Motor calc for European motor

[brother]

I haven't had to deal with foreign motors much. So I want to bounce off the calculation off of someone else since most of the US motors fit nicely into the NEC 2008 table 430.250.

this motor is a 5.5 HP 460volt 40degree c 1.15 service factor 7.3 nameplate amps. 3 phase AC.

Since there wasn't a 5.5 HP on the table, I used the 5 hp and 1/2 hp and added those full load currents together, 5 hp = 7.6 1/2hp=1.1 7.6+1.1= 8.7 flc

8.7 X 1.25= 10.875 minimum condutor size 8.7X 2.5= 21.75 shortcircuit/ground fault protection

Was it ok to just add the 1/2 hp and 5 hp together that way to get the actual flc since we dont have a 5.5 on our table, maybe Europe has that on their table? I was concerned ,probably for nothing, to be sure the math was right.

 

[Smart $]

I haven't had to deal with foreign motors much. So I want to bounce off the calculation off of someone else since most of the US motors fit nicely into the NEC 2008 table 430.250.

this motor is a 5.5 HP 460volt 40degree c 1.15 service factor 7.3 nameplate amps. 3 phase AC.

Since there wasn't a 5.5 HP on the table, I used the 5 hp and 1/2 hp and added those full load currents together, 5 hp = 7.6 1/2hp=1.1 7.6+1.1= 8.7 flc

8.7 X 1.25= 10.875 minimum condutor size 8.7X 2.5= 21.75 shortcircuit/ground fault protection

Was it ok to just add the 1/2 hp and 5 hp together that way to get the actual flc since we dont have a 5.5 on our table, maybe Europe has that on their table? I was concerned ,probably for nothing, to be sure the math was right.

Does the motor nameplate actually say the motor is 5.5HP?

There is no prescribed method for determining conductor or equipment ampacities or ratings where nameplate HP does not correspond with those listed in the Table.

There is a prescribed [general] method for determining conductor or equipment ampacities or ratings where the motor is marked in amperes, but not horsepower. 430.6(A)(1) states in part, "Where a motor is marked in amperes, but not horsepower, the horsepower rating shall be assumed to be that corresponding to the value given in Table 430.247, Table 430.248, Table 430.249, and Table 430.250, interpolated if necessary."

If you were to use the basic linear interpolation method with your 5.5HP rating and Table values below (5HP, 7.6A) and above (7?HP, 11A) you'd get:

(5.5-5)?(7?-5)?(11-7.6)+7.6=8.28A​

If you were to use the nameplate amperes (7.3A) and interpolate amperes to HP you can automatically see the motor would be rated less than 5HP.

 

[GeorgeB]

Does the motor nameplate actually say the motor is 5.5HP?

I've seen 4kW (5.5HP) before ...

If you were to use the basic linear interpolation method with your 5.5HP rating and Table values below (5HP, 7.6A) and above (7?HP, 11A) you'd get:

(5.5-5)?(7?-5)?(11-7.6)+7.6=8.28A​

Yes, pretty good; I'd have said that 5.5 was 110% of 5 and used 110% of 7.3 ... 8.0 or 8.1 ... in my decisions.

If you were to use the nameplate amperes (7.3A) and interpolate amperes to HP you can automatically see the motor would be rated less than 5HP.

I don't try to work backwards for this ... different motor constructions will have different efficiencies and power factors. Take what they give (ASSUMING SAME FREQUENCY) and make scaling corrections.

 

[Smart $]

I've seen 4kW (5.5HP) before ...

Are you saying the "(5.5HP)" was on the nameplate, or just the 4kW? If just the 4kW, then the NEC, under 430.6(A)(1), says you'd have to interpolate the nameplate amperes backwards to get HP. In the OP case, that would put the HP at less than 5.

Also, technically, 4kW is only 5.36 horsepower (assuming input)... and we all know motors are not 100% efficient.

Yes, pretty good; I'd have said that 5.5 was 110% of 5 and used 110% of 7.3 ... 8.0 or 8.1 ... in my decisions.

As I said, there is no prescribed method going from HP to amperes. So your method is just as valid as any other.

I don't try to work backwards for this ... different motor constructions will have different efficiencies and power factors. Take what they give (ASSUMING SAME FREQUENCY) and make scaling corrections.

If nameplate info' provides ampere data but not horsepower, you have to work backwards to get horsepower... but I don't see the point in doing so, because the Code don't use unconverted HP for anything, does it?

 

[brother]

Thanks for th info. I'm pretty sure the hp was 5.5 on the name plate, checked it twice.

So basically you're saying I just use the nameplate amps and go from there to hp??

750 watts in 1hp if I recall correctly.

 

[Smart $]

Thanks for th info. I'm pretty sure the hp was 5.5 on the name plate, checked it twice.

So basically you're saying I just use the nameplate amps and go from there to hp??

750 watts in 1hp if I recall correctly.

IMO, Article 430 Tables are based on NEMA designs... and using a non-NEMA design motor can be difficult, due to an intermixing of conventions.

For GF and SC protection, I would interpolate your motor as being a 5HP motor, since the Table amperes for a 5HP motor is higher than your motor's nameplate marking of 7.3A. 5.5HP namplate marking on a non-NEMA design motor could be the electrical input HP. IIRC, NEMA design motor HP is mechanical output. Input ? Efficiency = Output, and we know motors are not 100% efficient.

For OL protection, it would be safer to use nameplate amperes, at least initially.

Question: Does the nameplate indicate frequency? If a 50Hz motor, it will have a bearing on all your data.