Calculating 120V loads on 208/120V system

[MooFish]

I'm a system engineer responsible for specifying power quantities for my systems. There are a lot of ways that are accepted in my industry to do this that I know to be incorrect, so I'd like to hear from someone with more specific expertise.

The systems are comprised of a combination of 120V resistive loads, and 208V resistive and inductive loads, although the devices that have inductive loads actually operate at various voltages, their power supplies accept 208VAC. Most of the power available is 208/120VAC 60hz, usually in 400A increments.

What a lot of people do is add up the load P, divide by operating voltage and divide by 3, so if we had 10 500w loads at 120V and 5 at 208V, it would look something like

((5000/110)+(2500/203))/3 = 20A needed.

I can't imagine that's correct. What I have done based on advice and reading is divide the load P by 1.73, a relevant power factor and the supply voltage, so this:

(5000+2500)/(1.73*0.9*203) = 24A needed

But I can imagine that being different, also. What is really the way to do this?

Thanks,
Moo

 

[advidana]

You don't figure amps that way when you mix up single phase loads on 3 phase source. That is why panel schedules were invented. You need to balance your loads.

The math can get tricky but with a graphic presentation such as a panel schedule it is easy. Your calucations would be correct if the loads were balance in threes. For example if you had 9 500watt loads at 120V and 6 500 watt loads at 208V then it would be ok. It is sometimes used for estimates assuming that the loads will be balanced.

But you are right a lot of people told you wrong.

 

[MooFish]

So what if the system is out of balance, as it sometimes might be, how does the math change?

 

[sslobodan]

you are calculating 3+3+4 single phase load by taking 4 for all. It goes for cable calculation but not for consumption.

Your consumption is 10*500W/120V + (5*500W)/(1.73*208V)
This means one phase will have load of 24 A while rest 2 will have current of 20A. For a equipment choosing you should take the upper value and it goes for transformer load as well. Unbalance is your lose in system equipment choice because if you have totally unbalance system like that 10 loads on one phase than you would have to pick gear with 50A ratings instead 24 like you have now.

 

[rbulsara]

Moofish:

To answer your last post. If the load is unbalanced, say it is 20A, 30A and 50A in a, b and c phases respectively in a 208/120V system.

Total VA is sum of the loads in each of the three phases for either balanced or unbalanced system. Voltage per phase (L-N) is 120V so total VA will be 120V*20A + 120V*30A + 120V*50A = 12000 VA or W if the load is all resistive.

In such case as sslobodan indicated you would pick feeder breaker rated 3P-50A and would require a larger source transformer rated 50A. This again shows importance of keepiong the loads balanced.

 

[nightfox1925]

Since your 208V three phase loads are balanced, then your total current based on three phase loads will be

FLA 3phase = (5*500W)/(1.732*208) = 6.93A

For the 10 x 500W at 120V (Line-Neutral), You have to balance them among the phases, then

Phase A = 3 x 500W = 1500W/[(120V)*(0.9)] = 13.88A

Phase B = 3 x 500W = 13.88A

Phase C = 4 x 500W = 18.51A

Therefore the highest current per phase will be:

I = 18.51 + 6.93 = 25.44A use 35A or 40A MCCB

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If the 5 x 500W are single phase 208V loads, then balance them again among the two phases and consider the phase with the highest total current.





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