YY transformer with floating primary

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mbrooke

mbrooke

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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?
 
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
 
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
 
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?
 
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.
 
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Phil Corso said:
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
Click to expand...

Are you referring to per-unit values?
 
Bugman1400 said:
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.
Click to expand...


H0 is floating, X0 is grounded down.
 
mbrooke said:
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?
Click to expand...
Pretty close, depending on impedance.
 
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%
 
OK wow Thanks!!!!!!!!!!!!! :D:):cool:

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