Dear Members

I need help from YOU. We have a motor rated for 415V, 3Ph, 50Hz for Boiler FAN application. KW rating of the motor is 250KW operating through VFD. The power consumption is high (220KW) in the field for the same flow and pressure of the fan during testing. The power consumption of the motor is 155KW during testing.

Why there is increase in power consumption for the same flow and pressure of the Fan in the field. Is there any effect of VFD.

Please help me to find the solution

Pandi
Design Engineer

 

It is a common misconception that a VFD will decrease power consumption of a motor operating at near full load conditions. The VFD will actually increase power consumption due to voltage drops across the semiconductor components, creating heat.

 

You can forget about the effects of the VFD.

A Motor will draw the necessary current to drive the load, independent of the source, either VFD or the mains.

 

Dear Senthil Kumar,
Probably there was an error in your intial or field readings, The figures will not change that much.
Possible reasons
1.Was the clamp ammeter kept very near to motor? The stray magnetic field produced by big motors affect the clamp on ammeters.
2.Was the suction side restricted when taking these measurements or the suction opening different for both cases?
3.Or the polarity of any one of the the measuring instruments reversed?
That's all I can see.
Best regards,
Sekar

 

Power required for a fan or pump depends on the product of mass flow and pressure drop. If the density of the air youare handling is greater than design, you will have the same volumetric flow and pressure difference, but require a greater power to drive the fan. This would arise if the fan is rated to handle pre-heated air but is being tested on cold air.

Bruce.

 

How r u calculating the power consumption when vfd is used?
Please clarify...

A.K.D.

 

You cannot forget about the VFD if the motor is almost always running at full load. If on the other hand it is running a part load most of the time, the savings due to operating at a lower power rating are large enough to overcome the VFD losses. Remember that P = VI and if you have the usual 0.7 volt drop across a power device then on a 300 amp load the VFD will generate 210 watts of heat per pole or 630 watt for all three phases. This is why VFDs have heat sinks, fans and other cooling equipment and why NEMA 4 and 12 ones are so much more expensive. If you are almost always running at full load then there is probably no need for a VFD for energy conservation. The
process or equipment may require it for other reasons.

You get the same problem with solid state starters. In some cases on larger HP ratings they install bypass contactors that close after the motor is up to speed and take the solid state part out of service so they can reduce the losses thru the electronics.

James Bouchard

 

Bruce's reply is well taken and if it solved your dilemma then a word of thanks should be expressed. If not, following are my comments:

If the application is not a preheater and the problem not eliminated, and, If the wattmeter is installed via fixed wiring, then you may have a connection error. Or, there may have been a phase-sequence or a polarity error if a clamp-on wattmeter is used. Were measurements for Factory Acceptance Test (FAT) purposes taken with a fixed single-element meter.

Were onsite tests performed with a clamp-on wattmeter? If so, did the test consist of one measurement? Two? Or, three? Also, what is the motor's rated-current as shown on its nameplate?

Regards,
Phil Corso, PE
(Boca Raton, FL)

 

Thanks for phil corso's answer
The motor full load ct is 446Amps.the measurement was done using clamp on wattmeter suitable for single phase system We observed an unbalance in the current drawn by the motor.The power factors are 0.6,0.94,0.4 in R,Y,B phase respectively.

The test is going on I will come back after collecting more details.

Pandi

 

Further to my earlier reply to Pandi:

1) First sentence... end of sentence should read "... used simultaneously."

2) Although I recommended a procedure using the R and B leads, other connections are possible as long as polarity and phase sequence are properly observed.

Regards,
Phil Corso, PE
(Boca Raton, FL)

 

You should not use a single phase watt meter for three phase loads. I am very suspicious about power factors that vary from .4 to .94 on the same
load. Even more so for a VFD. Also you need a watt meter that will respond to the full frequency range provided by the VFD and will give accurate readings with a non sinusoidal wave form.

James Bouchard

 

A single-phase wattmeter is a perfectly legitimate method for accurately determining power or energy in a balanced or unbalanced three-phase load, regardless of voltage and/or current waveform. A graphical method is also available, based solely on voltage and current measurements.

Regarding the wattmeter approach, two methods are possible: the first is called 2-wattmeter method; the second is called the 3-wattmeter method. The latter is only used if one must determine the power in individual phases of an unbalanced 3-phase load.

My earlier Wed, Jul 3, 1:12pm response, illustrates a procedure for the 2-meter method. It also explains, although not fully, why PF readings are different.

If you would like a discussion on "The Mathematics of n-1 Wattmeters to Measure n-Wire Power", contact me.

Regards,
Phil Corso, PE
(Boca Raton, FL)

 

Replying to Pandi's observation:

Three measurements are taken only if the "neutral" is available, or three instruments are being used. If not, then take just two
measurements:

1st measurement... Phase leads from R to Y, with ammeter in R line.

2nd measurement... Phase leads from B to Y, with ammeter in B line.

NOTES(S):

1) Above requires that phase rotation be R-Y-B. Clamp-on amp window must be in the same direction in the R nd B phase leads, eg, from source thru window to load.

2) Also, even though you provided unbalanced PF readings, the difference may be related to meter connections.

3) True unbalance will show up in the line current measurements. If more than, say 10%, then VFD is cause of unbalance. Please provide line
current measurements.

If you want corresponding equations, please contact me.

Regards,
Phil Corso, PE
(Boca Raton, FL)

 

Replying to Phil corso.
We are back to office now.
We have taken the readings with 3ph clamp on meter and observed that the current readings are almost same (1Amps difference in R,Y,B phases ) and the power factor is 0.98 .The voltage fed to the motor from VFD is 406 (rated voltage is
415V).Whether the pf 0.98 is due to VFD,then how the VFD will improve the pf for induction motor from 0.85 at full load to .98 at full load.Please explain the concept of pf improvement with VFD.

Regards.
pspandi

 

Attn: P. S. Pandi:

I want to apologize for providing instructions via text. My goal is to illustrate how just two power measurements can be used to determine
power factor. This could refute the value of 0.98 that was obtained. (BTW, this method can also be used for unbalanced loads.)

Therefore, please forward your Fax #, and I will forward the proper procedure, including vector diagrams. This offer is also made to anyone else interested in this topic.

In reply to your query regarding the VFD and power factor... correction is only possible thru the application of capacitors for that purpose.
The VFD cannot delete the motor's magnetizing current requirement. I believe that the 2-wattmeter test will confirm this.

Regards,
Phil Corso, PE
(Boca Raton, FL - [email protected])

 

Hi Mr. Phil Corso,

We too are facing similar problem. A Hoist Motor (22 kw, 4 pole) is used through VFD to achieve 550 RPM; (frequency used around 23 to 24) and the current flow (amps) is becoming very high. Can we use 8 pole Motor instead of 4 pole so that frequency need not be so low. Kindly advise.

 

Subhendu... yes, it will certainly reduce the current, that is, if the replacement motor's output rating is equal to the existing motor's rating.

Caveat,
It will also halve the speed, and increase the available torque from the motor. Check to be sure that the crane's hoist requirement does not exceed its existing rating.

Regards,
Phil Corso (cepsicon [@] aol [.] com)

 

Subhendu... additional information:

The FLA of an 8-pole motor will probably be higher than that of the 4-pole motor. Thus, be sure that the motor's existing feeder-cable and protective device are adequate.

Regards,
Phil

 

...
>The voltage fed to the motor from VFD is 406 (rated voltage
>is 415V).Whether the pf 0.98 is due to VFD,then how the VFD
>will improve the pf for induction motor from 0.85 at full
>load to .98 at full load.Please explain the concept of pf
>improvement with VFD.

Think of a VFD is a NEW power source for your motor,. The input to the VFD from the AC line source is essentially the "raw material" from which the motor power is manufactured. The AC is rectified to DC and stored in the capacitors, which serve to smooth out the ripple caused by the diode bridge rectification. Once you establish the smooth steady DC bus, transistors on the inverter side of the VFD fire that DC in a PWM pattern that makes the motor act as if it is getting AC power again, but at the RMS voltage and frequency that you control with the VFD. So there is no longer as direct of a connection from the AC source to the AC motor.

In the process of rectifying the AC into DC, the current is drawn by the capacitors through the diodes to recharge them as they are depleted by the transistor firing into the motor. Since capacitors charge almost instantaneously, as the voltage increases within each sine wave (beyond the forward conduction voltage of the diode) the current is drawn almost immediately as well. So there is almost no lag in current compared to voltage, ergo the power factor is very close to unity, .98 as you observed (assuming you were measuring on the input side of the VFD).

On the other end of the VFD, going to the motor, the current through the INVERTER section does indeed include both the active current and the reactive current. When you read the current on the VFD display however, it will not integrate, it will only show the active current, and if intelligent enough, maybe the motor power factor. Therefore, when you compare the input current to the output current, it always appears as though the output current is higher than the input, because the output current shown is at the power factor of the motor, whereas the input current is at near unity power factor as it is drawn by the diode bridge. The difference will be representative of the motor power factor at the given loading, which is dynamic.

Since we do not know how the values you are discussing were attained, the concern you have is, as yet, unfounded. In other words this might just very well be a simple issue of not understanding that the values will differ from one side of the VFD to the other.

 

Hello,

When a Variable Frequency Drive (VFD) is used for fan applications, it can significantly reduce power consumption. VFDs control the speed of an AC induction motor by varying the frequency and voltage supplied. With a VFD, you can match the speed of the motor-driven equipment (like a fan) to the load requirement, meaning no energy is wasted. For fans, a small reduction in speed can lead to significant power savings, thanks to the fan affinity laws. However, to get an accurate measure of power consumption, one would need specifics such as motor efficiency, load profile, operating hours, etc.