single phase 3 phase 208volts loads

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murli8

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I am doing a load schedule where I have 208 loads 3 phasea nd single phase. There is some confusion in my mind.

LOAD 1----> 10 AMPS, 208 VOLTS Single phase
Load 2----> 15 amps, 208volts, single phase
Load 3-----> 20 amps, 208volts, single phase
load 3-----> 20 amps, 208volts, 3 phases
load 4------>10 amps, 208volts , 3 phases

size of main breaker ?

10 amps, 208v = 208 x 10 = 2080 VA
15 maps, 208v = 208 x 15 = 3120 VA
20 AMPS,208V = 208 X 20 =4160
20 amps, 208volts, 3 phases = 1.732 x 208 x 20 = 7205 VA
10 AMPS, 208V, 3 PHAsea = 3602 VA

TOTAL = 20167 VA
CURRENT = 20167 / 208 X 1.732 = 56 AMPS
125% = 1.25 X 56 = 70 AMPS
MAIN BREAKER = 75 AMPS

IS THIS OK?
How can I balance phase ? What will be curent in each leg or phase ?
thanks in advance
murli
 
I'd go for at least 100a for future additions; the main does not have a maximum limit.

You'll have to settle for the differences in current. The 3-ph loads are irrelevant. The singles will just have to be A-B, C-A, and B-C.

The highest amperage will be on the phase with the 15a and 20a 1-ph legs.

I envision the panel with the two 3-pole breakers on one side and the three 2-pole breakers on the other side.
 
Actually, using Larry's circuit layout you'll end up with;
A phase @ 60A
B phase @ 55A
C phase @ 65A

65A * 1.25 = 81.25 so a 90A protection will be necessary assuming that there are no motors in this system.
 
Larry has the right answer. Trying to pin down the _exact_ load in a panel is like trying to nail jello to a brick wall. You might be able to do it on paper for an exam, but in the real world it is better to provide a small excess of capacity, both for future expansion, and because this excess of capacity costs less than the time of figuring out the exact answer.

murli,

Your math is an approximation that assumes a perfectly balanced panel, and will thus underestimate the load on individual phases.

Snorks,

Your math is an approximation that ignores the necessary vector addition of loads between different phases, and will thus overestimate the load on the individual phases.

Both approaches approximate the load values with their nominal values; a '10A' load is probably really a 'somewhere between 9 and 11A load'.

-Jon
 
forgot to mention

forgot to mention

That this is a DELTA system used overseas. 208 DELTA. What will be the difference? Phase or line currents or total loads?
thanks
 
Hmm. Two 208V delta posts in one day. Is this a sign????

Delta or wye supply, it will make no difference for the calculation of the feeder amps, the math is the same. There may be a difference in the way transformer loading gets distributed, or how the system responds to harmonics, but the calculations above (and their approximate nature) do not change.

However the OVERSEAS part may make a difference for the rules that you are required to use for the calculation; are you under the NEC??? In particular, the 125% factor is not required in all circumstances by the NEC, and I bet that different national codes have different ways of applying similar factors.

-Jon
 
Usa

Usa

I am designing the tool for ASIA and am based in USA. But i was told that the phase currents and line currents are different for delta and wye system
So for my calculations above, will the feeder panel will have similar currents for the 2 systems?
 
This is a terminology issue. Just what does one mean by 'phase' and 'line' currents. I will try to ignore these terms in the below paragraph:

Consider the wires coming from the transformer and feeding this panel. The current on these wires will be essentially the same, no matter if the transformer is a 'wye', 'delta', 'open delta' or some other three phase source that produces the same voltages and phase angles. There may be some slight differences in how the system responds to 'harmonic' content from non-linear loads, but for purposes of this calculation they are the same.

Now look 'inside' the transformer, and measure the current flowing on the individual coils inside the box. Here 'wye' versus 'delta' does make a difference. In the case of the 'wye' connected transformer, the voltage created by an individual coil is lower (it is your line-neutral voltage) but the current carried is higher (each coil connects a single wire, thus carrying the full current of the wire). In the case of the 'delta' connected transformer, the voltage created by an individual coil is higher (the line-line voltage), but the current carried in each coil is lower (two coils supply each wire, and thus split the current).

So inside of a source or load, wye versus delta makes a difference. But on three wire runs between the loads, wye versus delta makes no difference.

When you say designing a tool, that implies a stand-alone device, not a premises wiring system. In the US, you would want to look at UL standards, not at the NEC, for proper selection and sizing of OCPD. For Asia, you probably have yet another set of standards that you need to follow. Since all of these rules are derived from the same physics, then you might use the NEC for _guidance_, but to actually build and sell something you will need to know the codes that have been legally adopted in the country that you are selling to.

-Jon
 
Panel Schedules

Panel Schedules

Thanks. But if you were to create a panel schedule, how will you show the vaious loads ? For example the phase current and line current on the schedules? This may be different for the delta or the wye system panel schedules?
 
Hi Winnie,

you wrote:
"Your math is an approximation that ignores the necessary vector addition of loads between different phases, and will thus overestimate the load on the individual phases."

You are getting the application of 3 phase vector application a little confused. When using the individual phase loads to calc a panel's requirement you only use the ampacity of the device. A 65A phase A load will require 65A of power regardless of the device's or panel's phase. At this point Amps are Amps.

When computing per balanced panel then you are right, you use the vector addtiton of the load to match the panel's phase to distribute the load evenly among the panel's three phases.

PanelA = device A * sqrt(3)

OR using the device's real power component;

A = watts / (panel voltage * sqrt(panel phase) * device EFF * device PF)

And, I've seen panel schedules in VA, kVA, kW and Amperes.... I do agree with kingpb that Apparent power is much more applicable, but until there is an established standard, any of the power units are being used. Along with that, a number of individuals are still trying to get amperes from apparent power by dividing voltage into apparent power. In an impedance based system this is wrong and will always be wrong. So maybe for these individuals the ampacity unit might also need to be correctly computed and displayed. I for one display the panel's summation in units of Amperes, Kilowatts and Kilovolt Amperes to avoid confusion.
 
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