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.

 

[jghrist]

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.

 

[lucino]

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

 

[moturcu]

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.

 

[ERaySir]

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

 

[jghrist]

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.

 

[moturcu]

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.

 

[lucino]

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

 

[jghrist]

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?

 

[Shortstub]

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.

 

[moturcu]

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.

 

[jghrist]

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

 

[jbartos]

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.

 

[jbartos]

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.\\\

 

[jghrist]

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.

 

[stevenal]

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.

 

[Shortstub]

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.

 

[Shortstub]

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!