Number of Circuits for a store
- Thread starter boku0003
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A lot of varibles... what is the 9 kw load ? Does it have HVAC ?
You only have to have 1 circuit that I know of, a sign.
If you have attic or crawl equipment, you would need lights and receptacles (a 2nd circuit ?)
Ridiculous as it seems, I can answer: 1 120v circuit ..for the required sign circuit
(sorry sparkey, it was such a strange question, I spoke too early)
You only have to have 1 circuit that I know of, a sign.
If you have attic or crawl equipment, you would need lights and receptacles (a 2nd circuit ?)
Ridiculous as it seems, I can answer: 1 120v circuit ..for the required sign circuit
(sorry sparkey, it was such a strange question, I spoke too early)
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How do you only get 1 circuit? It's a store, so you have to at least include 3VA per square foot per T220.12. I was not told what the 9kV load is. Do you use that, or use the calculated load (4000 square feet x 3VA) to compute the number of circuits? Do you multiply by 1.25 for continuous load as well?
If you use the calculated load, then 4000x3x1.25=15kVA, which would be 6.25=7 circuits (at 120Vx20A per circuit).
If you use the calculated load, then 4000x3x1.25=15kVA, which would be 6.25=7 circuits (at 120Vx20A per circuit).
120v
120v
The exam question (I took the Master test back in Feb, so I'm going off of my notes here...) said:
For a store that is 4000 square feet with a connected load of 9kVA, calculate the number of 120V/20A circuits needed.
What I explained in my last post is what I did for my answer.
120v
The exam question (I took the Master test back in Feb, so I'm going off of my notes here...) said:
For a store that is 4000 square feet with a connected load of 9kVA, calculate the number of 120V/20A circuits needed.
What I explained in my last post is what I did for my answer.
Shouldn't you take 4000 X 3 then divided by 2400 to get 5 circuits.
9kva would not be used because the connected load is less than the calculated 3va per square foot.
Where does the 1500 VA sign come into play? Only in a bank or office?
If you add the sign load per 220.14, then it would be 6 circuit.
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- Chapel Hill, NC
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- Retired Electrical Contractor
I don't agree with your answer if this is a store you must calculate for continuous load. So, you would be dividing by 1920 not 2400. This answer could change also by the size of the loads being connected. Suppose that each light, as we had in another thread, was 500 watts, then the connected load for each circuit could only be 1500 watts.
I believe the answer that they are looking for is RND(3 x 4000 x 1.25 /(120*20))=7 ? (same answer as Dennis gave you)
Of course, a better real world example would give you the store sign frontage,add ckts for cash reg's and equip, HVAC (larger of heat/ac), and have you add a 20A sign ckt , but taking a test means figuring out exactly what answer they want, not providing the most correct answer (jmshio)
Of course, a better real world example would give you the store sign frontage,add ckts for cash reg's and equip, HVAC (larger of heat/ac), and have you add a 20A sign ckt , but taking a test means figuring out exactly what answer they want, not providing the most correct answer (jmshio)
Ok, we would take 4000 x 3 = 12,000va x 1.25 (continuous load)= 15,000va
15,000va + 1500va (sign) = 16500va/ 2400 = 6.875 circuits or 7 circuits
Even if you take 9kva connected load at 1.25 that would be 11,250. 11,250 is still less than the 3va per square foot calculation. So, we still need to used the 3va per square foot calculation according to the nec instead of the connected load.
If a store is subject to 220.14 such as the with the bank and office, the we need to added the 1500va sign (1200va x 1.25).
120v x 180va = 2400, Why would you divided by 1920?
The question did not give each light per so many watts (for each light), only the connected load. Some other similar question may give you light per so many watts. Some question even give you the number of unit with the light amperage with the voltage. What ever that is given, you still have to calculated out the "connected" load. In this particular question, the connected load is stated and nothing else.
What am I missing here?
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- Chapel Hill, NC
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- Retired Electrical Contractor
Let's ignore the sign and look at what you did and what I did. They are the same. You multiplied 1.25 times the 12,000 and got 15000 and without the sign divide by 240 and you get 6.25 which means 7 circuits plus the sign
Now I took 80% of the 2400 watts allowed for a 20 amp circuit and you get 1920 watts. I then divided 12000 by 1920 and I get 6.25 or 7 circuits plus one for the sign.
So you used 15000/240 and I used 12000/1920--- they are the same.
As t the value of the lights-- I was making a point about what a bad question it was.
This is not a bad question. There are questions like this in Holt's books. These are exam practice questions, so they do not always give every little detail. Please note that I posted this (and others) in the "Exam Preperation" section of this forum, so there are always going to be details left out. Slavan found a question asked exactly like this online as well:
http://books.google.com/books?id=8c...ge&q=calculating store branch circuit&f=false
I agree with his calculation, take the bigger of the calculated and actual load, though I don't know where this is stated officially in the NEC:
15,000va + 1500va (sign) = 16500va/ 2400 = 6.875 circuits or 7 circuits
http://books.google.com/books?id=8c...ge&q=calculating store branch circuit&f=false
I agree with his calculation, take the bigger of the calculated and actual load, though I don't know where this is stated officially in the NEC:
15,000va + 1500va (sign) = 16500va/ 2400 = 6.875 circuits or 7 circuits
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- Lockport, IL
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- Retired Electrical Engineer
This is a bad question, and Mike Holt will not claim that his sample tests are perfect. Indeed, he would like to hear the news, if anyone finds an error in one of his books.
The reasons this one is a bad question are that (1) You don't determine circuits on the basis of connected load, but rather on the basis of calculated load, and (2) Giving you the square footage is not enough to allow you to calculate the total load.
I'm not claiming that Holt has bad info in his books. Just that he has practice questions that sometimes do not give every little real world detail. And that's good. It helps to isolate the concept.
Take a look at the link that I provided. It says that "The lighting load to be used in the branch-circuit calculations for determining the required number of circuits must be the larger of the values obtained by using one of the following: The actual Load or A minimum load in VA as specified in the NEC". Then it goes on to say "If the actual lighting load is known and if it exceeds the minimum determined by the watts per square foot basis, the actual load must be used because the NEC specifies that branch-circuit conductors shall have an ampacity not less than the max load to be served". That to me is just common sense. If your connected load is larger than your calculated, you go with that or you're going to have problems. Now in this problem, the actual connected load was less so you use the calculated.
However, even if this was not true, including it would not make it a bad problem. It would have been an example of extraneuos information which is very common on an exam. I could have included other details like "...this installation is in Wisconsin..." and it would be a legit exam question because it is designed to throw off the exam taker.
Take a look at the link that I provided. It says that "The lighting load to be used in the branch-circuit calculations for determining the required number of circuits must be the larger of the values obtained by using one of the following: The actual Load or A minimum load in VA as specified in the NEC". Then it goes on to say "If the actual lighting load is known and if it exceeds the minimum determined by the watts per square foot basis, the actual load must be used because the NEC specifies that branch-circuit conductors shall have an ampacity not less than the max load to be served". That to me is just common sense. If your connected load is larger than your calculated, you go with that or you're going to have problems. Now in this problem, the actual connected load was less so you use the calculated.
However, even if this was not true, including it would not make it a bad problem. It would have been an example of extraneuos information which is very common on an exam. I could have included other details like "...this installation is in Wisconsin..." and it would be a legit exam question because it is designed to throw off the exam taker.
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- Lockport, IL
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- Retired Electrical Engineer
I understand your point, and I agree that you have to take the larger of the calculated load and the installed load. But let me make my point a bit clearer. This is a bad question because it does not give you the amount of HVAC load, and the load related to the stove and dishwasher in the break room, and the load related to the point of sale registers, computers, and such, and the load related to the sump pump, the water heater, and the pumps serving hot and cold water, and any other loads. All they gave was square footage.
Now if the question asked how many circuits would be needed to handle the general lighting load, then it would have been a fair question. It did not, and thus it was not.
Now if the question asked how many circuits would be needed to handle the general lighting load, then it would have been a fair question. It did not, and thus it was not.
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
I will rephrase my response and say it is a terrible question. The connected load is what???? and does it or does it not relate to this problem? It obviously can't be the lighting load so I would have to assume it has no lighting load in the 9kw number.
So now do they want us to add the store window in there or are they looking for the load based solely on the sq. ft. Perhaps with a choice of answers one could make an educated guess but I still think this is a terrible question.
So now do they want us to add the store window in there or are they looking for the load based solely on the sq. ft. Perhaps with a choice of answers one could make an educated guess but I still think this is a terrible question.
this may indeed be a bad question, but unfortunately,I'm pretty sure that you will find questions like this on masters exams, so if you are planning on taking one of those tests, decide how you will answer the question now so that at least you won't waste your time at the time of the exam wondering how you should answer it.
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I agree!! I can post a ton of them from a few books written by well known teachers of the code. If one takes the time to pick up one (an exam prep book), you will see that most--at least the ones I have--exam preparation books are well written--explaining how to get to an answer (using all the details), then how to answer an exam question--deciphering the question without all the details.
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