Panel loading calc for motors

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Telaid

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I have a three phase motor/pump, 3 phase 208 volt rated at 2 HP. Im trying to determine panel loading for both continuous and non-continuous loading for purposes of balancing the loads on the phases in the panel. (kva on all the phases)

Ive been through ch 4 but cant find the answer. Can someone point me to the correct section in the NEC?

My interpretation for the continuous loading was to take
2hp* 745.6 = 1491 watts -> 1491/208 = 7.17 amps. using the value from table 430.250 i get FLC of 7.5 amps.

this gives me 1.56 kVA/phase; is this correct? what if this was a intermittent duty motor?

thanks
 
Telaid said:
I have a three phase motor/pump, 3 phase 208 volt rated at 2 HP. Im trying to determine panel loading for both continuous and non-continuous loading for purposes of balancing the loads on the phases in the panel. (kva on all the phases)
Click to expand...

a 3 phase motor has the same current on each leg already
 
Thanks bob, i understand that part; but where im confused is to the kva rating i should use. Im reading the code as saying i should calculate the load as 125% of the FLA of the motor. is this correct?

Then for non- continuous duty motors, is there a section that deals with derating for motors?
 
First you must calculate the breaker size to accomodate inrush square "d" says you could use a 15a breaker for your application but to reduce neusence tripping i would use a 20a. You then take your FLA for the 2hp/3ph motor. FLA=7.8 x Volts=208 x sq root (3) / 1000. That gives you kva. Divide by 3 and that will be your kva per phase. your kva is 2.81/3=.94kva per phase. to reverse engineer for other apps. take your kva for all 3 ph. and add them up. multiply that number by 1000. then divide by the single ph volts or the 3 ph volts. ie... 120v 1 ph = divide by 120 480v 3 ph = divide by 831.
 
Also, motor devices must always be computed as a continuous load. This is due to their inrush characteristic during startup.
 
codster said:
First you must calculate the breaker size to accomodate inrush square "d" says you could use a 15a breaker for your application but to reduce neusence tripping i would use a 20a. You then take your FLA for the 2hp/3ph motor. FLA=7.8 x Volts=208 x sq root (3) / 1000. That gives you kva. Divide by 3 and that will be your kva per phase. your kva is 2.81/3=.94kva per phase. to reverse engineer for other apps. take your kva for all 3 ph. and add them up. multiply that number by 1000. then divide by the single ph volts or the 3 ph volts. ie... 120v 1 ph = divide by 120 480v 3 ph = divide by 831.
Click to expand...
Umm.... your calculation used the wrong FLC value. For a 2HP 208V 3? motor the FLC is 7.5... not 7.8.

For kVA per phase on a wye system, its easier to just multiply FLC by Line to Neutral voltage:
  • 7.5A ? 120 V ? 1000V/kV = 0.9 kVA/Line
As proof:
208V = 120V ? √3
208V x √3 = (120V ? √3) ? √3 = 120V ? 3
208V x √3 ? 3 = (120V ? 3) ? 3 = 120V​
Substitute into the formula:
7.5A ? (208V x √3 ? 3) ? 1000V/kV =
7.5A ? 120 V ? 1000V/kV = 0.9 kVA/Line​

Multiply by 3 for system kVA:
  • 0.9 kVA/Line x 3 = 2.7 kVA, system
You can also use the abbreviated formula for delta systems. The Line to System Neutral voltage is the Line to Line voltage divided by √3. For example:
5HP 460V 3? motor on a 480V Delta system
Table 430.250 FLC = 7.6A
7.6A x (480V ? √3) ? 1000V/kV = 2.1 kVA/Line
2.1 kVA/Line x 3 = 6.3 kVA, system​
 
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