3 phase rectification

I have an issue solving this problem.

1. Does the firing angle below 30 degrees matter in 3 Phase rectification.

2. How do we know which type of connection is this. 3 diode, 6 diode between L-L or 6 diodes center tap  with star connection.

View attachment 9466

I'll try to point you in the right direction: 

First, it is rectified, so we are taking an AC input and getting a DC output.  The given equation is an instantaneous expression for the voltage, v(t), where 460 is the amplitude (peak) of the voltage waveform (no need to multiply/divide by sqrt2, it is already in terms of what is needed to solve).

Hope this helps!

Also, since the firing angle is at 0, the waveform is a purely sinusoidal input (it is not chopped at pi/6, as would be the case if it were at 30 degrees).

TNPE said:

I'll try to point you in the right direction: 

First, it is rectified, so we are taking an AC input and getting a DC output.  The given equation is an instantaneous expression for the voltage, v(t), where 460 is the amplitude (peak) of the voltage waveform (no need to multiply/divide by sqrt2, it is already in terms of what is needed to solve). 

Hope this helps!

Click to expand...

Thanks TN ; so far it is good, but in three phase rectification the dc outputs are different for different configurations i.e half wave, full wave, what to assume? The answer shown is 293V, which is for full wave rectification of single phase bridge rectifier, I suppose. Secondly in three phase the diodes get forward biased only after 30 degrees, how to account it.

TNPE said:

Also, since the firing angle is at 0, the waveform is a purely sinusoidal input (it is not chopped at pi/6, as would be the case if it were at 30 degrees).

Click to expand...

It is a full-wave rectifier. On the positive half-cycle of the input wave, current flows through the top SCR while the bottom SCR is reverse-biased. On the negative half-cycle of the input wave, current flows through the bottom SCR while the top SCR is reversed-biased.

A half-wave rectifier would only have one SCR...

That figure represents one phase of the three total, so the input would be better described as v(t) of the "A" phase. Just visualize two more of them; one across "B" phase and the other across "C" phase...this is just a simplified figure.

There are a couple of things you need to take note of for this particular problem: one, omega is 2 x pi x f; therefore, 377 = 60 Hz. Second, the firing angle of the SCR was given in the problem as 0 degrees.

So, to answer the question, you should use the "Average" formula from your PPI reference manual (sorry, I don't have it right in front of me to type it out here), but it is a definite integral evaluated from the firing angle of the SCR to 2 pi, but since the firing angle is said to be 0 degrees in the problem, you could just accept that the average is approximately 0.637 of the input voltage, so 460 x 0.637 = 293V.

If this question was reworded so that the firing angle was at some other value than 0 degrees, then you would need to evaluate the integral to get the right answer.

Hope this helps.

It should also be noted that the question states that rectification is accomplished with SCRs, not diodes. The symbol used in the schematic doesn't show the "gate" of the SCR for simplicity, which makes it look like a diode symbol.

It's the gate that controls the firing angle of the SCR, so the answer to your first question is: yes, firing angle less than 30 degrees does matter. The gate gets a voltage pulse that turns the SCR "on." Once the SCR turns on, it stays on until the current that is running through the SCR falls below a threshold known as the "holding current," so you can program a controller to turn the SCR on at 0-degrees, 5-degrees, 75-degrees...whatever you want the average voltage to be since the average voltage is:

, where the "0" point denoted in the definite integral would be the firing angle (in radians) where you want the SCR to turn on.

So here is an example of a schedule of firing angles vs. average voltages provided by a rectifier with the result of your problem denoted for firing angles other than 0-degrees.

0-degrees (0 radians): (2/pi)*Vp = 0.637*Vp = 293V

5-degrees (pi/36 radians): (1.996/pi)*Vp = 0.635*Vp = 292.1V

13-degrees (pi/13.846 radians): (1.974/pi)*Vp = 0.628*Vp = 288.9V

25-degrees (pi/7.2 radians): (1.906/pi)*Vp = 0.607*Vp = 279.2V

90-degrees (pi/2 radians): (1/pi)*Vp = 0.318*Vp = 146.3V

For some reason, the clipping didn't go through:

View attachment 9485

As @BigWheel said, the given expression is for one phase (call it A).  The subsequent phases would be the same with an argument of (wt-120) and (wt-240/or+120), respectively.

Remember, if you are asked for an average, you will almost always integrate. In some instances, integration may not be necessary, but I would complete the integral anyway to be certain.  This test will pique your intuition and challenge real-life applications, but it will present and quiz your theoretical understanding of ideal circumstances.  Solve the problem with what is given!

Thanks for your efforts @BigWheel and @TNPE but I am more than fully confused now. I made an integration and the answer 293 =2 Vm/Pi is for one phase fullwave rectification for 0 degree firing angle . Can someone post the integration done and show the solution. And regarding firing of an SCR I have this information so we need not worry about firing angle. According to me the answer should have been 3Xroot3X460/Pi=761.2 V. View attachment 9486

rg1 said:

Thanks for your efforts @BigWheel and @TNPE but I am more than fully confused now. I made an integration and the answer 293 =2 Vm/Pi is for one phase fullwave rectification for 0 degree firing angle . Can someone post the integration done and show the solution. And regarding firing of an SCR I have this information so we need not worry about firing angle. According to me the answer should have been 3Xroot3X460/Pi=761.2 V. View attachment 9486

Click to expand...

BigWheel wasn't exactly right...

The integral should be:

(1/T)S(sin(wt))dt --where S is the integral and evaluate this from 0 to pi.

Inherently, it is evaluated over one half cycle, since a full cycle would yield an average of 0 (i.e same curve area above and below and it would reduce to 0).  The average can be looked at as the infitesimal points along the curve (one half cycle) and divided by the period (pi).

Let's evaluate sin(t)dt :

-cos(t) evaluated from 0 to pi = 1+1=2

Put it all together and you have:

2/pi = 0.637

460(0.637)=293 Vavg

This holds true for any pure sinusoid evaluated over one half cycle with period T (and given your firing angle of 0 degrees).  

Stop overthinking it.  Look at a single phase and evaluate the avg. voltage for that phase.  This would hold true for any phase, but best to stay with A phase since it is assumed to be at 0 degrees (and the given equation says as much), unlike B or C phase.

Your approach with 460 is way off.  You're thinking about line voltage, when what you are given is peak voltage of 1-phase.  You are averaging the peak, not the line, RMS or any other value.  An instantaneous equation of voltage or current should immediately tell you that you're dealing with a peak value, unless, of course, you are told otherwise.  If that were the case, you should appropriately multiply by sqrt2 to get it in terms of peak.  

Also, another tidbit, the ratio of the RMS voltage to the average voltage for a pure sinsusoid is always 1.11.  Analytically, this would be {(Vp/sqrt2)/(2Vp/pi)} or pi/(2*sqrt2).

TNPE said:

BigWheel wasn't exactly right...

The integral should be:

(1/T)S(sin(wt))dt --where S is the integral and evaluate this from 0 to pi.

Click to expand...

@TNPE is correct, the Average Value formula I clipped from the internet did not include the 1/T term in the integral (it was included in the schedule I provided), and I erroneously stated that the integral was evaluated over 2pi, which is obviously wrong for the reason @TNPE stated for a sine wave and is also wrong for a fully rectified sine wave, as evaluating a fully rectified sine wave over 2pi would result in 2 times the average voltage, which is also wrong. I apologize for the confusion...

rg1 said:

Thanks for your efforts @BigWheel and @TNPE but I am more than fully confused now. I made an integration and the answer 293 =2 Vm/Pi is for one phase fullwave rectification for 0 degree firing angle . Can someone post the integration done and show the solution. And regarding firing of an SCR I have this information so we need not worry about firing angle. According to me the answer should have been 3Xroot3X460/Pi=761.2 V. View attachment 9486

Click to expand...

The "worked" integration you requested is below. Your average voltage will never exceed the peak voltage.

The firing angle is moot for this particular problem since you were told that it fires at 0-degrees, that is correct - my intent for elaborating on the firing angle portion of this discussion was my attempt to explain why the firing angle matters and how it affects your answer so that if you are thrown this curve ball during the exam, you'll know how to handle it. At the bottom of the attachment, I tried to graphically show why it matters.

As I post this, I'm reminded of a rule I learned a long time ago in college: "Never do math in public." I think I did this right, but if it's wrong please point out where I went off the rails and be especially cruel in your ridicule - I'll deserve it...

Thanks.

View attachment 9490

Pardon the typo in my previous post.... should read infinitesimal, not infitesimal. 

@BigWheel, the work was spot on and much more elaborate than I could show by manually entering via a phone.  I've tried to attach pictures and clips before, but the files have been too large and I'm never in a mood to compress/reformat.  

TNPE said:

Pardon the typo in my previous post.... should read infinitesimal, not infitesimal. 

@BigWheel, the work was spot on and much more elaborate than I could show by manually entering via a phone.  I've tried to attach pictures and clips before, but the files have been too large and I'm never in a mood to compress/reformat.  

Click to expand...

Thanks a ton TN for working hard to make the case clearer. The diagram is right, the maths is right but I not only feel, I am sure that  it is for one phase full wave. Three phase rectification output wave has a different waveform where it never reaches zero line. The integration there gives value of 3 * Vpeak(L-L)/Pi. I too give you all rights for correction. I request you to google three phase rectifier waveform. I am not able to attach it for now. 

rg1 said:

Thanks a ton TN for working hard to make the case clearer. The diagram is right, the maths is right but I not only feel, I am sure that  it is for one phase full wave. Three phase rectification output wave has a different waveform where it never reaches zero line. The integration there gives value of 3 * Vpeak(L-L)/Pi. I too give you all rights for correction. I request you to google three phase rectifier waveform. I am not able to attach it for now. 

Click to expand...

My apologies ,Thanks to both TN and BW (Bigwheel)

Actually, on second thought, my initial approach may be incorrect.  I'm inclined to think your approach may be the way to go.  

TNPE said:

Actually, on second thought, my initial approach may be incorrect.  I'm inclined to think your approach may be the way to go.  

Click to expand...

Thats good. 

View attachment 9493

rg1 said:

I have an issue solving this problem.

1. Does the firing angle below 30 degrees matter in 3 Phase rectification.

2. How do we know which type of connection is this. 3 diode, 6 diode between L-L or 6 diodes center tap  with star connection.

View attachment 9466

Click to expand...

Yeah, I think it depends on how you interpret the question IMO. I completely missed, ignored, or otherwise overlooked where the root cause of the confusion was when @rg1 posted this question, which I think could be summarized as, "The problem statement says it's a 3 phase rectifier, but only one of the phases is represented. Do I really solve this as a 3-phase system, or do I solve the single-phase system shown?"

I may have completely misrepresented what rg1 was thinking, so I apologize if that's the case and I also apologize for not paying attention to his real question and assuming that I had all the answers.

In any case, I looked at the diagram and approached it from a "Let's solve for the one phase they have represented in the figure." My instincts were confirmed when rg1 said that the answer provided matched what I calculated based on my chosen approach - it basically confirmed my thoughts that I understood what the author was asking for.

Again...I could be way off...

I do still think that the way that @TNPE and I suggested is the correct approach because I feel that was the author's intent, especially since the author's solution does match the solutions we arrived at with our approaches.

HOWEVER, I agree that @rg1 is right regarding the correct way to analyze a real 3-phase rectifier circuit, and I'll be the first to admit that when I saw @rg1's follow-up questions and posts, I fashioned my right hand into a pistol, pointed it at the screen, let the hammer fall, and thought to myself, "this guy...he's way off," but I guess I should have pointed this mimed pistol to my own head and let the hammer fall.

All kidding aside - the author's answer suggests that he was looking for the solution of only one of the phases of a complete three-phase system. I think most people would agree that the questions posed in the actual PE exam are screened and will leave only enough ambiguity in the real question to see if you're paying attention.

The PE is not as ambiguous as the above situation.  Yes, it will force you to use some intuition and make some assumptions, but it alleviates ambiguity to that extent.  To be honest, the PE is more about concepts, concepts, concepts!  Not how well do you integrate and "guess" what I'm exactly asking for (integral/differential calculus is covered on the FE -- no need for that here, but is fair game).   This exam with punch you until you feel like biting its ear off, but concepts are much more heavily presented than pure analytics and mathematic applications.  @BigWheel can attest to that, at least I would guess he can (I wasn't at this administration so I can't say with certainty what was presented, but from my administration and others before me, I have gathered the exam is almost always presented heavily on concepts).  To be perfectly honest with you, while not giving too much away, I would advise you to spend a lot of time on motors and all applications associated with them (analytical and especially conceptual).  That's as far as I'm willing to go with regards to the concepts.  Good luck!

TNPE said:

The PE is not as ambiguous as the above situation.,,the exam is almost always presented heavily on concepts.

Click to expand...

Yup - can you see through the problem well enough to think your way around it? If I don't pass this time around, I'll for damned sure know how to pass the next time around in October.

That's about as far as I feel comfortable confirming.

Thanks @TNPE and @BigWheel My Power electronics was not that good in school except for the concepts and so I thought I must not be wrong. I knew you guys were missing the point and I tell  you can not add up single phases for a three phase solution. You have to know the exact  output waveform and then take average or rms.

1. I do not know exactly how deep I should study this subject of Power Electronics for PE. Also

2. I would like you guys to just wrestle with my last question on synchronous motor. The answer takes Xd*Id as internal generated voltage while actually it is V+/-Xd*Id (Generator/Motor)

I wish you guys all the best for upcoming result.