NCEES Question #510

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

TakiTaki

Member
Joined
Aug 31, 2019
Messages
8
Reaction score
2
A 50-hp, 460-V, 3-phase induction motor is served by a single feeder conduit... and it is rated for continuous duty... Starter running overload protection is available in 5-A increments... the initial maximum sized protection device (amperes) that can be applied is most nearly:

This might be a very dumb question but where is the current 65 [A] coming from? I understand the solution and I know why we need to multiply it by 1.15 and round down to get the final answer but maybe I am missing a step here. Any help is appreciated.

Answer from NCEES: 65(1.15) 74.75 [A] using 430.32(A)(1)

Round down and it equals 70 [A]

 
To find the maximum sized protection device, you need to know the motor nameplate full load current.  The problem does not provide this, so you have to look it up based on the information they give you.

Use Table 430.250 (for 3 phase motors) to get the full load current at 50HP, 460V to get 65A.

 
Last edited by a moderator:
A 50-hp, 460-V, 3-phase induction motor is served by a single feeder conduit... and it is rated for continuous duty... Starter running overload protection is available in 5-A increments... the initial maximum sized protection device (amperes) that can be applied is most nearly:

This might be a very dumb question but where is the current 65 [A] coming from? I understand the solution and I know why we need to multiply it by 1.15 and round down to get the final answer but maybe I am missing a step here. Any help is appreciated.

Answer from NCEES: 65(1.15) 74.75 [A] using 430.32(A)(1)

Round down and it equals 70 [A]
To be honest, this is a question I always wonder if NCEES made a mistake on. The NEC is pretty clear on using the name plate full load amps to size separate overload protection. The NEC is also very clear that motor FLC table values are for sizing OCPD and conductors. I guess since NCEES left out the nameplate FLA, you are forced to use the NEC FLC instead, but I still think it is a slight error. 

 
To be honest, this is a question I always wonder if NCEES made a mistake on. The NEC is pretty clear on using the name plate full load amps to size separate overload protection. The NEC is also very clear that motor FLC table values are for sizing OCPD and conductors. I guess since NCEES left out the nameplate FLA, you are forced to use the NEC FLC instead, but I still think it is a slight error. 
I think this question may issued considering the permissions under article 440.41 (A) :"In case the motor controller is rated in horsepower but is without one or both of the foregoing current ratings, equivalent currents shall be determined from the ratings as follows. Table 430.248, Table 430.249, and Table 430.250 shall be used to determine the equivalent full-load
current rating. Table 430.251(A) and Table 430.251(B) shall be used to determine the equivalent locked-rotor current ratings.
", and article 430.83(D). 

 
I think this question may issued considering the permissions under article 440.41 (A) :"In case the motor controller is rated in horsepower but is without one or both of the foregoing current ratings, equivalent currents shall be determined from the ratings as follows. Table 430.248, Table 430.249, and Table 430.250 shall be used to determine the equivalent full-load
current rating. Table 430.251(A) and Table 430.251(B) shall be used to determine the equivalent locked-rotor current ratings.
", and article 430.83(D). 


Hi @Mohammed Kamel,

For me, the issue is that the NCEES® solution appears to be using the FLC from the NEC® Table 430.245 for a 50 Hp, 460 V, 3 phase motor even though they do not mention this table in the solution. 

Using the Table 430.245 FLC value to size the over load directly contradicts NEC® 430.32(A)(1) which states to size separate overload protection as a "percent of the motor nameplate full-load current rating"

The nameplate full load current rating is the FLC stamped on the motor nameplate from the manufacturer, which is usually a slightly different value than the NEC FLC table values. 

If you dig into Article 430 in the NEC, the FLC tables (430.247-250) are used for selecting the FLC of the motor for sizing OCPDs and Conductors, and the nameplate full load current value is used for sizing overloads. If you have it, the handbook edition of the code book spells this out even further. 

The nameplate full load current rating is never given in the problem, and NCEES® does not mention that they are using the FLC rating from Table 430.245 based on the motor ratings in the problem. 

Also, careful with 440.41, this section appears to only apply to motor controllers that are specifically for motor compressors, and does not appear to cover overloads until 440.52(A)(1).

 
Does anyone know why we need to multiply the current by 1.15 and round down to get the final answer? I do not understand the rounding down since the question asks for the maximum sized protection device that can be applied
 
Does anyone know why we need to multiply the current by 1.15 and round down to get the final answer? I do not understand the rounding down since the question asks for the maximum sized protection device that can be applied

The 1.15 FLC multiplier comes from NEC® 430.32(A)(1) Continuous-Duty Motors Separate Overload Device.
 

Latest posts

Back
Top