208v load vs 120v load 3

[itsjackma]

a contractor measured 3 1p-208v circuit breakers' loads to be 1A, 2A, 3A. the total load of all 3 breakers is 1248VA. if we are to change these loads into 1P-120V circuit breakers phase A becomes 3+1=4A, B becomes 1+2A=3A, C becomes 2+5A=, for a total of 12A. how did the load all of a sudden become 1440VA? the load VA cannot just magically increase when we change from 208V single phase to 120V single phase right? thank you all for your help.

 

[enverd]

Generally at higher voltage 208V or 480 and additional phases your load will go down because it is distributed between A-B or A-C or B-C or A-B-C etc.. vs running on one phase at 120V.
The watts or energy consumed is dictated by the type of load and its specifications.
So you really can't just take a load that is running at 208V and calculate it at 120V to prove that the Watts will not change.

1. You need to make sure that the existing equipment can run at 120V?
Remember this is a 208V, 1phase circuit feeding a load that is designated to run at just that and not at 120V.
2. If the manufacturer however lists that you can run that load at 120V 1phase configuration than you need to look at the cut sheet and figure out what the amp draw is at that voltage.
It should be higher, and with that the watts will be higher because the same piece of equipment will need more energy to perform the same task at smaller voltage because it gets less "juice".

One example (maybe not so great but still an example) that shows that the higher the voltage the smaller the amp draw can be seen in NEC 430.248 or .249 or .250 tables

 

[waross]

phase A becomes 3+1=4A, B becomes 1+2A=3A, C becomes 2+5A=, for a total of 12A. how did the load all of a sudden become 1440VA?

NO. 3A + 1A does NOT equal 4 Amps when the phase angles of the 1 Amp current and the 3 Amp current are different. You must use a little more than simple arithmetic to add currents from different phases.
Try some vector sketches.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jmbelectrical]

The load absolutely can change when applied with a different voltage. Recall that P = (V^2)/R. Suppose that the load is purely resistive and is 5Ω. At 120V, the load would be (120^2)/5 = 2,880W. At 208V, it would be (208^2)/5 = 8,653W. Having said that, unless we're talking about a universal voltage electronic lighting ballast, a power supply of some sort, or a special motor, very few devices can safely operate over such a wide voltage range. In the preceding example, you can see that, at 208V, the load dissipates three times as much heat as it does at 120V. Are you certain that your 208V, single-phase loads can operate at 120V?

In AC systems, voltages and currents are vector quantities - that is, having both a magnitude and a phase angle. As such, you cannot use basic artihmetic to compute voltages, currents, and other values of interest.

Let's define three voltages, Vab, Vbc, and Vca as follows:
Vab = 208∠30° V
Vbc = 208∠-90° V
Vca = 208∠150° V

Let's also assume that Vab serves your 1A load, Vbc serves your 2A load, and Vca serves your 3A load. For the sake of simplicity, we can also assume that your loads are purely resistive (Meaning that the current will have the same phase angle as the voltage). The resultant currents on each phase are as follows:
Ia = 1∠30° + 3∠150° = 2.65∠130° A
Ib = 1∠30° + 2∠-90° = 1.732∠-60° A
Ic = 2∠-90° + 3∠150° = 2.65∠-169° A

A common mistake that people make is to assume that, once you know the current on each phase, you can simply multiply each current by the phase-to-neutral voltage to obtain the load for each phase. Doing that, in this case, would yield (2.65 * 120) + (1.732 * 120) + (2.65 * 120) = 844 VA. Rather, as you indicated your original post, the load is (208 * 1) + (208 * 2) + (208 * 3) = 1,248 VA.

You should take a look at [URL unfurl="true"]http://en.wikipedia.org/wiki/Euclidean_vector[/url] when you have some free time.