# PE Power Question for the Week

### #1

Posted 02 January 2012 - 02:40 AM

The Question For The Week will help those taking the Power PE exam.

The way it works is:

1. Every Monday a new question is provided on the "Question For The Week" tab on http://www.spinupexams.com/ no multiple choices will be provided on Monday's. Attempt to solve the problem without any multiple choices available.

2. On Wednesday's the multiple choices will be provided.

3. On Friday's the complete solution will be provided

4. This will be repeated every week with a new sample exam question.

Enjoy the Question For The Week. Let us know of any improvements you would like to see for the "Question For The Week"

Good Luck on your studies!

Joan

### #2

Posted 02 January 2012 - 09:29 PM

### #3

Posted 04 January 2012 - 12:12 AM

Spin-Up is now providing the "PE Power Question for the Week" http://www.spinupexams.com/

The Question For The Week will help those taking the Power PE exam.

The way it works is:

1. Every Monday a new question is provided on the "Question For The Week" tab on http://www.spinupexams.com/ no multiple choices will be provided on Monday's. Attempt to solve the problem without any multiple choices available.

2. On Wednesday's the multiple choices will be provided.

3. On Friday's the complete solution will be provided

4. This will be repeated every week with a new sample exam question.

Enjoy the Question For The Week. Let us know of any improvements you would like to see for the "Question For The Week"

Good Luck on your studies!

Joan

I like the question for the week, but can the solution be posted at the same time as the question. It makes it more convenient that way. I prefer not to have to wait until Friday.

### #4

Posted 04 January 2012 - 12:58 AM

Can you keep the prior week's question up also, incase I forget to look one week.

### #5

Posted 04 January 2012 - 02:37 AM

The solution has been uploaded for the QFTW. Both Question and Solution will be uploaded at the same time.

The question and solution from the prior week will be made available.

Let me know what other changes you would like.

http://www.spinupexams.com/

Joan

**Edited by spinup, 04 January 2012 - 02:38 AM.**

### #6

Posted 04 January 2012 - 04:14 AM

Thank You for the feedback.

The solution has been uploaded for the QFTW. Both Question and Solution will be uploaded at the same time.

The question and solution from the prior week will be made available.

Let me know what other changes you would like.

http://www.spinupexams.com/

Joan

Spin-up, regarding the answer for that question....since the 480V Line to line was given as 0 degrees....V phase lags or leads the line voltage by 30 degrees depending on the sequence....so the voltage could be 277<-30 or 277<+30....i don't have my references with me...but i'm sure the 30 degrees needs to be added to the equation, that's something that needs to be checked. A problem that NCEES added the 30 degree difference is in one of the NCEES practice problems I don't remember the problem #. I took all my references to my dads home and leave it there for good i didn't want to touch a book in a loonggg timee. But if some one have their references please clarify. I don't remember if it was the voltage or the current that the 30 degrees needs to be added.

### #7

Posted 04 January 2012 - 07:45 AM

Thank You for the feedback.

The solution has been uploaded for the QFTW. Both Question and Solution will be uploaded at the same time.

The question and solution from the prior week will be made available.

Let me know what other changes you would like.

http://www.spinupexams.com/

Joan

Spin-up, regarding the answer for that question....since the 480V Line to line was given as 0 degrees....V phase lags or leads the line voltage by 30 degrees depending on the sequence....so the voltage could be 277<-30 or 277<+30....i don't have my references with me...but i'm sure the 30 degrees needs to be added to the equation, that's something that needs to be checked. A problem that NCEES added the 30 degree difference is in one of the NCEES practice problems I don't remember the problem #. I took all my references to my dads home and leave it there for good i didn't want to touch a book in a loonggg timee. But if some one have their references please clarify. I don't remember if it was the voltage or the current that the 30 degrees needs to be added.

There is either a +/- 30 deg added when converting from Line-Line to Line-Neutral. if I remember correctly in a balanced a-b-c sequence there is a -30 deg going from line-line to line-neutral. For a a-c-b sequence we add +30 deg going from line-line to line-neutral. In Spin-Up's problem the angle is not given for the voltage nor indicated if it is a positive or negative sequence. But that is not needed since PF is relative between the voltage angle and current angle. Since the problem states the load has a 0.75 lagging PF we know that the angle is 41.4 deg. Since current lags voltage, we get the minus sign. Looking at spin-ups solution, it does not show the angle for voltage, but I do not think it is needed since we already know the angle between current and voltage. If we knew the current instead, but not the power factor, then knowing the voltage angle is critical because we need it to find the delta between the two angles to solve for pf. In my opinion, the answer spin-up has is correct. What does others think.

It is amazing that I know this stuff better now two months after finishing studying. Not sure if I needed time for the info to sink in or I have more confidence now after knowing that I passed the exam.

In the NCEES problem, i remember that we needed to use the angle because we converted from a delta configuration to like a virtual neutal to simplify the problem and then converting back. i think it was problem NCEES 111.

### #8

Posted 04 January 2012 - 06:23 PM

Thank You for the feedback.

The solution has been uploaded for the QFTW. Both Question and Solution will be uploaded at the same time.

The question and solution from the prior week will be made available.

Let me know what other changes you would like.

http://www.spinupexams.com/

Joan

Spin-up, regarding the answer for that question....since the 480V Line to line was given as 0 degrees....V phase lags or leads the line voltage by 30 degrees depending on the sequence....so the voltage could be 277<-30 or 277<+30....i don't have my references with me...but i'm sure the 30 degrees needs to be added to the equation, that's something that needs to be checked. A problem that NCEES added the 30 degree difference is in one of the NCEES practice problems I don't remember the problem #. I took all my references to my dads home and leave it there for good i didn't want to touch a book in a loonggg timee. But if some one have their references please clarify. I don't remember if it was the voltage or the current that the 30 degrees needs to be added.

There is either a +/- 30 deg added when converting from Line-Line to Line-Neutral. if I remember correctly in a balanced a-b-c sequence there is a -30 deg going from line-line to line-neutral. For a a-c-b sequence we add +30 deg going from line-line to line-neutral. In Spin-Up's problem the angle is not given for the voltage nor indicated if it is a positive or negative sequence. But that is not needed since PF is relative between the voltage angle and current angle. Since the problem states the load has a 0.75 lagging PF we know that the angle is 41.4 deg. Since current lags voltage, we get the minus sign. Looking at spin-ups solution, it does not show the angle for voltage, but I do not think it is needed since we already know the angle between current and voltage. If we knew the current instead, but not the power factor, then knowing the voltage angle is critical because we need it to find the delta between the two angles to solve for pf. In my opinion, the answer spin-up has is correct. What does others think.

It is amazing that I know this stuff better now two months after finishing studying. Not sure if I needed time for the info to sink in or I have more confidence now after knowing that I passed the exam.

In the NCEES problem, i remember that we needed to use the angle because we converted from a delta configuration to like a virtual neutal to simplify the problem and then converting back. i think it was problem NCEES 111.

I disagree (with EEVA I think) , the answer provided is correct ONLY IF we assume V

_{an}angle is 0. It would also be a reasonable assumption to assume that V

_{AB}is at an angle of 0 and then the solution given is incorrect. If you are giving a solution with an angle it has to be respect to something. A simple addition of "Assume positive phase sequence and angle of V

_{an}= 0 would straighten it up.

### #9

Posted 04 January 2012 - 09:46 PM

### #10

Posted 04 January 2012 - 10:47 PM

I believe DK PE is on the right track. Line-line voltages lag line-neutral voltages by 30 degrees, this is true.

Think you need to be more specific... the line to line V

_{AB}LEADS V

_{an}by 30 degrees, not lags, correct? I assume positive phase sequence.

The current lags that reference by 41.4 degrees.

You are correct, the problem would be correct if written "what is the current?" and added "(assume the voltage across the load is the reference at 0°)".

### #11

Posted 05 January 2012 - 12:26 AM

As to whether the question is written wrong... well, that's hard to say. It doesn't specifically say that Va is at 0 degrees, but I've always understood that to be a common assumption.

### #12

Posted 05 January 2012 - 01:48 AM

A little vagueness is sometimes a good thing, it stirs up discussion and creates a learning environment. Sometimes vagueness is not good.

We appreciate the feedback.

Joan

**Edited by spinup, 05 January 2012 - 01:54 AM.**

### #13

Posted 08 January 2012 - 07:37 PM

http://www.spinupexams.com/

Joan

### #14

Posted 15 January 2012 - 01:50 AM

Starting current is equal to locked rotor current only for DOL (direct on-line) stating method. In other case, in starting lower voltage is applied, so starting current is lower than locked rotor current.

Thanks,

### #15

Posted 15 January 2012 - 02:48 PM

About this week NEC - Design B motor question:(determination motor HP)

Starting current is equal to locked rotor current only for DOL (direct on-line) stating method. In other case, in starting lower voltage is applied, so starting current is lower than locked rotor current.

Thanks,

Insaf, Thanks for your feedback. I will forward your comment on.

Joan

### #16

Posted 15 January 2012 - 02:52 PM

http://www.spinupexams.com/

Joan

### #17

Posted 18 January 2012 - 05:21 PM

### #18

Posted 18 January 2012 - 06:18 PM

Maybe take another peek at delta/wye transformation relationship when the leg impedances are not equal...Should the answer to the QFTW be found using Z-WYE = Z-DELTA / 3 for each leg?

### #19

Posted 18 January 2012 - 07:21 PM

Should the answer to the QFTW be found using Z-WYE = Z-DELTA / 3 for each leg?

The formula applies for balanced load only! The question has an unbalanced load.

### #20

Posted 20 January 2012 - 12:36 AM

### #21

Posted 22 January 2012 - 02:47 PM

http://www.spinupexams.com/

Many people make the mistake of skipping this week's topic during their studies. These are easy points if this topic is studied a little. Don't leave these points on the table. It can make the difference between passing and needing to retake it.

Spin-Up's 5 Sample Exams Power PE book is available at the introductory price of $99.99

http://www.spinupexa...php?pr=Products

Joan

### #22

Posted 22 January 2012 - 05:17 PM

### #23

Posted 29 January 2012 - 02:41 PM

The topic this week is real power.

http://www.spinupexams.com/

Joan

### #24

Posted 06 March 2012 - 09:06 PM

A question on Week 2 problem:

The 3-phase motor is fed from the 3-phase 4600V (phase-to-phase) generator via a 10:1 delta-wye transformer.

On primary side, the phase-to-phase voltage is 4600V --> on secondary side, the phase-to-neutral voltage is 460V.

Should the motor voltage equal to the phase-to-phase voltage of the secondary side of the transformer, which is 460V*sqrt(3) = 797V?

### #25

Posted 06 March 2012 - 10:24 PM

Spin-up,

A question on Week 2 problem:

The 3-phase motor is fed from the 3-phase 4600V (phase-to-phase) generator via a 10:1 delta-wye transformer.

On primary side, the phase-to-phase voltage is 4600V --> on secondary side, the phase-to-neutral voltage is 460V.

Should the motor voltage equal to the phase-to-phase voltage of the secondary side of the transformer, which is 460V*sqrt(3) = 797V?

The three phase motor voltage is 460 V. It is correct as shown in the solution.

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