Cram for the PE Sample Test 1 - Problem #13 - Page 2 - Power Exam Sub Forum - Engineer Boards
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# Cram for the PE Sample Test 1 - Problem #13

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Hello ... i am actually was working on the same problem set and was wondering if you could help with the following question?

Attached above is the question/answer.... i am having a hard time understanding where this formula comes from or how to derive it.

Also how would it change if its 3 phase, more than 2-pulse?, or not a fully controlled rectifier? I know .. a loaded question lol. Any help would be appreciated.

Thank you!

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On 1/19/2020 at 5:16 PM, Kris said:

Hello ... i am actually was working on the same problem set and was wondering if you could help with the following question?

Attached above is the question/answer.... i am having a hard time understanding where this formula comes from or how to derive it.

Also how would it change if its 3 phase, more than 2-pulse?, or not a fully controlled rectifier? I know .. a loaded question lol. Any help would be appreciated.

Thank you!

To find the average value of a waveform we just integrate over the area of interest (finding the sum) and then divide by the period (finding the average). For the stuff below, we're integrating over the shaded area (in the case of a full wave rectifier, from theta to T/2). T=2π, w=2π/T, Vpk=Vrms.

The final formula is a tad bit different from what was used in the solution, but if you notice there are a few terms they could've canceled out.

EDIT: Typed this up for clarity.

Edited by Chattaneer PE

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The 2 pulse is simply because there are two thyristors that need to be turned on at the same time at any point.

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Useful formulas for this: https://www.electronics-tutorials.ws/power/single-phase-rectification.html

Edited by Chattaneer PE

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@Kris :) The formula can be found on page 211, equation 8.6 of this book.

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I drew two diagrams below; one for before the conductor break occurred and one for after. In the diagram before the break, the phase and neutral conductors are connected directly to the load. In the diagram after the break, the phase conductor is connected directly to the enclosure, and the neutral conductor connection is irrelevant (but it's still connected to the load). We don't need to worry about the neutral even though it's still connected to the load, because there is no power to the load since the phase conductor broke off. We only need to worry about the return path for current through the load's enclosure. So in the diagram after the break, the load inside the enclosure is drawn not connected. All current will flow through the enclosure. Since the enclosure is connected to ground, and the source's neutral is connected to ground, this will complete the circuit and will be the path for current. Once we calculate the current in the circuit, we can calculate the voltage at the enclosure by finding the resistance from enclosure-to-ground and multiplying it by the current.

Edited by Chattaneer PE

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This is all great stuff!  Thanks everyone for posting (both the questions and answers).

These qualitative questions are great.  I try, when I'm reviewing the more math-relative (quantitative) problems, how the NCEES might change it or ask a qualitative question on the same kind of problem.  For example, as Chattaneer wrote, losing the hot side of a WYE circuit.  What's the overall effect?  How does it affect the current/voltage though the other legs?  How is the load affect (before vs after)?

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That'd be a nice zap if someone touched that enclosure without the 10ohm neutral grounding resistor installed!

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Hello, we’re working on the same problem set, but from book #2 now. We’ve got a question for problem #19 below.

The solution says that there is no neutral in the wye-wye transformer to carry harmonics. However, going back to the question.... we don’t see a mention of a wye with neutral. For problems like these, do we just assume no neutral?

Also, if there was a neutral, does this help to mitigate the harmonics?

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I drew two diagrams below; one for before the conductor break occurred and one for after. In the diagram before the break, the phase and neutral conductors are connected directly to the load. In the diagram after the break, the phase conductor is connected directly to the enclosure, and the neutral conductor connection is irrelevant (but it's still connected to the load). We don't need to worry about the neutral even though it's still connected to the load, because there is no power to the load since the phase conductor broke off. We only need to worry about the return path for current through the load's enclosure. So in the diagram after the break, the load inside the enclosure is drawn not connected. All current will flow through the enclosure. Since the enclosure is connected to ground, and the source's neutral is connected to ground, this will complete the circuit and will be the path for current. Once we calculate the current in the circuit, we can calculate the voltage at the enclosure by finding the resistance from enclosure-to-ground and multiplying it by the current.

Question on this one - Why is there a 0.1 ohm resistance in the equipment grounding conductor? Didn’t see it as a given in the problem - the problem specifically states “A single-phase load is connected phase-to-neutral and is feed from the source with each conductor, including the equipment grounding conductor, having a total resistance of 0.1 ohm”

It says “total resistance”... so why does the hot have a resistance of 0.1 ohm, and why does the EGC have a resistance of 0.1 ohm?

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12 hours ago, wiliki said:

Question on this one - Why is there a 0.1 ohm resistance in the equipment grounding conductor? Didn’t see it as a given in the problem - the problem specifically states “A single-phase load is connected phase-to-neutral and is feed from the source with each conductor, including the equipment grounding conductor, having a total resistance of 0.1 ohm”

It says “total resistance”... so why does the hot have a resistance of 0.1 ohm, and why does the EGC have a resistance of 0.1 ohm?

I think it makes more sense with a different wording of the question:

A single-phase load is connected phase-to-neutral. Each conductor has a total resistance of 0.1 ohms, including the equipment grounding conductor, .

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13 hours ago, wiliki said:

Hello, we’re working on the same problem set, but from book #2 now. We’ve got a question for problem #19 below.

The solution says that there is no neutral in the wye-wye transformer to carry harmonics. However, going back to the question.... we don’t see a mention of a wye with neutral. For problems like these, do we just assume no neutral?

Also, if there was a neutral, does this help to mitigate the harmonics?

I honestly don't understand what he's trying to get at with this question.

We know delta-wye transformers are useful because the delta winding allows triplen harmonics to be contained within the transformer, reducing the harmonics on the wye side. Maybe he's assuming with a wye-wye transformer the harmonics are not contained within the transformer and are reproduced on the secondary side. I don't understand his solution of "because there is no neutral." Maybe he's implying that if the transformer had a neutral conductor the harmonics would flow in the neutral. However, I can't think of an instance where a wye connection would want to be used without a neutral.

I don't like the question, but at least we can eliminated the a, b, and c answers.

Edited by Chattaneer PE
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1 hour ago, Chattaneer PE said:

I honestly don't understand what he's trying to get at with this question.

We know delta-wye transformers are useful because the delta winding allows triplen harmonics to be contained within the transformer, reducing the harmonics on the wye side. Maybe he's assuming with a wye-wye transformer the harmonics are not contained within the transformer and are reproduced on the secondary side. I don't understand his solution of "because there is no neutral." Maybe he's implying that if the transformer had a neutral conductor the harmonics would flow in the neutral. However, I can't think of an instance where a wye connection would want to be used without a neutral.

I don't like the question, but at least we can eliminated the a, b, and c answers.

These are the type of practice exam questions I run into once in a while that I just can't seem to quite get.  My first thought was, is it not safe to assume the Wye has a neutral?  I don't remember seeing any kind of sample or practice exam problem showing a Wye without one.  I've seen one example of a transformer bank arranged in delta, and you lose one and it asks you how it can be connected (open delta).

But examples like the one above I feel like I just have to move on, so I simply print it out with everyone's comments and put it in my binder and tag it in the rare event it might show up on the actual exam.

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8 hours ago, Chattaneer PE said:

I honestly don't understand what he's trying to get at with this question.

We know delta-wye transformers are useful because the delta winding allows triplen harmonics to be contained within the transformer, reducing the harmonics on the wye side. Maybe he's assuming with a wye-wye transformer the harmonics are not contained within the transformer and are reproduced on the secondary side. I don't understand his solution of "because there is no neutral." Maybe he's implying that if the transformer had a neutral conductor the harmonics would flow in the neutral. However, I can't think of an instance where a wye connection would want to be used without a neutral.

I don't like the question, but at least we can eliminated the a, b, and c answers.

I agree that he’s trying to get at the harmonics reduction with a delta-wye transformer, but is doing it backwards by giving the example of the wye-wye. I do not understand the “there is no neutral” comment either. In the delta-wye, you don’t have a neutral on the delta side and create a separately derived system on the wye side, but that is not what the solution seems to be saying.

I don't mind the question, but the solution explanation leaves a lot to be desired.

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Another problem here, problem #21. We're at loss with the solution, as this is the first time we're seeing a relation between generator curves, to R-X diagrams, and to a distance relay. Does anyone understand the solution or is able to provide more clarity?

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5 minutes ago, wiliki said:

Another problem here, problem #21. We're at loss with the solution, as this is the first time we're seeing a relation between generator curves, to R-X diagrams, and to a distance relay. Does anyone understand the solution or is able to provide more clarity?

I'll check some of my protection books at home to see if they cover anything like this. This seems like a very specific question for protection engineers.

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This is from Elmore's Protective Relaying Theory & Applications, sponsored by ABB.

The KLF & KLF-1 relays are specific microprocessor relays from ABB, circa 2004.

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Also if you have 4th edition of Blackburn, check out page 262-266.

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Thanks guys, that helps a lot 👍 @Chattaneer PE @LyceeFruit PE

We're gonna have to purchase more books on protection over here. Heard about Blackburn, but have yet to look into it. Please let me know if you have any other good recommendations for protection references.

My approach

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On 1/23/2020 at 8:28 PM, wiliki said:

Question on this one - Why is there a 0.1 ohm resistance in the equipment grounding conductor? Didn’t see it as a given in the problem - the problem specifically states “A single-phase load is connected phase-to-neutral and is feed from the source with each conductor, including the equipment grounding conductor, having a total resistance of 0.1 ohm”

It says “total resistance”... so why does the hot have a resistance of 0.1 ohm, and why does the EGC have a resistance of 0.1 ohm?

NGR's are required by law in mining to limit frame V to <40

what is the N-G V of the ungrounded phases during a gnd fault?

hint: what is the N V elevated to?

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On 2/5/2020 at 3:41 PM, wiliki said:

Another problem here, problem #21. We're at loss with the solution, as this is the first time we're seeing a relation between generator curves, to R-X diagrams, and to a distance relay. Does anyone understand the solution or is able to provide more clarity?

I posted my answer to this question here:

If you have any other questions feel free to contact me.

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On 1/3/2020 at 8:50 PM, wiliki said:

Hey guys, I know this is a simple problem, but can anyone provide their feedback on attaining the answer to this? The solutions manual says it's (D), but we had initially assumed the answer was (C) due to no ground.

I believe letter C is still a proper connection. It still protect the load. MOVs are also used in DC applications ( electronics and semiconductors).

Edited by BebeshKing

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On 1/9/2020 at 3:52 PM, Chattaneer PE said:

Maybe this helps?

Hi @Chattaneer PE, what if the connection will be High side wye and Low side delta? will the result for positive and negative sequence current change?

Will it become:

Ia1(wye) =Ia1 <-30

Ia2(wye) = Ia2 <30

Iao=0

???

I know the result will still be the same (1/square root of 3) since we are just adding the 3 components. But I am just curious with the result each components.

Thanks,

Edited by BebeshKing

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9 hours ago, BebeshKing said:

Hi @Chattaneer PE, what if the connection will be High side wye and Low side delta? will the result for positive and negative sequence current change?

Will it become:

Ia1(wye) =Ia1 <-30

Ia2(wye) = Ia2 <30

Iao=0

???

I know the result will still be the same (1/square root of 3) since we are just adding the 3 components. But I am just curious with the result each components.

Thanks,

Yes it is dependent on connection (and technically phase sequence: abc vs acb, but most problems are assumed abc, read up on DAB vs DAC transformers).

The phase shift is either plus or minus 30 degrees from primary to secondary for positive sequence voltages and currents (essentially, positive sequence follows the phase shift of the transformer). Negative sequence shifts in the opposite direction from positive.

These two posts are worth reading, though they're not talking about symmetrical components:

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