Chiller plant power factor

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raiderUM

Member
Location
Ohio
Hello all,

I have some interesting info to share and I hope someone can help me figure it out. We have a new chiller plant that feeds 5 buildings for there A/C. The chillers are always on and can never be shut off because they have oil heaters inside of them.

When the Chillers are chilling water the PF is almost at unity (.99) but when the chillers are idle the PF is very low (.12) I'll post a document with all the information that I have. I am getting all my info from E-Mon. We do not pay a PF correction charge but we do pay a KVA demand charge, so the closer we can get KVA to KW obviously our PF will be closer to unity.

I have physical gone to the chillers and used my amp meter to check the load on one of the two chillers. I had 45 amps on one of the three parallel legs on A phase. So, 45*1.73*480 which is 75 KVA for both Chillers idle.

Can anyone help me figure out what is going on here:?

View attachment doc2.doc
View attachment doc3.doc
 
R

rcwilson

Senior Member
Location
Redmond, WA
Readings look funny with 0 on phase C. Could be a metering problem?

Power factor is just the ratio of watts to volt-amps. When a motor or transformer is energized it takes vars to maintain the magnetic field in the motor winding or transformer core. These excitation vars are present at no load. When the chiller unloads and is idle, the var losses do not go down but the kw does, dropping the pf ratio.

More information on the size of the equipment and its kW and amps at full load would help in analyzing if this is an issue.
 
Jraef

Jraef

Moderator, OTD
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Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
rcwilson said:
Readings look funny with 0 on phase C. Could be a metering problem?
Click to expand...
C phase CT is backward perhaps?

Power factor is just the ratio of watts to volt-amps. When a motor or transformer is energized it takes vars to maintain the magnetic field in the motor winding or transformer core. These excitation vars are present at no load. When the chiller unloads and is idle, the var losses do not go down but the kw does, dropping the pf ratio.

More information on the size of the equipment and its kW and amps at full load would help in analyzing if this is an issue.
Click to expand...

To further this idea, what you are seeing is not all that unusual. But remember that since PF is very low in idle, so is your kW. So most likely your kVA, although technically still higher than you might expect at a low kW value, is still probably significantly lower than any peak you are going to have when loaded. In other words even though it looks bad by the strict numbers, it is not likely doing you any harm as far as kVA demand charges go. But like he says, you need more information to truly evaluate it.

By the way, your use pattern appears to be one that could probably benefit from a VFD on the chiller motor, have you considered it? Chillers are not simple VFD applications because of the possibility of creating surging in the chiller, but most of the chiller mfrs and service companies now have canned chiller control + VFD packages that take advantage of the VFD's capability to reduce energy use. I've seen some that pay for themselves in under a year.
 
R

raiderUM

Member
Location
Ohio
rcwilson said:
Readings look funny with 0 on phase C. Could be a metering problem?

Power factor is just the ratio of watts to volt-amps. When a motor or transformer is energized it takes vars to maintain the magnetic field in the motor winding or transformer core. These excitation vars are present at no load. When the chiller unloads and is idle, the var losses do not go down but the kw does, dropping the pf ratio.

More information on the size of the equipment and its kW and amps at full load would help in analyzing if this is an issue.
Click to expand...

This is what my data looks like when just one of the two chillers are running. There are Capacitors already inside the unit itself. I also attached a pic of the schematics.
IMAG0457.jpg
View attachment Doc4.doc
 
R

rcwilson

Senior Member
Location
Redmond, WA
I"m guessing that the low power factor is actually a leading power factor due to kVAR contribution of the harmonic filter capacitors.

I don't have time to run the numbers and see what kVAR 3 phase-to-ground connected 47 mfd capacitors could supply and compare that to your no load amp readings. Try it and you may find your answer.

It looks like your system does have a VFD on the main motor.
 
Besoeker

Besoeker

Senior Member
Location
UK
rcwilson said:
I"m guessing that the low power factor is actually a leading power factor due to kVAR contribution of the harmonic filter capacitors.
I don't have time to run the numbers and see what kVAR 3 phase-to-ground connected 47 mfd capacitors could supply and compare that to your no load amp readings. Try it and you may find your answer.
Click to expand...
Interesting thought.
The drawing calls it a filter cap assy.
We don't know what the inductor value is but the capacitors appear to be 46uF and connected in star. So, ignoring the inductor for the moment, each capacitor would have 277V across it.
Not enough for the 45A.
 
R

raiderUM

Member
Location
Ohio
wouldnt my formula be this:

480^2*2*Pi*60*46*10^-6/1000 = 4 kVAR per cap
There are 27 caps, so 4*27=108 kVAR
I=kvar/1.73*V = 130 amps

I must not be doing something correct? 130amps is a long ways away from 300 amp fuses that they have per leg

Would it be safe to say that the Negative Phase angle on the EMON chart means I have LEADING PF??
 
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Besoeker

Besoeker

Senior Member
Location
UK
raiderUM said:
wouldnt my formula be this:

480^2*2*Pi*60*46*10^-6/1000 = 4 kVAR per cap
There are 27 caps, so 4*27=108 kVAR
I=kvar/1.73*V = 130 amps

I must not be doing something correct? 130amps is a long ways away from 300 amp fuses that they have per leg

Would it be safe to say that the Negative Phase angle on the EMON chart means I have LEADING PF??
Click to expand...
Not sure where you get the 27 capacitors but that could account for the 45A at idle.

Xc =1/ωC

For 46uF at 60Hz that gives Xc about 58 ohms. Since the circuit is star connected, each capacitor would take around 5A. (277/58).
If there are 27 capacitors that would be nine in each leg.
And that would account for the 45A.
 
R

raiderUM

Member
Location
Ohio
Besoeker said:
Not sure where you get the 27 capacitors but that could account for the 45A at idle.

Xc =1/ωC

For 46uF at 60Hz that gives Xc about 58 ohms. Since the circuit is star connected, each capacitor would take around 5A. (277/58).
If there are 27 capacitors that would be nine in each leg.
And that would account for the 45A.
Click to expand...

Yes there are 9 caps per leg but, this is only for one chiller. I have two identical chillers in this building so I'm not sure???
 
GeorgeB

GeorgeB

ElectroHydraulics engineer (retired)
Location
Greenville SC
Occupation
Retired
raiderUM said:
... but when the chillers are idle the PF is very low (.12) ...
Click to expand...
It would not be a bad approximation for a motor to assume its PF uncoupled is 5%, and 20% at 20% load, linear between. I've previously posted an old Reliance (now Baldor) curve set ... a good GENERAL approximation, and am doing so again. This sheds lots of light on motors ...
 

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R

raiderUM

Member
Location
Ohio
Besoeker said:
Then, as per post #8, that could account for the 45A and the dreadful power factor.

46 uF
9 per leg
414 uF
277 V
60 Hz
376.99ω
0.16 ωC
6.41 Xc
43.23A
Click to expand...

Ok so we are getting the same numbers like I had in post #7. And if I have 43.3 amps per leg, that equals 130 amps per chiller, and I have two chillers that are identical, so a total of around 260amps of load in the building from the capacitors.

So I can only see this as a LEADING PF not a LAGGING PF??? Which is why I have a negative phase angle on the Emon display? Sorry for the delayed response, Thank you!
 
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