Motor overload protection device

[saw]

Hi there,

Question to the one who are expert in NEC! to calculate the size of overload protection device for motor, we use the currents listed on table 430-247 through 430-251 plus additional percentage of the currents depends on what type of motors we work with. What I don't understand is the gray portion of the notes on end of 430-32((A)(1) in NEC 2011 that it says:

"Motors are required to be protected from overloads. To portect a motor from an overload, the motor nameplate full-load current is used to select the overload protection rather than the full-load current values from Tables 430-248 through 430-250, which are used to design the feeder and branch circuit wrining. "

Does that mean we use the full load current of the motor not the NEC tables?

Thank you in advanced for your help

 

[knight1fox3]

In general practical application, when you don't know the actual motor FLA, the NEC tables are a good conservative substitute. However, when you have the motor nameplate information, that should be used when calculating overload protection.

 

[saw]

In NCEES problem 510 there is enough information to find what the FLA is but the book still used the NEC value. Is that because the problem did not specifically mentioned the amount of the full load current?

 

[knight1fox3]

In NCEES problem 510 there is enough information to find what the FLA is but the book still used the NEC value. Is that because the problem did not specifically mentioned the amount of the full load current?

I would assume so, yes. Again in practical real-world application, calculated full-load current can differ from actual test load current. It's best to go with the figure that is most accurate.

 

[saw]

Thank you

 

[cupojoe PE PMP]

In NCEES problem 510 there is enough information to find what the FLA is but the book still used the NEC value. Is that because the problem did not specifically mentioned the amount of the full load current?

I would assume so, yes. Again in practical real-world application, calculated full-load current can differ from actual test load current. It's best to go with the figure that is most accurate.

I haven't actually read your problem, but 430.6(A)(1) requires the use of the FLC listed in the tables NOT the FLA on the motor name place. It's a common mistake in these problems and in the real world. With some exceptions, the NEC requires the FLC to be used for nearly every sizing application, including motor protection, and cable ampacity. I agree with KF that in real-world situations most people use the motor nameplate FLA, but may be incorrect. You will also notice that almost universally the table FLCs work out to be slightly higher than A=HP*746/v, I've often thought this is because they are being slightly conservative and assuming the voltage will run low (thus amps may run high). In my experience, this true in many cases.