Soft Starter With Energy Saving

[SGK2007]

I came across a live deom of a softstarter with energy saving feature for avarying load application. The voltage across the motor was varying in relation to the load on the motor. Eg for a 40HP motor, at NL current of 13A, the voltage measured ws 200V and at 50A, it was 380V. The voltage kept on varying. It did show up energy saing on the energy meter. My question is what is this technology and how reliable is this?

[marke]

Variations of this concept have been around for around 30 Years.

The original technology was developed by Frank Nola of NASA and worked by reducing the voltage applied to the motor when the power factor of the motor dropped.

 

By reducing the voltage, you reduce the iron losses in the motor.

If the load current is greater than the magnetizing current, you wil increase the motor current with reducing voltage and so you willincrease the copper loss.

 

The true energy saving is only going to occur when the load current is less than the magnetizing current and is a portion of the iron loss of the motor.

 

On modern motors, the iron loss is very small relative to the motor rating and so the potential to save significant energy is very small.

 

On small single phase motors, the iron losses are much higher and better results wil be achieved.

 

There are a number of threads on this topic on this forum, plus there is a aper at http://www.LMPhotonics.com/energy.htm

 

Best regards,

Mark

[SGK2007]

Hi Marke

 

I understand what you are saying.

 

I am not talking about the energy saving in the terms of iron losse of motor. I am not talking about motor efficiency here. but "KWH".

 

Will "V x I" factor obviously not reduce for a 200V as compared to 400V applied across the motor by 50%? Depending upon how much time such a condition remains, the energy consumption will be 50% lower than what would be if I dont reduce the voltage?

[marke]

I am not talking about the energy saving in the terms of iron losse of motor. I am not talking about motor efficiency here. but "KWH".

 

Actually, you are. You can only reduce the KWHr if you reduce the KW.

A power flow of 1KW for one hour equals one KWHr

 

As you can not reduce KWHr with out reducing KW unless you reduce the run time, you do need to look at where the KW are reduced and how much.

 

If the motor is very efficient, there is very little energy to save and therefore very little KWHr to be saved.

 

For example, if you reduce the iron loss by 500W, you will save 0.5KWHr for every hour that you reduce this amount of power

 

Will "V x I" factor obviously not reduce for a 200V as compared to 400V applied across the motor by 50%? Depending upon how much time such a condition remains, the energy consumption will be 50% lower than what would be if I dont reduce the voltage?

 

Power equals V x I x pf where pf is the power factor and this changes.

On a lightly loaded motor, the power factor is low and increases as you reduce the voltage. It is a very important part of the equation!!

 

On a motor that is partly loaded, as you reduce the voltage, the load component of the current will increase, so half voltage does not mean half power. This would only be true if the current stayed the same and the power factor stayed the same. This wil not happen with an induction motor.

If your motor is operaing at say 90% efficiency and you reduce the voltage, the power input to the motor can not drop much, as the motor can not "generate" power.

I have seen many claims where the motor is operating at half load and it is claimed that the magic device will save 50% energy, but the motor may be operating at 87% efficiency at half load. The only way it can save 50% power is to make everything operate atgreater than 100% efficiency and that is not possible.

 

Best regards,

Mark.

[PowerQualityDoctor]

When the motor is loaded by more than 50% its efficiency is indeed high and cannot be improved. However, when the motor is loaded by less than 50% the efficiency is reduced. When you reduce the voltage you reduce the maximum power of the motor. For example, a 100HP motor loaded with 40HP (30kW) is loaded by 40%, which means efficiency of around 80%. If you reduce the voltage to 80% from nominal, the maximum power is also reduced by 0.7*0.7=0.49, which means it operates as 49HP motor and the new loading factor is 80%. At this loading, the efficiency is as rated. This means around 10%-15% saving. The problem lies in the harmonics. As soft starters create harmonics due to chopping of the waveform, they cause additional losses and failures. As a result, you will have kWh saving on the motor but less or no saving on the main service (harmonics affect the efficiency of all loads). There are modern implementations of this concept using sinusoidal waveforms, so you gain only the savings without the "side effects" of harmonics. Unlike the soft starter based solutions, they provide real saving. Note that it depends on the motor loading - in some applications they provide great value while in others they are useless.