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goodguy
10-03-2007, 02:11 PM
W:) ondering if anybody can help, I have a 50 hp motor running a belt driven compressor. FLA is 47.1A at 575v 60hz. SF of 1.25 and a efficiency of 75%.
could anyone tell me how to figure out the watts used at full load conditions. Ultimately I would like to figure out the power factor for the whole plant. If anyone can shed some light on this it would be much appreciated.
thanks

Brian_UK
10-03-2007, 06:13 PM
Ho Goodguy, not my area really but doing a Google on power factors brings up a fair selection...

Here's one to start with......
http://www.angelfire.com/pa/baconbacon/page2.html

NoNickName
10-03-2007, 07:44 PM
Unless you know the RMS drawn current, you are unable to know the power factor. Supposing the FLA is the actual running current:
50hp = 36.75kW
36.75/47.1/1.73 = 0.451 - very bad power factor.

Electrocoolman
10-03-2007, 09:19 PM
Hi Goodguy,

The only way of measuring the power factor is with a proper KVA power meter (as opposed to an electricity meter which will only register kWh). You will need to measure both current and voltage.
A power meter should actually give you a direct reading of the power factor. If you are considering the whole site then it would be worth renting one in or possibly purchasing one?

Motors can have a power factor anywhere between 0.1 up to 0.9 depending upon the loading. Lightly loaded motors having a poor power factor.

I guess the figures you quote are purely those from the motor rating plate.

A motor is normally rated at a pf of 0.8 lagging, thus in your case:

50 hp = 50 x 746 = 37.300 kW ( = P) [1 hp = 746 W]

P= E x I x 1.732 x pf
= 575 x 47.1 x 1.732 x pf
= 46.907 x pf (kW)
thus
pf = 37.300 / 46.907
= 0.795

I expect that your actual p.f will be 0.4 to 0.7 as generally when designing, the 'safety factors' push the motor rating up to the next standard size motor, so in reality your motor will not be fully loaded, and hence will not be at its most efficient loading.

Have you investigated VF Drives, as these can vastly improve efficiency and reduce losses.
Energy useage these days is a 'hot topic'.

try looking at www.carbontrust.co.uk (http://www.carbontrust.co.uk)
and search for 'motors' GPG002 is intresting.
Hope this helps.

ECM.

goodguy
11-03-2007, 02:37 AM
Thank you all for your knowledge and web links. But one further question, at what point do you start taking corrective actions fo poor power factor?

Peter_1
11-03-2007, 04:09 PM
Hi Goodguy,

The only way of measuring the power factor is with a proper KVA power meter (as opposed to an electricity meter which will only register kWh). You will need to measure both current and voltage.
A power meter should actually give you a direct reading of the power factor. If you are considering the whole site then it would be worth renting one in or possibly purchasing one?

Motors can have a power factor anywhere between 0.1 up to 0.9 depending upon the loading. Lightly loaded motors having a poor power factor.

I guess the figures you quote are purely those from the motor rating plate.

A motor is normally rated at a pf of 0.8 lagging, thus in your case:

50 hp = 50 x 746 = 37.300 kW ( = P) [1 hp = 746 W]

P= E x I x 1.732 x pf
= 575 x 47.1 x 1.732 x pf
= 46.907 x pf (kW)
thus
pf = 37.300 / 46.907
= 0.795

I expect that your actual p.f will be 0.4 to 0.7 as generally when designing, the 'safety factors' push the motor rating up to the next standard size motor, so in reality your motor will not be fully loaded, and hence will not be at its most efficient loading.

Have you investigated VF Drives, as these can vastly improve efficiency and reduce losses.
Energy useage these days is a 'hot topic'.

try looking at www.carbontrust.co.uk (http://www.carbontrust.co.uk)
and search for 'motors' GPG002 is intresting.
Hope this helps.

ECM.

Electrocoolman, nice explanation.
Power factor could also be increased with an automatic switching capacity battery.
I thought the net voltage was 600 V in Canada?

goodguy
11-03-2007, 05:31 PM
peter1, your right, net voltage is closer to 600v and have seen it around 620v in normal circumstances.
It seems thought, alot of nameplates are only listed 575v, in my experience anyway. Is this common?

Peter_1
12-03-2007, 08:22 AM
try looking at www.carbontrust.co.uk (http://www.carbontrust.co.uk)
and search for 'motors' GPG002 is intresting.
.

This link is easier http://www.carbontrust.co.uk/Publications/publicationdetail.htm?productid=GPG002&metaNoCache=1
Do also once a search on refrigeration on this website

Robearbam
03-04-2007, 06:11 PM
Thank you all for your knowledge and web links. But one further question, at what point do you start taking corrective actions fo poor power factor?

Well, if it gets too low I'm sure the power company would let you know. We had capacitors installed in our plant many years ago because of a low power factor. I know now we have a meter where power comes in and tells you what the power factor is. The lower the power factor the cheaper power is for you. Inductive reactance (motors) are responsible for this.

Peter_1
03-04-2007, 07:13 PM
I can speak for Belgium: <0.92