Yes, I do plan to expand on these videos in the near future. Unfortunately, I can only do so as time allows. At the moment, I'm not sure how soon it will be. Thanks for your interest!
The service amps is 503. You would have to go to table 240.6(A) to find the standard breaker size. You can’t use the standard 500 amp breaker because your calculated load is over that. The next standard size is a 600 amp breaker. Go to table 310.16 and go to the 75 degree Celsius column. Find the wire size that has an ampacity of 600 amps or more. The service wire size will be 1500 kcmil Copper or parallel runs of 500 kcmil Aluminum.
Excellent explanation. Thank you so much for your time. Really helpful to see how you went through it line by line. I tend to always think I am missing something since the code is a bit convoluted trying to cover all scenarios. Thanks again.
I've been banging my head against load calc for the past couple weeks and I have to say that your video, BY FAR, is the best, most thorough, in depth, precise tutorial on how to do commercial load calc (or any load calc for that matter) on TH-cam. It's not only incredibly thorough, your layout is beautiful. Seriously. I've gone from rewriting & condensing notes in 220, trying to figure it out, to making a spreadsheet or load calc calculator, to using the Line Method with a shit ton of notes, to finally landing on your most excellent video and yours is the best. You cover something that only ONE other guy mentions, which is continuous load, and you even explain WHY you're using it where you're using it. We are all so lucky to have you. One question. I'm preparing to take my test, which, sadly, is still 2017. Do we assume continuous load is included in Table 220.12? I ask, because the continuous load note at the bottom of Table 220.12 in the 2020 NEC is not in the 2017 NEC, and I've scoured the interwebs for an answer and nobody mentions it.
Hey, thank you so much for the incredibly kind words! Sorry, I've been away for a bit and the comments piled up. As for T220.12, you'll notice that the values are different between the 2017 and 2020. That's because they added in the continuous load factor in 2020. So, on a 2017 exam, you will need to account for it yourself, if necessary. In my experience, most exam questions aren't detailed enough to expect you to factor that in, but just read it carefully. When in doubt, try it both ways and see which result fits an answer option. Sometimes that works. ;)
Please correct me if I am wrong. When you apply the demand of table 220.56 you got a total less then the 2 biggest appliances so you should use what it said on 220.56 “however in not case shall the feeder or service calculated load less than the sum of the larges 2 kitchen equipment load “ please explain me that I am not correcting you I just want learn.
It sounds like you may be getting the total kitchen appliance load confused with the total service load. The second paragraph in 220.56 it referring to the total feeder or service calculated load. So, in this example, your two largest kitchen appliances combined would have to equal more than 181,135 VA in order for this sentence to come into play. That sentence is mainly intended for a feeder to a kitchen equipment panel. They want to make sure the feeder can handle the two biggest appliances at full rating. Hopefully that helps! Let me know if I misunderstood your question.
Since landscape lighting isn't specified under article 220, it would have to go under Appliances. It falls under the definitions of "Appliances" and "Utilization Equipment". However, don't forget to add 25% for being a continuous load!
Do you have a retail calculation crash course version? Similar to your dwelling videos with multiple examples they're great for studying also when is the program coming out so we can pay you!
No, I haven't made such a video yet, but I will plan to do so as part of my 2023 series. Thanks for the request. I really appreciate the sentiment to support my efforts! I just haven't had time to develop a product yet that I felt was high enough quality. But it is still in the works. I apologize for it taking so much longer than I originally expected. For now, please just spread the word on forums and such. That's the best way to help the channel at the moment. Thanks for your kind words!
@@simply.electrical these videos have helped me pass around 30 exams especially the Texas Masters which was a nightmare. i will spread the word on your channel.
@@ericpolk9004 That's awesome! Glad you found this information helpful. It's always great to hear positive feedback. I really appreciate your help getting the word out! The Texas Master exam was definitely the toughest state test I've taken. When I passed it, the gals that were administering it were surprised. They said it had been quite a while since they had seen anyone pass it on their first attempt. That was in 2022. I was fortunate to have the advantage of 27 years of experience and five other state licenses under my belt. Even so, it gave me a run for my money!
I am studying for my Washington master exam and your videos have been extremely helpful. I think I have watched all of them and am much more confident taking my test tomorrow. I am wondering about taking your show window, track lighting, and sign loads at 125%? I understand that they would be continuous loads but what if they don't use the show window? In annex D example D3 (store building) they have a sign and show window in the calculation at 100% not 125%. What would you do on a exam?
Those would fall under 220.14(A). Part III of 220 does not provide any demand factors for commercial laundry loads, so they would be calculated at nameplate value.
Optional calcs only apply to occupancies covered in part IV (220.80-88): residential (220.82-83), multifamily (220.84-85), schools (220.86), existing loads (220.87), and new restaurants (220.88).
I see you were able to make sense of the NEC example. I'm not a big fan of the examples in the Annexes, simply because I don't think they do a very good job of explaining themselves. It leaves you scratching your head more often than not, IMO. That said, they are helpful if you have no other resources to draw on. Thanks for the comments!
Quick Question as i see this on almost every test and i cant seem to get a good answer... #1 can you install class 2 and 3 circuits with class 1 in the same raceway with a barrier if the insulation for all circuits is rated for 600v? #2 what if the same test question doesen't mention a barrier? only the 600v rating? i believe the answer is no in that case?
See 725.136, particularly (B). The insulation rating doesn't appear to be an issue here, either way. But there are many other factors to consider - see parts (C) - (I). With the questions being written as they are above, I would say Yes to #1, and No to #2. Because we can't make assumptions regarding information that we aren't provided. Rather, we presume that (C) - (I) do not apply.
One of the answers is if there is 600v insulation I still think without a barrier the answer needs to be no It is hard to guess what they’re thinking Thanks again
@@ericpolk9004 They will often throw in extra needless information just to try to trip you up by making you overthink it. They want to make sure you really know your stuff. That may be why the insulation is mentioned. I'm not aware of any place where the insulation rating matters for Class 1 vs Class 2/3 wiring.
That's correct. You'll notice that the note for Table 220.12 was added in the 2020 NEC, stating the inclusion of the 125%. And you can see that the values have all been adjusted accordingly. They did a whole re-vamping of the table in 2020.
Hey Seth, sorry for not responding to you sooner! I was away for a bit. You must use the standard method for a duplex --- BUT --- 220.85 allows you to calculate it under the optional method AS A TRIPLEX and take the lesser of the two totals. I'll put together a duplex video soon. Here are my Multifamily videos: Multifamily Optional Crash Course - th-cam.com/video/zRKbEu2zNWM/w-d-xo.html Multifamily Optional Deep Dive - th-cam.com/video/CzIiZwBkNNU/w-d-xo.html Multifamily Standard Crash Course - th-cam.com/video/0Ryf2QlmO-k/w-d-xo.html Multifamily Standard Deep Dive - th-cam.com/video/laZ4THu15Hs/w-d-xo.html Let me know if you have any questions. And thanks for checking out my channel! James
That's correct. You'll notice that the note for Table 220.12 was added in the 2020 NEC, stating the inclusion of the 125%. And you can see that the values have all been adjusted accordingly. They did a whole re-vamping of the table in 2020.
I had a question pop up that i got wrong any advice see below: Step 7 - A/C Load The total VA load for the AC should be calculated as follows: 24 A × 240 = 5,760 VA This load is included at 100% in this question shouldn't it be 125 % since it is the largest motor? This is a Store
Did they tell you that the A/C load was indeed the largest load of the building? Because on a test, you are not expected to assume. They will give you that kind of pertinent data. Some people add the extra 25% during a different step, so could they have perhaps done that? Hard to know without seeing the whole problem. Was this a test, or maybe a practice test, or worksheet?
@@simply.electrical it was the only motor in the calculation but taken at 100 and not 125 seems wrong No mention of heat or any other motors Practice Exam
@@ericpolk9004 One thing to note about practice tests: Many of them are imperfect. They can be written by well-intentioned folks who are trying to help, but they do have some mistakes sometimes. I'm not saying that is the case here, but it's a possibility. I'm not sure where your practice test came from. Can you reach out to them to see how they arrive at their answer? The state exams tend to be more vetted and accurate. Often times, on a state exam, the calc problems will be very abbreviated. They are not looking for a whole service load calc most of the time. So you won't treat it quite the same way. You just need to give them what they ask for.
I recently took a load calculation exam on the 2020 NEC. One of the questions gave me a two tennant manufacturing facility. One of the tennants had (5) 60 amp transformer arc welders at a 10% duty cycle. How does that load get calculated? Article 630 has a duty cycle table that I don't know how to read, and 630.12 talks about 200% conductor ampacity for the OCPD. For the exam I took it as (5) 60 amp loads with no demand factor, but I'm certain that isn't right. Can you offer some insight?
Welders throw a lot of folks for a loop. And the question they gave you on the exam was tricky because they gave you intentionally misleading information. The duty cycle has no bearing on calculating the load for a GROUP of welders. So that bit of info was meant to be a curve ball. Table 630.11(A) only applies to an individual welder. For a GROUP of welders, apply the percentages specified in 630.11(B). Add the values together to determine the minimum CONDUCTOR ampacity (60+60+51+42+36=249A). This would be your MINIMUM load value for welders in your calc. You would plug this number into the table under Other Loads [220.14(A)] in the table from the video. Side note FYI: Once you have the conductor ampacity, see 630.12(B) if you want to find the max OVERCURRENT PROTECTION rating for one or more welders. Take 200% of the conductor ampacity to get your MAXIMUM load value in amps (249x200%=498A). This would apply to your branch circuit (or subfeeder) OCPD -- not your service. So this would NOT be part of your calc for the exam question. This is how I interpret and apply article 630. Hope this helps!
@@simply.electrical I had that same question and your explanation helps a lot. I'm positive I can get this next time because I know I didn't get it this last time
@@lukethesithlord Yes, one of the frustrating things about the state exams is the extra (irrelevant) information they often stick in there to throw you off. But they want to make sure you really know your stuff, so I can't really blame them. :) Good luck on your next, and hopefully final, attempt!
I had a question come up where it was single family dwelling using ( Optional Method ) but the calc said the dwelling was running on a generator and not utility does that change the calc at all?
No, a service load calc simply establishes the minimum load of the building. It makes no difference what the power source is. It could be powered by a generator, solar panels, wind turbine, exercise bicycle/magneto combo, or the Tesseract - doesn't matter. However, whatever it is, pay attention to the voltage because that tells you how to find amps. The state exams will often times put in extra information that has no bearing on the problem - just to throw you off. They're trying to make sure you really know your stuff.
I had a question on an exam recently Below i feel like the answer is 9000 va ? 30 x 200 = 6000 x 1.25 is 9000? i'm confused on the annex D Store Example it say coutinous duty loads but only accounts for 6000 VA? If a 4800 sq. ft. store has 30 ft. of show window, a total of 80 duplex receptacles, service of 120/240 V, single phase 3-wire, and actual connected lighting load of 9120 VA, what is the total calculated lighting load for the show windows in VA? A: 5000 B: 6000 C: 7000 D: 9000 Any Input would be appreciated
Total calculated lighting load for 30 feet of show window would be 7500 VA. So none of those answers are correct. 30 x 200 = 6000 x 125% = 7500 VA The example in Annex D3 shows that to be true (about 1/4 of the way down the left column of page 818). Now, the SIMPLE calculation is 30 x 200 = 6000. But the TOTAL calculated load would be including the continuous load multiplier (1.25). Therefore, 7500 VA. I'm not sure where this exam is from, but it appears to have either left out critical info (e.g. the show window lights are on a timer, and therefore not continuous), or it is otherwise inaccurate. If I was forced to answer this question with the given answers, I would choose B, 6000. Because the other answers have no relation to the problem whatsoever. But I think there is something amiss here.
@@simply.electrical ok but it's not 6000 because they are saying it's 6000? is there any world where i don't add 125% to signs and show windows in a store calculation?
@@simply.electrical the Exam is in Maryland ME. the question is what is the full calc but then asks you to say what the load is for show windows only. so it's either 6000 or 7500 in this case ? they also say things like the store is 4800 sq ft and the "countinous actual connected lighting load is 7500VA" My math is 4800 x 1.9 = 9120VA and 7500 x 1.25% is 9375VA i would use the larger of the two. when they use the phrase countinous actual connected lighting load are they already adding 1.25% to the 7500VA? Another confusing part is when they say what is the General light ing load of the store i assume we are excluding the show windows,track,and signs?
@@ericpolk9004 Typically, "actual" (aka "connected") load is PRIOR to applying any correction factors, demand, etc. "Calculated" load would be AFTER all adjustments are made, and would include the extra 25% for continuous loads. It sounds to me like they might be asking for the show window lighting load value BEFORE factoring in the extra. Some people add up all of their loads first and then go back and apply any additional factors to respective values at the end. So perhaps that is what they are expecting. As far as "general lighting" goes, look at my table in the video and notice the 5th row in the far left column. It's called "Demand Load". But it is also the "General Lighting Load" (220.42). So this does not include the specialty lighting such as show window, track, etc.
@@ericpolk9004 Only if those loads are prevented from operating for more than 3 hours at a time (like on a timer). Otherwise, they are continuous loads and must be calculated as such. My hunch is that they are asking for the pre-adjustment connected value - not the full calculated total. Therefore 6000VA would be correct. I apologize if I'm not answering your questions adequately. Without seeing the complete exam question, I kinda have to make some presumptions on what they are asking for.
I’m studying for the Oregon General Supervisor exam (which it brutal by the way), but maybe this question isn’t the best example because it is so generic. They didn’t include equipment voltages. What if the heat was (3) 15000W, 1PH, 208V loads? Can you please do a an example on a more challenging question, like a restaurant without just wattages and a mixture of 1ph/3ph loads?
Yes, Oregon is tough. When I took mine 20 years ago, they had trick questions on the exam. Some questions had no correct answer and others had more than one correct answer. They provided us with scratch paper so we could explain which answers we chose and why. Like you said... BRUTAL. If they don't include the equipment voltages, there are two possibilities: either the voltage isn't necessary to know for the answer, or there is enough information given in the question to determine the voltage via Ohm's Law. An Ohm's wheel chart is super helpful, if you don't have one. ohmslawcalculator.com/ohms-law-wheel If you have (3) 15,000w heating loads in commercial, just add them together (assuming they are not noncoincident with larger loads), regardless of the voltage or phasing. If you have a load calc question that has several 1ph and 3ph loads, go to Table 430.248 and Table 430.250, respectively, to determine the FLC. Multiply the FLC by the voltage of each motor to get your wattage (don't forget your 3ph multiplier - 1.732). They must either provide the voltage or else provide the means to attain the voltage within the question. Otherwise, no solution is possible. Hope that helps at least a little. Let me know if I misunderstood any of your questions. 🙂
Hey Nicolae, thanks for visiting my channel. You're the second request for a duplex, so I'm putting one together now. I'll hopefully have it posted in a day or two. - James
Hey David, I don't monetize my videos. I don't have near enough subscribers to monetize (need 1000 min). This must be TH-cam doing that on their own. I'm sorry to hear that there are a lot of ads. When I watch them, I don't see any ads, but that's probably because it's my channel. But thanks for watching my video. Sorry about the ads!
At this point, I'm just trying to build the channel. If you would be so kind as to spread the word to anyone you think might be interested, I would be so grateful! Likes, subs, and comments help a ton. Thanks!
amazing video, can you continue this video and help to find feeder size and also designing branch circuit for such system ?
Yes, I do plan to expand on these videos in the near future. Unfortunately, I can only do so as time allows. At the moment, I'm not sure how soon it will be. Thanks for your interest!
@@simply.electrical Thank you
The service amps is 503. You would have to go to table 240.6(A) to find the standard breaker size. You can’t use the standard 500 amp breaker because your calculated load is over that. The next standard size is a 600 amp breaker. Go to table 310.16 and go to the 75 degree Celsius column. Find the wire size that has an ampacity of 600 amps or more. The service wire size will be 1500 kcmil Copper or parallel runs of 500 kcmil Aluminum.
Excellent explanation. Thank you so much for your time. Really helpful to see how you went through it line by line. I tend to always think I am missing something since the code is a bit convoluted trying to cover all scenarios. Thanks again.
You're welcome! No doubt, the code is anything but straight-forward. Glad you found this helpful. Thanks for the comment!
I've been banging my head against load calc for the past couple weeks and I have to say that your video, BY FAR, is the best, most thorough, in depth, precise tutorial on how to do commercial load calc (or any load calc for that matter) on TH-cam. It's not only incredibly thorough, your layout is beautiful. Seriously. I've gone from rewriting & condensing notes in 220, trying to figure it out, to making a spreadsheet or load calc calculator, to using the Line Method with a shit ton of notes, to finally landing on your most excellent video and yours is the best. You cover something that only ONE other guy mentions, which is continuous load, and you even explain WHY you're using it where you're using it.
We are all so lucky to have you.
One question. I'm preparing to take my test, which, sadly, is still 2017. Do we assume continuous load is included in Table 220.12? I ask, because the continuous load note at the bottom of Table 220.12 in the 2020 NEC is not in the 2017 NEC, and I've scoured the interwebs for an answer and nobody mentions it.
Hey, thank you so much for the incredibly kind words! Sorry, I've been away for a bit and the comments piled up.
As for T220.12, you'll notice that the values are different between the 2017 and 2020. That's because they added in the continuous load factor in 2020. So, on a 2017 exam, you will need to account for it yourself, if necessary. In my experience, most exam questions aren't detailed enough to expect you to factor that in, but just read it carefully. When in doubt, try it both ways and see which result fits an answer option. Sometimes that works. ;)
@@simply.electrical Thanks. And, I passed on the first try.
@@icevariable9600 That's awesome! Congrats!
Please correct me if I am wrong. When you apply the demand of table 220.56 you got a total less then the 2 biggest appliances so you should use what it said on 220.56 “however in not case shall the feeder or service calculated load less than the sum of the larges 2 kitchen equipment load “ please explain me that I am not correcting you I just want learn.
It sounds like you may be getting the total kitchen appliance load confused with the total service load. The second paragraph in 220.56 it referring to the total feeder or service calculated load. So, in this example, your two largest kitchen appliances combined would have to equal more than 181,135 VA in order for this sentence to come into play.
That sentence is mainly intended for a feeder to a kitchen equipment panel. They want to make sure the feeder can handle the two biggest appliances at full rating.
Hopefully that helps! Let me know if I misunderstood your question.
@simply.electrical thanks for your answer . Please make more videos base on the 2020 nec. And your videos are the best. Thanks
Hello, where shall I put landscape lighting in my commercial building?
Since landscape lighting isn't specified under article 220, it would have to go under Appliances. It falls under the definitions of "Appliances" and "Utilization Equipment". However, don't forget to add 25% for being a continuous load!
What about washer , dryer load in commercial building like hotel, motel? Also in appliences?@@simply.electrical
Do you have a retail calculation crash course version?
Similar to your dwelling videos with multiple examples they're great for studying also when is the program coming out so we can pay you!
No, I haven't made such a video yet, but I will plan to do so as part of my 2023 series. Thanks for the request.
I really appreciate the sentiment to support my efforts! I just haven't had time to develop a product yet that I felt was high enough quality. But it is still in the works. I apologize for it taking so much longer than I originally expected.
For now, please just spread the word on forums and such. That's the best way to help the channel at the moment. Thanks for your kind words!
@@simply.electrical these videos have helped me pass around 30 exams especially the Texas Masters which was a nightmare.
i will spread the word on your channel.
@@ericpolk9004 That's awesome! Glad you found this information helpful. It's always great to hear positive feedback. I really appreciate your help getting the word out!
The Texas Master exam was definitely the toughest state test I've taken. When I passed it, the gals that were administering it were surprised. They said it had been quite a while since they had seen anyone pass it on their first attempt. That was in 2022. I was fortunate to have the advantage of 27 years of experience and five other state licenses under my belt. Even so, it gave me a run for my money!
Thank You for sharing this is great informatiom.
You are very welcome!
Excellent video😍
Thanks!
UAU thank you for your quick reaction on may answer it is unexpected . thank you very much!!!!!!!
You bet. Check it out here:
Duplex Standard Calc: th-cam.com/video/9kiYYCs3m5Q/w-d-xo.html
Great video.
Thanks!
I am studying for my Washington master exam and your videos have been extremely helpful. I think I have watched all of them and am much more confident taking my test tomorrow. I am wondering about taking your show window, track lighting, and sign loads at 125%? I understand that they would be continuous loads but what if they don't use the show window? In annex D example D3 (store building) they have a sign and show window in the calculation at 100% not 125%. What would you do on a exam?
See comment below...
Hello, What about washer , dryer load in commercial building like hotel, motel? Just one per whole building.
Those would fall under 220.14(A). Part III of 220 does not provide any demand factors for commercial laundry loads, so they would be calculated at nameplate value.
is there an optional method for this calculation?
Optional calcs only apply to occupancies covered in part IV (220.80-88): residential (220.82-83), multifamily (220.84-85), schools (220.86), existing loads (220.87), and new restaurants (220.88).
never mind 125% is right thanks for the great videos :)
I see you were able to make sense of the NEC example. I'm not a big fan of the examples in the Annexes, simply because I don't think they do a very good job of explaining themselves. It leaves you scratching your head more often than not, IMO. That said, they are helpful if you have no other resources to draw on.
Thanks for the comments!
Quick Question as i see this on almost every test and i cant seem to get a good answer...
#1 can you install class 2 and 3 circuits with class 1 in the same raceway with a barrier if the insulation for all circuits is rated for 600v?
#2 what if the same test question doesen't mention a barrier? only the 600v rating? i believe the answer is no in that case?
See 725.136, particularly (B).
The insulation rating doesn't appear to be an issue here, either way. But there are many other factors to consider - see parts (C) - (I).
With the questions being written as they are above, I would say Yes to #1, and No to #2. Because we can't make assumptions regarding information that we aren't provided. Rather, we presume that (C) - (I) do not apply.
One of the answers is if there is 600v insulation I still think without a barrier the answer needs to be no
It is hard to guess what they’re thinking
Thanks again
@@ericpolk9004 They will often throw in extra needless information just to try to trip you up by making you overthink it. They want to make sure you really know your stuff. That may be why the insulation is mentioned. I'm not aware of any place where the insulation rating matters for Class 1 vs Class 2/3 wiring.
so in the 2017 NEC the 125 percent need to be added?
That's correct. You'll notice that the note for Table 220.12 was added in the 2020 NEC, stating the inclusion of the 125%. And you can see that the values have all been adjusted accordingly. They did a whole re-vamping of the table in 2020.
Can you do a duplex ? I had 2 questions on test for a duplex standard calc and optional calc for a duplex
Hey Seth, sorry for not responding to you sooner! I was away for a bit.
You must use the standard method for a duplex --- BUT --- 220.85 allows you to calculate it under the optional method AS A TRIPLEX and take the lesser of the two totals. I'll put together a duplex video soon.
Here are my Multifamily videos:
Multifamily Optional Crash Course - th-cam.com/video/zRKbEu2zNWM/w-d-xo.html
Multifamily Optional Deep Dive - th-cam.com/video/CzIiZwBkNNU/w-d-xo.html
Multifamily Standard Crash Course - th-cam.com/video/0Ryf2QlmO-k/w-d-xo.html
Multifamily Standard Deep Dive - th-cam.com/video/laZ4THu15Hs/w-d-xo.html
Let me know if you have any questions. And thanks for checking out my channel!
James
Here it is! Duplex Standard Calc: th-cam.com/video/9kiYYCs3m5Q/w-d-xo.html
@ericpolk9004
0 seconds ago
so in the 2017 NEC the 125 percent need to be added? to the general lighting load?
That's correct. You'll notice that the note for Table 220.12 was added in the 2020 NEC, stating the inclusion of the 125%. And you can see that the values have all been adjusted accordingly. They did a whole re-vamping of the table in 2020.
Thanks i appreciate your channel I'm up too 15 masters now and you have helped tremendously @@simply.electrical
I had a question pop up that i got wrong any advice see below:
Step 7 - A/C Load
The total VA load for the AC should be calculated as follows:
24 A × 240 = 5,760 VA
This load is included at 100% in this question shouldn't it be 125 % since it is the largest motor?
This is a Store
Did they tell you that the A/C load was indeed the largest load of the building? Because on a test, you are not expected to assume. They will give you that kind of pertinent data.
Some people add the extra 25% during a different step, so could they have perhaps done that? Hard to know without seeing the whole problem.
Was this a test, or maybe a practice test, or worksheet?
@@simply.electrical it was the only motor in the calculation but taken at 100 and not 125 seems wrong
No mention of heat or any other motors
Practice Exam
@@ericpolk9004 One thing to note about practice tests: Many of them are imperfect. They can be written by well-intentioned folks who are trying to help, but they do have some mistakes sometimes. I'm not saying that is the case here, but it's a possibility. I'm not sure where your practice test came from. Can you reach out to them to see how they arrive at their answer?
The state exams tend to be more vetted and accurate. Often times, on a state exam, the calc problems will be very abbreviated. They are not looking for a whole service load calc most of the time. So you won't treat it quite the same way. You just need to give them what they ask for.
I recently took a load calculation exam on the 2020 NEC. One of the questions gave me a two tennant manufacturing facility. One of the tennants had (5) 60 amp transformer arc welders at a 10% duty cycle. How does that load get calculated? Article 630 has a duty cycle table that I don't know how to read, and 630.12 talks about 200% conductor ampacity for the OCPD.
For the exam I took it as (5) 60 amp loads with no demand factor, but I'm certain that isn't right. Can you offer some insight?
Welders throw a lot of folks for a loop. And the question they gave you on the exam was tricky because they gave you intentionally misleading information. The duty cycle has no bearing on calculating the load for a GROUP of welders. So that bit of info was meant to be a curve ball. Table 630.11(A) only applies to an individual welder.
For a GROUP of welders, apply the percentages specified in 630.11(B). Add the values together to determine the minimum CONDUCTOR ampacity (60+60+51+42+36=249A). This would be your MINIMUM load value for welders in your calc. You would plug this number into the table under Other Loads [220.14(A)] in the table from the video.
Side note FYI: Once you have the conductor ampacity, see 630.12(B) if you want to find the max OVERCURRENT PROTECTION rating for one or more welders. Take 200% of the conductor ampacity to get your MAXIMUM load value in amps (249x200%=498A). This would apply to your branch circuit (or subfeeder) OCPD -- not your service. So this would NOT be part of your calc for the exam question.
This is how I interpret and apply article 630. Hope this helps!
@@simply.electrical I had that same question and your explanation helps a lot. I'm positive I can get this next time because I know I didn't get it this last time
@@lukethesithlord Yes, one of the frustrating things about the state exams is the extra (irrelevant) information they often stick in there to throw you off. But they want to make sure you really know your stuff, so I can't really blame them. :) Good luck on your next, and hopefully final, attempt!
great information
Glad it was helpful!
I had a question come up where it was single family dwelling using ( Optional Method ) but the calc said the dwelling was running on a
generator and not utility does that change the calc at all?
No, a service load calc simply establishes the minimum load of the building. It makes no difference what the power source is. It could be powered by a generator, solar panels, wind turbine, exercise bicycle/magneto combo, or the Tesseract - doesn't matter. However, whatever it is, pay attention to the voltage because that tells you how to find amps.
The state exams will often times put in extra information that has no bearing on the problem - just to throw you off. They're trying to make sure you really know your stuff.
I had a question on an exam recently Below i feel like the answer is 9000 va ?
30 x 200 = 6000 x 1.25 is 9000?
i'm confused on the annex D Store Example it say coutinous duty loads but only accounts for 6000 VA?
If a 4800 sq. ft. store has 30 ft. of show window, a total of 80 duplex receptacles, service of 120/240 V, single phase 3-wire, and actual connected lighting load of 9120 VA, what is the total calculated lighting load for the show windows in VA?
A: 5000
B: 6000
C: 7000
D: 9000
Any Input would be appreciated
Total calculated lighting load for 30 feet of show window would be 7500 VA. So none of those answers are correct.
30 x 200 = 6000 x 125% = 7500 VA
The example in Annex D3 shows that to be true (about 1/4 of the way down the left column of page 818).
Now, the SIMPLE calculation is 30 x 200 = 6000. But the TOTAL calculated load would be including the continuous load multiplier (1.25). Therefore, 7500 VA.
I'm not sure where this exam is from, but it appears to have either left out critical info (e.g. the show window lights are on a timer, and therefore not continuous), or it is otherwise inaccurate.
If I was forced to answer this question with the given answers, I would choose B, 6000. Because the other answers have no relation to the problem whatsoever. But I think there is something amiss here.
@@simply.electrical ok but it's not 6000 because they are saying it's 6000?
is there any world where i don't add 125% to signs and show windows in a store calculation?
@@simply.electrical the Exam is in Maryland ME.
the question is what is the full calc but then asks you to say what the load is for show windows only.
so it's either 6000 or 7500 in this case ?
they also say things like the store is 4800 sq ft and the "countinous actual connected lighting load is 7500VA"
My math is 4800 x 1.9 = 9120VA and 7500 x 1.25% is 9375VA i would use the larger of the two. when they use the phrase countinous actual connected lighting load are they already adding 1.25% to the 7500VA?
Another confusing part is when they say what is the General light ing load of the store i assume we are excluding the show windows,track,and signs?
@@ericpolk9004 Typically, "actual" (aka "connected") load is PRIOR to applying any correction factors, demand, etc. "Calculated" load would be AFTER all adjustments are made, and would include the extra 25% for continuous loads. It sounds to me like they might be asking for the show window lighting load value BEFORE factoring in the extra. Some people add up all of their loads first and then go back and apply any additional factors to respective values at the end. So perhaps that is what they are expecting.
As far as "general lighting" goes, look at my table in the video and notice the 5th row in the far left column. It's called "Demand Load". But it is also the "General Lighting Load" (220.42). So this does not include the specialty lighting such as show window, track, etc.
@@ericpolk9004 Only if those loads are prevented from operating for more than 3 hours at a time (like on a timer). Otherwise, they are continuous loads and must be calculated as such.
My hunch is that they are asking for the pre-adjustment connected value - not the full calculated total. Therefore 6000VA would be correct.
I apologize if I'm not answering your questions adequately. Without seeing the complete exam question, I kinda have to make some presumptions on what they are asking for.
Very nice, thank you
Thanks, you bet!
I’m studying for the Oregon General Supervisor exam (which it brutal by the way), but maybe this question isn’t the best example because it is so generic. They didn’t include equipment voltages. What if the heat was (3) 15000W, 1PH, 208V loads?
Can you please do a an example on a more challenging question, like a restaurant without just wattages and a mixture of 1ph/3ph loads?
Yes, Oregon is tough. When I took mine 20 years ago, they had trick questions on the exam. Some questions had no correct answer and others had more than one correct answer. They provided us with scratch paper so we could explain which answers we chose and why.
Like you said... BRUTAL.
If they don't include the equipment voltages, there are two possibilities: either the voltage isn't necessary to know for the answer, or there is enough information given in the question to determine the voltage via Ohm's Law. An Ohm's wheel chart is super helpful, if you don't have one.
ohmslawcalculator.com/ohms-law-wheel
If you have (3) 15,000w heating loads in commercial, just add them together (assuming they are not noncoincident with larger loads), regardless of the voltage or phasing.
If you have a load calc question that has several 1ph and 3ph loads, go to Table 430.248 and Table 430.250, respectively, to determine the FLC. Multiply the FLC by the voltage of each motor to get your wattage (don't forget your 3ph multiplier - 1.732). They must either provide the voltage or else provide the means to attain the voltage within the question. Otherwise, no solution is possible.
Hope that helps at least a little. Let me know if I misunderstood any of your questions. 🙂
PLEAS do duplex! Thank you.
Hey Nicolae, thanks for visiting my channel. You're the second request for a duplex, so I'm putting one together now. I'll hopefully have it posted in a day or two. - James
Here it is! Duplex Standard Calc: th-cam.com/video/9kiYYCs3m5Q/w-d-xo.html
You’re a fantastic teacher. Your lesson is too polluted with ads though. I understand monetizing your content but this was ridiculous.
Hey David,
I don't monetize my videos. I don't have near enough subscribers to monetize (need 1000 min). This must be TH-cam doing that on their own. I'm sorry to hear that there are a lot of ads. When I watch them, I don't see any ads, but that's probably because it's my channel.
But thanks for watching my video. Sorry about the ads!
I didn't get a single ad when I watched this video.
how to monetize your channel ?
At this point, I'm just trying to build the channel. If you would be so kind as to spread the word to anyone you think might be interested, I would be so grateful! Likes, subs, and comments help a ton.
Thanks!
Why didn’t you balance the loads between phases?
Thanks for the question. Before I address it (I don't want to assume), which specific loads are you inquiring about?
Sorry, I posted my response in the wrong area.