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Good day to everybody. Presently I
10

Good day to everybody. Presently I

Good day to everybody. Presently I

(OP)
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.

RE: Good day to everybody. Presently I

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

RE: Good day to everybody. Presently I

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

RE: Good day to everybody. Presently I

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

RE: Good day to everybody. Presently I

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.

RE: Good day to everybody. Presently I

(OP)
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.

RE: Good day to everybody. Presently I

(OP)
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

RE: Good day to everybody. Presently I

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.

RE: Good day to everybody. Presently I

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

RE: Good day to everybody. Presently I

3
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.

RE: Good day to everybody. Presently I

(OP)
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!!!!

RE: Good day to everybody. Presently I

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

RE: Good day to everybody. Presently I

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

RE: Good day to everybody. Presently I

(OP)
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

RE: Good day to everybody. Presently I

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.

RE: Good day to everybody. Presently I

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

RE: Good day to everybody. Presently I

(OP)
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

RE: Good day to everybody. Presently I

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.

RE: Good day to everybody. Presently I

(OP)
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 below:(load 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
        

RE: Good day to everybody. Presently I

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).

RE: Good day to everybody. Presently I

In jbartos answer, each phase would count as a line segment, so you could use the actual current Ik in that phase.  In this case, since you are adding up three losses for each line section, you don't use the multiplier 3.

The line segments to use are those from the generators to the capacitor location.  Beyond the capacitors, the current does not change (I2=I1) so the losses do not change.  If the losses don't change, you don't have to calculate them because you are only after the difference.  The closer to the load that you place the capacitors, the more loss reduction you will have.  Placing the capacitors at the generators will result in no savings in conductor losses.

What to do about losses in the generator?  I guess you could find a resistance value to use, but it might be better to get efficiency vs power factor curves and determine loss reduction from the change in efficiency.  I don't know if the change in efficiency is directly related to the I²R losses in the generator.

RE: Good day to everybody. Presently I

(OP)
Hello guys,

Jbartos:
Its a clerical error Va=461.9
Your Ploss calculation is alien to me specially I do not know what value is specific resistance of each segment.
Where shall I got it? Can I compute it? If it is computed, can I use the meter reading data or do i need some other data that is still needed to be verified?
Thanks.


Jhgrist:
Your comments makes my initial plan to suggest that capacitor bank to correct power factor be installed nearest to the load, more meaningful.
However as I understand it I can only count the losses between the generator and the capacitor bank as losses eliminated thereby made into the basis of fuel savings at the generator consumption. By this I shoud need to consider its distance.Is it a correct assumption?
Thanks.

Lucino

RE: Good day to everybody. Presently I

Lucino
your total load side power is 390.8kW. On the other hand, you mentioned average system system power is 385.2kW. I do not know where you measured this last one but it conflicts with total load side power.

RE: Good day to everybody. Presently I

Lucino:

-  Cooling lowers Generator losses.
-  Higher system Voltage lowers I2R losses. (motors require constant power higher V lowers I. Be careful SCRs and IGBTs may fail prematurely(variable speed drives soft starts.))
-  Check your generator sets’ efficiency curves to make sure your operating at the most cost effective point.  85-110%? If operating >100% ensure proper service factors and cooling is available.

The only concern I have if your running 3 units at 85% and a cap bank installation brings them down to 78% you may not realize a savings because you may be still using the same fuel quantities.

Best Regards

Ray Micallef, P. Eng.
Power Generation/Utility Industry

RE: Good day to everybody. Presently I

Yes, you do need to consider distance.  The conductor resistance is found in tables on an ohms/km (or ohms/1000' or ohms/mile) basis.  To get the resistance for your loss calculations you need to multiply this times the distance.

Be aware that there are other considerations than cost of losses.  As ERaySir noted, you need to consider the effect on variable speed drives and soft starters.  You need to consider the effect on the generators if you overcompensate.  You need to consider cost of capacitor controls.  Also capacitor switching transients.

RE: Good day to everybody. Presently I

I guess he does not need to wory about R calculation. As far as I understand he has measurement for load and generator side power (and line currents). The difference would give us line copper losses. I he wishes he can find the line resistance as R=power difference/(3*(Iav)^2)
where Iav is the average of the 3 line currents.

However, generator side loss reduction analysis is another story. But it will certainly be copper + core losses reduction in generator as well.

RE: Good day to everybody. Presently I

(OP)
Hello guys,

EraySir:
I am still consolidating data.
Our generated voltage average is 463 volts while the user side only needs 440 volts. My data is not complete yet. Any progress will be posted.

Jghrist:
Seems that I have no choice but to approximate distance of loads from generator side. My approximation is 200 meters but I have yet to verify sizes so that I can use resistance values from tables for the purpose.

Moturcu:
You are right. With a lot of inputs from you guys I  am now confident of determining the losses. As it would progress I will make a post to keep everyone updated.

Regards.

Lucino

RE: Good day to everybody. Presently I

Finding the resistance by using the loss difference between two points on the distribution system is not a very accurate solution.  If you have 200 m of 500 mm² copper cable, losses would be about 8 kW with 530A current.  This is only about 2% of the load.  If the measurement error (meter error plus CT error) is 1% on each end, what kind of accuracy can you expect in subtracting the two readings?

RE: Good day to everybody. Presently I

Caveat:

Never install a PF correction capacitor that offsets more than, say 90%, of the no-load reactive-draw of the motor.  Over-voltage may be produced.

RE: Good day to everybody. Presently I

jghrist
Yo are somewhat right. But I am not satisfied how you come up with 8kW. In an almost 400kW low voltage distribution system, 8kW loss is too small.

RE: Good day to everybody. Presently I

moturcu,
Resistance r = 0.015 ohm/1000' from NEC Table 9 for 1000 kcmil (approx 500 mm²) copper in PVC conduit.
Current I = 530A
Distance d = 200 m = 656 ft
Loss per ø = I²·r·d = 530²·0.015·656/1000 = 2764 W
Total loss = 3·2764/1000 = 8.3 kW

RE: Good day to everybody. Presently I

Suggestion to lucino (Electrical) Jun 4, 2003 marked ///\\\
Jbartos:
Its a clerical error Va=461.9
Your Ploss calculation is alien to me specially I do not know what value is specific resistance of each segment.
///If the cable sizes, or generator parameters are not available , it would be difficult to obtain RI**2 losses since the formula includes R of those items.\\\
Where shall I got it?
///Perhaps, field surveys could reveal some data, one line diagram others, wiring diagram another, physical drawing would reveal lengths of cables, etc. This is a very routine task to those who have already performed this type of tasks.\\\
 Can I compute it?
///Yes.\\\
 If it is computed, can I use the meter reading data or do i need some other data that is still needed to be verified?
///Documentation mentioned above would be helpful and save some legwork.\\\
Thanks.

RE: Good day to everybody. Presently I

Suggestion to jghrist (Electrical) Jun 4, 2003 marked ///\\\
In jbartos answer, each phase would count as a line segment, so you could use the actual current Ik in that phase.
///Yes, indeed. Additionally, a single phase circuit has a neutral. This one counts too in RI**2. A little refreshment of basic electrical circuitry would be helpful.\\\
  In this case, since you are adding up three losses for each line section, you don't use the multiplier 3.
///I am not aware of posting any multiplier of 3. Please, would you clarify your statement?\\\
The line segments to use are those from the generators to the capacitor location.
///Most certainly.\\\
  Beyond the capacitors, the current does not change (I2=I1) so the losses do not change.
///Is this really true that I2=I1? Please, clarify it. It appears that the voltage will increase at the point where the capacitor is connected and downstream since there is smaller voltage drop upstream. The higher voltage will cause higher current drawn by loads, e.g. resistors. Then obviously, the losses RI**2 will increase in those segments too since the current to loads is higher.\\\
  If the losses don't change, you don't have to calculate them because you are only after the difference.
///This needs clarifications as far as unchanged losses are concerned.\\\
  The closer to the load that you place the capacitors, the more loss reduction you will have.
///Correction. Closer to the inductive load or a load that has the inductive component.\\\
  Placing the capacitors at the generators will result in no savings in conductor losses.
///What about reduction of losses in the generator and assumption that the voltage remains unchanged at the point of capacitor connection? Please, clarify.\\\

What to do about losses in the generator?  I guess you could find a resistance value to use, but it might be better to get efficiency vs power factor curves and determine loss reduction from the change in efficiency.  I don't know if the change in efficiency is directly related to the I²R losses in the generator.
///The generator electrical equivalent circuit may help. Even here in this equivalent circuit, RI**2 is applicable.\\\

RE: Good day to everybody. Presently I

jbartos said:
Additionally, a single phase circuit has a neutral. This one counts too in RI**2. A little refreshment of basic electrical circuitry would be helpful.
The facts stated by the initial poster indicated a fairly balanced load, so neutral current would not contribute much to losses and the loss reduction in the neutral from the addition of a balanced capacitor bank would be even less.  That last sentence is a little nasty isn't it?

I am not aware of posting any multiplier of 3. Please, would you clarify your statement?
This was to clarify that the factor of 3 posted by others was not necessary as you posed the solution.

Is this really true that I2=I1? Please, clarify it.
Any change in load current beyond the capacitor location caused by a voltage increase would be minor and difficult to determine.  If the load were resistive as you say, the current would increase a little, but if the load were motors, the current would decrease.  I don't think the added complication is worth it.

This needs clarifications as far as unchanged losses are concerned.
If you know that the losses in the line beyond the capacitor bank do not change, then you don't have to calculate them because you are after the loss reduction, not the total value of losses.

The generator electrical equivalent circuit may help. Even here in this equivalent circuit, RI**2 is applicable.
But it might be easier to find an efficiency vs pf curve that to get the parameters for the equivalent circuit.

RE: Good day to everybody. Presently I

I agree; nasty.
Once we got past basic electrical circuits we learned that everything we learned in that class and every other engineering class is based on a series of simplifying assumptions. Jgrist's treatment also ignored the facts that all circuits are really distributed and non-linear. Shame on him for making an in-solvable problem solvable.

Actually I would further simplify the problem in this manner: Utilities are interested in pf correction due to the long lines involved. Industrial customers of utilities are interested in pf correction due to the penalties imposed by the utilities (although many decide it's easier to pay the penalty). Factories using on-site generation should look elsewhere to increase efficiency.

RE: Good day to everybody. Presently I

Adding fuel to the fire, here is a simple, albeit useful method to calculate total conductor losses in a 3-ph, 3-wire system, without having to know cable R and X.  Conductor material, length, and size, are all that are required.:

  P = K x A^2 x L / S, in Watts.

For metric applications:
  K = Resistivity in ohm-mmq/mt, 51.2E-03
  A = Current in Amperes.
  L = Circuit length, in meters.
  S = Cable size, in mmq (or mm^2).

For AWG or MCM applications:
  K = Resistivity in ohm-kcmil/ft, 41.1E-03.
  A = Current in Amperes.
  L = Circuit length in feet.
  S = Cable size, in kcmil (was MCM).

Constraints/Limits/Multipliers:
  a) Above for copper conductors.
  b) Multiplier, 3, resulting in total loss is included.
  c) Multiplier for aluminum is 1.6 to 1.7.
  d) Conductor temperature is 20 deg C. For other temperatures add 0.4% per deg C, above 20. Subtract 0.4% per deg C below 20.
  e) Multiplier for 1-phase, 2-wire is 0.667.
  f) Multiplier for 2-phase, 4-wire is 1.333.
  g) Assumes resistivity to size ratio is constant. Of course, it isn't, especially for very large conductors.
  h) Capacitance and skin-effects are not included. If necessary and calculation method is not available to you, contact me.
 

  

RE: Good day to everybody. Presently I

Oops (the most feared word in an operating room),

K for AWG or MCM case should have been 31.1E-03, not 41.1E-03!

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