Power calculation for 3 phase motor with VFD

[Mya44]

I'm sure this has already been answered somewhere but I can't find it...if someone could point me in the right direction?

I have a 480/3/60 motor with a VFD powering a blower. Motor is designed for 900rpm but we're using the VFD to turn it down to 800rpm for normal operating and 300rpm as an absolute minimum.
How do I calculate the power consumption for this? I'm not sure how the VFD affects this. Is there an efficiency or something for it that needs to be taken into account?

Thanks!

 

[davidbeach]

Fan curves should tell you the require horsepower at the different speeds. This will be the shaft power required of the motor. Then you have the motor and VFD losses to add in and that's your power required by the VFD. Meter that and you can calculate your system efficiency.

 

[itsmoked]

If possible you can turn off everything BUT the motor and read your power meter after a minute or two. That will give you the actual energy that you pay for. Energy/Time = Power

You can count the little disk rotations if you have one on the meter. They represent a specific amount of energy.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[waross]

Fyrther to itsmoked's comments;
On the face of your KWHr meter you should see a parameter called kh or kH. Some typical values are 3.6 and 7.2. Other values are possible. This is the energy in Watt hours of one turn of the meter disk.
I often record the times for 10 consecutive turns. This indicates whether the load is steady. 3.6 /(time per turn in seconds) = KW demand or draw.
[3600 seconds in an hour. Divide 3600 by 1000 to convert from Watts to kilowatts draw.]

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jraef]

How accurate do you need to be? Most VFDs on the market now will provide you with the kW output and if you look at their technical data, they should give a part-load efficiency percentage of the VFD itself. Those numbers are pretty close.

So for example if you have a 100HP fan motor running at 80% speed and the VFD says the output is 38kW and the VFD tech data says the VFD is 90% efficient at 1/2 speed, then a pretty close estimate of your throughput power consumption will be around 42kW (probably a little better than that because the efficiency of the drive will be better than that 90% number since you are not really at 1/2 speed).

"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[Mya44]

Thank you everyone for the excellent suggestions. I should have been a little clearer in my post though...this equipment is not actually running yet so I can't physically measure the power consumption. I think what my customer is after is an estimate of what to expect so they can size their equipment so what jraef is suggesting should be fine.

Just to make sure I understand clearly though. Basically what the grid will see is the power draw from the VFD only, not the motor itself, correct? So if I have a max from the VFD of 664 kVA then, assuming PF=1 as it should be running close to unity then that would be 664kW also? And this should be what my client needs to size the supply equipment.

 

[Skogsgurra]

"what the grid will see is the power draw from the VFD only, not the motor itself, correct?"

No. Not correct. As far from truth as you can get.

The motor will consume power. The VFD modifies the speed at which the power is consumed. It also draws a few percent for its own use (aka VFD loss).

Look at the fan specification to see what it needs in terms of HP or kW. Divide by assumed (or known) efficiency. I would use something like .8 to be on the safe side.

If your VFD draws distorted current from the grid (most still do) then use the next higher fuse size, conductor size etcetera.

Gunnar Englund

http://www.gke.org

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

 

[jraef]

I think what my customer is after is an estimate of what to expect so they can size their equipment...

Ahh then... For the porpoises of selecting feeder equipment in the US on a VFD circuit, it's actually somewhat simpler. Determine the motor size, select a typical VFD to control it, and per the National Electric Code, you MUST size the feeder conductors for 125% of the VFD MAXIMUM NAMEPLATE CURRENT RATING, not what the motor is going to draw. Then you size the feeder circuit to protect those conductors.

"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[apuck]

Don't forget to look at the following: a) Verify that the conditions the blower inlet was designed for are accurate, and that a worst case for design conditions is properly established. A fan taking atmospheric air can be -20F in winter, as well as 105F in summer, will have different motor draws (at a given fan speed) due to a change in gas density. If essentially a constant inlet gas density (this fan is inside a building), then the motor bhp will follow standard fan law, whereby BHP2=BHP1x(RPM2/RPM1)^3. If at design point, the fan vendors states the motor will draw 10 bhp at 900 rpm, then at 300 rpm, BHP2=0.37 bhp.

b) Fan curves are based on a specific fixed (i.e. buyer stated) inlet condition. If the inlet side process changes, or you have a complex duct system, suction or discharge, you also may request a set of fan curves from the fan supplier with your preferred operating speeds stated.

c) Verify that your fan design point is based on valid analysis of the system (i.e. pressure drops at the designated flow rate thru all duct, fittings, dampers, entrance/exit losses from/into larger or open chambers, etc), since this will determine where on the curve you will actually operate. Oversized duct, shorter runs, fewer elbows, etc than originally planned, and you will operate further out on the fan curve - thus handling greater flowrate, which will consume greater power.

Also be careful with motor SF and inverter nameplate ratings. A 10hp inverter operating a 10hp motor having a 1.15 service factor cannot access this safety factor, since the inverter is not provided with a matching SF. If you want access to this motor SF, jump to a 15 hp drive. Un-matching drives are not uncommon, and can be used for a future upgrade in motor hp if the process can potentially change down the road.

Hope this is helpful and not overly complicated, but simple questions often do not have simple answers. - Al

"Beware of the man who won't be bothered with details".