Power Factor/%Eff.

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feez

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I'm new to the instruction field. I've got almost twelve years in the trade with a wide variety of experience but I've never taught before, so it's been a bit of a challenge to teach myself what I'm teaching my students. I never attended trade school or anything, just went and took the MD Masters test. That means I didn't get the benefit of all that formal math training.
Here's the point to all that: In showing my students RLC circuits we found an interesting phenomenon. The power factor of a series RLC ckt. is equal to R/Z. Easy enough. But the only way our textbooks show to find PF in a parallel RLC ckt. is that it's (PF) the cosine of the phase angle. Now, I don't have access to 'scopes, or even protractors and scale rulers to do vector diagrams (the school's pretty cheap, I must say, and it's a brand new program. I'm the guinea pig.), so I needed another way to find PF in a para. RLC ckt. %Efficiency = True power over TP + Losses (R losses, in phase). TP= Apparent power (E x I) x PF, right? Right.
So, I was told that the only true power in a RLC para. ckt. are the R losses. The problem is that this always results in a 50% EFF. Example: TP = 200w (100v x 2a @ R), TP over TP + Losses (200w @ R) = .5, 50% every time. I figured this must be the wrong calculation. But then the night instr. (electronics guy) said that no, that's just a characteristic of those type of ckt.'s. OK. Good enough for me.
Then the next day my boss invades my class with another friend (also electronics guy). This guy demands to see how I'm doing my calculations. I patiently explain them and am told that I'm doing it all wrong. You have to take Ix over Ir, then the inverse tangent of that answer (which will be your phase angle), then the cosine of that. Sounds sensible enough, despite his angry attitude. Just one problem. It still comes out as 50% EFF, no matter what the PF is.
To shorten this up, suffice to say that four or five different people have assaulted me over the last few days, all with different methods of calculating PF in a RLC parallel ckt., all of which, surprisingly, work. And all of which end up giving a 50% EFF.
What the hell's going on? Is this just a characteristic of these ckt.s or am I nuts? Or both?
Can someone please explain this?
 

bphgravity

Senior Member
Location
Florida
Re: Power Factor/%Eff.

Hello Feez,

Hang in there, it gets easier each year. After the 100th time you perform this type of calculation, you will be doing it in your sleep.

I may be showing you a another method to this equation, but I assure you that you will not get 50% for every one.

The first thing that must be done in a parallel RLC circuit is to find net reactive current (X). Then total current must be found by the square root of the sum of the square of R and the square of X. Next you use Ohm's law again to find the power of the circuit. Power Factor is the watts/va.

Lets do an example! We have a 30-ohm resistor, 30-ohm capacitor, and a 20-ohm coil in parallel at 120-volts.

Using Ohm's Law: R=4A, L=6A, C=4A.

Net Reactive current: 6A - 4A = 2A

Total Current: Square root of 4^2 + 2^2 = 4.47A.

Find Power: Watts = 120V x 4A (R) = 480W
VA = 120V x 4.47A (Z) = 536.4va

pF: 480W / 536.4va = 89% or lag anlge of 27 degrees.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Re: Power Factor/%Eff.

I can see that you are trying to learn, trying to understand, and that is commendable. But reading your description, I conclude that you are using terms incorrectly, or at least imprecisely. Let me start by clarifying several of your terms.

%Efficiency = True power over TP + Losses (R losses, in phase). TP= Apparent power (E x I) x PF, right? Right.
Let?s start there. There is no need to talk about efficiency. It is not related to power factor in any way.
I was told that the only true power in a RLC parallel circuit are the R losses.
You are using the word ?losses? incorrectly. A person would build an RLC circuit for some purpose: to do some work, or to perform some task. The three parallel elements would have some function related to the desired task. The power that they use is the normal price you pay, in order to get the task done. But there are also ?losses? to consider. You lose power in the wires that connect the three elements. The wires don?t perform the task, but you cannot connect the three elements without them, so they are necessary. If there is any power expended within the wires (the so called ?I-squared-R? losses), and there is always some, it causes the efficiency to be less than 100%. There are also losses associated with the development of an electric field within the capacitive (?C?) element, and also losses associated with the development of a magnetic field within the inductive (?L?) element

What you are doing wrong in your calculation is to treat the power expended by the resistive element (the ?R?) as though it were losses. It is not! It is the useful work for which the RLC circuit was designed. You have not given us any information related to the actual ?losses? within the circuit. So let me suppose that in order to put 200 watts through the circuit, you experience ?losses? within the wires (and within the electric and magnetic fields) of 5 watts. Then, Efficiency equals (200) divided by (200 plus 5), or 97.6%.

The bottom line is (1) You can?t calculate efficiency, because you don?t know the losses, and (2) You don?t need to calculated efficiency anyway.
PF in a parallel RLC circuit is the cosine of the phase angle.
That is its definition.
The power factor of a series RLC circuit is equal to R/Z.
This is another way of saying the same thing. Draw a right triangle as follows. First draw a horizontal line, and label it ?R.? At the right endpoint of the first line, draw a line straight up, and label it ?X.? Complete the triangle by drawing a line from the top of the vertical line to the left endpoint of the horizontal line, and label this third side ?Z.? The angle formed at the left endpoint corresponds to the phase angle. The cosine of that angle is defined as R/Z.
You have to take Ix over Ir, then the inverse tangent of that answer (which will be your phase angle), then the cosine of that.
This is yet another way of saying the same thing, but this one is a bit trickier to use. Back to the triangle I described above. If you divide X/R, and then take the inverse tangent, you get the phase angle. Next take the cosine of that angle, and you get the power factor.
Now, I don't have access to 'scopes, or even protractors and scale rulers to do vector diagrams
You don?t need them. Just draw the triangle I described above.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Re: Power Factor/%Eff.

Bryan gives a good example. In fact, he raises a point that I had intended to raise, and therefore has saved me the trouble. Before you can do any calculation, you must first figure out what information is given. In your example, what is given is the watts of the resistive element, no information about the other two, and no information about losses. That is not enough information to calculate either power factor or efficiency. In Bryan?s example, he gives you the value of resistance of each of the three elements. That is enough to calculate power factor, as Bryan has shown.

His example appears to assume no losses, and therefore a 100% efficiency. That is probably the most common approach to an RLC circuit. You generally don?t get into efficiency issues until you deal with transformers.

I have one caution to offer:
Originally posted by bphgravity:
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">net reactive current (X).</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">R=4A, L=6A, C=4A.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
Using the letters X, R, L, and C for various values of current is not the usual naming convention. It is more common to use ?I? for current, and use subscripts to distinguish currents in various elements. In order to limit possible confusion, I might suggest the terms Ix, Ir, Il, and Ic.
 

bphgravity

Senior Member
Location
Florida
Re: Power Factor/%Eff.

Good call Charlie.

I realize after reading my post how that could be confusing. "X" is definitely not a value of current but a value of reactance. XL, XC, and Z are figured in Ohms. IL, IC, IZ, and IR would have been better values to use for current figures.

Thanks. :cool:
 

feez

Member
Re: Power Factor/%Eff.

bphgravity- Find Power: Watts = 120V x 4A (R) = 480W
VA = 120V x 4.47A (Z) = 536.4va


That I got. The problem is indeed in the terms, part of it anyway. VA would be the apparent power, Er x Ir would we the true power in the efficiency equation, no need for the PF value, correct? Just divide VA into W.


CharlieB-
Let?s start there. There is no need to talk about efficiency. It is not related to power factor in any way.


Why does our textbook mention it, then? Tp over Tp + L: does that deal with something else other than RLC ckt's? Maybe I'm taking it personally, or maybe I should re-read my own post (could be I wasn't clear) but the problem was that this wasn't "my" calculation, they were all given to me, either by a text or a fellow instructor. In a way I'm kind of proud of where I've gotten through nothing but the skill I've taught myself, and pride is rotten. So I guess what bugged me wasn't the math or the terms (I understand the difference btwn. Z and R, and the R of wire) but the fact that each and every person I approached, without fail, seemed to indicate that I was at fault.
I'll be the first one to admit that I'm probably too sensitive, so please, don't think I'm not grateful for your replies. Like the Chinese proverb: I hear; I forget I see; I remember I do; I learn.


many thanx,
feez
 

Ed MacLaren

Senior Member
Re: Power Factor/%Eff.

feez,

A motor is the most common type of load device for which both power factor and efficiency would normally be calculated.

I use this sketch to illustrate the basic concepts.

Example 1 shows a common exam question that requires the calculation of the motor current at a specified power factor and efficiency.

Example 2 is a typical power factor correction calculation.

Ed

Power4.gif
 

charlie b

Moderator
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Location
Lockport, IL
Occupation
Retired Electrical Engineer
Re: Power Factor/%Eff.

To feez : I certainly meant no offense. If I knew a person?s background and level of experience, then I can more easily understand the nature of his question, and can focus my answer on the one or two minor points that he might be getting confused. But in a forum like this, I can only guess how much a person already knows, on the basis of the way the question was worded.
Originally posted by feez: Tp over Tp + L: does that deal with something else other than RLC ckt's?
Yes. It applies in the mechanical world as well. The efficiency of a pump or a steam turbine uses the same equation. My point was that the resistive value within the RLC circuit is not the ?losses? that you use in the efficiency equation. That is the reason you kept getting an incorrect value of 50% efficiency. You used the wrong value of ?L.?

It is worth noting that the RLC model of a circuit is intended to be used in describing EVERY SINGLE CIRCUIT in the electrical world. In saying this, I distinguish ?electronic? from ?electrical,? since circuits that include amplifiers or that change the current into something other than a pure sine wave cannot be modeled as RLC circuits. But everything an electrician will install, from transformers, to panels, to lights, to fans, to tanning beds, to elevators, to transmission lines, can be modeled as a part of an RLC circuit. That is why that type of circuit is taught in the electrical fundamentals classes.
Why does our textbook mention it (i.e., efficiency) then?
I don?t know, as I haven?t read your textbook. My point is that if you are looking for power factor, it is not necessary to first find the efficiency. You can certainly find both, if you are given enough information, but the two are not related to each other. Power factor is one of the most difficult concepts in electrical science. It has no analogy whatsoever in the mechanical sciences.
 

roger

Moderator
Staff member
Location
Fl
Occupation
Retired Electrician
Re: Power Factor/%Eff.

Charlie b, (my post here is not directly related to this thread, or anyone in it, so forgive me)
Originally posted by charlie b:
To feez : I certainly meant no offense. If I knew a person?s background and level of experience, then I can more easily understand the nature of his question, and can focus my answer on the one or two minor points that he might be getting confused.
I know this has come up before and wouldn't necessarily be acurate, but why was some of this information eliminated from the members profiles?

I know there would be misrepresentation (as there is now) even with this available, but if a person over represented themselve, they would only be doing themselve harm in that the answers would probably be over their head.

I guess the bottom line is, what was the reason for eliminating this info?

Roger
 

Ed MacLaren

Senior Member
Re: Power Factor/%Eff.

Power factor is one of the most difficult concepts in electrical science
An analogy that I have often used to explain power factor can be described in terms of the old method of door-to-door milk delivery in the days when they used reusable glass milk bottles.

The delivery function required the delivery person to transport, from the dairy to the consumer?s house, a total load consisting of milk and glass.
Of this total, only the milk component was paid for and consumed by the customer. The glass, although essential to the operation, was stored at the customer?s home, washed and returned to the dairy on the next delivery day.

The function of an AC electrical circuit is to deliver energy from a source, through the circuit conductors to a load, in this example, a motor.
The energy consumed ( actually converted by the motor into output horsepower and heat losses) is represented by the true power. It is the product of the in-phase current times the voltage, and is measured in watts. This is paid for by the consumer, and is analogous to the milk in the comparison.

The energy required to create the motor?s magnetic flux is not converted (consumed), it is stored in the magnetic field and returned to the source during the next cycle. It is represented by the reactive power, is the product of the out-of-phase current times the voltage, and is measured in vars (volt-amps reactive). It would be analogous to the glass in the comparison.

The total energy input to the motor, the product of the total current times the voltage is called the apparent power. It is measured in va (volt-amps), and would be analogous to the total load carried into the home by the delivery person, milk and glass, in the comparison.

Ed

[ May 09, 2004, 01:23 PM: Message edited by: Ed MacLaren ]
 
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