The first step in problem solving is to define the problem.
The second step is to identify the source(s) of the problem.
The third step is to devise a solution or compromise to wholly or partially eliminate the problem.
It is often unproductive to skip the second step and propose a solution without first identifying the source of the problem.
Step two:-
Possible reasons for a large voltage drop at a panel.
Metering errors; Not likely if the customer is complaining about voltage dips. If there are no noticeable voltage dips, but the monitoring equipment is showing large dips youmay have a metering issue.
Feeders from the transformer to the panel:- This is a good place to look for voltage drops. The conductors may be arranged in such a way as to cause an unacceptably high impedance. some things to look for:
Feeders with insufficient ampacity.
Feeders too long for the unit impedance.
Feeder conductors arranged in such a way as to develop excessive inductive reactance. This may be something such as a duct bank with the conductors of each phase contained in separate ducts rather than being distributed so that each duct has an equal number of conductors from each phase with the neutral equally distributed among the ducts.
Measure the voltage drop on each phase conductor from the panel to the transformer.
Loose connections. Loose connections with a single conductor per phase will cause quite small voltage drops. More than a few volts drop across a connection carrying load current will develop so much heat energy that catastrophic failure will soon occur.
At 25% load you have about 1000 amps. With only a 1% voltage drop (6 Volts)across a loose connection, there will be 6000 Watts of heat developed. That much heat generated by a loose connection will quickly develop into catastrophic failure.
However, with multiple conductors per phase, poor connections on some conductors throws the load on the remaining conductors. This may cause excessive voltage drops.
Transformer regulation; Possible. Voltage drop in a transformer is caused by a combination of reactance (X) and resistance (R). (Hence X/R ratios)
At full load and unity power factor the voltage regulation is mainly caused by the resistance of the transformer windings. AS the power factor drops, more and more reactive voltage drop is developed. The full available short circuit current is only developed when the only resistance in the circuit is the transformer resistance.
What is the per/unit (%) regulation of the transformer?
Do you know the per unit (%) impedance of the transformer AND the X/R ratio?
This will give us the ability to suggest how much voltage drop to expect in the transformer.
Primary voltage drop:-
You may have a "soft" primary feed. Do you have access to primary CTs to check the primary voltages?
Many of the newer electronic kWHr meters have the ability to store a wide range of information. This may include phase voltages, phase amperages, peak voltages, peak currents, power factor, peak kW, peak KVA, peak KVAR, in addition kWHrs, KVARHrs, and KVAHrs. 15 minute intervals are common.
I have had good cooperation from some utilities regarding metering information.
The utility may be able to send out a meter man to download the information directly.
The utility may be able to send out a meter man to program the meter to provide the information in the future.
The utility may be able to send out a meter man to install a battery and program the meter.
The utility may be willing to install an electronic meter with recording capability.
You will probably be billed for some of these services. At the most, being charged for the supply and installation of a plug in meter with the wiring, Potential Transformers and Current Transformers already in place and paid for is by far the most economical way to acquire valuable data.
Your problem may be a combination of primary voltage drop and secondary drop. See if you can get more information and we can provide more assistance.
In regards to the neutral current;
The average amps draining on ground wirer towards the transformer is 50 A, wile the average amp on main neuter is jus 6 A.
I don't quite understand "main neuter is jus 6 A." Not the place to guess a meaning or make an assumption. Canyou explain a little more please?
Bill
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"Why not the best?"
Jimmy Carter