validation

[cjbouchard@bergco.com]

Hello all my stupid question is to validate a Ongoing debate 120/208 v
we have 6 20 amp 120v circuits drawing 16 amps each 1920w x6 =96 amps correct !
we also have a environmental control unit 3 phase 208 5500w divided by 360 =15.27 amps +96 amps total load would be 111.27 = 125 amp main ,am i thinking about this right or should I be taking the total watts of both 1920 and 5500 adding together total watts 1920+5500 divided by 360 for total ,Load on the 120/208 service My thought is you have to separate the 3 phase loads form the single phase loads,

 

[infinity]

Hello all my stupid question is to validate a Ongoing debate 120/208 v
we have 6-20 amp 120v circuits drawing 16 amps each, 1920w x6 =96 amps correct !
we also have a environmental control unit 3 phase 208 5500w divided by 360 =15.27 amps +96 amps total load would be 111.27 = 125 amp main ,am i thinking about this right or should I be taking the total watts of both 1920 and 5500 adding together total watts 1920+5500 divided by 360 for total ,Load on the 120/208 service My thought is you have to separate the 3 phase loads form the single phase loads,

I'll let an engineer handle this calculation but I will give you something to think about, the bold assumes that all 6-120 volt loads are on the same phase.

 

[kwired]

Like infinity said it looks like you possibly have all the 120 volt load on one phase. If so then yes one phase will be loaded to your 111.27 amps the other two will only be loaded to 30.5 amps.

If you balance the 120 volt loads across all three phases then they all see about 47 amps.

Next thing to consider is if the loads are continuous. Your minimum conductor ampacity and minimum overcurrent protection both require at least 100% of non continuous load plus 125% of continuous load. You could need up to 139 amp conductor and 150 amp overcurrent protection if all the 120 volt load is on one phase.

 

[Frank DuVal]

??? " 3 phase 208 5500w divided by 360 =15.27 amps " I don't see how this works, but you got the right answer!

5500/208/1.73=15.28 Amps per phase.

You would normally place two of the 20 amp 120 volt circuits on each phase, giving a total load of:

16 + 16 + 15.28 = 47.28 amps per phase.

47.28 x 208 / 1.73 = 5684 watts total load per phase, 47.28 x 208 x 1.73 = 17.013 kVA total three phase. If I used actual square root of three instead of 1.73, the figures would be exact. 5677.8 / phase, 17.033 kVA.

 

[david luchini]

??? " 3 phase 208 5500w divided by 360 =15.27 amps " I don't see how this works, but you got the right answer!

208*1.732=360

 

[infinity]

208*1.732=360

Yup, a common shortcut that I tend to forget more than I remember.

 

[kwired]

Yup, a common shortcut that I tend to forget more than I remember.

depening on how much accuracy you need as it is still a rounded off figure. 360.25 is closer to actual

 

[retirede]

depening on how much accuracy you need as it is still a rounded off figure. 360.25 is closer to actual

Technically, you can’t have more significant digits in the result than you have in either operand.

 

[LarryFine]

Technically, you can’t have more significant digits in the result than you have in either operand.

Next thing you're gonna tell us is that the ID of a conduit can't exceed the OD.

 

[synchro]

It may be obvious but if each L-N voltage is exactly 120V and each line current is ILINE, then the total KVA from the three line outputs is exactly 3 x 120V x ILINE = 360V x ILINE. Of course in this case the line-to-line voltage is √3 x 120V = 207.85.... volts. But this is 208V to 3 significant figures (the issue that retirede brought up) which is more than precise enough for most applications.