You very well may be correct, you definitely know a lot more about electrical code than me and I have so much appreciation for all of your videos, this is a point where I think it's not so cut and dry. Luckily using the 90 or 75 degree column usually won't change to answer enough to through things off drastically.
Hello, in the example they give in the code book on page 763 they use the 90 degree column to get the ampacity of the 12 AWG wire. Does rule 4-006 2) a) not apply here since there is no termination temperature given?? im confused here.
Great Channel!! Doing CSA(!!!) exam prep course... one question for voltage drop does not give termination temps(for residential baseboard heaters) conductor is NMD90-..my question.. do I calculate conductor ampacity using 90 degree,assume 75 degree termination, or go by 4-006(2) which tells me unmarked equipment under 100a is at 60.... CSA tells me 90
Thank you for the support! The assumption used to be 90 Based off previous code, but now using 4-006 2) as you mentioned, if the load is 100A or less you would use 60 if the equipment termination temperatures are unknown.
Great! I just missed in your explanation why we have to use the row "85-90°C" and not "75°C". I know you stressed the following note: "Use insulation temperature ONLY!" but Table D3 refers to insulated conductors. I am confused. Thanks for posting. I mean it.
We would use it due to insulation being 90° C rated. Everywhere else, 4-006 wins and we use the lower temps. Thank you for the encouragement!
Table D3 is out of range for crane and hoist since we use table 58 and we can run more amp in wire (ex: 32amp in a 14 awg), what is the reference than?
Hi Schuler, I feel I am about to ask a dumb question, but here goes.
Is there a D3 calculation formula for 3 phase conductor runs?
Definitely a good question. For three phase, there is a permission in some jurisdictions to increase lengths on D3 by a factor of 1.155.
@@schulerruler no kidding eh? Thank you. The question was bugging me trying to find anything in the CEC. Look forward to your next video.
how did you get conductor insulation temperature? 90 c
Great video, Question, How do you calculate VD on cables larger than 4/0? seeing how Table D3 doesnt go beyond that size?
I came across that today in class. It's in D3 note 6. Let me know if that helps
Hi, I am trying to find the Solution for the Red Seal Question : What is the minimum size of RW90 copper conductors required to feed a 230 V, 5 hp, single-phase motor located 60 m from the distributor centre, without exceeding a 3% voltage drop?
I cannot find the correct answer. I am getting #6 AWG but the answer is #10 AWG. Can you please help me finding the answer for it. ?
You'd need to know the current drawn by the motor. 5HP x 746 = 3730W. But that's the output power. You would need to know the rated % efficiency of the motor to find the input watts. Then you can divide watts by volts to find the current drawn (input) of the motor. After that you would just follow the advice in part 2 of his video series.
In this case we are permitted to use table 44 to obtain the FLA rating of the motor.
What would you do if dealing with 3 phase, 600 volt service?
You will also need to use the multiplier for a three phase calculation of 1.155.
my brain isnt wrinkly enough
Volt drop calculations will help put some wrinkles in there!