First, while neutral currents may be conducted on ground paths, neutral currents are different than ground currents.
The load of the 15 KVA transformer will cause a voltage drop across phases A and B.
A single phase load across two phases will also cause a neutral current to flow. That is a neutral current, not a ground current.
Second, consider the motor.
The motor will be developing substantially equal back EMFs on all three phases.
Single phase load (at unity power factor for simplicity of explanation) across two phases of a transformer will a primary neutral current equal to the load current divided by the transformer ratio.
The load current is about 31 Amps. with a 14.88 kV primary that will be about 1 amp of primary neutral current.



Another mitigation will be the voltage drops in the supply conductors, including the neutral conductor.
These voltage drops will tend to mitigate the neutral current.
The calculations?
Each effect interacts with the other effects, the transformer impedances (absolute and directed, not PU), the conductor impedances, the motor impedance.
The load on the motor will have an effect on the PU unbalances.
The transformer X/R ratios must be considered as must the phase angles of the various voltage drops, for a rigorous solution of the various currents.
BOTTOM LINE
A simple solution:
15000 KVA /480 Volts = 31 Amps.
With no other effects, the primary neutral current with a 14.88 kV primary will be about 1 Amp.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

" #1. we have is a 500 kVA three phase transformer with wye secondary, 480/277V. There is one three-phase 250 Hp motor, and one single-phase 15 kVA transformer, 480V - 240/120V. The 15 kVA transformer is connected across phases A and B. .....#2. ... have been told that the current imbalance caused by the single phase transformer will result in ground currents circulating at the main transformer".

I have the following opinion for your consideration:
1. It does not matter whether the three-phase transformer wye secondary neutral is grounded or resistor/impedance grounded (or floating); the result is the same. There is NO ground currents circuiting at the main transformer.
2. Line A = Line B will be having higher current than Line C. However, the vector sum of Line 1,2 and 3 is zero. There is NO ground currents circuiting at the main transformer.
Che Kuan Yau (Singapore)

All,
I am very clear that there are no ground or neutral currents circulating. There isn't even a neutral connection at any of the loads. I'm not asking for clarification on the circuit analysis. What I'm looking for is a bombproof way to explain this to someone. I have presented vector equations showing this but so far I have made no difference. I was hoping for a video or paper that shows this calculation.

Thanks

By definition, if all you have are phase-phase connections there is nothing in the zero-sequence network. Not sure why anybody would make a video of the definitionally obvious.

I’ll see your silver lining and raise you two black clouds. - Protection Operations

Quote (eeprom)

...What I'm looking for is a bombproof way to explain this to someone...

Consider reversing the scenario - ask "someone" to explain to you the cause of the alleged circulating ground currents.

davidbeach,
I agree. I just need something more convincing than vector equations.

480 Volts almost always implies a 277/480 Volt wye connection.
Look at two 277 Volt transformers connected for 480 Volts.
Connect a load across the two transformers.
Each transformer will supply 277 Volts.
The current trough each transformer will be at 50% leading on one phase and at 50% lagging on the other phase.
The current at the common point will be equal to the load current.
If a third transformer is connected to form a three phase wye, not a delta, there will not be any current in the third transformer/phase.
There will still be a current in what is now a neutral conductor.
The three phase currents will not sum to zero with a single phase load across two phases of a wye transformer.
The three phase currents plus the neutral current will sum to zero.
With a delta primary, the three primary phase currents will sum to zero.
With a wye primary, the three primary phase currents plus the neutral current will sum to zero.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

Maybe it is now clear why I'm looking for a foolproof way to explain this. What is obvious to me is not obvious to everyone. I'm not even sure we are all referring to the same thing when we say neutral currents.

There will be no current flowing in any neutral conductor. There are no devices connected to a neutral. And there will be no residual ground current at the main transformer. And the sum of the currents from the main transformer will be zero.

There may be a tiny imbalance due to the fact that the service transformer has impedance, and loading phases a and b slightly more than c will result in a slightly reduced voltage on a and b. But that amount of imbalance is normal and will always be there due to utility line variations.

Perhaps the principle of superposition might help (hopefully the engineer will be familiar with this principle, otherwise you have other problems)

Line to line load - no neutral current (no neutral connection so no neutral current can flow)

Balanced 3 phase load - no neutral current (current sums to zero)


Add them together - still no neutral current.

Wcasetharman,
That is an excellent idea. Thank you and merry christmas.

If I get the chance I’ll build a little model in ATP (alternate transients program) and post the results. My co-workers and I have used ATP a few times to validate our hand calculations.
Merry Christmas to you as well.