Energy Efficinecy

only if your paying by the amp. Most of us pay by the watt. P=I*E

380v * 100 amps = 38000 watts

400v * 95 amps = 38000 watts

There is a modest increase in power loss (less efficiency) from insulation leakage currents that will happen by raising the supply voltage. Without knowing what intelligent power supplies or variable frequency drives you have for these unspecified motors, it is impossible to say if this will be more efficient, less efficient or just making your equipment more susceptible to catastrophic failure from a positive power surge.

I think that you should change this transformer wiring to the higher voltage. There's several electricians waiting for your call to repair your facility.

The secondary voltage you are referring to is the open circuit voltage. When the transformer will be loaded fully , 400V will get reduced to approximately 380V due to voltage regulation of transformer. If the % regulation of transformer is 5% then the secondary voltage (x ) at full load will be as under:

5= (400-x) *100 /x i.e. 105 x = 40000 or x = 380.95 V

When you increase voltage above "standard", you begin to "kill" your electronic lamp ballasts- life will diminish by about the square of difference. So, at 110% of normal, ballast life will be 1/(1.1x1.1) or about 80%- 12 years instead of 15.

Also- motor failure triggered by an occasional voltage spike causing phase-phase failure in winding through insulation. Maybe 10% higher losses- up to 20% with high-effic motors due to thinner insulation and more copper.

Your electronic controls will also be at higher risk of spike failure.

Secondary transformers will not be able to handle rated power flow because their cores get "loaded" by the higher voltage input.

Bottom line- it is not worth it. You may not see it directly, but you WILL be losing money. A site we visited several years ago, with the same thought process, was losing over US$100,000 a year due to this issue. They re-tapped the transformers.

You have mentioned the word 'active energy' & told that the current is reduced by 5%.

Current is not the active energy.

Infact, if you accurately measure the active energy in KWH unit, you will find something different.

On reducing the voltage by 5%, you will find the reduction in motor losses ( magnetic losses & Copper losses ) check the loss curve of motor given by manufacturer

I agree with you that core loss or magnetic loss will get reduced if the voltage is less. But current will correspondingly increase as Power = √3 V I Cos ø. Hence copper loss will be more.

That all depend upon the nature of load.

He has mentioned for Pumps. - Contant head - so, constant power out put - so, current will also reduce little . This can be verified practically - or verified with the head /power curve of pump + loss/voltage curve of motor.

In fact, I had verified the reduction in losses with voltage on shop floor.

Instead of acting on the voltage level , which itself is not, as said by others, a good solution, you should investigate (amongst others) :

Proper lighting management and may be lighting automation

Installation of VSD to optimize motors usage

Installation of Power Factor Correction and harmonics filtering to optimize losses and tariffs .

But energy efficiency methods is a vast topic, which cannot be adressed in a simple post reply. My advise is to consult some specialist blogs, forums, or manufacturers websites.