Cram for the PE Sample Test 1 - Problem #13

[Orchid PE]

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.

 

[LyceeFruit PE]

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.

 

[LyceeFruit PE]

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

 

[wiliki]

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. 

 

[Dude99]

My approach

 

[Dude99]

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

An add on ?

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

hint: what is the N V elevated to?

 

[Cram For The PE]

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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I posted my answer to this question here:

http://cramforthepe.com/index.php/2020/02/10/question-about-loss-of-excitation-protection/

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

 

[BebeshKing PE]

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. 

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

 

[BebeshKing PE]

Maybe this helps?

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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, 

 

[Orchid PE]

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:

 

[BebeshKing PE]

Hi, I'm just going to use this thread for some of my Cram Vol.1 practice exam questions instead of creating a new one.

For problem #2, 

If the efficiency lost is 20%, is that mean that the charging efficiency will be 80%?. And that being said, should the answer be

Time of charge= capacity/(charge rate current x efficiency) = 100Ah/(40A x 0.8) = 3.125 hours.???

[SIZE=11pt]Not sure where the 1.2 came from because 1/0.8 is 1.25, though.Could someone please tell me if my solution is correct?[/SIZE]

[SIZE=11pt]Thanks,[/SIZE]

 

[Orchid PE]

Hi, I'm just going to use this thread for some of my Cram Vol.1 practice exam questions instead of creating a new one.

For problem #2, 

If the efficiency lost is 20%, is that mean that the charging efficiency will be 80%?. And that being said, should the answer be

Time of charge= capacity/(charge rate current x efficiency) = 100Ah/(40A x 0.8) = 3.125 hours.???

[SIZE=11pt]Not sure where the 1.2 came from because 1/0.8 is 1.25, though.Could someone please tell me if my solution is correct?[/SIZE]

[SIZE=11pt]Thanks,[/SIZE]

If there was no efficiency loss, the battery would take 100% of the time to charge (100 / 40 = 2.5hr).

But since there is an efficiency loss of 20%, that means it will take an extra 20% of the time to charge. So 100% + 20% = 120% (or 1.2). We know to do this because an efficiency loss will increase the time it takes to charge the battery.

If, for whatever reason, we had an efficiency increase of 20%, we would multiply by 80% (because the total time to charge would be reduced).

 

[Cram For The PE]

Hi, I'm just going to use this thread for some of my Cram Vol.1 practice exam questions instead of creating a new one.

For problem #2, 

If the efficiency lost is 20%, is that mean that the charging efficiency will be 80%?. And that being said, should the answer be

Time of charge= capacity/(charge rate current x efficiency) = 100Ah/(40A x 0.8) = 3.125 hours.???

[SIZE=11pt]Not sure where the 1.2 came from because 1/0.8 is 1.25, though.Could someone please tell me if my solution is correct?[/SIZE]

[SIZE=11pt]Thanks,[/SIZE]

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View attachment 16487

A previous individual had the same question a while back. I answered it here:

http://cramforthepe.com/index.php/2019/07/27/batteries/

 

[Dude99]

hey @Chattaneer PE I've got another problem here from the Cram for PE test. I probably need to refresh myself on symmetrical components fault analysis after this...  

Why is the j0.07 reactance not included in the zero-sequence portion of the circuit? Also, is it standard to always assume V = 1 pu? 

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This is a good example: THINK before grabbing the pencil and calculator.

can't be 0 unless it is the faulted line or inf bus no and no Z.

can't be 13800 unless NO Z except a hi NGR.

same for 12090, nrg must be >>>> X1

that leaves 6990, which makes sense since L-N is ~7970
 

imo the point of a question like this is to make you think about basics and give exercise on SC's, but I personally spend time on it during testing.  No if the range of answers was: 6900, 6990, 7100, 7600, maybe.  It can be done quickly using only Z and arrive at 6900.  Don't always jump into the most complicated method, think basic concepts first.