Unrelated question: With 220.83(C) optional method I am having a bit of confusion with classifying heating types into the largest of the 6 options. I have see “central heat” “ furnace” “heat pump” and “space heaters” all applied to option (4) at 65%. Are there any other heating types you would include in this category? Also can you give an example of a heating type that would fall into category (6) of this section.
Quick Question on a Motor What is the full-load running current rating of a 208-volt, 3-phase, 50 hp, squirrel-cage, continuous-duty, ac motor? A: 130 B: 143 C: 162 D: 195 in the code book 430.250 i get 143 amps x 1.25 is 178.75 since it is continous duty I feel like i'm going crazy...
They are trying to trick you. It is a continuous duty motor. But that is just its rating. That does NOT mean it is being used in a continuous duty application. In other words, it is not necessarily being used for more than three hours at a time. But besides that, they are not asking about conductor ampacity. So Part II of 430 does not apply. 430.6(A) tells us to use the appropriate table to determine FLC of motor applications (except for overload protection), which you did correctly (143A). Side note: Also, the inclusion of 'squirrel-cage' is meant to throw you off as well, tempting you to go to T430.52. Doesn't matter. They want to make sure you really know the terminology, and which tables to use, and when not to use them. Answer: B. 143
I have a question on Dryers: Given: you have a (10) apartments each with 4500 VA Dryers what is the total load of the dryers? The test question says nothing about standard or optional... do i use nameplate 4500 VA and 220.54 or do i use 5000 and 220.54 or do i use namplate and 220.84 i notice some test don't say standard or optional some do. So do we assume standard?
The general idea for any load calc is to find the minimum load required. So if they don't tell you which method to use, generally go with whichever gives you the lowest value. That said, I can't speak for every testing agency out there. They may have different expectations in different jurisdictions. I'm going off my own experience and what I've heard from many others. 220.54 states that no dryer can be calculated at less than 5000VA (standard method). So in your case, you cannot use the nameplate when applying T220.54. You must use 5000VA for each dryer. This would be: 5000VA x 10 = 50,000VA x 50% = 25,000VA. Now, it would be very awkward to try to use the optional method here for a couple reasons. This method assumes an entire service calc - not just the dryers. This is because the values are cumulative and interdependent. In other words, you can't mix and match methods. You are not given enough info to merit using the optional method, IMO. But let's say you did use the optional: 4500VA x 10 = 45,000VA First 10,000VA @ 100% = 10,000VA Remainder (35,000VA) @ 40% = 14,000VA 14,000VA + 10,000VA = 24,000VA So we see this would be a lower value when compared to the standard method. But my gut tells me the optional method should not be used here unless they direct you to do so. And in my experience, that has always been the case. I'd be interested to hear other people's input on this. I can see an argument being made both ways on this.
For this example would it be 350OCPD with 400KCMIL phase conductors and 1/0 GEC (GEC size can vary based on type this value is from 250.66)
Unrelated question:
With 220.83(C) optional method I am having a bit of confusion with classifying heating types into the largest of the 6 options. I have see “central heat” “ furnace” “heat pump” and “space heaters” all applied to option (4) at 65%. Are there any other heating types you would include in this category? Also can you give an example of a heating type that would fall into category (6) of this section.
Quick Question on a Motor
What is the full-load running current rating of a 208-volt, 3-phase, 50 hp, squirrel-cage, continuous-duty, ac motor?
A: 130
B: 143
C: 162
D: 195
in the code book 430.250
i get 143 amps x 1.25 is 178.75 since it is continous duty
I feel like i'm going crazy...
They are trying to trick you. It is a continuous duty motor. But that is just its rating. That does NOT mean it is being used in a continuous duty application. In other words, it is not necessarily being used for more than three hours at a time.
But besides that, they are not asking about conductor ampacity. So Part II of 430 does not apply. 430.6(A) tells us to use the appropriate table to determine FLC of motor applications (except for overload protection), which you did correctly (143A).
Side note: Also, the inclusion of 'squirrel-cage' is meant to throw you off as well, tempting you to go to T430.52. Doesn't matter. They want to make sure you really know the terminology, and which tables to use, and when not to use them.
Answer: B. 143
I have a question on Dryers:
Given: you have a (10) apartments each with 4500 VA Dryers what is the total load of the dryers?
The test question says nothing about standard or optional... do i use nameplate 4500 VA and 220.54 or do i use 5000 and 220.54 or do i use namplate and 220.84
i notice some test don't say standard or optional some do. So do we assume standard?
The general idea for any load calc is to find the minimum load required. So if they don't tell you which method to use, generally go with whichever gives you the lowest value.
That said, I can't speak for every testing agency out there. They may have different expectations in different jurisdictions. I'm going off my own experience and what I've heard from many others.
220.54 states that no dryer can be calculated at less than 5000VA (standard method). So in your case, you cannot use the nameplate when applying T220.54. You must use 5000VA for each dryer. This would be:
5000VA x 10 = 50,000VA x 50% = 25,000VA.
Now, it would be very awkward to try to use the optional method here for a couple reasons. This method assumes an entire service calc - not just the dryers. This is because the values are cumulative and interdependent. In other words, you can't mix and match methods. You are not given enough info to merit using the optional method, IMO.
But let's say you did use the optional:
4500VA x 10 = 45,000VA
First 10,000VA @ 100% = 10,000VA
Remainder (35,000VA) @ 40% = 14,000VA
14,000VA + 10,000VA = 24,000VA
So we see this would be a lower value when compared to the standard method.
But my gut tells me the optional method should not be used here unless they direct you to do so. And in my experience, that has always been the case. I'd be interested to hear other people's input on this. I can see an argument being made both ways on this.
@@simply.electrical so 4500 va and 220.54? i feel like they are trying to trick me as both answers were there.
@@simply.electrical I didn't see the rest of your answer thanks agian on the test i went with 5000 VA X 50%