Correcting the PF on a Three Phase Motor

Ok I think I have got it, please correct me if I am wrong.....

P_(kW) = √3 × PF × I_(A) × V_{L-L (V)} / 1000

P_(kW) = 1.73 × 0.43 × 59 × 353 / 1000

15.493 kW

kVA x 0.85 = kW

kW/0.85 = kVA

15.493 / 0.85 = 18.23 kVA

I think I am doing something wrong here...

The voltage is at 360 as the capacitors should stabilize the 353 a bit? And the PF I am hoping for is 0.95.

P_(kW) = √3 × PF × I_(A) × V_{L-L (V)} / 1000

P_(kW) = 1.73 × 0.43 × 59 × 353 / 1000

15.493 kW

kVA x 0.85 = kW

kW/0.85 = kVA

15.493 / 0.85 = 18.23 kVA

Monthly Cost = 15.493 x 24 x 30 x $0.28

$3,123.39

P_(kW) = √3 × PF × I_(A) × V_{L-L (V)} / 1000

P_(kW) = 1.73 × 0.95 × 50.15 × 360 / 1000

29.671 kW

kVA x 0.85 = kW

kW/0.85 = kVA

29.671/ 0.85 = 34.91 kVA

Monthly Cost = 29.671 x 24 x 30 x $0.28

$5,981.67

If the feeder cable to this motor is running hot, then it is undersized.

Has the motor been properly matched to the mechanical load on its shaft? The figures presented suggest it is well oversized for the job.

What is the economic value of the proposed higher power factor? Is the facility being billed a penalty charge on low power factor?

Hi PWSlack,

If I correct the PF on all our motors this should save money as we won't get penalized by the utility company?

- Rob

Please answer the questions with some facts, and not with a question.

This pretty much sums it up in a nutshell, I know that our company does get penalized for low PF from the utility company:

"One very important aspect of improving quality of supply is the control of power factor.

Low power factor means poor electrical efficiency. The lower the power factor, the

higher the apparent power drawn from the distribution network. This means that the

supply company must install larger generation capacity, larger size transmission lines

and cables, transformers and other distribution system devices, which otherwise would

not be necessary. This results in a much higher capital expenditures and operating costs

for the Electricity Supply Company, which in many cases is passed on to the consumer

in the form of higher tariff rates.

This is the main reason behind why the Electricity Supply Companies in modern

economies demand reduction of the reactive load in their networks via improvement of

the power factor. In most cases, special reactive current tariffs penalize consumers for

poor power factors."

"The following is an example to illustrate the savings by installing power factor correction

equipment.

Let us assume that the penalty is 37.57 cents per day per kVAr, for the kVAr necessary

to improve the power factor to 0.95 lagging.

1000 kW load at a PF of 0.75 = 882 kVAr

1000 kW load at pf of 0.95 = 329 kVAr

The extra kVAr drawn from the supply is:

882 kVAr - 329 kVAr = 553 kVAr

553 x 37.57 = $207.76 penalty per day

Assume the Power factor correction unit will cost installed about $60 per kVAr

Therefore 600 kVAr x $60 = $36,000

Payback period = $36'000 divided by $207.76 = 173 days or about five to six months.

In about 5 to 6 months the cost for power factor correction is recovered and any further

penalties are avoided for the life of the electrical installation.

Any further savings then become profits that add to the company's bottom line.

Note: This example is for illustration purposes only. Actual savings will vary from

installation to installation."

Then the pf correction capacitor assembly needs to be immediately downstream of the tariff meter to look after the complete facility, and not on individual motors.

Why has this motor been oversized? Why is it driving a mechanical load that is far smaller than 50HP? Indeed, what is the load?

Hi PWSlack,

This motor is used as a chiller pump (Main circulation pump of chilled water with a head of 203 feet), they replaced the old chiller with a new one and no adjustment was made to the motor size.

Thank you for your input, much appreciated!

- Rob

Putting capacitors on each motor is just plain silly. PFC equipment saves very little in terms of kWh; they are there to reduce low pf penalty charges.

Correct the entire facility instead. Contact a specialist manufacturer or two, and get each of them in to work-up a detailed proposal for an automatic power factor correction unit, by which it is meant that the unit itself determines the correct amount of capacitance to add at any moment depending on what the facility as a whole is doing at any moment. If the proposal has a favourable ROCE calculation in terms of poor pf penalty charges saved, and the utility bills received will display the financial information that can be used to work out the savings, then buy it. If it is adverse, then don't worry, and sleep easy in the knowledge that the exercise has been valuable for learning and applicable elsewhere.

<unsubscribes>

Thank you for your input PWSlack, it is much appreciated!

I was going to suggest installing a smaller motor. This would have the advantage of reducing actual power consumption as well as increasing the PF.
However, after reviewing the data you provided, things don't add up. This is a 208-230/460V, 60Hz motor, but you measure 353V.

* What is your nominal supply voltage and frequency?

* What is the facility overall PF?

* Is this motor connected as 208-230 or 460?

* Can you provide a pump curve or data on its power requirements?This system needs to be understood better before attempting a 'fix'.

Hi Bigg,

Thank you for the response, the nominal voltage is meant to be 460 @ 60Hz, I do not know the entire facilities PF yet as we are still going to install a verification meter. The motor is supposed to be connected as 460VAC (According to the drawings, but does not seem to be), I am unable to give you the pump curve as it is an old system and there is no data available on it. I am not sure if this has been rewired as I am only coming across it now.

This is what my smart meter reads:

Voltage1 = 353.6

Voltage2 = 0

Voltage3 = 488

Current1 = 58.6

Current2 = 54.8

Current3 = 57.6

Watts1 = 10758

Watts2 = 0

Watts3 = 28194

PF1 = L0.43

PF2 = 0

PF3 = 1.00

Thanks for your help here, it is much appreciated!

To me those numbers look very wrong in all major aspects.

What are you referencing each phase to to get the individual voltage and amp readings?

Hi Tcmtech,

I am in another country at the moment, I am getting these readings off a smart meter that one of my guys installed, I will look at this when I get back. Thank you for your response here, much appreciated.

It seems that your meter is not correctly connected with CT and PT and it is showing incorrect readings. Your actual load may differ. I suggested 15 kw motor based on your load calculation given in your first post.

Yes, something is wrong there, I will need to check this all when I get back. Thank you for all your input here, much appreciated.

Have you thought about investing in an inverter?

Reading the thread seems to suggested that the system hasn't been sized very well, which lends itself to the implementation of an inverter. This will dramatically improve the PF and, also give you the ability to tune the speed of the pump to the flow requirements of the cooling process. This too will give a dramatic financial saving and, should pay back the investment in the inverter in a relatively short period. Get somebody in to work out the cost saving. You'll kill two birds with one stone.

Hi Dodgy Solenoid,

This may be a plan, however I am looking into buying a spare motor for each pump for preventative maintenance purposes anyway so it may be more cost effective to just replace the whole pump as this system is extremely old (30 years).

Thank you for your advise on this!

Thank you JRaef,

You have provided me with extremely useful information here, much appreciated!!!!

Connect a datalogger to the mainswitchboard & record all parameters(V,I,pf,Hz, harmonics, kVa,kVAr,kW etc)for the entire working hours(8 or 24) in a typical day,from which you can size the capacitor bank

Thank you Pnaban, will do!

You are using over sized motor for the job. Better to use 15 kw motor for the job. PF will improve automatically.

If you are using over sized pump also and running the system in throtteled conditions, you can use 11 kw motor with lower rpm and save a lot of power.

Practically you need not require to use any capacitor to improve power factor here. No any significant saving could be achieved by adding a capacitor at 15 kw load.

Hi Satendrakumart,

Thank you for the response. Please may you tell me how you got to a motor size of 15 kW. This motor is used for circulating chilled water throughout the building, it is an end suction pump. The details are as follows (Sorry this is all I have as there is no manual on this):

Head: 203 Feet

Flow: ? USGPM

Pressure: ? PSI

Here are the new chiller details:

SMARDT

MODEL# SWA080-2EG101-2A2AN

SERIAL# FF0010H244Q1201

VOLTAGE: 460

FREQUENCY: 60

PHASES: 3

MIN. DISCONNECT SIZE: 400 AMPS

SHORT CIRCUIT CURRENT: 35kA RMS

NOMINAL CAPACITY COND: 8.35 PSIG

NOMINAL CAPACITY EVAP: 6.65 PSIG

HIGH SIDE DESIGN PRESSURE: 191 PSIG

LOW SIDE DESIGN PRESSURE: 191 PSIG

COMPRESSORTYPE:

TT350-G10-1-HL-A-O-NC

F.L.A: 160A

L.R.A: 176A

COND FLOWRATE: 713.6 USGPM

EVAP FLOWRATE: 572.5 USGPM

REFRIGERANT TYPE: R134A

REFRIGERANT CHARGE: 619 LBS

EMPTY WEIGHT: 7882 LBS

Here are the details of the chiller that used to be there:

YORK CODEPAK

LIQUID CHILLING SYSTEM

UNIT MODEL: YT C3 C3 B2 - CJ E

REFRIG. DWP, PSIG: 15 (COOLER) 15 (CLG.COND)

LIQUID DWP, PSIG: 150 (COOLER) 150 (CLG. COND)

NO. OF PASSES: 3 (COOLER) 2 (CLG. COND)

REFRIGERANT: 11

REFRIGERANT CHARGE 730 LBS

VOLTS 480

PHASE 3

HERTZ: 60

MIN. CIRCUIT AMPACITY: 297

MAX. DUAL ELEMENT FUSE AMPS: 400

MAZ. CIRCUIT BREAKER AMPS: 400

COMPRESSOR -

MODEL: YDTJ-76

CODE: LC

SERIAL# YHXM577640

OIL PUMP H.P: 3/4

VOLTS: 460

F.L.A: 1.75

OIL HEATER F.L.A: 8

OIL HEATER VOLTS: 115

CONTROL SUPPLY: 115 VOLTS 15 AMPS

CHILLER STARTING CURRENT: 780 AMPS

105% OVERLOAD TRIP: 245 AMPS (F.L.A: 233 AMPS)

I suggested 15 kw motor based on actual load of the system as given by you:

P_(kW) = √3 × PF × I_(A) × V_{L-L (V)} / 1000

P_(kW) = 1.73 × 0.43 × 59 × 353 / 1000

15.493 kW

I faced similar situation, where a 30 KW motor was being used and was running at 50% load with throttled inlet.

I replaced the 30 kw motor 1470 rpm with 11 kw 980 rpm motor successfully.

Hi Satendrakumart,

Thanks for the reply, this chiller is at about 70% demand with the figures I gave you, I will have to calculate this at full load, however it gets better as we are going to install another "New Chiller" and run it in parallel as on hot day the one cannot keep up.

Would it really be less expensive to change a motor than to change a capacitor?

Thank you for your input Bigg, much appreciated!

I tried to send you all the cooling tower and heat exchanger details but I get an error?

Anyway, thanks again!

It sounds as if you have a corner-grounded system, most likely not intentional. Collectively, the given numbers don't compute. I think you need to sort out the voltage imbalance before worrying about PF.

Hi Tornado,

Thank you for your advise here, I am going to look into this.

Take care!

I'm surprised at the 203 feet of head, whether on the evaporator or condenser side. I may have missed it, but I don't recall seeing the refrigerating capacity (tons or Btu/h) of this system. That would at least give some ballpark numbers to work with.

Hi Tornado,

Sorry it is a 250 Ton chiller unit. Thanks for your responses here.

Hi JRaef,

Thank you for your input. You are right, this is a complex calculation that I am not a specialist in, however I like to challenge myself, as you can see by earlier posts, I am not looking for hand outs as I do attempt to do all calculations myself, I did study a lot of this but it has been a few years since I have had to put it all into practice.

I do appreciate your sound advice and I will be getting BMS pros in next month, their flight has already been booked, I am just looking at possible issues and their solutions to help them out in their groundwork, there are many issues that are not correct at the facility that I manage, that I will probably share with you guys as time goes on because I do appreciate every ones valuable input.

Once again, thank you and I look forward from hearing from you in future posts, take care.

-Rob

The KVA value you calculated in point 3 should be 36.07 not 39.07.

Energy meter is not correctly connected and the following reading is false:

Voltage: 353 V

Current: 59 A

P.F: 0.43

Above data is misguiding all of us.

Let him measure again the actual load (voltage, current and power factor)