Good day to everybody. Presently I 10

[lucino]

Good day to everybody.
Presently I am part of an operations improvement team.
In one of my studies I observed that power factor is below industry standard of not less than 95%.
Since the factory is on a 100% generation of its electricity requirements by means of Three Diesel Engine driven generators, I had estimated a substantial savings if power factor is to be corrected say to 97%.
These savings will be a result of lesser fuel for the same amount of kilowatt-hour produced(I hope I am correct).

However, I am not so sure how to correctly refer to the savings...Is it by power loss due to the low power factor or by KVA vs. fuel ratio.

If by power loss, how to compute it?
If by KVA/fuel ratio how to justify it and support the energy savings line of reasoning?

I am calling everybody to clarify.
Thanks.

 

[smms]

Qc = Pel x (powerfactor[old] - powerfactor[new])

 

[lucino]

smms thank you.
However your reply is so short.
Please explain further.

 

[smms]

Hi Lucino,
e.g. if you want to compensate from a powerfactor of 0,9 to 0,95 and you have an electric motor with a mechanical rating of 400kW with an efficiency of 95%, you need
QC = 400/0,95 x (tan(invcos(0,90)) - tan(invcos(0,95)) = 65kvar

 

[alehman]

Improving power factor will reduce fuel consumption because of lower system I^2R losses. This will have to be calculated based on your generator, other system components and circuitry. If you are asking about the benefit for going from 95% to 97% power factor, I would think the savings would be very small.

Be careful with capacitors on a small generator systems. Your generator may have voltage control problems if the load drops which will cause the generator to see leading power factor.

 

[lucino]

Hi smms.
You are very helpful. My computations for power factor capacitor is confirmed. But this is not what I am soliciting.
I am asking the proper technical justification in computing for fuel savings that can be realized after the correcting capacitor bank is installed.
Thanks.

 

[lucino]

Good day alehman.
So it should be system losses in the form of I2R.
I have already tried some trial calculations but still I am very doubtful of the proper presentation.What I did was to compute for the system current at existing power factor by means of voltage and KVA relationship. Then computed for the sytem resistance as in the sample below:

Initial Situation-Average Data:
kWh : 310
Power Factor : 0.77
Generator Fuel ratio : 0.28 liter / kWh
Fuel Cost : P12.43/liter

Computation:
@ Initial power factor = 0.77 and kWh constant

Power = Voltage x Current x Power factor x Square Root of Three
P = VI cos Ø √3
I1 = P ÷ V cos Ø √3
= 310,000 ÷ (440) (0.77) (1.732)
= 310,000 ÷ 586.80
= 528.29 amperes

@ Desired power factor = 0.97 and kWh constant

Power = Voltage x Current x Power factor x Square Root of Three
P = VI cos Ø √3
I2 = P ÷ V cos Ø √3
= 310,000 ÷ (440) (0.97) (1.732)
= 310,000 ÷ 739.22
= 419.36 amperes

Difference in Amperes:
ID = I1 – I2
ID = 528.29 – 419.36
= 108.93 Amperes

Power loss:
PL = ID2 R √3
Where:
PL = Power loss
ID2 = Differential current after improving power factor from 0.92 to 0.97
R = Resistive load due to heating at lower power factor.
√3 = Multiplying factor in a three phase system
To compute for R:
R = Voltage ÷ Differential Current
= V ÷ ID
= 440 ÷ 108.93
= 4.04 Ohms
= 4.04 Ω
Then:
PL = ID2 R √3
= (108.93) 2 (4.08) (1.732)
= 83,850 watts
= 83.85 kW
Energy savings computation:
ES =(PL) (Time)
Where:
ES = Energy savings
PL = Power loss
Time = Operating hour in one year
= (24 hour/day) (7 days/week) (48 weeks/year)
= 8,064 hour/year
Then:
ES = 83.85 kW (8,064 hour/year)
= 676,166 kWh/year

Computation for Fuel savings:
FS = (ES) (FR)
Where
FS = Fuel savings
ES = Energy savings in one year
FR = Fuel ratio (in liter/kWh)-fuel consumption of generator.

Then:
FS = (ES) (FR)
= (676,166 kWh/year) (0.28 liter/kWh)
= 189,326.48 liter/year


Computation for Peso savings:
SM =(FS) (FC)

Where:
SM = Savings in monetary value
FS = Fuel savings per year
FC = Fuel cost in peso per liter


Then:
SM =(FS) (FC)
= (189,326.48 liter/year) (12.43 peso/liter)
= 2,353,328.15 pesos/year

Is this sample computation correct?
Please comment.
Thanks

 

[sparky1976]

Dear Lucino,

Please see the the thread "power factor correction from generation stand point" (I donot know to put a link here, try by seacrh). You will find my first calculation is wrong and Bigamp gave the correction.

Basically agreed with Alehman,
See the IEEE Red Book Chapter 8, Power Factor and related cosideration.

Long story short, You will reduce the current on your cable xfmr etc that will "extend" life expentancy, reduce the system loss (this will be easily metered), reduce the generator loss (only calculated from naufacturer curve).

In My case (I almost had the same case, 120 MW system 0.89 PF want to increase to 0.95 PF) I cannot economically justifed the expenses to install capacitor, the payback time is to long 7 to 10 years.

Good luck hope you can justified the expenses.

 

[lucino]

Yamin thanks.
I will try to search it and feedback in due time.
Regards
Lucino

 

[jghrist]

Some problems with your calculations:

1. Loss reduction is (I1²-I2²)·R. This is not the same as (I1-I2)²·R.

2. The R that is important is the resistance between the generators and the point where the pf correction is applied plus the resistance of the generator. You appear to be trying to calculate the load resistance, although I can't figure out your method.

Calculation of loss reduction by capacitors can be simplified by realizing that the loss reduction depends only on the resistance, the capacitor current, and the reactive load current.

Watts loss reduction = WLR = 2·IC·R·IX - IC²·R
where IC is the capacitor current and IX is the reactive load current.

Note that the load current term is not squared, so you can use average load current to get average loss reduction. The capacitor current does not change with the load. This is a large benefit to this approach when dealing with varying loads.

See Calculation of Loss Reduction by Capacitors, Victor J. Farmer, Electrical World, Oct. 29, 1956 for derivation of the above.

 

[lucino]

Yamin: I had already found the thread.This is most helpful.Thank you.

Jghrist:Thank you. Seems this will what make my calculation more believable.

Alehman: Thank you

Smms: Thank you.

All of you guys is given a star!!!!

 

[DanDel]

lucino, just a question; are you certain 0.28 liters of fuel is used for 1 kWh?

 

[jghrist]

We have a client with 1800 kW diesel generators that use about 0.5 liters per kWh.

 

[lucino]

DanDel, Good morning.
Yes I am certain. In fact this is an average of the data I personally and religiously record every hour for successive 24 hours without sleeping.

Average Fuel consumption:243 liters

Average energy produce :861kW-Hr
Therefore the fuel ratio = 243liter/861 kW-Hr.
= 0.28 liter/kW-Hr.
Thanks for the inquiry.
Makes my day energetic.

Lucino

 

[moturcu]

Hello Lucio
I think following correction should be made about your calculation


1_) in loss equation multiplier should be 3 not square root of 3

2_)You should change the loss equation as
P=3*(Iold2-Inew2)*R


3_) Your resistance estimation method should be changed. I does not seem to be reasonable. You should use manufacturer data of each element along the way.

 

[GusD]

HI lucino

I don't know of any distribuition system providing PF correction that will go to such the level of correction as you plan.
Your perceived gains by running at higher power factor may very well be offset by the possibility of overcorrection.
If you control close to 97% PF there is a good possibility of overcorrecting due to changing loads.The reason for the 95% Pf industry standard is to take that into account.
System Load changes are anavoidable and the 95% Industry standard seems to take that possibility into account.
The only way to know for sure is to monitor loads and PF over a period of time and see if you have room for real improvement without a high price.

GusD

 

[lucino]

Moturcu:
Please expound on item no.1 & 2... Why I have to multiply by 3 and not square root of three. Is it not that 3 phase system calculations involved square root of three?
Thanks. I would be indebted.


GusD:
Yes. I agree with you. Actually the 97% is only for kVAR calculations.In actuality, if this project would push through, the PF controller will be set to 95%.

Good day.
Lucino

 

[moturcu]

Hi Lucino:

In a single phase system, copper loss is I2R. For 3 phase system you should multiply this by 3. Square root of 3 will be in play only when you deal with line to line voltages in 3 phase system.

for the second item: (Ia2-Ib2) is not equal to (Ia-Ib)2

Moturcu.

 

[lucino]

Moturcu,
Good day.
Yes, now I understand.
But our system is a three phase 440 Vac system supplied by three diesel driven generators with one generator having a capacity of 1800KVA.
How can I compute for the copper losses in a single line so that I can multiuply it by 3.
Our actual data as belowload side metering)

Va=416.9, Vb=462.2, Vc=460.8
Ia=532.2, Ib=543.3, Ic=524.5
kWa=134.2, kWb=132.3, kWc=124.3
PFa=0.89, PFb=0.9, PFc=0.91

Also from the meter which I assume the average for the system.

kW=385.2, PF=0.89, Hz=60

Please comment how these data can be use to compute for the line copper losses.
Thank you for your time.
Regards

Lucino

 

[jbartos]

Comment to the previous posting: Va=416.9V appears to be too low.
To compute losses RI**2, sum line segments, with a specific resistance Rk and an associated current in this line Ik
Then, perform summing
Ploss=Sum Rk x Ik**2 from k=1 (i.e. the first line segment) to k=K (i.e. the last segment equal to K, e.g. 90).