Oregon Supervisor Test....

[pjholguin]

If you pass not more than your score...at that time the Electrical Personnel would sit down with you if it wasn't a total blow out and review primarily part 2...not so much part 1. But that has been ~20 years and I will never forget a sign circuit...damn! LOL!
GOOD LUCK...may the NEC be with you!

Yeah theres a 2 week wait . Ill look into that . Im not even sure what info they give you about the results.

 

[Dennis Alwon]

Dennis, the min. lighting load for warehouses is 1/4 watt/sf.

250000 x 0.25 = 62,500. First 12,500 @100% rest at 50% (T. 220.42) 62,500 - 12,500 = 50,000. 50,000 X .5 = 25,000. 25,000 + 12500 = 37,500 watts lighting load.

That being said, I think the answer to the first question is "none of the above."

Thanks I was using va/sq. meter--- doh. not being careful

 

[kwired]

When they give loads in KW, they expect you to treat it as KVA.

I say if they are giving an exam for a professional license they need to show some professionalism in the way the questions are presented. If they give you a resistive load in kW that is easily converted to kVA, but to just give you something non specific in kW means you will at least need to know power factor to get accurate kVA to proceed with further calculations


I don't know how much of what you posted was exactly what was printed on your exam and what is just your recollection, but assuming what you posted is what was actually on the exam I insert my additional questions in red necessary to be able to consider answering some of those questions.

Just took my Supervisor's test in Bend, OR C.O.C.C, I wanted to help others who are taking this test. There were quite a few questions in article 605 and 820 in t he first portion of the test. In the second portion of the test I was given an industrial plant calculation that was all related to each other in this 12 question portion. I feel the first part of the test was easier, however this second part was very challenging so here it goes my questions:

1] What is the demand on a feeder for a 500x500 warehouse with 120 volt lighting and 500 general purpose receptacles. The service is 208/120 3 phase ? Here are the answer choices I was given
A.265
B.356 Are these figures Amps, VA, kVA, dollars spent on monthly electric bill????
C.251
D.229

2] What is the fault current available at the secondary terminals 1.8 impedance 500 KVA 480, 208/120 3 phase transforme Details on the supply to the transformer? or are we assuming infinite bus ability for the purpose of this question?

3] Minimum size conductor in THWN per phase feeding 1200 amp 480 volt 3 phase load 1800 ft from source minimum 3% voltage drop?Single conductor, parallel conductors? If parallel how many sets of parallels? Copper, aluminum?

This is the calculation from the whole part 2 of the exam:

480/277
535 KW lighting load power factor?, or give me value in kVA please
2-250 HP 480 V motors
1- 480x208Y 120V 3 phase transformer that feeds:
5-10HP 208V 3 phase motors
3-25HP 208V 3 phase motors
17-20 amp punch press receptacles voltage, number of phases? better yet actual load details of the machines, chances are they do not draw 20 amps.
8-30 amp receptacle voltage, number of phases? better yet actual connected load 30 amp receptacle could only have 10 amps of load connected to it
250-general purpose receptacles most will assume 15/20 amp 120 volt receptacles, but it would be nice if it were made clear that is what is intended.


I can't remember all the questions, this is just what is fresh in my head but all related to this. Anyone that can help me with this info I have here and do a regular calc? I was thrown off by the fact it was KW instead of VA... and the fact that there is receptacles listed separately from general purpose. How will I add in the 17 punch press receptacles and the 8/30 ampswr As you can see I have similar questions - the exam was poorly written IMO. Like I said before it is sad when they give you a professional exam for a professional license, and they are not professional with the questions

Thanks for your time and help!

 

[highlakeselectric]

Question 1 was word for word even the answers. The large calc with the motors and transformer was all of the info given but there were like 9 . questions on that one and i could not remember all the choices but basically it was just do a full calc and it would answer them. 2 was word for word . Thats what it seemed like the test was not asking if your capable of doing a calc but if you can quess what there asking you to do. Anyone wanna try doing the one with the punch press recepts and the 30 amp receptacles

 

[Jon Coulimore]

1] What is the demand on a feeder for a 500x500 warehouse with 120 volt lighting and 500 general purpose receptacles. The service is 208/120 3 phase ? Here are the answer choices I was given
A.265
B.356
C.251
D.229

500X500 = 250,000 x .25(Table 220.12 Warehouse)=62,500va X 1.25 (125% for continuous use) = 78,125va
500 x 180va (Receptacles 220.14(I))=90,000va
78,125 + 90,000= 168,125va
220.44 says we can apply the receptacles to the demands of Table 220.42
Table 220.42 Warehouse(Storage) states First 12,500va at 100% and the Remainder at 50%

168,125va -12,500va(first at 100%) =155,625va
155,625va x .5 = 77,812.5va
12,500va(first at 100%) + 77,812.5va(remainder at 50%) = 90,312.5 / 360.256 (208 X 1.732)=250.689 amps
C is the correct answer 250.689 round up to 251

 

[Jon Coulimore]

Available Fault current

Available Fault current

2] What is the fault current available at the secondary terminals 1.8 impedance 500 KVA 480, 208/120 3 phase transformer?

The formula is: Secondary amps divided by the % of the transformers Impedance
500,000 / 360.256 (208 x 1.732) = 1387.90 amps
1387.9 / .018 (1.8%) = 77,105.55 of Available Fault Current at the secondary terminals

 

[Byron C.]

[h=2]Supervisors Test[/h]

Just took my Supervisor's test in Bend, OR C.O.C.C, I wanted to help others who are taking this test. There were quite a few questions in article 605 and 820 in t he first portion of the test. In the second portion of the test I was given an industrial plant calculation that was all related to each other in this 12 question portion. I feel the first part of the test was easier, however this second part was very challenging so here it goes my questions:

1] What is the demand on a feeder for a 500x500 warehouse with 120 volt lighting and 500 general purpose receptacles. The service is 208/120 3 phase ? Here are the answer choices I was given
A.265
B.356
C.251
D.229




I thought I would give you my 2 cents on this question. Here goes.
demand on a feeder - For a feeder there is no continuous load. Ampacity is the maximum current in amperes, that a conductor can carry continuously under the conditions of use without exceeding its temperature.
There is no mention of what the demand would be on the OCPD, therefor we don't have to include the 125% from 215.3. The question is simply asking what the demand on the feeder is.

220.12 in regards to floor area - 500 x 500 = 250,000 sq. ft.
Table 220.12 - Warehouse 1/4 Volt-Amperes / Square Foot
250,000 sq. ft. x 1/4 VA = 62,500 Load
220.42 General Lighting - Apply Table 220.42 to branch circuit load calculated for general lighting illumination.
Table 220.42

Warehouse Option 1:
First 12,500 VA @ 100%
Remainder at 50%
62,500 VA - 12,500 VA = 50,000 VA
50,000 x 50% = 25,000 + 12,500 = 37,500

Receptacles
500 general purpose receptacles
220.14(I), 220.44 & Table 220.44
500 receptacles x 180 VA = 90,000
First 10,000 @ 100%
Remainder @ 50%
90,000 - 10,000 = 80,000
80,000 x 50% = 40,000 + 10,000 = 50,000

37,500 + 50,000 = 87,500 / (208 x 1.732) = 242-A So it is not Option 1

Warehouse Option 2:
First 12,500 VA @ 100%
Remainder at 50%
This time include the receptacles permitted by 220.44 to Table 22.42 instead of 220.44

62,500 VA + Receptacles (500 x 180)
62,500 VA + 90,000 VA = 152,500
152,500 - 12,500 VA = 140,000 VA
140,00 x 50% = 70,000 + 12,500 = 82,500
82,500 / (208 x 1.732) = 229-A Option 2 it is!

D. is the correct answer

​

 

[Bugman1400]

2] What is the fault current available at the secondary terminals 1.8 impedance 500 KVA 480, 208/120 3 phase transformer?

The formula is: Secondary amps divided by the % of the transformers Impedance
500,000 / 360.256 (208 x 1.732) = 1387.90 amps
1387.9 / .018 (1.8%) = 77,105.55 of Available Fault Current at the secondary terminals

Does this assume an infinite bus on the primary side? If so, where is that advised in the code book?

 

[highlakeselectric]

Anybody try there hand at this!!
Warehouse with
480/277 3 phase
535 kw lighting load
With A transformer that feeds
5-10 hp 208 3ph motors
3-25 hp 208 3ph motors
17-20 amp punch press receptacles
8-30 amp receptacles
250 general purpose receptacles
Just to figure the loads


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[kwired]

Anybody try there hand at this!!
Warehouse with
480/277 3 phase
535 kw lighting load
With A transformer that feeds
5-10 hp 208 3ph motors
3-25 hp 208 3ph motors
17-20 amp punch press receptacles
8-30 amp receptacles
250 general purpose receptacles
Just to figure the loads


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Do you have an assumed correct answer or are you just throwing that one out there to see what answers you get?

IMO regardless of what NEC may say, in reality the maximum expected load could be sum of those items (some of which are not really that specific of what the load is) or some items may never draw their full load rating and/or some items possibly never run at same time as another item.

General purpose receptacles with no designated load are likely to be calculated for NEC at 180 volt amps per receptacle.
The 10 and 25 HP motors - could be figured at sum of table values at end of 430, but is possible some may never run while others do - then you only need to account for highest load at any given time.

20 amp punch press receptacles - do the presses draw 20 amps - probably not - same for mentioned 30 amp receptacles, the actual load on them is likely less then 30. Then there is the possibility that whatever plugs into those receptacles is only one utilization equipment but is portable and has several places it may be able to be plugged into.

All that said, IMO not enough details have been given.

 

[highlakeselectric]

Got a few calc questions anyone help?
Copy center in a large building with other retail spaces
Copy center 50x100 120 240 volt single phase flourescent lights heat and ac not on at same time
200' 120 v fixed multioutlet assembly intermittent use
11-16.7 amp 120 v copy machines
1-240 v 54.5 amp a/c
1-15kw 240 v electric heat
4-20 amp 120 volt dedicated computer circuits
2-12 amp 120 volt color copiers
1-120 volt 1200 watt laminator

So here goes the questions asked with answers
Minimum demand in amps for feeder 323 300 276 375

Minimum nuetral load in amps for feeder 276 300 242 256

Minimum overcurrent device? Did not get the choices but what do you guys think ?


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[highlakeselectric]

That last question was half of the last part of supervisors test


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[highlakeselectric]

Anyone ? I could not get any of the answers listed . Im unsure on which part of the receptacles listed could be derated


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[kwired]

Anyone ? I could not get any of the answers listed . Im unsure on which part of the receptacles listed could be derated


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I come up with 375 once but now can't figure out how to get it again, may have been an error and just was luck to match one of their answers:roll: But took a little thinking on what they maybe intended to require. This space is not too absolute as to which general lighting load should apply or if it should have receptacle loads treated like "banks and office buildings".

Also just what are we supposed to do with 4-20 amp dedicated computer circuits? We either need a specific load or number of receptacles or it means nothing.

Only thing I see that could have a demand factor applied is the multioutlet assembly - but there is not enough of it to get to the level where demand factor can apply. Needs to be over 10kVA and we only have 7200.

 

[highlakeselectric]

That was my problem they never give enough info to determine how to calculate these questions so i cant figure out how to study for them


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[kwired]

That was my problem they never give enough info to determine how to calculate these questions so i cant figure out how to study for them


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I think that was a little too difficult of a situation with many unknowns as it was presented to use for a test question myself. One other thing that I thought of but still didn't work out to their answers was that this described application likely requires at least 1200 VA from 220.14(F) for the circuit required in 600.5(A). There are other loads that typically may be there but were not mentioned like general receptacle loads other then the multioutlet assembly, and how many outlets (or a specific load level) for the 20 amp computer circuits- so still a poor question IMO. They needed to use an example that didn't have so many questionable variables, or clearly state which of those variables they intend to apply.

 

[highlakeselectric]

Any one else wanna try this ? Har to understand how the want recpt loads calculated


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[highlakeselectric]

Anybody else wanna try this ? Still cant come up with any of the answers they provided


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[joshua davis]

warehouse lighting is continous duty

warehouse lighting is continous duty

I think you are correct about the receptacle load (the above was just for lighting load), but "D" isn't correct. None of them are.

The demand would be...

Lighting: 250,000sf x 1/4va/sf = 62,500VA.
Lighting demand: 12,500va @ 100% + remaining at 50% = 37,500va.

Receptacle: 500 rec @ 180VA/rec = 90,000VA.
Receptacle demand: 10,000 @ 100% + remaining at 50% = 50,000va.

Total demand = 37,500va + 50,000va = 87,500VA which equals 243 Amps @ 208/120V.

But 243A wasn't one of the selections - "none of the above."

250,000*.25*1.25=78,125
500*180=90,000
90,000+78,125=168,125
As per table 220.42 12,500@100
155,625@50=77,813
77,813+12,500=90,313
90,313?360.25=250.6
As per 220.5 (b) 251 the answer was there

 

[kwired]

250,000*.25*1.25=78,125
500*180=90,000
90,000+78,125=168,125
As per table 220.42 12,500@100
155,625@50=77,813
77,813+12,500=90,313
90,313?360.25=250.6
As per 220.5 (b) 251 the answer was there

I don't see that you need to apply 1.25 factor to the general lighting load from 220.12. if you take that out you come up with 229 amps - also one of the possible answers listed - but IMO still not correct answer because 220.42 only applies to "that portion of the total branch-circuit load calculated for general illumination" but you applied it to the receptacle outlets plus the lighting, I think David's calculation was correct.