Load calculation for large flat storage

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ctaylo360860

ctaylo360860

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I’m going to be installing a service for a large flat corn storage bunk. I’m trying to do a load calc and have a few questions on this one. The bunk will have 8-10hp 3 phase motors operating on 480volts There is also a conveyer and a drive over that need to be ran to fill the bunk.

The conveyer has 1-30hp conveyer motor, 2- 2hp front and rear axle motors, all are 480v 3ph motors, it also has 1- 1/2hp hood actuator @ 115volt 1 ph, and 1-2hp 230v 1ph hydraulic motor. And also 2 400watt lights

the drive over has a 1-15hp conveyer motor, and 1-2hp hydraulic motor with a 30 minute duty rating, and 2-400 watt lights

first question I have is I’m planning on running my conveyer and driver over off one disconnect and breaker. How do I figure the load of these two pieces of equipment? Would I find my largest motor and multiply 1.25x whatever fla 4.30.250 gives me for a 30hp? Then find my continuous load x 1.25 + non continuous load. Find the sum of al these loads and then size my main disconnect ocpd using 240.6.?

How do I figure the load on the hood actuator? Would I use single phase motor chart in 430, or use the nameplate rating?

to size my ocpd and my disconnect for my conveyer/drive over would I find the largest motor on the equipment, use the fla found in the in 430.250 “3phase motor flat char” and multiply by 1.25? Then add the continuous loadx1.25 + the non continuous load. Side my ocpd using 240.6?
 
Feeder conductors: I would typically use 1.25 * the largest single load plus the remaining. Use the charts.

The largest load being all the motors that start at one time. A second or two can make a difference.

The panel builder you selected should have this on their label.
We are looking at a piece of equipment that asks for and will get a 70 amp breaker with #4 SO cord. They list the FLA.

Real world...it could run on 50 amp or less with #6. Big difference if you are the one dragging that cord.
 
Using the charts in 430, I get 112.45 ampers. The existing conveyer has a 125a breaker that feeds a 100a disconnect with 2/0 aluminum, and then they have #4 sow chord feeding the control box. Doesn’t seem right… 5A30EC69-849D-47E7-825D-8EF8366948F3.jpeg
 
ptonsparky said:
Feeder conductors: I would typically use 1.25 * the largest single load plus the remaining. Use the charts.

The largest load being all the motors that start at one time. A second or two can make a difference.

The panel builder you selected should have this on their label.
We are looking at a piece of equipment that asks for and will get a 70 amp breaker with #4 SO cord. They list the FLA.

Real world...it could run on 50 amp or less with #6. Big difference if you are the one dragging that cord.
Click to expand...
So the largest load should be the 30hp. The motors shouldn’t really ever start at the same time. The 30hp and 15hp should be the only two that run together with the exception of them adjusting the small hood actuator. All the other motors are to move the conveyer, or to set the height of drive over. so the only other two motors that would Likely run at the same time would be the 2-2hp front and rear axle motors… I just sent email to control builder. WhT should they have on there label?
 
ptonsparky said:
Feeder conductors: I would typically use 1.25 * the largest single load plus the remaining. Use the charts.

The largest load being all the motors that start at one time. A second or two can make a difference.

The panel builder you selected should have this on their label.
We are looking at a piece of equipment that asks for and will get a 70 amp breaker with #4 SO cord. They list the FLA.

Real world...it could run on 50 amp or less with #6. Big difference if you are the one dragging that cord.
Click to expand...
Here is what the controls look like for the drive over. Currently they have the drive over fed off a 30atwistlock plug and receptacle. Female twist lock is set on the pump panel and is fed from 30a fuse block. When I do the new controls what would be an easy way that I can hardwire the controls for the drive over, or should I wire a pin and sleeve into the new controls so that the drive over can be moved when needed unplugged and not uninstalled/ reinstalled by me every time they move it…
 

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How about making it simpler by adding 125% of all continuous loads (equipment that runs 3 or more hours) and 100% of the non-continuous loads (to get the minimum feeder size NEC 215.2) and then comparing it with your other methods?
 
topgone said:
How about making it simpler by adding 125% of all continuous loads (equipment that runs 3 or more hours) and 100% of the non-continuous loads (to get the minimum feeder size NEC 215.2) and then comparing it with your Other methods
Click to expand...
That is what I have done. Look at post #4 of added in the lighting load after
 
ctaylo360860 said:
That is what I have done. Look at post #4 of added in the lighting load after
Click to expand...
You are adding motor currents of different voltages! I've got 90.96A total without the lights (using 430.250 table). I see that there are some errors also when you read out from the table: 2HP at 460V is 3.4A, not 4.8!
 
topgone said:
You are adding motor currents of different voltages! I've got 90.96A total without the lights (using 430.250 table). I see that there are some errors also when you read out from the table: 2HP at 460V is 3.4A, not 4.8!
Click to expand...
I’m using 250.248 for single phase flc motors. The single phase motors will be fed of some type of step down transformer inside the control box…. I didn’t happen to see anything in 240.23 that states I shouldn’t add motor currents with different voltages…. I re did my math on paper and am going to post it shortly…. Has to re figure 3.4 instead of 4.8amps…
 
So for my total load calculation I get 256.55 amps. For my total drive over/ conveyer I get 116.5amps.
 

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That’s what I’ve been using, except in 2020 book it’s 430.23. My controls aren’t really interlocked… the drive over controls are in separate controls than the conveyer controls…. Guys use the 240v hydraulic ram motor, rear axle front axle to drive big conveyer into place. Once conveyer is in place they bring the drive over set it where it’s needed and plug it into the conveyer controls.they can now start the conveyer motor. Then walk over to drive over controls where they can start the driver over conveyer. At that point they can adjust the hydraulic motor on drive over so trucks can dump corn. They can also turn on the lights or adjust the hood actuator…
 
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.
 
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