Let’s take a look at how Revit computes the current on each phase in various scenarios, and see how this compares to a current meter.

## 277V/1ph/10A load on a 480/277v 3-phase panel

The load for the circuit in VA shows 2770 VA (10A * 277V). Additionally, the Total Load shows 2770 VA, and the Total Amps as 10A. The current on the circuit wire is 10A, and the current on the ‘bus’, as indicated by the Total Amps is also 10A.. as is the current on the feeder. This is shown diagramatically below.

In this case, the voltage phase-to-neutral on that bus/phase is 277V, and thus, the current going through it is 10A... since the current isn't divided onto multiple paths, and since the voltage is constant, the current is constant.

When the phase column values are shown as Current, the 10A load is shown for the circuit. Nothing else has changed, so the Total Load is still 2770 VA, and the Total Amps is 10A.

One anomaly you can see is that the Total Conn: shows 3.33 A. Why is this? This value is simply taking the Total Conn Load (2770 VA) and dividing by the 3-phase voltage (480*sqrt(3)), which is quite common in engineering practice. 2770 VA / 480*sqrt(3) = 3.33A. This is one of the reasons that the per-phase totals show the way they do, it is sort of a reality check. In reality, the load on the bus is 10A, not 3.3A. If one were dealing with 10x larger numbers, say 100A vs. 33.3A, one would need to select a panel with busses rated at 100A… no less (i.e., 60A). Otherwise, the bus would be under-rated. Additionally, the wires would need to be sized for 100A, not something smaller, like 40A as might be interpreted if one only looked at the Total Load / 3-phase voltage. Typically, of course, one is dealing with balanced loads, where the disparity between the current per phase and the total current are closer to the same.

## 480V/1ph/10A load on a 480/277v 3-phase panel

The load for each phase of the circuit shows 2400 VA, for a total of 4800 VA. (10A * 480V = 4800 VA). Additionally, the Total Load shows 2400 VA per phase, with a Total Conn Load of 4800 VA. In this case, again, the total amps on each phase is 10A, even though the load has shifted from 2770 VA on one phase to 2400 VA on two phases. Again, the current on the circuit wire is 10A, and the current on the ‘bus’, as indicated by the Total Amps is also 10A. The only thing that can change the current is if the voltage changes. In this case, the voltage phase-to-phase on the busses is 480, and thus, the current going through it is 10A.

When the phase column values are shown as Current, the 10A load is shown for the circuit as expected. Nothing else has changed, so the Total Load is still 4800 VA, and the Total Amps is 10A per phase.

The Total Conn: is computed as 4800 VA / 480*sqrt(3) V = 5.77A. Again, the reality check of the per-phase loads confirms that in the case the loads were 10x, we need to size the bus of the panel and feeder wires based on the per-phase load of 100A, not a Total Conn: load of 57.7A.

## 480V/3ph/10A load on a 480/277v 3-phase panel

For a 3-phase balanced load, the load on all three phases are the same. In this case, 2770 VA per phase. There is still 10A on each wire, and 10A on each bus/phase. Further, the total connected load is 10A.

Showing the per-phase loads in Current (amps) reveals the same results. All phases of the circuit have 10A.

The Total Conn: is computed 8310 VA / 480*sqrt(3) V = 10.00A. In this balanced case, it doesn’t really matter if we refer to the per-phase load or the total connected load. Since in the balanced Current Phase A = Current Phase B = Current Phase C = Total Current.

## Unbalanced loads on a 480/277v 3-phase panel

In this scenario, I am starting with the phase column values showing in units of Amps. This is where things may not ‘add-up’, but there is a very logical explanation derived from the discussion above, and he theory behind 3-phase power.

In this case, we have the same three loads, but in a different configuration, with them all connected at the same time:

- Circuit 2,4,6: 480V/3ph/10A load
- Circuit 8: 277V/1ph/10A load
- Circuit 10,12: 480V/1ph/10A load

As you can see, Phase B and Phase C each show two loads at 10A each…. However, the phase B and C Total Aps is 18.66A! How can this be? Let’s take a look at the per-phase loads in units of VA:

Here, we can more clearly see that the loads aren’t simply 10A, but rather are made up of different types of loads at varying voltages. As such, the math behind it is a bit more complicated than simply dividing by the phase-to-neutral or 3-phase-to-phase voltages.

If we look simply at the 3-phase load total of 15880 VA / 480*sqrt(3) we get the reported 19.10A, consistent with everything sated so far.

If we were to remove 370VA from Phase A, we would have Phase A = Phase B = Phase C = 5170 VA. Then, take the total load divided by the voltage: 5170 VA * 3 / 480*sqrt(3) = 18.66A. Not-so-coincidentally, this is the current we see on Phase B and Phase C in the schedule, and is what we would measure on the feeders to the panel in an installation.

Further, if we took this difference of 370VA, and divided it by the unbalanced load (which in this cases is at 277V/1p), we would have 370VA / 277A = 1.34A. Not coincidentally, if we add this 1.34A to the balanced three-phase-current of 18.66A on Phase A, we get 20.00A on Phase A… again, the current one would measure on the Phase A feeder.

Since the load is not balanced, naturally, the current on Phase A will be slightly larger than the reported Total Conn current across all three phases… similarly, Phase B and C are slightly smaller.

In a more realistic scenario, of course, there would be de-rating on the loads, so the value you use to size your bus bars and feeders is not solely dependent on the Total Connected current. For most practical purposes, where systems are almost balanced, one can use the Total Est. Demand current as the basis of sizing the panel bus bars and feeders. However, in the case you have an unbalanced system Revit provides the detail on a per-phase basis to show you how the connected load and current on the phases tabulate. If you only looked at the Total Est Demand, there are scenarios where you’d get into trouble, and therefore, in some cases, the additional detail is necessary.

In a real installation, you are un-likely to really see the design load or current, or even the voltage. There will be voltage drop, which will vary per phase depending on the loading, and of course, the loads per phase will vary depending on the type of load and their utilization.

However, all the same theory presented here will apply.

Here we have the readout from a 3-phase current meter (the numbers in the black circles correspond to numbers in the next image). As you can see on the top row, right column, the voltage is 93.697, though, due to unbalanced loads and other variations, the voltage varies on each phase. However, if you average the per-phase voltages, you get precisely this value. (In Revit, we don’t have the ability to account for voltage fluctuations on per phase… but, typically, designers don’t do this anyway.)

As you can see in the green VA row, the total VA is a sum of the VA from the three phases.

If you take the VA, and divide by the voltage in the same column, you get the current shown in that column (e.g., for phase A: 89.231 VA / 95.626 V = 0.9331 A).

If you average the values in the blue row (average of 0.9331, 0.8483, and 0.9884) you get the reported 3-phase current value: 0.9233 A. This is almost the same value as taking the total VA, and divide by the voltage * 3 (this is showing the line-to-neutral average, and thus, we need an extra sqrt(3)), you will get very close to the reported 3-phase current:

260.29 VA / (93.697 V * sqrt(3) * sqrt(3) ) = 0.9260 A (a difference of less than 0.3%)

So, how do we test this against what Revit reports as the per-phase current and the total current? Quite simply, create a device with an unbalanced load as shown above, and connect it to a 162.29 / 93.70 V/3Ph panel (93.70 * sqrt(3) = 162.29), and check out the panel schedule.

The top image is from the Revit Panel Schedule… the bottom portion is computed in Excel to show step by step where the values come from. The values here have been rounded since Revit only stores two decimal places on power (VA) and voltage (V), so the results will vary a bit from the meter photo.

Let’s see where these numbers come from.

(1) shows the original entered loads. The spreadsheet shows how the load is divided across each phase and what voltage. There are three parts to this load, a 3-phase L-L part, a 1-phase L-L part (phases A and C), and a 1-phase L-L part (phase C). The current from each load part is tabulated in the column marked (2). The currents are then shown on the appropriate phases, and summed (3).

(4) shows the total VA load divided by the 3-phase voltage. (5) shows the average across all phases. These numbers are slightly different because a portion of the load is at 1-phase L-L. Since the voltage across each phase in the meter photo varies, the results aren’t exactly the same.

However, if the loads were balanced, and the voltage was consistent, which is the ideal design scenario, the following truism holds.

Current Phase A = Current Phase B = Current Phase C = Total Current

Or, alternatively for the unbalanced condition:

Total Current = Average (Current Phase A , Current Phase B , Current Phase C)

Revit shows the ‘Total Conn’ as VA divided by the 3-phase voltage, since this is how most engineers compute simplify the calculation. It is also very easy to compute the Average across the phases. And, in the event you need it, you also have the connected current per phase.

Knowing what the demand per phase (after demand factor values are applied) is a bit more complex because internally Revit isn’t keeping track of the load type per phase (and most engineers don’t either). However, you could estimate the values by computing the ratio between the Total Est Demand and the Total Conn (or the Average 3-ph Conn Cur).. then, multiply this ratio by the per-phase current.

If you happen to have any photos of such a meter at 480/277 (instead of my example at 162/94), I would be happy to update this example.

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