480V to 208V - KVA transformation?

[hrc]

Given a 3p 480V source, transforming to 208 Y 3phase (sorta the normal commercial type PDU thing), if I have 750KVa on the 480V side, does this mean...

250KvA per phase, on the 480V side
520A per phase, on the 480V side

and then on the output side of the transformer

1201A available on each leg of the 208?

 

[aolalde]

A 3 phase transformer (or any phases) modifies the system voltage, not the power.

The primary is 750 kVA, 480 Volts
I line= 750*1000 / (1.732*480) = 902 Amperes

The secondary is 750 kVA, 208 Volts
I line= 750*1000 / (1.732*208) = 2081.8 Amperes

I line is the current you are able to draw from each line for single or three phase loads, if that figure is exceeded the transformer is overloaded.

 

[hrc]

*pounds head on table*

duh...

forgot about the 1.73 factor

thanks!!!

 

[hrc]

oh wait...the 2081.8A is per phase? In other words, this is feeding a PDU where I have all 3 phases available. If I run a power supply using 2 of the 3 phases, is it that each leg can provide this amount of current? Then again, I suppose it doesnt matter because the main breaker is 225A rated, or allowable 180A on each leg.

 

[DougMSOE]

hrc,
While this is not directed at you (as I have seen this elsewhere in the web site) it is something that needs to be understood (please read, I made the same assumption in school). It is not amps per phase but rather the maximum (RMS) offered at a terminal.
Another way of saying this is if a transformer is "rated" at 100 amperes per phase, this does not mean that the total current from the transformer is 300 amperes.
After all when performing the calculation of KVA we use Vline-to-line and line current as well as SQRT(3).
Have a great day!
Doug

 

[aolalde]

Yes the 2081.8 Amps are per line.

See it the following way:

If the secondary were wye connected and 208 Volts line to line, the voltage line to neutral is 208/1.732=120 Volts.

If a single phase is loaded with 2081.8 amperes the power per phase is P=120*2081.8= 250,000 Watts. Note that in this example one phase is from one line to the neutral, when the three phases are evenly loaded the current in the neutral balances to zero.

The ideal for the transformer is a balanced three-phase load, in real life with single-phase loads the transformer could work unbalanced into the current rating limits.

 

[jbartos]

Suggestions to hrc (Electrical) Mar 11, 2004 marked ///\\Given a 3p 480V source, transforming to 208 Y 3phase (sorta the normal commercial type PDU thing), if I have 750KVa on the 480V side, does this mean...

250KvA per phase, on the 480V side
///Yes\\520A per phase, on the 480V side
///No. It is not clear whether you have 1) 480V 3phase 3-wire system or 2) 480V/277V 3phase 4-wire system.
1) Each phase will be allowed to draw such a current as to not overload the primary transformer winding.
Iline=750000VA/(sqrt3 x 480V)=902A
However, it is not possible to connect the load across the two terminals that would draw 480V x 902A=432960VA since only 250000VA or I1phase=250000VA/480V=521A is allowed not to overload the winding and transformer. Please notice that 902A/521A=1.731~sqrt3, which is the design basis for the delta winding current when delta current is 521A and line current is 902A. Similar reasoning holds true for:
2) The line current is again 902A for three phase load. For the single phase load:
I1phase=250000VA/277V=902A for the line to neutral voltage, and
VA1phase=480V x 902A=432960VA
which is higher VA than for the delta connection. Please notice that the two phase load allowed on wye is:
VAa=Van x Ia = 277V x 902A~250000VA
VAb=Vab x Ib = 277V x 902A~250000VA
VAab=VAa + VAb=500000VA\\
and then on the output side of the transformer
1201A available on each leg of the 208?

///The similar reasoning holds true for the transformer output side:
It is not clear whether you have 1) 208V 3phase 3-wire system or 2) 208V/120V 3phase 4-wire system.
1) Each phase will be allowed to draw such a current as to not overload the secondary transformer winding.
Iline=750000VA/(sqrt3 x 208V)=2082A
However, it is not possible to connect the load across the two terminals that would draw 208V x 2082A=433056VA since only 250000VA or I1phase=250000VA/208V=1202A is allowed not to overload the winding and transformer. Please notice that 2082A/1202A=1.732~sqrt3, which is the design basis for the delta winding current when delta current is 1202A and line current is 2082A. Similar reasoning holds true for:
2) The line current is again 2082A for three phase load. For the single phase load:
I1phase=250000VA/120V=2083A for the line to neutral voltage, and
VA1phase=208V x 2083A=433264VA
which is higher VA than for the delta connection. Please notice that the two phase load allowed on wye is:
VAa=Van x Ia = 120V x 2083A~250000VA
VAb=Vab x Ib = 120V x 2083A~250000VA
VAab=VAa + VAb=500000VA\\\

 

[hrc]

Interesting data...

Jbartos asked some additional questions so here is some more detail

The input is a 480V delta configuration, while the output is a 208wye configuration. This is actually a large PDU type unit, where the transformer and the panel board is all in one box. The panelboard has a main breaker rated 225A, so per NEC only 180A are allowed to be run thru it.

The final assumption that I am making is that given a design of this PDU (I don't have any more data on it)has the 225A breaker, that across any two poles, I can draw up to 180A, for a total draw of 540A continious from the box, across all three phases. Is this correct?

 

[aolalde]

No.

You are driving 180 amperes per line of a three-phase circuit.

the power in kVA=1.732*208*180/1000 = 64.85 kVA

If you add the three line currents ( 0, 120 & 240 deg out of phase), the result is (0) zero amperes.

 

[hrc]

BUT....

PhaseA PhaseB PhaseC
\ / \ /
208V 208V

And of course PhaseAC is also 208

So if you have a 225A 3pole breaker, allowing 180A per pole, each of the phase to phase can pull 180A, with no cancellation because the load see's it similar to single phase...correct?

 

[rbulsara]

It is true that you can have a 3 phase load (for example a motor or a 3 phase heater) with no neutral connection, and you can draw 180A balanced current in 'each' line of 208V 3-phase system.

In one way it is equivalent of 540A (180A x3) but at 120V ( you always take per phase voltage is 120)giving you total power of 540*120/1000=64.85 kva, which is same as 3 phase power at 208V at 180 amps in each line given by 1.732*208V*180A/1000=64.85kvA

 

[rbulsara]

Also refer to thread238-89630

 

[jbartos]

Suggestion to hrc (Electrical) Mar 16, 2004 marked ///\\BUT....

PhaseA PhaseB PhaseC
\ / \ /
208V 208V

And of course PhaseAC is also 208

So if you have a 225A 3pole breaker, allowing 180A per pole, each of the phase to phase can pull 180A, with no cancellation because the load see's it similar to single phase...correct?
///180A currents do cancel at the neutral point on the secondary winding for balanced wye or delta load. They are added vectorially, 180Amps+180/_120°Aamps + 180/_240°Amps=0+j0, Amps at the neutral point of wye or at the ficticious neutral point of the load delta winding.
However, if only current magnitude 180A is considered (no angles), then in scalar form, 120V x 180A + 120V x 180A + 120V x 180A=120Vx(180A+180A+180A)=120Vx3x180A=120Vx540A=64800VA.\\\

 

[rbulsara]

It should also be noted that current cancelling (or not cancelling) in the neutral has nothing to do with power consumed in each phase or total thereof.

Curernt flow in neutral only indicates that the load is unbalanced(assuming all linear loads) or certain harmonic current are present (typical of non-linear loads).

The total 3-phase power is still the total of power in each of the three phases. Only in case of a balanced 3-phase load, the special relationship of 1.732*V*L applies.

 

[aolalde]

The current of a three phase circuit with balanced load could be calculated with the formulas above.
If voltages and currents are not balanced approximated results are obtained using the three phase average values.
To get an accurate result the resolution of the symmetrical components is required which certainly must consider the phase angle between phases.

Ia = Ia1+ Ia2 + Io

were Ia = phase A current
Ia1=phase A positive sequence current
Ia2=phase A negative sequence current
Io= zero sequence current

 

[jbartos]

Suggestion to Mar 16, 2004 lbursara posting:
Consider the Power Vector U consisting of the active power, P, reactive power, Q, and distortion power, D, i.e.
U=iP+jQ=kD
U is in VAs
P is Watts
Q is VARs
D is VAs
Reference; IEEE Std 100-2000 Dictionary