reducing voltage to three-phase induction motors

[magoo2]

When you reduce voltage to an induction motor, how much energy savings will result?

To my way of thinking, supply voltages are a usually little higher than motor nameplate voltages (like 480 V supply with 460 V motors, or 4160 V supply with 4000 V motors). Looks like about 4% higher.

There will be some voltage drop to the motors, but probably not 4%, maybe 2%. This means that the nominal voltage at the motor is 2% high. If I reduce the voltage (through transformer taps) such that I get rated voltage at the motor, I should save some energy.

I'm trying to find how low can I go with voltage and still save energy. I'm assuming that generally the motor will not be fully loaded.

Are there any good references or links dealing with this topic?

 

[SteveWag]

Reducing voltage will increase losses, decrease efficiency, shorten motor life and cost you more in general.

Steve

 

[dpc]

Motors are pretty much constant kVA loads - voltage goes down, current goes up. There may be a sweet spot of efficiency at a slightly lower voltage due to reduced core losses, but in general, I don't think lower voltage to an induction motor is going to save much energy. Different story for resistive loads, of course. When voltage gets too low, the motors can become overloaded.

 

[magoo2]

Hey dpc - I like that term - sweet spot. It certainly fits here.

 

[Skogsgurra]

magoo2,

Your thinking has some validity with some loads. Low loaded motors usually do not need all the excitation current available to maintain good efficiency. That is why some frequency inverters have a "square law" mode where V/Hz is reduced when lowering the speed. Works quite well with fans.

Also, a vector drive usually does not have the nominal V/Hz number when running lightly loaded. Simply because the motor doesn't need the full voltage. If load is increased, the voltage goes up. This sometimes confuses newbies - and old-timers as well.

But, usually, you will not gain much by doing such an adjustment. And very often you lose due to higher slip.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Marke]

This is the basic concept behind the Nola energy saving algorithm. - Reduce the voltage ans save energy when the motor is operating at reduced load.
There is quite a lot of information around on this, but essentially, if you look at the motor, you have two major sources of loss. The iron loss and the copper loss.
If you reduce the voltage, the iron loss will reduce and the magnetizing current will reduce. The work current will increase.
If the total current is very close to the magnetizing current, then a reduction in voltage will reduce the current and the iron loss. This gives you a reduction in the coper loss and the iron loss. If the current is much higher than the magnetizing current, then any reduction in the magnetizing current will be lost by the gain in work current. This can result in a copper loss gain higher than the iron loss saved so you get a net loss, not a gain.

This really is only relevant if the motor is over fluxed. If the motor is operating in it's normal operating range, there will be little or no gain and probably a loss.
- Have a look at the magnitude of the iron loss as a percentage of the motor rating. - pretty small on a large machine.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[electricuwe]

When discussing this topic it is very important to consider the size of the motor. What might be a good approch for fractional horse power motors is totally stupid for motors running from medium voltage.

As mentioned by Marke you will find a lot of further information if you take a look on all the discussions on the Nola approach which is basic operating principle of all the "energy savers" not capable to change the frequency.

 

[gepman]

Refer to NEMA MG-1 to see how induction motors should behave with varying input conditons. For motors with over 50% loading there is not much savings in lowering the voltage. The EASA (Electrical Apparatus Service Association) Electrical Engineering Pocket Handbook has a nice graph of what happens on page 20 of my year 2000 version. It is adapted from the above source. Following is some information from US Motorsat

http://www.usmotors.com/products/ProFacts/1-133.htm

Effects of Variation of Voltage and Frequency Upon the Performance of Induction Motors:

Induction motors are at times operated on circuits of voltage or frequency other than those for which the motors are rated. Under such conditions, the performance of the motor will vary from the rating. The following is a brief statement of some operating results caused by small variations of voltage and frequency and is indicative of the general changes produced by such variations in operating conditions.

With a 10 percent increase or decrease in voltage from that given on the nameplate, the heating at rated horsepower load may increase. Such operation for extended periods of time may accelerate the deterioration of the insulation system.

In a motor of normal characteristics at full rated horsepower load, a 10 percent increase of voltage above that given on the nameplate would usually result in a decided lowering in power factor. A 10 percent decrease of voltage below that given on the nameplate would usually give an increase in power factor.

The locked-rotor and breakdown torque will be proportional to the square of the voltage applied.

An increase of 10 percent in voltage will result in a decrease of slip of about 17 percent, while a reduction of 10 percent will increase the slip about 21 percent. Thus, if the slip at rated voltage were 5 percent, it would be increased to 6.05 percent if the voltage were reduced 10 percent.

A frequency higher than the rated frequency usually improves the power factor but decreases locked-rotor torque and increases the speed and friction and windage loss. At a frequency lower than the rated frequency, the speed is decreased, locked-rotor torque is increased, and power factor is decreased. For certain kinds of motor load, such as in textile mills, close frequency regulation is essential.

If variation in both voltage and frequency occur simultaneously, the effect will be superimposed. Thus, if the voltage is high and the frequency low, the locked-rotor torque will be greatly increased, but the power factor will be decreased and the temperature rise increased with normal load.

The foregoing facts apply particularly to general-purpose motors. They may not always be true in connection with special-purpose motors, built for a particular purpose, or as applied to very small motors.

 

[aolalde]

Typical figures for an induction motor 400 HP, 4 poles are; Eff = 94.5% and PF = 87.5%. This means the input power is 400/. 945 = 423.28 HP and the total losses = 423.28 – 400 = 23.28 HP (17.367 kW)
If the efficiency should be increased to 100% (zero losses, ideal motor) the saving is 17.367 kW

When the efficiency reaches its maximum value; Lfix (the Friction+Windage loss + The Magnetic core loss) = L (I) (Stator I^2*R + Rotor I^2*R + SLL).
Assuming the maximum efficiency happens at full load, Lfix = L (I) = (17.367)/2 = 8.684 kW

The magnetic core loss and the stator I^2R loss could be up to 70% of each group. (8.684*0.7= 6.078 kW)

Voltage reduction will reduce the core loss drastically but will increase the current in the windings. Assuming 15% voltage reduction the new values approach:

Core loss2 = (.85)^2 * 6.078 = 4.391 for a saving of (6.078 – 4.391 = 1.707 kW)
Stator I^2R2 = 1.15^2*6.078 = 8.038 for an increased loss of ( 8.038 – 6.078 = 1.960 kW)

And the net result is an increase in losses around 0.250 kW

This means, unless you modify the motor construction parameters, changing the line voltage will have not significant improve on the efficiency.

 

[magoo2]

aolalde
I think your conclusions are based on a fully loaded motor. I'd be interested how the numbers shake out if you assume a lighter load like 75% of nameplate.

 

[aolalde]

At 75% load assuming the power factor will be reduced around 10 points or PF = 86.5 % at full voltage and due to the flux reduction it keeps the 87.5 with reduced voltage and 75% load and no change in efficiency ( very optimistic).

At full voltage, the current I1= (0.746* 400*.75) / (1.73 V*.865*Eff)
With 85% voltage; I2=(0.746*400*.75) /(1.73 *.85V*.875*Eff)

Comparing these currents; I1 / I2 = 1.73*0.85V*.875 *Eff/ ( 1.73*V*.865*Eff) =0.856

That means, at 75% load the line current at full voltage is still smaller (85.6%) in spite of any improvement on the power factor due to the voltage reduction to 85%V.

If we repeat the loss calculation process above, we see that any gain while reducing the core loss by reducing the voltage is masked by the winding loss due to the current increase with the voltage reduction.

 

[jraef]

The quick and dirty rule of thumb used for years on the Nola voltage reduction devices is that tangible savings from voltage reduction happens at around 50% loading. Any more than that and aolalde's observations begin to hold true. Somewhere between 50% and 100% loading there may be very slight savings, but it's rarely worth the other consequences of the Nola controller (harmonics, heat losses in the controller etc.) The big question then is always, if it is only loaded to 50%, why is it even running?

 

[gepman]

There is a very good discussion on the Nola Energy Saving Algorithm at

http://www.lmphotonics.com/energy.htm

 

[magoo2]

The other point to keep in mind is that we're not talking about a voltage change of 10% or 15%. 5% is probably the maximum voltage reduction that could be achieved and still meet the range of allowable voltage regulation per ANSI C84.1.