Oregon Supervisor Test....

[highlakeselectric]

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

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

3] Minimum size conductor in THWN per phase feeding 1200 amp 480 volt 3 phase load 1800 ft from source minimum 3% voltage drop?

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

480/277
535 KW lighting load
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
8-30 amp receptacles
250-general purpose receptacles


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 amps?

Thanks for your time and help!

 

[pjholguin]

Just some inputs, I remember when I took my General Supervisor test(1995)...it doesn't sound like it has changed. Yes, the second part is tough(that's why there are only 2 questions), but if you know how calculate loads and size of electrical components...you shouldn't have any issues except running out of time. Be sure to know codes pertaining to Electrical calculation questions. I made sure I had my NEC Handbook, the handbook has a lot of good information. I hope all went well on your exam. Let us know how you did.

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

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

3] Minimum size conductor in THWN per phase feeding 1200 amp 480 volt 3 phase load 1800 ft from source minimum 3% voltage drop?

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

480/277
535 KW lighting load
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
8-30 amp receptacles
250-general purpose receptacles


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 amps?

Thanks for your time and help!

 

[highlakeselectric]

Thanks im pretty sure i passed the first part easily but the second i was thrown off. The way they worded the questions was unlike how i studied calcs . Thanks for the info im going to keep studying calcs till i know them in and out. If anyone out there can help me answer the ones i posted it would be a great help thanks..

 

[highlakeselectric]

......

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How would I calculate this question if it's in KW and I have no power factor for the main second part of the calculation portion? IE 535 KW lighting load as posted above, I have searched my calc books and still am lost. Thanks

 

[Dennis Alwon]

I see the lighting load as 3 watts/sq.ft. 500x500= 250000 x 3 = 750,000. First 12,500 @100% rest at 50% (T. 220.42) 750,000 - 12,500 = 737,500 . 737,500 X .5 = 368,750.

368,750 + 12500 = 381,250 watts lighting load.

 

[david luchini]

I see the lighting load as 3 watts/sq.ft. 500x500= 250000 x 3 = 750,000. First 12,500 @100% rest at 50% (T. 220.42) 750,000 - 12,500 = 737,500 . 737,500 X .5 = 368,750.

368,750 + 12500 = 381,250 watts lighting load.

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."

 

[highlakeselectric]

That is what i read in the code .75 va x 250000 then add the recepts at 180 right? And do recepts get reduced by 220.44 deman factors? Im lost kind of. Ps thanks for everyone helping me!

 

[augie47]

I see the lighting load as 3 watts/sq.ft. 500x500= 250000 x 3 = 750,000. First 12,500 @100% rest at 50% (T. 220.42) 750,000 - 12,500 = 737,500 . 737,500 X .5 = 368,750.

368,750 + 12500 = 381,250 watts lighting load.

warehouse, Dennis... believe that's 1/4w ft?
for #2 I got 77k
#3 1200 amp load.. 1 conductor per phase 3%.. I don't think they make a condcuctor that large.

 

[augie47]

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."

david, I definitely may be mistaken, but for non-dwelling I don't think receptacle load is included in the lighting demand and must be added at 180va.

edit..forgot to include demand on recept when I obtained answerr

 

[david luchini]

david, I definitely may be mistaken, but for non-dwelling I don't think receptacle load is included in the lighting demand and must be added at 180va. If I'm correct the answer is then "D", 229

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."

 

[highlakeselectric]

Thanks david i agree with your last post the asnswers i posted were the only choices. It seemed like a bunch of trick questions or the state calculates load differently then everyone else. Thanks again

 

[david luchini]

Thanks david i agree with your last post the asnswers i posted were the only choices. It seemed like a bunch of trick questions or the state calculates load differently then everyone else. Thanks again

The answer they are looking for is A) 265....You should get extra credit for figuring out how they get their answer.

To calculate max available short circuit current through a transformer, you would multiply the rated secondary current by 100, then divide by the percent impedance. So 1387.86A * 100 / 1.8 = 77,103 amps available.

How would I calculate this question if it's in KW and I have no power factor for the main second part of the calculation portion? IE 535 KW lighting load as posted above, I have searched my calc books and still am lost. Thanks

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

 

[augie47]

FWIW My answer for #1 is A
500 x 500 = X 1/4 = 62500 watts
continuous load = 78125
1st 12500 @ 100% 12500
remainder at 50% 32812


Recept
500 x 180 = 90000
1st 10k at 100% 10000
bal at 50% 40000

Total 95312 w
@ 208/120 phase = 264

For #1 my answer would be A

 

[highlakeselectric]

Thank you for that info . That will help me on the next one ps for anyone reading this and might be testing almost all choices for answers on the calc portion were feeders in multiple pipes with derating and all that so you have to know that as well parallel feeds

 

[david luchini]

FWIW My answer for #1 is A
500 x 500 = X 1/4 = 62500 watts
continuous load = 78125
1st 12500 @ 100% 12500
remainder at 50% 32812


Recept
500 x 180 = 90000
1st 10k at 100% 10000
bal at 50% 40000

Total 95312 w
@ 208/120 phase = 264

For #1 my answer would be A

Extra credit for you, Gus :happyyes:

264A is still the wrong answer, though. "Continuous loads" don't have anything to do with demand factors.

For instance, if the feeder for that lighting and receptacle load was in a raceway with another feeder such that there were 6 ccc's, and the feeder was 300mcm THHN, with an adjusted ampacity of 256A (320*0.8), would that feeder be adequate to supply the demand load?

 

[augie47]

I agree, but apparently the guy who wrote the test didn't

 

[david luchini]

I agree, but apparently the guy who wrote the test didn't

:thumbsup:

Makes you wonder what's the point of the test, if the test answers aren't even right.:happysad:

 

[augie47]

You might be able to prove me wrong, but as I recall, there is no specific wording in the Code that states "commercial lighting shall be considered a continuous load", yet it occurs so often in testing and calculating.
With the new energy codes and the increased use in occupancy sensors it seems less likely today that lighting loads would be continuous.
On the other hand, with the new energy codes prohibiting any oversizing of HVAC units it seems possible that during extreme weather the units may actually run a full load greater than 3 hours but we never consider them continuous
Add into that the requirement that water heater storage type branch circuits are required to be continuous... never saw one that was
Strange

 

[pjholguin]

That's great...do you still have to wait 2 weeks for test results? Maybe, call for your test results. Once, you have your results and if part 2 is as you say contact the Electrical BCD and ask if you can sit down with the Chief or Assistant Chief to review that potion of the test.

Thanks im pretty sure i passed the first part easily but the second i was thrown off. The way they worded the questions was unlike how i studied calcs . Thanks for the info im going to keep studying calcs till i know them in and out. If anyone out there can help me answer the ones i posted it would be a great help thanks..

 

[highlakeselectric]

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