Help on a few stumping questions

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Ben M

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Texas
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Electrician
I believe I did this right for voltage drop but just in case. 1. A pole mounted light is 2000 watts, 240 volt single phase and mounted 100 feet away fed with #8 awg wire. What is the voltage at the pole? I got 238.6volts after getting around 1.47 for the drop and minus that from 240v.
2. what is the minimum size inverse time breaker supplying 10- 15hp 3/0 motors 460 volt? Wasn’t sure how to do this one.
3. what size thhn 75 degrees Celsius would feed 2 ac units at 30 amps each?
4. what is the daily power usage in kilowatt hours of 1200 fluorescent lights each at 277 V 1.2 A and on continuous for 14 hours per day. I did 398,880 but the other answers were 6980, 55,084, 4654. I had no clue how to do this one.
Thank you for your help!
 
When dealing with study/test questions we like you to show what you feel is the answer (and basics as to how it was derived0 and many here will
be glad to help you.
This as opposed to simply giving you an answer.
 
A few hints.

1. #8 wire has about 0.6282 Ohms/1000feet. You have 200 feet of wire in the circuit. You can calculate the current and then determine the voltage drop in the wire.

2 and 3 are pretty straightforward if you look at the motor feeder requirements in the code.

4 is kind of a trick question. I don't think there is enough information since it asks for kw-hrs but does not give the power factor.
 
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augie47 said:
When dealing with study/test questions we like you to show what you feel is the answer (and basics as to how it was derived0 and many here will
be glad to help you.
This as opposed to simply giving you an answer.
Click to expand...
If we teach you that you can have other people do your work for you, you are becoming a manager, not an electrician.

That then means we will have to despise you…
 
Please check the answers you show for #4. Did you foul up one of them?
.
I took a stab at #4 and think I did it correctly-- I don't think this type of question is interested in power factor. That's next week's question!
How did you get 398,880 ? Doesn't match my answer!
 
PaulMmn said:
Please check the answers you show for #4. Did you foul up one of them?
.
I took a stab at #4 and think I did it correctly-- I don't think this type of question is interested in power factor. That's next week's question!
How did you get 398,880 ? Doesn't match my answer!
Click to expand...
It says nothing about power factor but mentions it is a fluorescent light load and they tend to have low pf so no way to calculate kw-hrs.
 
PaulMmn said:
Please check the answers you show for #4. Did you foul up one of them?
.
I took a stab at #4 and think I did it correctly-- I don't think this type of question is interested in power factor. That's next week's question!
How did you get 398,880 ? Doesn't match my answer!
Click to expand...
277*1.2*14 = 4654 watt hours per day per lamp if PF is one.
 
petersonra said:
277*1.2*14 = 4654 watt hours per day per lamp if PF is one.
Click to expand...
Extended to 1200 lamps, those calcs have the same digits as one of the answers... almost. That's why I asked if he mis-typed one of the suggested answers for #4. And I match your calcs for a single lamp.
 
Ben M said:
4. what is the daily power usage in kilowatt hours of 1200 fluorescent lights each at 277 V 1.2 A and on continuous for 14 hours per day. I did 398,880 but the other answers were 6980, 55,084, 4654. I had no clue how to do this one.
Click to expand...

Hint #1, remember to multiply power by time, when it asks for kW-hrs, or energy. The dash in kW-hrs, and lack of a "per" in its pronunciation, means units are multiplied.
Hint #2, what does kilo mean? Adapt accordingly.

I got an answer consistent with a plausible misprint of your answers.
 
Kwh would be kw (I X E = P) x hours
I get one of the "listed" answers other than yours....
 
It appears as though the OP hasn't been back after post #4.
 
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