Floating voltage calculation for three-phase ungrounded wye

E

EC Dan

Senior Member
Location
Florida
Occupation
E&C Manager
I have a proprietary heater that has three heating elements wired in a wye configuration, each requiring phase voltage of 277 V. Initially I was planning on using a neutral connection at the center point, however running a 4-wire circuit may not be possible or advisable, so I'm trying to evaluate how bad the voltage rise could get in the center with different resistance tolerances for each element. It seems like a simple application of KCL, but I'm hoping someone could double-check my work to make sure I'm not over-simplifying this. Here's what I got (with examples values for resistance):

V1: 277<0 V
V2: 277<120 V
V3: 277<-120 V
V0: x (center point voltage)

R1: 10 Ohms
R2: 9 Ohms
R3: 11 Ohms

Via KCL:

(277<0 - x)/10 + (277<120 - x)/9 + (277<-120 - x)/11 = 0
x = 16.07<93.3 V

Thanks!
 
If the 480 V system does not trip on a ground fault - typical for a high-res ground, then you have to assume that the voltage on one phase could be 480 V in the event of a ground fault somewhere in the system. If one phase is grounded, the voltage to ground on the two remaining phases goes to 480 V until the fault is cleared. Transient voltages could exceed 480 V.
 
EC Dan said:
I have a proprietary heater that has three heating elements wired in a wye configuration, each requiring phase voltage of 277 V. Initially I was planning on using a neutral connection at the center point, however running a 4-wire circuit may not be possible or advisable, so I'm trying to evaluate how bad the voltage rise could get in the center with different resistance tolerances for each element. It seems like a simple application of KCL, but I'm hoping someone could double-check my work to make sure I'm not over-simplifying this. Here's what I got (with examples values for resistance):

V1: 277<0 V
V2: 277<120 V
V3: 277<-120 V
V0: x (center point voltage)

R1: 10 Ohms
R2: 9 Ohms
R3: 11 Ohms

Via KCL:

(277<0 - x)/10 + (277<120 - x)/9 + (277<-120 - x)/11 = 0
x = 16.07<93.3 V

Thanks!
Click to expand...
That looks correct to me.

The current through each resistor is delta V / resistance.

The sum of currents on the floating neutral is 0.

I plugged your numbers in and got within rounding of zero.
 
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