Star Delta Star Converter for Press Machines

Sounds like you need a lesson in REAL motor energy issues.

An AC induction motor draws energy in proportion to the applied load, meaning it only draws as much energy from the source as is necessary to do the work it is designed to do. So right there it is INHERENTLY "energy saving" without doing anything. This fact is the main thing that is exploited by people trying to sell useless or marginal "energy saving" schemes for them. They try to take advantage that most people perceive that a motor is using maximum energy all the time. It just is not true. So right now, with no "energy saver" scheme, when your press is running at 40% load, the motor is ALREADY drawing 40% of the energy it was at full load.

With one very small exception; motor losses. Of the energy that the motor consumes from the line, not all of it is converted to useful work. Some of that energy is lost n MAKING the motor work in the first place. So when you read a motor's efficiency value, that is what this refers to. If the motor says that it is 90% efficient (as many new high efficiency motors are now), then that means that at full load, 10% of the energy it consumes never makes it to the load. These losses can be further broken down into two rough categories and then again into 4 more refined categories. The two rough categories are fixed losses and variable losses. The fixed losses are those that remain constant without regard to load, and the variable losses themselves vary with motor load.

Starting with the variable losses, these are resistance (I²R losses in the conductors) and stray load losses (flux related losses in the stator due to minor manufacturing defects). Of these, because loading affects current, the I²R losses themselves are already load dependent and therefore lower current means lower I²R losses in the copper. I²R losses represent roughly 65% of the total losses in a motor, so right there you again ALREADY have "energy savings" even among the losses just from lower loading on the motor. The stray load losses are only controllable in the motor manufacturing process, but represent a small percentage of the total, roughly 7-10%, and nothing can be done about them after the fact.

The fixed losses are Iron losses (the energy used to magnetize the core and the eddy current losses in the laminations), friction and windage (the cooling fan and bearings). There is really not much that can be done after the fact to further reduce the friction and windage losses, but they too only a small fraction of the total losses (windage and friction together represent roughly 10% of the total). So the so-called "iron losses" representing roughly from 15 to 20% of the total losses, are the only thing that can be affected from the outside.

The iron losses are related to the applied voltage, so if you decrease the voltage, you can decrease these losses. But the relationship is not linear, because not ALL of the iron losses will vary with voltage. Running a motor in Star that is designed to run in Delta, will change the power configuration to effectively reduce the voltage across the windings to 58% of normal. So consequently, the iron losses can never be reduced to anything less than 58% of the 20% of the total losses, which themselves are already a small percent of the total energy. Extrapolate that out to see what it gets you, using a motor that is even an inefficient one at 85% efficiency. So you have a motor that already has only 15% losses, then of those, only 20% can be affected, so that's 3% of the total, and of those, only 58% maximum can be affected by reducing the voltage, so you are down to 1.74% reduction in the energy use. But remember the first statement, the energy use is ALREADY at 40% because your load is at 40%. So your NET energy savings, as a percent of full power rating of the motor, are 40% of 1.74%, or 0.7%. That means if you have a 100kW motor, you are going to save 7kW by running it in Star.

Not insignificant you say? Well, the problem is the risk / reward ratio. When you run a motor in Star, as said you reduce the voltage to 58% of design. Since torque is related to the square of the applied voltage, you now have 33% of peak torque capability! So from the outset, your 40% load reduction is already a myth, most likely based on an outdated assumption that all motors are over sized by 20%, something that machine designers stopped doing 20 years ago. So your 100kW motor is effectively now a 33kW motor and if your load is 40kW, or even 35kW, then you are going to OVERLOAD the motor. So keep that in mind, you really cannot use the Star connection safely unless you have a load reduction of 67% or greater (33% load, not 40%). That then also means that your savings is not as much either: you are only going to realize 0.574% of savings at the risk of overloading your motor, PLUS that 0.574% savings is ONLY going to happen during the time periods that the machine in running at 33% capacity. If that is only a fraction of the time, make sure you factor that fraction into the fractions of fractions of fractions that already exist. THEN look at the fact that if the machine is running at 33% capacity, that means that the revenue it generates is ALREADY lower than it used to be, so your energy cost PU of production is now actually going to go BACK UP, regardless of the savings.

Lastly, add to that the fact that you have added 3X more components to the system to affect this savings; more contactors to wear out and maintain, electronics that must be kept cool, control circuits that can go awry, and you have probably decreased your MTBF (Mean Time Between Failures) 4 fold. So for example that means if you had a failure of one sort or another once every 4 years, you can now expect a failure of some sort once every year.

Factor the cost of the added down time risk and that 0.574% of energy savings for a fraction of the operating time doesn't look so good.