Maximizing Savings: Power Factor Correction Homework for Improved Efficiency

In summary, the conversation discusses the concept of power factor and its impact on energy costs. The solution involves calculating the real power using the complex power formula and then finding the angle between real power and apparent power. By reducing this angle, the cost can be reduced. However, more information is needed to determine the exact amount of money being lost due to poor power factor.
  • #1
raiderUM
32
0

Homework Statement





Homework Equations


If I want to figure out how much money we are losing because of poor power factor and the information I have is Power Factor 60% and total KWH I can take 60% of my total and then subtract that from my total then multiple the subtracted number by my current rate and it will tell me how much money we are throwing away because of poor power factor?


The Attempt at a Solution

 
Physics news on Phys.org
  • #2
there is not enough info here to calculate anything

Power factor is (real power/apparent power) it is a measure of how much power is real and is calculated as a cosine of the phase angle.

power companies charge for volt amps so the worse your power factor is then the more you will pay. i guess an example would help !

lets say you own a great big motor 20 kw your power factor was 0.7 lagging and the price per unit is 1 euro per kilowatt hour. you want to run your motor for one hour.

because your power factor is bad you are paying for allmost 29 Kw when a good power factor would be considered as 22 Kw
 
  • #3
commelion said:
there is not enough info here to calculate anything

Power factor is (real power/apparent power) it is a measure of how much power is real and is calculated as a cosine of the phase angle.

power companies charge for volt amps so the worse your power factor is then the more you will pay. i guess an example would help !

lets say you own a great big motor 20 kw your power factor was 0.7 lagging and the price per unit is 1 euro per kilowatt hour. you want to run your motor for one hour.

because your power factor is bad you are paying for allmost 29 Kw when a good power factor would be considered as 22 Kw

It doesn't make sense to say they charge your more for a worse power factor because you are simply getting less power. Why would they charge you more? It would make more sense for them to charge you more for a better power factor because you are getting more power.

If they charge you per amount of real power, you'd want to get your power factor as close to unity as possible. (For maximum power that is)

You can always tweak your power factor by adding either a capacitor or an inductor bank. (It will either consume or supply VAr's)
 
Last edited:
  • #4
raiderUM said:

Homework Statement





Homework Equations


If I want to figure out how much money we are losing because of poor power factor and the information I have is Power Factor 60% and total KWH I can take 60% of my total and then subtract that from my total then multiple the subtracted number by my current rate and it will tell me how much money we are throwing away because of poor power factor?


The Attempt at a Solution


If you want to figure out how the power factor is affecting your intake of real power you need the complex power, S. (This is in volt-amps)

Then your real power is going to be,

[tex]P = Scos(\theta)[/tex]

Where the cosine portion is your powerfactor,

[tex]PF = cos(\theta)[/tex]
 
  • #5
ok this the all the information that I have:
KW 98
KVA 130
So PF is 75%

That is all three phases added together.
The voltage is 283 per phase and amps is 472 for all three phases

Lets say I only pay 8 cents a Killowatt.. how do I figure out how much money I am losing? I guess that is the confusing part for me. Do I need for info?
 
  • #6
jegues said:
It doesn't make sense to say they charge your more for a worse power factor because you are simply getting less power. Why would they charge you more? It would make more sense for them to charge you more for a better power factor because you are getting more power.

If they charge you per amount of real power, you'd want to get your power factor as close to unity as possible. (For maximum power that is)

You can always tweak your power factor by adding either a capacitor or an inductor bank. (It will either consume or supply VAr's)

it has nothing got to do with the consumer getting more or less power man !

the whole idea behind powerfactor is the simple fact that it costs more for the supply company to generate to customers with a bad power factor i.e the need more volt amps on the transmission line.
 
  • #7
raiderUM said:
ok this the all the information that I have:
KW 98
KVA 130
So PF is 75%

That is all three phases added together.
The voltage is 283 per phase and amps is 472 for all three phases

Lets say I only pay 8 cents a Killowatt.. how do I figure out how much money I am losing? I guess that is the confusing part for me. Do I need for info?

your soloution ! [i suspect this is homework so you can figure this out yourself]

your power factor is correct

now do an arccos on this value, this is the angle between your real power and apparent power. when you draw this your apparent power phasor will be the hypotenuse of a right angled triangle. straight away you can see that this line is longer that your real power ?
(by the way its normal to put real power on your x axis)

and that's it... is reduce the angle reduce the cost good luck !
 
  • #8
commelion said:
your soloution ! [i suspect this is homework so you can figure this out yourself]

your power factor is correct

now do an arccos on this value, this is the angle between your real power and apparent power. when you draw this your apparent power phasor will be the hypotenuse of a right angled triangle. straight away you can see that this line is longer that your real power ?
(by the way its normal to put real power on your x axis)

and that's it... is reduce the angle reduce the cost good luck !

This is not my homework. I am just curious as to how much money I can tell my boss we could save by installing capacitors at our poor power factor locations. I know the phase angle, its 40 degrees. I have an energy system that let's me see all this real time. I'm sorry if my question comes off confusing because you think this is my homework. I'm just looking for some better examples of how to quickly calculate how much money we are throwing away using the Data I stated in my previous post.
 
  • #9
not homework then ea !

Power factor cannot be solved on this forum properly, because it needs a detailed study of plant and equipment and how it might be connected. I don’t recommend telling your boss anything unless you completely correct, as it is normal that supply companies dump huge fines on to customers with bad power factor, so he should know if he does not. That said I’ll use your example to demonstrate but you will need to fill in some blanks yourself ok?

Your power factor angle loos like being 41.1 degrees (you don't mention lagging or leading so I am going with the persumption of lagging)

p = 98
s = 130

and most importantly Q = 85.46 kvars lagging (take out your trigo book to see where this comes from)

you need capacitors which are rated in kvars to reduce Q (ie capacitors/inductors don't use real power but the create reactance or kvars) this is just the opposite of your problem.

ive allready told you how to calculate the cost of your bad power factor, if you are indeed curious then you should not have a problem turning the handle on this from here.

good luck
 
  • #10
commelion said:
there is not enough info here to calculate anything
That's true. Not enough information.

power companies charge for volt amps
Wrong. Power companies charge domestic consumers for the actual power used.

It would make a lot of sense (i.e., make more money for the power company) if they were to charge according to kVA but they don't, and I think we should not encourage them to consider doing otherwise. :smile:

Power authorities prefer that you have a power factor close to 1, and will probably pressure big companies to take steps to do so, but for small domestic consumers, there is generally no incentive to consider powerfactor correction.

EDIT: Just noted where raiderUM indicates his employer is a medium commercial user of power. In that case, it is just possible that the power company may be imposing a charge for kVA. It would make sense to have that written into the supply contract, as an incentive for a good powerfactor. But we can't know unless OP is able to find out.

So to put it starkly, the boss may possibly be paying for 130 units of supply, when he need only be paying for 98 units. This is 32% higher than if the pf were unity. You can only know by looking at the account, or reading over the contract, whether the power supplier is imposing a penalty for a poor pf.

raiderUM said:
Lets say I only pay 8 cents a Killowatt.. how do I figure out how much money I am losing?
If you really are paying per kilowatt, then you are losing no money. You are paying only for the actual power you use. On the other hand, if you were paying per kilovolt-amp then there is a potential saving of 32% to be made if you fixed your poor powerfactor to make kW ≈ kVA. You need to scrutinise the quarterly account to see whether powerfactor is somehow being factored into the bill.
 

Related to Maximizing Savings: Power Factor Correction Homework for Improved Efficiency

1. What is power factor correction?

Power factor correction is a technique used to improve the power factor of an electrical system. It involves adjusting the electrical load to match the supply voltage, thereby reducing the amount of reactive power needed and improving the overall efficiency of the system.

2. Why is power factor correction important?

Power factor correction is important because it helps to optimize the use of electricity and reduce energy waste. A low power factor can lead to higher energy bills and can also cause strain on the electrical system, resulting in equipment damage and downtime.

3. How is power factor correction achieved?

Power factor correction can be achieved through various methods, such as adding capacitors to the electrical system or using specialized equipment like power factor correction controllers. These methods help to reduce the amount of reactive power and bring the power factor closer to unity.

4. What are the benefits of power factor correction?

The benefits of power factor correction include improved energy efficiency, reduced energy costs, increased reliability of the electrical system, and extended lifespan of equipment. It also helps to meet regulatory requirements and reduces the carbon footprint of a facility.

5. Can power factor correction be applied to all electrical systems?

Power factor correction can be applied to most electrical systems, including industrial, commercial, and residential systems. However, the specific approach and equipment used may vary depending on the type and size of the system. It is best to consult with a professional to determine the most suitable power factor correction solution for a specific system.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Electrical Engineering
Replies
8
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
16
Views
2K
Replies
12
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
2K
Replies
77
Views
992
Back
Top