Load calculation for large flat storage

[ctaylo360860]

IMHO, the best for you to do is add the kVAs of the loads as classified (continuous and non-continuous).
Be sure to express the loads properly in kVA, assume 0.8 PF for 3-phase motors and 0.7 PF for single-phase motors (if you can't have those pf figures from the equipment nameplates).
For the load distribution to be correct, divide the kVA that you can come up with by 3 for three-phase and place the kVA per phase in each of the line loads. The single phase motors and load should be placed in either phase once.
My calcs here tell me you only have a total demand of 117.4 kVA for a total demand amps of 141.2A.

I’ve wrote all my motors and ocpd in a diagram. I have used the methods from ecmag that reference the nec
I found all my flc using 430.250/430.248.
I wrote the nameplate sizes down and sized my ol protection for each motor using 430.32
I sized my branch circuits using 430.22. I took the flc given in 430.250/430.248 and multiplied by 125%
I used 310.16 to reference branch circuit conductor size.
I then sized my ocpd’s for my branch circuits using 430.52 this should be done by control builder I’m guessing. I used inverse time breaker, but not sure if I need to use the instantaneous Column because I’m guessing the builder will use instantaneous breaker in combo with a motor controller.
From there I used 430.63 and 430.24
I took largest motor load 40ax1.25=50a
I added the continuous loadx1.25=26.25
I added the non motor loadx1.25=8.3
I added the sum of the full load current rating dog all other motor loads (3.4x3)+9.8+12=32
Giving me a total load of 116.55amps
I can now size my feeders and feeder protection
I used 400.6 and 310.16
I’m planning on running either 1/0 or 2/0 aluminum from feeder protection to a 200a disconnect
I will use #2 -4c type w so chord from the disconnect to the conveyer control panel breaker. It should be good for 154a t 75 degree

now for feeder protection I use 430.63 and 430.62 rating I’d the motor feeder short circuit ground fault protection device is determined by adding the largest branch circuit short circuit and ground fault protective device for any motor supplied by the feeder to the sum of the full load currents of all other motors supplied by that feeder. I took my largest branch circuit 100a+21+3.4+3.4+3.4+9.8+12= 159.667 I can’t have a breaker larger than 159.667 using 240.6 I can go down to a 150a breaker for my feeder protection. Would I be allowed to go down even further say a 125a breaker? I’m wondering because that is how existing conveyer/drive over is on…

 

[ctaylo360860]

Here is the article I read to come up with the methods I used for my calculations.

StackPath

 

[ctaylo360860]

IMHO, the best for you to do is add the kVAs of the loads as classified (continuous and non-continuous).
Be sure to express the loads properly in kVA, assume 0.8 PF for 3-phase motors and 0.7 PF for single-phase motors (if you can't have those pf figures from the equipment nameplates).
For the load distribution to be correct, divide the kVA that you can come up with by 3 for three-phase and place the kVA per phase in each of the line loads. The single phase motors and load should be placed in either phase once.
My calcs here tell me you only have a total demand of 117.4 kVA for a total demand amps of 141.2A.

here is a diagram of what i have done, it should show all specs of motors and code articles used. I don't know if i need to add the single phase motor/ lighting loads to my 480v feeder caluclation? mabye i need to just add the load that the step down transformer will pull supplying the 1 phase motor/ lighting loads?

 

[ctaylo360860]

Look at 430.24 (1) thru (4).

Then down to exception #3.

looked into what you said a little more and something clicked! thanks for all the help. this is what i have designed as of now... the only thing i'm not sure i have correct is the sizing of my ocpd for the feeders, and the main breaker inside the control panel...