Saving Energy with Drives

It is not saved in that case - you have actually discovered the HOLY GRAIL - that is an over-unity device / free energy / .........

You have broken the laws of physics - and the nay-sayers said it couldn't be done.

I would keep this very quiet, guard the machine, reverse engineer it, and then patent it. You will be a millionaire.

(or check your numbers again )

lspraba; are you reading the amps with a true RMS meter or the read out on the drive? perry

So what is the 'true' power usage of a VSD? Using an RMS meter on the incoming supply? or the VSD's display?

Geoff

The difference in current comes from the displacement power factor.

The drive inlet is almost unity while the motor's is probably 0.8.

You are using more energy in this case. Approximately 2-5% more.

BUT, you save when operating at lower speed. The exact quantity depends on the type of load.

if your set is working continiously. how woderful you are buying a set. no one can match it.

conservation of energy was broken big , big bigly. we will have enough energy sources.

http://www.jacobson-engineering.com/carriage-drives.shtml

Nice elegant translation of what Lspraba attempted to describe. You get a GA from me.

Thank you redfred--I have ran into a bunch of different anomalies working with drives and this is one that not only resolves needing dynamic braking, but turns a problem against itself and uses the enertia instead of wasting it---I think that it is interesting.

The article states '95%' efficient electronics but the energy consumed will be largely the mechanical losses in the machine itself, these cost energy in acceleration and reduce 'payback' on deceleration.

Another interesting strategy is when using several VSD's, some motoring some braking, as in mill/rotor applications, is to connect via THE dc LINK returning energy from one to the other. A neat trick but the braking VFD has to be returning less energy than the others are outputting.

Lspraba,

The best way to check power consumed is by using power meter at input side.

Voltage/currrent measurements done at output of VSD will not be correct given the nature of waveform at the output.

Best way is to put powermeter at input and measure power

a)when using VSd

b) Run the motor direct on line.

compare....

cheers.

Hi Lspraba,

Your machine must be a disciple of the great Indian Magician Dr. PC Sarkar who can produce "Water of India" from an empty jar any time during his show . Otherwise it cannot happen. Check up your meters there is some thing wrong.

Electrical engineering follow certain rules .If your transformer is a simple isolation transformer it can only effect loss of energy and not saving. It cannot do any magic. The unit you may be talking about can be VFD (Variable Frequency Drive) used for energy conservation of motor drives. Please check up .

_{"The input current is 1.2A}

_{But the output current is 2.4A}

_{So how power is Saved ?}

_{Kindly explain"}

You discrepancy is due to measurement error. If you are measuring the output current with a basic ammeter, it will not give you accurate information because of the high harmonics on the PWM side of the VFD. In fact, your input current will be higher than the output current if the voltage is exactly the same because of switching losses in the VFD.

Dear LS praba,

By your narration, First of all you does not require any VFD for your application. since you are running the motor as normal, 415v,50hz,

By law, speed (N)=(120 * F)/P

and Power is directly proprtional to N quebe.

Your current reading may wrong. jsut measure current at insput side at 50 Hz and at 25 Hz, you feel the saving. Running at same speed , you never get any savings. your VFD will be dead invsetment.

Regards

Keshavamurthy

Low Motor Current:

Oh, the other common reason for such a low motor current at 415V when the load is very low is that the motor is connected for 660V, so you could have 415V on the 660V connection and achieve low current.

If this is the case, DO NOT apply much load or the motor will stall, the current may not be very high but the efficiency loss may result in motor burnout!!

Metering:

A normal CT makes a good estimate of motor side current. Supply, (input) current is more tricky as it has a high (low order) harmonic content due to the inverter being a rectifier load on the mains. High frequency response true RMS or Power analyser is best. This waveshape improves using a choke hence the input RMS current reduction at the same power.

Dear Malcolmk,

Thank you very much for clearing things to me.

AC Choke on the supply side improves the waveshape of the current and reduces its RMS value

This is the answer that i need. The drive unit is having AC choke in the input side and DC choke at the output side due to which the clamp meter was showing lesser current in the input side. So i need to measure with a real RMS meter.

But at the output side there was no problem to measure with the normal clamp meter. I have measured motor current in drive mode & DOL mode. Both are same.

That is why it was showing higher current than the input side. Please confirm whatever i have written is correct or not?

Thank you very much to all, who replied to this question. I have learnt many things from your answers.

Dear Lspraba,

the output choke (AC) is used for other reasons to reduce dV/dt, normally to prevent 'nuisance' overcurrent tripping if you have a long motor cable length (High capacitance). This does not normally affect your current clamp readings.

The supply side AC input choke 'flattens' out the current shape thereby reducing RMS current - the power is basically the same. Whether the electricity bill is reduced depends on your kW and kVAr metering details.

The DC Choke I mentioned is an inverter option which fits to the positive side of the DC link, see inverter instruction manual for details if you want.

Hope this helps

Thank you very much Mr. Malcolmk

Looks to be the most feasible answer of secondary higher than primary other than the instrument (or typing error - ie the primary values and secondary values transposed)

Hi..

I'm working with Schneider..You can only save energy when ur motor is running on low speed with Drive, If on full speed no chance to same energy..

Means u have to reduce frequency of VFD to save power.. maximum possible savings are with FAN/Pump/HVAC application where u can save 35-60% based on ur application..

You can check out this Schneider software for energy savings calculations.. with Schneider VFD's..

http://www.download.schneider-electric.com/C12571B9003C86E1/all/C125713F005265E2C12572EC0026F59E?OpenDocument&L=EN

'You can check out this Schneider software for energy savings calculations.. with Schneider VFD's..'

In my opinion, you can only save energy in a process by increasing efficiency or by a reduction in the process itself.

In the case of Fan/pump/hvac the key is to reduce the speed to meet the process requirement rather than wasting energy in turbulence in pipework, dampers or throttling valves etc.

To suggest that an energy calculation is only valid using the software estimate is only applicable with that supply VSD is erroneous and mistaken, any efficient way of changing speed would effect the same saving. Many suppliers offer this type of estimate, independent models offer similar.

VSD's are a flexible, practical way of achieving speed reduction and hence power reduction but keep your focus on the main aspect in how this is achieved. Of course, the implication of introducing VSD also entails VSD losses itself , EMC considerations and associated supply side harmonics being introduced to the power supply.

It would be interesting to install and calibrate CTs on the line side and meter your amps and see how the speed affects them and compare it to the HMI. We have a few big drives that run up around 80 hertz and the higher speed seems to give the machine more momentum and less torque--I may have to take an amp clamp and see what the motor is doing--Interesting.

Running 'supersynchronous' above the motor rated voltage/frequency normally effects underfluxing and motor current may reduce at lighter loads accordingly. If the motor is pretty loaded and you increase frequency (assuming voltage stays the same at max output) and so machine power is increased, motor amps will rise.

If your supply voltage is simiilar to motor rating (i.e. 'correct') and your inverter control is excellent at all frequencies, you should see motor load torque quite accurately on the inverter. This torque and your different speeds will tell you how much you have in hand for speed increase over rated motor speed and also what increase in power or torque this requires. If your motor's big enough you can then contemplate the speed up.

e.g. 400V 4p 90kW motor 160AFLC, No Load c.40A so 100A is approx 50% load etc.

Been a long day, hope this is understandable, I think I know what I mean!

CT's are OK on motor side, supply is more difficult, I normally use a power analyser (expensive)to see RMS A, pf, and harmonics or voltage distortion on input. A cheap 1000:1 CT straight into my mA input (multimeter) is OK on the output side! A good quality true RMS clamp should be fine, some clamps vastly over-read because of the harmonics in my experience. A DC choke is most cost effective way to reduce input amps, then AC choke on supply side.

Yes, to a degree--There is a lot more reading I need to do fully understand this. You have dedicated a lot of time here. Thank You-- K T

Dear L.S.Prabha,

What you are measuring at the out put is not correct. The out put of VFDs never pure sine wave. Also you can not measure voltage and frequency. If you are running the motor at less load than motor rating then you will be definitely saving power.