Feeder Calculation to temporary MCC

[mull982]

We are needing to turn a piece of equipment that will require temporary power to some lubrication pumps. This is a new construction project so power is not yet avaliable in the area, so we will need to get power from an existing MCC and run a feeder circuit to a small "portable MCC" arrangement which will have two small starters and two breakers. The following are my loads wich will be fed from the "portable MCC"

All loads are continuous and are rated at 480V:

Motor 1: 24.9A
Motor 2: 24.9A
Heater 1: 1.83A
Heater 2: 1.83A

Adding these 4 loads I come up with 53.46A and taking 125% of the largest motor gives me 66.8A as the connected load off of the portable MCC. Rounding this number up I figure that I will at least need a feeder with an ampacity of 67A to be able to feed this portable MCC. Since the 480V existing MCC being used to feed the portable MCC only has an 100A breaker avaliable I figured that I would have to run a #2 feeder off of the 100A breaker to feed the portable MCC. This would also allow extra capacity on the feeder. Voltage drop will not be an issue.

Does it seem like I have taken the correct steps and made the correct calculations for this application. Do you agree with my design choices?

 

[charlie b]

I don't follow your math, but your design choices seem appropriate.

Adding 24.9, 2.49, 25% of 24.9, 1.83, and 1.83 gives me 59.78, not 66.8.

Also, where did you get the value of 24.9? We generally start with the motor HP, and look up the value of current in Table 430.250, instead of taking current values off the motor nameplate.

 

[charlie b]

Let me add that I am assuming the current value for the heaters (1.83 amps each) has the 125% built in (i.e., for their being a continuous load). But if not, adding that in still does not bring the total to 66.8. :-?

 

[charlie b]

OK. I see it now. You added an extra 25% of everything.

(24.9 plus 24.9 plus 1.83 plus 1.83) times 1.25 equals 66.8

You didn't need to add 25% to the second motor.

 

[mull982]

OK. I see it now. You added an extra 25% of everything.

(24.9 plus 24.9 plus 1.83 plus 1.83) times 1.25 equals 66.8

You didn't need to add 25% to the second motor.

Yes this is what I did, added all of the loads and then added 25% of the largest motor. You are saying that I do not need to add this 25% of the largest motor onto the total load? Do I only add this 25% onto the largest motor and then add all of the loads? Should I have added added a 25% load onto my heaters since they are continuous loads?

Also, where did you get the value of 24.9? We generally start with the motor HP, and look up the value of current in Table 430.250, instead of taking current values off the motor nameplate.

The motor outputs were not given in hp but were given in kw (german motors) with an associated FLA rating.

 

[charlie b]

Yes this is what I did, added all of the loads and then added 25% of the largest motor.

That you did. But afterwards you also added 25% of the second motor, 25% of the first heater, and 25% of the second heater. In other words, you took all the loads (24.9 + 24.9 + 1.83 + 1.83), a total of 53.46, and added 25% (13.36) to that, and wound up with your 66.8. You could have omitted the 25% of the second motor. That would have given you a total of 60.6.

 

[mull982]

That you did. But afterwards you also added 25% of the second motor, 25% of the first heater, and 25% of the second heater. In other words, you took all the loads (24.9 + 24.9 + 1.83 + 1.83), a total of 53.46, and added 25% (13.36) to that, and wound up with your 66.8. You could have omitted the 25% of the second motor. That would have given you a total of 60.6.

Ok I follow what you are saying with only adding 25% to only one of the motors. The 1.83 value for the heaters did not account for 25% for the continuous loads, so I guess I have to add 25% for each heater (or 25% to the combined heater load) since they are a continuous load. Doing this I come up with a new total of 60.6A!

So I guess in summary you take 125% of your largest motor only, and 125% of your total connected continuous load to come up with your total connected load.