Originally Posted By: pabernathy This post was automatically imported from our archived forum.
If you are refering to the 5HP motor question…no power factor needed in that equation…if no factor is listed per the NEC you have to assume ONE ( listed per code )…or none is needed in the equation…but 12 AWG is WAY too small…talk about a fire hazzard…lol
Remember NEC exams will not ask you anything to be factored in other than information given.
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Originally Posted By: tallen This post was automatically imported from our archived forum.
pabernathy wrote:
ok...here is one for you...very simple...
A 5hp motor is located 90 feet from a 120/240v panelboard. What size conductor should be used if the motor nameplate indicated 52 ampere at 115volts? Terminal rated for 75 degree C.
In the field type question....give it a shot ole sparky !
I did it what about you???
-- I have put the past behind me,
where , however, it now sits, making rude remarks.
Originally Posted By: pabernathy This post was automatically imported from our archived forum.
I know the answer…post yours…FIRST and how you came up with it fella…Old Sparky
Question reading 101....I did not ask you for the breaker size....I asked for the conductor size....
You stated : 45 amp min breaker at 100 %....not what I asked for in the question......
First thing we teach....read the question slowly and understand what the proctor is asking...lol
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Originally Posted By: pabernathy This post was automatically imported from our archived forum.
Aw heck…I am getting tired as it is past midnight here…
Here is the answer for all you electrical wiz people who know this....
CM= 2 x K x I x D
_____________
VD ( Voltage Dropped )
K= 12.9 Ohms, copper ( we always ASSUME copper in NEC calculations if not mentioned. )
I = 52 amperes at 115 volts
D = 90 feet
VD = 120 volts x 0.03= 3.6 volts
CM = 2 x 12.9 x 52 x 90
_______________( divided )
3.6 volts
= 33,540 CM ( Chapter 9, table 8 of the NEC )
Use a # 4 THHN.........be happy.....![icon_smile.gif](upload://b6iczyK1ETUUqRUc4PAkX83GF2O.gif)
-- Paul W. Abernathy- NACHI Certified
Electrical Service Specialists
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Originally Posted By: pabernathy This post was automatically imported from our archived forum.
Oh crap…I only need one more post to reach 100…so I will answer my other questions so I can get 100…yeeehhhaaaaaa ( it is late…sorry )
What is the current in ampere of a three-phase, 18KW, 208 volt load?
Easy one.... W
______
E x 1.732 ( square root of 3 )
Answer is : 49.96...rounding rule....up..up..up...50 Amps
or
What is the power loss in watts for a conductor tha carries 24 ampere and has a voltage drop of 7.2 volts?
Ok.....easy as well.....everyday electrician us it...
Question..what are you looking for? Power right....
P=ExI so E = 7.2 volts and I = 24 amps so....
Answer is 172.8 Watts lost per hour
Night all....
-- Paul W. Abernathy- NACHI Certified
Electrical Service Specialists
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Originally Posted By: pabernathy This post was automatically imported from our archived forum.
I know Jeff… me and Mr. Allen got carried away at a Pi@@ing contest…lol…he removed his question which I still dont know if it is right or wrong…lol…so then I posted some everyday field electrician math questions and well…I got carried away…lol
Mr. Allen was a good sport about it........it was fun !
-- Paul W. Abernathy- NACHI Certified
Electrical Service Specialists
Licensed Master Electrician
Electrical Contractor
President of NACHI Central Virginia Chapter
NEC Instructor
Moderator @ Doityourself.com
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Originally Posted By: pabernathy This post was automatically imported from our archived forum.
So I will leave with ONE additional EASY electrical calculation for everyone to answer while they read my long…winded…boring post…lol
Q : The true power of a single phase, 2.1 KVA load with a power factor of 91 percent is _________.
heck I will make it multiple choice since this is NOT a question I came up with...lol...but I know how to calculate it.
a.) 2.1kW b.) 1.91 kW c.) 1.75kW d.) 1,911 kW
Ok all you HI sparkies....fire away.....
EDIT # 1 - It is 9:55AM still not attempts to answer this question.
-- Paul W. Abernathy- NACHI Certified
Electrical Service Specialists
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Originally Posted By: pabernathy This post was automatically imported from our archived forum.
Award goes to :jwortham
The correct answer is : 1.91kW
-- Paul W. Abernathy- NACHI Certified
Electrical Service Specialists
Licensed Master Electrician
Electrical Contractor
President of NACHI Central Virginia Chapter
NEC Instructor
Moderator @ Doityourself.com
Visit our website- www.electrical-ess.com