Load Calc on 3 Phase 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 3 Phase, 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 (208-5 for voltage drop), so this:

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

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

Thanks in advance,

Moo

 

[infinity]

You would need to convert all of the loads to volt-amps and then calculate the total. If you're looking for the smallest possible three phase ampacity you would need to balance the loads as closely as possible.

 

[bob]

.

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 (208-5 for voltage drop), so this:

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

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

Thanks in advance,

Moo

Your calculation is correct if the load is balanced across the 3 phases. Use 208 volts for the calculation and not 203. Once you calculate the amps you can select the conductor size .Voltage drop is another problem. After you calculate the amps and conductor size then you calculate the voltage drop. If VD is too high you need to increase the conductor size.

 

[Smart $]

Your calculation is correct if the load is balanced across the 3 phases...

I believe his question is, paraphrased: How do I calculate when the loads are not balanced across the 3 phases?

IMO, he should (simplest method) balance the loads as best possible and add phantom loads to the lesser loaded phases to balance it out completely...

...or (more accurate method) balance as best possible then vectorially sum the current per line and use the load on the line having the highest amount current.

 

[bob]

I believe his question is, paraphrased: How do I calculate when the loads are not balanced across the 3 phases?

IMO, he should (simplest method) balance the loads as best possible and add phantom loads to the lesser loaded phases to balance it out completely...

...or (more accurate method) balance as best possible then vectorially sum the current per line and use the load on the line having the highest amount current.

When the phase loads are balanced, you only need to multiply the loads by 1.73 to get the line currents. When the loads are not balanced you must convert the phase loads to a vector format and then add them together
to get the line current. Its just a little more complex because you have to keep track of the phase angles in order the get the + and- signs right.

 

[ramsy]

Here's an unbalanced formula using VA for A & B below.

I=√(A?+B?+A*B)/Vln

I don't believe Power Factor is used for resistive loads.

However, for inductive motor loads, I modify this by I-Rise

I=√(A?+B?+A*B)/Vln/I-Rise

I-Rise = 1-(1-PF)*(%Motor Load)

I believe rising current for Motor Power Factor losses, I/Pf, can be defined by (%Motor Load) above.

For inductive motor loads, I'm not sure how additional I-Rise occurs with circuit impedance or voltage drop, given ambient and temperature rise (T2) of conductors (R2).

 

[steve66]

Here's an unbalanced formula using VA for A & B below.

I=√(A?+B?+A*B)/Vln

I don't believe Power Factor is used for resistive loads.

However, for inductive motor loads, I modify this by I-Rise

I=√(A?+B?+A*B)/Vln/I-Rise

I-Rise = 1-(1-PF)*(%Motor Load)

I believe rising current for Motor Power Factor losses, I/Pf, can be defined by (%Motor Load) above.

For inductive motor loads, I'm not sure how additional I-Rise occurs with circuit impedance or voltage drop, given ambient and temperature rise (T2) of conductors (R2).

I'm not sure what A and B are supposed to be. Are you sure this is the right formula for a three phase system? I seem to recall seeing a similar formula posted here that had three currents - A, B and C.

Steve

 

[ramsy]

I believe this unbalanced formula is limited to loads across two specific phases at a time, but A & B & C should be interchangeable. To get all three at once you would need 3 formulas.

Phase Formula
(Ac) I=√(A?+C?+A*C)/Vline/I_rise
(Ba) I=√(A?+B?+A*B)/Vline/I_rise
(Cb) I=√(B?+C?+B*C)/Vline/I_rise