VFD eff, power factor and harmonics 10

[BigInch]

Yes, I've drifted into water deeper than usual.

VFDs are said to have an efficiency of apx. 95%, so
with a 100 brake HP pump, I should therefore add power. For a 100 HP (74.5 kW) rated 94% eff motor, I need 79.3 kW) input power to the motor and I apparently need 83.4 kW supplied to the VFD.

First, is the VFD efficiency factor of 0.95 used to account for the chopped up current delivered from the VFD to the motor instead of the usual AC sine waves?

After reading for a couple of days now, I am still confused as to if poor power factors of some VFDs affect power consumption costs. I think I understand that power factor from VFD to pump is close to unity, so they must be talking about power factor of the supply to the VFD. If this is true, does this result in extra power costs? How much extra? Does the poorer power factor vary with produced frequency, or is it constant and does it occur only during some events or is it constant through time (given a constant speed at the pump)?

VFD heat loss.

A VFD manual says that it loses 2310 W of heat. Its controlling the same 100 HP 480 V motor above. (Only 0.6% of rated motor power?). Is this heat loss constant, or only at full rated motor speed? It would seem like VFD eff should be less at reduced frequencies and reduced motor speeds. If so, what function describes the VFD efficiency vs the output frequency (as a function of % of full rated motor speed?).

Harmonics.

Apparently harmonics affect the power factor and the power bill too. True? How much power is lost by harmonics? If a filter is added to reduce harmonics, is power consumed by the filter (how much?) and is the harmonic power loss correspondingly reduced or do both power losses add together?

Is the typical 95% VFD efficiency accounting for poorer power factor and harmonic losses, or is it only considering the loss by chopping up the sine wave and these others are additional?

Much Thanks as always.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[itsmoked]

1) The VFD efficiency is not 95% 'period'. It is whatever it is stated to be - in this case 2310W verses the load.

The losses are caused by the switching and a VFD does a ton of switching. During this switching the actual devices get hot as they have voltage losses and current is passing thru them. So V_loss x I = P_loss.

Added to this loss is also the energy used to run the actual controls and the fan(s).

Keith Cress
kcress -

http://www.flaminsystems.com

 

[CJCPE]

I would expect that a 100 Hp AFD should have a little higher efficiency than 95%. I would use 96.5% for estimating purposes, but I believe that some manufacturers are advertising 98%. The AFD efficiency should take into account the losses in the AFD but not in the motor. The waveform causes some additional losses in the motor, but probably on the order of 1/2% of the motor’s output power. Remember that if the motor is 95% efficient, it has 5% losses and an increase to 5.5% losses would be a 10% increase. The VFD losses are mostly the output transistor losses. The input rectifier is probably responsible for the second largest component of the losses. Other losses occur in the DC bus filter, the logic and gate/base driver circuitry and the cooling fans.

If the VFD has input and/or output reactors or filters, those losses would not be listed as part of the standard drive losses, but would need to be considered as part of the system losses. I don’t believe that those losses would be much more than 1% for input and output combined.

The drive manufacturer’s listing of 2310W of losses would not include input and output reactors or filters unless they are furnished as part of the standard VFD. They would probably assume a 95% efficient motor. A 95% efficient 100 Hp motor would require 78.5 kW adding 2310W to that would make the input power 80.8 kW and the efficiency 97.1%. I don’t think that is unrealistic. The losses listed for the VFD would be for full load, full speed motor operation. At reduced speed and/or torque, the VFD losses would be less. Some VFD losses, such as logic and fan losses, are constant regardless of motor load. Some losses are proportional to motor current and others are proportional to motor power.

The VFD has a high displacement power factor (.95 or so) at the input, but a lower total power factor due to the harmonic content. Harmonic content is lower for VFDs connected to power systems that have higher source impedance (lower short circuit current capacity). Harmonic content varies with motor load. Input harmonic losses occur in the power distribution system and in the VFD input rectifier. Adding an input harmonic filter reduces harmonic losses, but adds losses due to the extra components that carry the total VFD input current.

The power factor at the output of the VFD is about the same as the motor power factor would be if it was connected directly to the power line. The VFD supplies the reactive portion of the motor current from the DC bus capacitors. That means that low motor power factor increases the DC bus cap losses and the transistor losses but not the input rectifier losses.

 

[itsmoked]

I'm doing this in pieces since it is a long one. I don't want anyone else also answering the same exact questions.

2)PF. Yes they are referring to the PF that the VFD presents to the power supply. No one much cares on the other side and it is also close to unity anyway.

Power factor is how far apart the peak current waveform is away from the peak voltage waveform. (one description) The farther apart they are the more physical current the source must provide. NOT more energy just current. No one has to burn more coal or fuel to provide this added current but all the equipment between the power source and the load has to bigger to carry this extra current. (There are secondary issues like added voltage drop caused by these larger currents that DO actually consume more energy.)

Power companies don't like having to have this bigger equipment even though they can't charge for it just via the end users kW usage. Recently power companies decided 'We aren't gonna take this any more!' So they now check the PF of their bigger customers and add a penalty to their energy costs based on the PF. The worst the PF (farther from 1) the steeper the penalties.

This structure makes it usually worthwhile to the end user to do things that 'correct' their PF. Capital outlays that actually pay back.

With respect to your VFD it can present a lower PF to the supply. Some utilities don't yet charge for PF or more likely, one load out in the boonies may not be considered for a PF penalty. But wise thinking would still have you considering this.

Because of this issue a lot of VFD makers have converted to PFC correction in their front ends. This actually forces the current and voltage waveforms into closer alignment. So often VFDs will provide a high PF and avoid the issue entirely.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[itsmoked]

I think we covered it.. If not let us know Big.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[ozmosis]

Another point worth considering is looking at the efficiency loss and determining whether the 'loss' can be utilised, thus cancelling some of the losses.
As an example, we are currently designing a system in India that will, according to the architect, be the greenest building in Asia Pacific. They have investigated long and hard the impact of each components efficiency losses and looked at utilising the heat productively. As pointed out in above posts, the main loss with a VFD is heat and this can be anywhere from 2~4% of the drive's rating. Using this heat from drives with a total capacity of 960kW means approx 25kw in heat can be re-used. The design of our drive and others on the market allow for back-channel ventilation and therefore the majority of heat dissipated can be focused into areas that need the energy.
Losses with VFD's are high when the losses are actual 'lost' and even moreso when you have to 'add' energy (cooling) in a plantroom to ensure the losses are not creating more problems with ambient temperatures.

 

[geefer]

I think the first thing to point out here is that if you require the driven load to run at 100% speed and at 100% absorbed power why fit a VFD at all. A cheaper and better option for such an application would be a soft start with by pass to run as DOL. If speed control is required then as the previous poster outlined the amount of losses greatly depend on the required speed, By utilizing the cube law for torque as the speed drops the torque requirement drops by its cube hence current required. If power factor correction is an issue then by fitting an active front end VFD unity power factor can be obtained . You could also if you wished run at a leading power factor. This setup will also give you very good Harmonics Mitigation. There is a trade off however slightly higher power electronics losses in the order of 3.5 to 4.5 %

 

[DickDV]

There are some errors in this string that should be cleared up. First, modern VFD's have about 2% losses but the 2% is LOAD KW, not motor rating or drive rating.

A 100hp drive at idle looses only about 100W which is primarily for the cooling fan. Some modern drives even go so far as to shut off the cooling fan when it isn't needed so the 100W may be even less at idle. A 500hp 460V drive will typically loose 1500W again, largely for the fans which may or may not be running.

As for input harmonics and distortion, I don't see where this affects drive losses much. It may result in higher losses in transformers and supply wiring but not in the drive.

Everything I've ever read or experienced indicates that drive heat losses are almost entirely dependent on output load.

 

[BigInch]

Thanks to all so far!

I've been reading and digesting these great comments and now understand power factor implications,but I'm still a bit confused about VFD efficiency, especially since I read this,

"Ask the Clearinghouse"
found on this page,

http://www.eere.energy.gov/industry/bestpractices/energymatters/full_issue.cfm/volume=19

There is a table of VFD efficiency values which I still can't get my head around in light of the above comments, which I understand to mean VFD eff is a constant and does not vary with output frequency.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[CJCPE]

We tend to focus on efficiency at full load and full speed because that is the most expensive operating point and the point at which maximum losses occur. That is the benchmark figure that is most easily found in the manufacturer's literature. When comparing two AFDs, it is usually assumed that the unit that is most efficient at full load and full speed will be most efficient at other operating points as well. If one unit has a feature like modulating the cooling fan speed, that assumption is probably not very accurate.

If the speed (frequency) is 100% but the torque is less than 100%, efficiency is less than it is at full torque because the output power (torque X speed) is less while losses are not reduced by the same percentage. Some loss components may not be reduced at all.

If the torque is 100% but the speed (frequency) is less than 100%, the same thing is true.

With a fan or centrifugal pump, the torque is reduced in proportion to speed squared whenever the speed is reduced. The output power is reduced in proportion to speed cubed. That means that the efficiency goes down very quickly as the speed (frequency) is reduced.

 

[itsmoked]

To add: The losses are the fixed losses of the controls and any fan. With the motor stopped you have the lowest losses but not zero losses. You, of course, actually have 0% efficiency as nothing is being done but power is being consumed.

Remember my first response equation;
V_loss x I = P_loss

That I is the big 'loss' driver as the V_loss value doesn't change much at all.

Don't need full motor HP? Then you are not going to have the maximum I, hence less P_loss.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[BigInch]

Now it makes sense.
Thanks again all.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain