Question about 3-phase panel load

[Asaras]

so I’m dealing with a 3-phase, 4W, 120/208 volt panel with a main breaker that’s rated for 100A and has about 20,000 VAs loaded on it. That means, the total amperage should be about 55.6 amps. So on two different circuits (both different) I’m planning adding two heating elements, both of which consume about 250 watts each. So my idea was that each heater draws about 2.1 amps (they’re being supplied with 120VAC) for a total of 4.2 amps added to the circuit. So when I add the 500 to the panel load, it totals 20,500, and after doing the load calculation for it again, it totals to about 57 amps. Why? Wouldn’t it total to around 59.8 Amps based on what I calculated from the heaters?

The formula I’m using for calculating the load amps is I = Total power/(sqrt(3)*208).

 

[HEYDOG]

so I’m dealing with a 3-phase, 4W, 120/208 volt panel with a main breaker that’s rated for 100A and has about 20,000 VAs loaded on it. That means, the total amperage should be about 55.6 amps. So on two different circuits (both different) I’m planning adding two heating elements, both of which consume about 250 watts each. So my idea was that each heater draws about 2.1 amps (they’re being supplied with 120VAC) for a total of 4.2 amps added to the circuit. So when I add the 500 to the panel load, it totals 20,500, and after doing the load calculation for it again, it totals to about 57 amps. Why? Wouldn’t it total to around 59.8 Amps based on what I calculated from the heaters?

The formula I’m using for calculating the load amps is I = Total power/(sqrt(3)*208).

Because you’re dividing it by 360. The extra load that you are adding is 120 volts so it will be hooked to one phase and the neutral (grounded conductor)So the extra amperage will only add to that one phase!

 

[Asaras]

Because you’re dividing it by 360. The extra load that you are adding is 120 volts so it will be hooked to one phase and the neutral (grounded conductor)So the extra amperage will only add to that one phase!

I’m still a bit confused; the extra load is added to a single phase, but I wouldn’t that still add to the total load calculation anyway? So like 250/120 is 2.1, and so the single phase has an extra 2.1 amps on it, but the total still increases by 2.1amps?

 

[kwired]

Is your existing 20kVA equally balanced across all three phases? If general use lighting and receptacles are there chances are it is not.

To some extent one needs to do a load calculation for each line (@120 volts) and sum up total on each line, and majority of time they won't be equal other than in cases where you only serve fixed and/or balanced loads.

Three phase motor - will load each line equally unless something is wrong with it.
208 volt single phase loads - if you have multiple identical ones and connect them with equal number of connections to all three lines you can balance them. Same with 120 volt single phase loads. 120 volt single phase loads tend to be what ends up being more difficult to balance often times though. Or any single phase loads that are intermittent, random, etc. in operation.

 

[Asaras]

Is your existing 20kVA equally balanced across all three phases? If general use lighting and receptacles are there chances are it is not.

To some extent one needs to do a load calculation for each line (@120 volts) and sum up total on each line, and majority of time they won't be equal other than in cases where you only serve fixed and/or balanced loads.

Three phase motor - will load each line equally unless something is wrong with it.
208 volt single phase loads - if you have multiple identical ones and connect them with equal number of connections to all three lines you can balance them. Same with 120 volt single phase loads. 120 volt single phase loads tend to be what ends up being more difficult to balance often times though. Or any single phase loads that are intermittent, random, etc. in operation.

Yeah, it’s a lighting/receptacle panel and they’re not balanced. It was already configured like that when I started working on it.

 

[Carultch]

The formula I’m using for calculating the load amps is I = Total power/(sqrt(3)*208).

This is only valid, if you have perfect balancing of phase currents, which generally, you should aim to do when you have control over the breaker layout. Alternatively, you can express this formula as I = Power/(3*120V).

Depending on what mixture of loads you have, a different formula would apply to add up the current. The easiest case is when all your loads are either line-to-neutral, or 3-phase loads that have no reason to be unbalanced. A-phase line current = sum of 1-pole A-phase loads + sum of 3-phase loads. Likewise for B & C phases.

The harder case is when you have 2-pole loads, and there's a square root formula to determine the equivalent line currents from a mixture of 2-pole loads staggered among the phases. It doesn't sound like this applies to you.

 

[winnie]

@Carultch has it. The _formula_ @Asaras is using assumes perfectly balanced loads and is an approximation for real world loading.

Add a single unbalanced load and the error in the approximation becomes apparent.

Jon

 

[Asaras]

This is only valid, if you have perfect balancing of phase currents, which generally, you should aim to do when you have control over the breaker layout. Alternatively, you can express this formula as I = Power/(3*120V).

Depending on what mixture of loads you have, a different formula would apply to add up the current. The easiest case is when all your loads are either line-to-neutral, or 3-phase loads that have no reason to be unbalanced. A-phase line current = sum of 1-pole A-phase loads + sum of 3-phase loads. Likewise for B & C phases.

The harder case is when you have 2-pole loads, and there's a square root formula to determine the equivalent line currents from a mixture of 2-pole loads staggered among the phases. It doesn't sound like this applies to you.

There are a couple of them that are two pole breakers. So from my understanding though, you’re saying that I’ll have to calculate each breakers current by doing P/120 and then getting the current that way?

 

[winnie]

How accurate to you want/need to be?

For most purposes the balanced panel approach is good enough.

The next more accurate approach is to assign load VA to each panel bus, then calculate each bus separately but using balanced load maths.

Even more accurate is to add load currents using vector math to accommodate the different load phase angles.

Then you can add currents using vector math for each harmonic component of the loads.

Each level of complexity adds lots more work but makes the calculation more accurate.

However prior to going down that path you need to decide how much accuracy you need. IMHO you are far better off understanding how the basic equation is simply an approximation, understanding how that approximation fails, and then using it knowing the limits.

Jon

 

[Asaras]

I get now what you guys are saying. Thank you all!!!!