YY transformer with floating primary

[mbrooke]

Ive got a hypothetical question (what if scenario) that's bugging me...


What would happen if a phase to ground fault took place on the secondary of a 12.47 to 480 volt YY transformer while the primary neutral (HO) was left floating? What would the vector shift look like?

 

[mivey]

Suppose we have
A:277<0d
B: 277<-120d
C: 227<120d
AB: 480<30d
BC: 480<-90d
CA: 480<150d

then for a A-N fault we have:
A:0<0d
B: 480<-150d
C: 480<150d
AB: 480<30d
BC: 480<-90d
CA: 480<150d

 

[mivey]

And H0 will be at 7200 volts, FWIW.

 

[JRW 70]

On the load (480V) I see they are still 120d
apart, so if only 480V loads this could run for
awhile, correct?

JR

 

[Phil Corso]

Mbrooke,

One of the primary lines would carry 2/3 of the (reflected) secondary fault-current! The other two primary lines would each carry 1/3! The result would be a severe neutral-shift in the primary circuit, and eventual failure!

A similar situation occurs when if the secondary loads are unbalanced or one-phase is overloaded with respect to the other two.

Regards, Phil Corso

 

[mbrooke]

So the 480 volt loads will see no change? My understanding is that if C phase went to ground, then the primary will act like one coil is shorted circuited, so its like the H3 is connected to H0?

 

[Bugman1400]

You didn't say if H0 and X0 use the same bushing. Even if H0 is floating, if X0 is a separate bushing and is solidly grounded then, a ground fault on the secondary side does not cause a neutral shift on the secondary side. For just a floating primary, a fault or load imbalance could cause the neutral to shift from center and result in a L-L voltage being applied to the xfmr terminals that are rated for only L-N voltage. this would probably result in a xfmr failure.

 

[Bugman1400]

Mbrooke,

One of the primary lines would carry 2/3 of the (reflected) secondary fault-current! The other two primary lines would each carry 1/3! The result would be a severe neutral-shift in the primary circuit, and eventual failure!

A similar situation occurs when if the secondary loads are unbalanced or one-phase is overloaded with respect to the other two.

Regards, Phil Corso

Are you referring to per-unit values?

 

[Phil Corso]

Bugman...

Reur per-unit query. Do you want V & A magnitudes?

 

[Bugman1400]

Bugman...

Reur per-unit query. Do you want V & A magnitudes?

That would be great!

 

[Phil Corso]

Bugman,

What are the system parameters!

Phil

 

[mbrooke]

You didn't say if H0 and X0 use the same bushing. Even if H0 is floating, if X0 is a separate bushing and is solidly grounded then, a ground fault on the secondary side does not cause a neutral shift on the secondary side. For just a floating primary, a fault or load imbalance could cause the neutral to shift from center and result in a L-L voltage being applied to the xfmr terminals that are rated for only L-N voltage. this would probably result in a xfmr failure.

H0 is floating, X0 is grounded down.

 

[mbrooke]

Anyone know?

 

[mivey]

On the load (480V) I see they are still 120d
apart, so if only 480V loads this could run for
awhile, correct?

JR

If the system would stay online.

 

[mivey]

So the 480 volt loads will see no change? My understanding is that if C phase went to ground, then the primary will act like one coil is shorted circuited, so its like the H3 is connected to H0?

Pretty close, depending on impedance.

 

[mivey]

Here is an example with voltages and currents, note the neutral to ground voltage after the fault:

Primary impedance: 4607 amps 3-phase, 3990 amps 2-phase, 3455 amps 1-phase with X/R = 10
Secondary impedance: 1012 amps 3-phase, 876.3 amps 2-phase, 54 amps 1-phase with X/R = 0.17
30 kVA transformer (three 10 kVA units)
Load = 15 kW, 2 kvar at 480 volts wye connected at transformer terminals

PRE-FAULT:

At the source:

VAN = 7199<0
VBN = 7199<-120
VCN = 7199<120
VAB = 12470<29.99
VBC = 12470<-90.01
VCA = 12470<150.00

IA = 0.7009<171.2
IB = 0.7009<51.18
IC = 0.7009<-68.82

SA = 4987+j773.4
SB = 4987+j773.5
SC = 4987+j773.6
P = 14.96 kW
Q = 2.320 kvar
S = 15.14 kVA
pf = 98.82%


At the load:

Van = 275.7<-0.99
Vbn = 275.7<-121.00
Vcn = 275.7<119.00
Vab = 477.6<29.01
Vbc = 477.6<-90.99
Vca = 477.6<149.00

Ia = 18.11<-8.584
Ib = 18.11<-128.6
Ic = 18.11<111.4

Sa = 4949+j659.9
Sb = 4949+j659.9
Sc = 4949+j659.9
P = 14.85 kW
Q = 1.980 kvar
S = 14.98 kVA
pf = 99.12%



Now introduce a bolted secondary c-n fault: Ic = 54.03<109.2

At the source:

VAN = 12180<-31.2
VBN = 12550<-91.73
VCN = 381.2<-168.9
VAB = 12470<29.99
VBC = 12470<-90.02
VCA = 12470<150.00

VAG = 7198<-0.01
VBG = 7200<-120
VCG = 7199<120
VNG = 7085<117.1

IA = 1.186<140
IB = 1.222<79.45
IC = 2.080<-70.78

SA = 14280+j2215
SB = 15150+j2350
SC = 112.1+j784.8
P = 29.54 kW
Q = 5.349 kvar
S = 30.02 kVA
pf = 98.40%


At the load:

Van = 466.6<-32.18
Vbn = 480.6<-92.71
Vcn = 0<0
Vab = 477.6<29.01
Vbc = 480.6<-92.71
Vca = 466.6<147.8

Ia = 30.64<-39.77
Ib = 31.57<-100.3
Ic = 0<0

Sa = 14170+j1890
Sb = 15040+j2005
Sc = 0+j0
P = 29.21 kW
Q = 3.895 kvar
S = 29.47 kVA
pf = 99.12%

 

[mbrooke]

OK wow Thanks!!!!!!!!!!!!!

This has no phase to neutral loads, so the above will work, correct? (I just skimmed it over so please forgive me)

 

[Phil Corso]

MBrooke...

Your analysis of the pre-fault case appears to be based on the primary-windings connected in a Wye-configuration. But, the mid-point is connected to the source neutral, i.e., 12kV source!

Phil Corso

 

[Phil Corso]

Sorry, MBrooke...

I was really addressing Mivey's calc!

Phil

 

[mivey]

Sorry, MBrooke...

I was really addressing Mivey's calc!

Phil

You could be right. Windings should have been paralleled. I'll check later.