NCEES Question #536 - Book's answer is correct

I am posting this as a correction to what I said before, and also for anyone who who needs the proof using the full expression:

Question:

Feeder supplying a balanced load. For a fixed current of 50A, which of the following statements concerning the phase-to neutral voltage drop at the receiving end of the feeder is most correct?

(A) Voltage drop is relatively independent of the power factor.

( Voltage drop will be largest for a unity load power factor.

© Voltage drop will be largest for a zero load power factor

(D) Voltage drop will be larger for a lagging load power factor of 0.707 than either a unity load power factor or a zero load power factor.

The answer given in the book is (D), and proves this using the approximate equation: R*cosphi + X* sin phi

The complete expression for the voltage drop assuming a LAGGING load pf phi is,

Phasor (Vs) - Phasor (Vr) = |I| {(R*cos phi + X* sin phi)+j (X*cos phi - R* sin phi)}

R = X = 0.05 and this simplifies the equation to

Voltage drop Vd = 50*0.05 (cos phi + sin phi) + j (cos phi - sin phi)

For phi = 0, Vd = 2.5 (1+j), |Vd| = 2.5

For phi = 90, Vd = 2.5 (1-j), |Vd| = 2.5

For phi = 45, Vd = 2.5 (2*0.707), |Vd| = 3.535

You will get the same answers if you use the approximate equation too. I was not just comfortable with making the decison based on the approximate equation. That's all.

tobeeepe said:

I am posting this as a correction to what I said before, and also for anyone who who needs the proof using the full expression:
Question:

Feeder supplying a balanced load. For a fixed current of 50A, which of the following statements concerning the phase-to neutral voltage drop at the receiving end of the feeder is most correct?

(A) Voltage drop is relatively independent of the power factor.

( Voltage drop will be largest for a unity load power factor.

© Voltage drop will be largest for a zero load power factor

(D) Voltage drop will be larger for a lagging load power factor of 0.707 than either a unity load power factor or a zero load power factor.

The answer given in the book is (D), and proves this using the approximate equation: R*cosphi + X* sin phi

The complete expression for the voltage drop assuming a LAGGING load pf phi is,

Phasor (Vs) - Phasor (Vr) = |I| {(R*cos phi + X* sin phi)+j (X*cos phi - R* sin phi)}

R = X = 0.05 and this simplifies the equation to

Voltage drop Vd = 50*0.05 (cos phi + sin phi) + j (cos phi - sin phi)

For phi = 0, Vd = 2.5 (1+j), |Vd| = 2.5

For phi = 90, Vd = 2.5 (1-j), |Vd| = 2.5

For phi = 45, Vd = 2.5 (2*0.707), |Vd| = 3.535

You will get the same answers if you use the approximate equation too. I was not just comfortable with making the decison based on the approximate equation. That's all.

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Sorry guys. I did a mistake in my complex calculations. (Still getting used to my calculator!!!)

These are the correct answers and this seems to suggest that my earlier position was correct. that is the voltage drop is independent of the pf.

For phi = 0, Vd = 2.5 (1+j), |Vd| = 3.535

For phi = 90, Vd = 2.5 (1-j), |Vd| = 3.535

For phi = 45, Vd = 2.5 (2*0.707), |Vd| = 3.535

Sorry again for the confusion, but I am back to the questioning of book's answer to this question....

You need to supply the entire question. I believe you're approaching the problem wrong, which is why you're confused.

536 Consider a long 3-phase feeder with the following phase conductor properties:
Reactance XL=0.050 Ohms

AC Resistance R=0.050 Ohms

The feeder is supplying a balanced load. For a fixed current of 50A, which of the following statements concerning the phase-to-neutral voltage drop at the receiving end of the feeder is most correct?

( A ) Voltage drop is relatively independent of the power factor.

( B ) Voltage drop will be largest for a unity load power factor.

( C ) Voltage drop will be largest for a zero load power factor

( D ) Voltage drop will be larger for a lagging load power factor of 0.707 than either a unity load power factor or a zero load power factor.

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R=X is the cable resistance, with a PF =.707. The problems states that the load is 50A, and asks us to relate the voltage drop when load is 50A/PF=1, 50A/PF=0 and 50A/PF=.707. So, I=50A will never change- that's the peak current. The only thing that changes is the power factor. R and X are given; phi is phi of the load's power factor.
I like to use the formula from NEC for this since I'm used to it and it's easy; V=I(Xsin + Rcos). It's good enough for most power purposes, as the load values we use are also always approximations and never represent the true waveform.

unity power factor, PF=1, phi = 0, I=50A:

V=I(X*0 + R*1) = I * 1 * R = 2.5V

zero load power factor, PF=0, phi = 90, I=50A

V= I(X*1 + R*0) = I * 1 * X = 2.5V

PF=.707 lagging, phi = 45, (R=X), I=50A

V=I(X*(.707) + R*(.707)) = I * 1.414 * R = 3.5V

Honestly, you don't even need to calculate it out if you understand what's happening- this cable has reactance equal to resistance- its maximum impedance occurs at 45 degrees. Therefore, if peak current flow ALSO occurs at 45 degrees, that's where you'll see the greatest voltage drop. The answer is (D).

Be careful- a LOT of the questions on the PE exam are like this, which is to say that they're almost trivial if you actually understand the material, or can alternately be brute-forced if you know your basics and have good references. Nearly every single question is extremely easy if you know how to solve it! If it seems too hard or takes you longer than 6 minutes, you're probably doing it the wrong way, or at the very least, the hard way. The PE exam is testing your mettle and skills as an engineer- and it's an excellent test for that!

grover said:

You need to supply the entire question. I believe you're approaching the problem wrong, which is why you're confused.R=X is the cable resistance, with a PF =.707. The problems states that the load is 50A, and asks us to relate the voltage drop when load is 50A/PF=1, 50A/PF=0 and 50A/PF=.707. So, I=50A will never change- that's the peak current. The only thing that changes is the power factor. R and X are given; phi is phi of the load's power factor.

I like to use the formula from NEC for this since I'm used to it and it's easy; V=I(Xsin + Rcos). It's good enough for most power purposes, as the load values we use are also always approximations and never represent the true waveform.

unity power factor, PF=1, phi = 0, I=50A:

V=I(X*0 + R*1) = I * 1 * R = 2.5V

zero load power factor, PF=0, phi = 90, I=50A

V= I(X*1 + R*0) = I * 1 * X = 2.5V

PF=.707 lagging, phi = 45, (R=X), I=50A

V=I(X*(.707) + R*(.707)) = I * 1.414 * R = 3.5V

Honestly, you don't even need to calculate it out if you understand what's happening- this cable has reactance equal to resistance- its maximum impedance occurs at 45 degrees. Therefore, if peak current flow ALSO occurs at 45 degrees, that's where you'll see the greatest voltage drop. The answer is (D).

Be careful- a LOT of the questions on the PE exam are like this, which is to say that they're almost trivial if you actually understand the material, or can alternately be brute-forced if you know your basics and have good references. Nearly every single question is extremely easy if you know how to solve it! If it seems too hard or takes you longer than 6 minutes, you're probably doing it the wrong way, or at the very least, the hard way. The PE exam is testing your mettle and skills as an engineer- and it's an excellent test for that!

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I agree that if you use the APROXIMATE formula, (D) is correct. But the QUESTION doesn't ask you to use the approximate formula. I am using the very basics, i.e. magniute doesn't change with the angle. And it took me less than 5 minutes to select (A), without doing any calculations. And later, when I found book's answer I was surprised and checked again using the EXACT formula and found again that (A) is the correct answer, not (D). Here it is again:

For phi = 0, Vd = 2.5 (1+j), |Vd| = 3.535

For phi = 90, Vd = 2.5 (1-j), |Vd| = 3.535

For phi = 45, Vd = 2.5 (2*0.707), |Vd| = 3.535

My point is that, book's solution is selecting the WRONG answer based on the ERROR in the APPROXIMATION.

Any approximaton comes with an error. An I think we engineers should be aware of the errors involved in approximations and should not make wrong decisions based on the ERROR in the APPORXIMATION.

I took the ECC module, so I'm not going to even attempt to enter the fray here. However, I am darn glad I never had to choose from a group of answers where one choice was "The voltage drop is RELATIVELY independent of the power factor."

For crying out loud, what kind of answer selection is that on a technical exam? What does relatively even mean? To me, everyone's concept of relatively is relative.

benbo said:

I took the ECC module, so I'm not going to even attempt to enter the fray here. However, I am darn glad I never had to choose from a group of answers where one choice was "The voltage drop is RELATIVELY independent of the power factor."
For crying out loud, what kind of answer selection is that on a technical exam? What does relatively even mean? To me, everyone's concept of relatively is relative.

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Well, the question is for power engineers and the term 'relative' was not that much of an issue for me. But I had no doubt that the other three were wrong since the voltage drop is independent of the power factor, for this question of 'constant current'.

I think you have your math wrong. Common sense says the answer is D, the approximate equation says the answer is D and NCEES says the answer is D. If your "exact" equation is giving a completely different answer, then I suspect there is an error in your approach- either in your math, or the way you set up the equations. The difference between "exact" and "approximate" should amount to no more than a rounding error- since we only need 2 significant digits of accuracy in voltage drop calculations, the approximate equation and "exact" equation should be in complete agreement.

The complete expression for the voltage drop assuming a LAGGING load pf phi is, Phasor (Vs) - Phasor (Vr) = |I| {(R*cos phi + X* sin phi)+j (X*cos phi - R* sin phi)}

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Where did you get that equation from? It looks wrong- there shouldn't be an imaginary component to voltage drop.

I'm going to use theta instead of phi because I can't recall the symbol codes for phi off the top of my head:

Breaking down your equation, are these the two phasors you're adding together?

Vs=IRcosø+jIXsinø

Vr=IXsinø+jIRcosø

I'm pretty sure both of those equations are wrong. I'll give derivation of the "exact" equation a shot:

Impendance phasor Z=0.05 + j0.05=0.0707<45.

Current phasor I=50<ø.

Voltage drop is Vd=IZ=(0.0707<45) * (50<ø)

ø=0, 90, 45:

Vd= (0.0707<45) * (50<0) = 3.5<45 = 2.5 (unity)

Vd= (0.0707<45) * (50<90) = 3.5<135 = 2.5 (zero)

Vd= (0.0707<45) * (50<45) = 3.5<90 = 3.5 (0.707 lagging)

grover said:

I think you have your math wrong. Common sense says the answer is D, the approximate equation says the answer is D and NCEES says the answer is D. If your "exact" equation is giving a completely different answer, then I suspect there is an error in your approach- either in your math, or the way you set up the equations. The difference between "exact" and "approximate" should amount to no more than a rounding error- since we only need 2 significant digits of accuracy in voltage drop calculations, the approximate equation and "exact" equation should be in complete agreement.Where did you get that equation from? Any why are you multiplying by the scaler of I and not the phasor representation of I?

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I proved the equation by myself. Here are the steps:

1. Draw the receiving end voltage vector Vr as the refernce (horizontal) like this: ->

2. Draw the lagging current vector I

3. The lagging pf angle phi is the angle between them. Both vectors start from the same point. Counter clockwise positive, so I is phi degrees lagging by Vr. (I is ponting roughly south east, since V is pointing east)

4. Then draw the vector IR starting from the end point of V, parallel to I (pointing roughly south east)

5. Then draw vector jIX starting from the end point of IR, but 90 degrees counter-clockwise. (jIX is pointing roughly north east)

6. Then draw the supply voltage vector Vs from the start point of Vr to end point of jIX.

7. The voltage drop vector Vd = Vs-Vr is the vector drawn from the END point of Vr to END point of Vs

8. Magnitude of the voltage drop is the length of this vector Vd.

9. Phasor Vd = Phasor (Vs) - Phasor (Vr) = |I| {(R*cos phi + X* sin phi)+j (X*cos phi - R* sin phi)}

10. Help in proving the above: Real part is the horizontal length of this vecor. Reactive part that stats with j is the vertical length of this vector.

11. Note: phi is the angle between Vr and I (answer to your question: Why are you multiplying by the scaler of I and not the phasor representation of I?)

12. Now apply the different pf angles and you get these answers:

For phi = 0, Vd = 2.5 (1+j), |Vd| = 3.535

For phi = 90, Vd = 2.5 (1-j), |Vd| = 3.535

For phi = 45, Vd = 2.5 (2*0.707), |Vd| = 3.535

Pl. let me know if this helps. I was in teaching before and I enjoy doing this!

tobeeepe said:

For phi = 0, Vd = 2.5 (1+j), |Vd| = 3.535For phi = 90, Vd = 2.5 (1-j), |Vd| = 3.535

For phi = 45, Vd = 2.5 (2*0.707), |Vd| = 3.535

Pl. let me know if this helps. I was in teaching before and I enjoy doing this!

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For voltage drop, I think you have to use only the real component, though. In fact, if you drop the imaginary component from your expressions, you get the right answer.

grover said:

For voltage drop, I think you have to use only the real component, though. In fact, if you drop the imaginary component from your expressions, you get the right answer.

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Why is the real portion called APPROXIMATION if that is the case?

Remember any approximation has an eror. In other words, when you 'drop' something from the exact equation, you don't get a 'right' answer. It is an approximate answer, which means it has an ERROR. You are trying to justify the solution in the book, but not trying to look at the problem correctly.

What do you do when you go to the exam (if you don't get the same problem of course)?

Anyway, I appreciate your effort in trying to solve the problem. Keep up your good work! I have seen that you are providing very good input to the group, for example, the CT question with the diagram etc. I think we had a good discussion on this topic. For me, learning is a life time process and I like to learn even a little bit at a time and trying to help others learn too, as you do.

Thanks again and we may end up discussing more problems like this in future!

tobeeepe said:

Why is the real portion called APPROXIMATION if that is the case?Remember any approximation has an eror. In other words, when you 'drop' something from the exact equation, you don't get a 'right' answer. It is an approximate answer, which means it has an ERROR. You are trying to justify the solution in the book, but not trying to look at the problem correctly.

What do you do when you go to the exam (if you don't get the same problem of course)?

Anyway, I appreciate your effort in trying to solve the problem. Keep up your good work! I have seen that you are providing very good input to the group, for example, the CT question with the diagram etc. I think we had a good discussion on this topic. For me, learning is a life time process and I like to learn even a little bit at a time and trying to help others learn too, as you do.

Thanks again and we may end up discussing more problems like this in future!

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I kept this discussion with some of my coleagues at work also and finally found the correct explaination of the problem and the book's answer:

What the question is referring to here is the DIFFERENCE in voltage MAGNITUDES at the sending and receiving end. There is NO exact equation for this. There is only an approximate equation. But the error of the approximate equation is ZERO when R=X and the pf = 0.707. This is a special case. So, book's answer D is correct.

The exact equation I derived is for the VECTOR difference, but not for the MAGNITUDE difference.

Grover, I think this is probably what you were trying to tell..... Thanks for your effort!

Hi guys,

I am a new member and this is my first post on this forum. I tied to post it as “New topic” but I got an error massage.

My question is Kaplan Sample exam book, NEC question #1.25, how the conduit size chosen is 2 inch? I have no issue with step 1 and step 2, I don’t understand step 3 and how they used table 4 to determine the conduit size. Can any body help me with an answer?

Thanks,

GAZOO said:

Hi guys,
I am a new member and this is my first post on this forum. I tied to post it as “New topic” but I got an error massage.

My question is Kaplan Sample exam book, NEC question #1.25, how the conduit size chosen is 2 inch? I have no issue with step 1 and step 2, I don’t understand step 3 and how they used table 4 to determine the conduit size. Can any body help me with an answer?

Thanks,

Click to expand...

I don't have Kaplan with me. Can you post the problem?

Good pick with Kaplan. You are on the right track.

A building with 120/208v, 3 phase, 4 wire wye, computed load = 645Amp, parallele 6 sets of #1 (THHN AL) per phase, What size IMC conduit is required for each run (of six)?

GAZOO said:

A building with 120/208v, 3 phase, 4 wire wye, computed load = 645Amp, parallele 6 sets of #1 (THHN AL) per phase, What size IMC conduit is required for each run (of six)?

Click to expand...

Have to go thru different tables in the NEC. As far as I can remember they are in Chapter 9. There is a table with the areas of the different conductors. I am not sure right now, will have to check my NEC, but on Chapter 9, look at tables 4 and 5. Look also at Table 8(not sure about this one). I will check my NEC later today but in the meantime check that. I think Table 5 will give you the areas in sq in of the conductors. Find the area for #1 AL THHN and take this value 6 times.

Then at Table 4(or 8...not sure now) look for the conduit size the will be big enough to have this cables.

If this is not enough help let me know. I will try to be more specific. Again, will check the NEC to see if I am not giving you bad info.

Good luck!!!!!

Luis said:

Have to go thru different tables in the NEC. As far as I can remember they are in Chapter 9. There is a table with the areas of the different conductors. I am not sure right now, will have to check my NEC, but on Chapter 9, look at tables 4 and 5. Look also at Table 8(not sure about this one). I will check my NEC later today but in the meantime check that. I think Table 5 will give you the areas in sq in of the conductors. Find the area for #1 AL THHN and take this value 6 times.
Then at Table 4(or 8...not sure now) look for the conduit size the will be big enough to have this cables.

If this is not enough help let me know. I will try to be more specific. Again, will check the NEC to see if I am not giving you bad info.

Good luck!!!!!

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Shouldn't you be using the Table C.4(A) at the end? What does the answer say?

tobeeepe said:

Shouldn't you be using the Table C.4(A) at the end? What does the answer say?

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Another attempt following up on Louis's help:

Table 5 says THHN #1 is 100.8 mm2, so 6x100.8 = 604.8 mm2. Table 4 for IMC over 2 wires (last column) next higher to 600 is 937 mm2. This is Trade size 2. Am I correct?

OK this what I am calculating:

from ch9 table 5 for #1 THHN AL the area = 0.1562 * 6= 0.9372.

now this is my question, I take this value to table4 under IMC which column I should be using?

should I use:

(Total area 100%) column which in this case the closest value is 0.959 and conduit size will be #1.

OR (Over 2 wires 40%) column?

The answer in the book is:

( NEC table 4, Chapter 9

9372 sqr in = 2" IMC) ?!

tobeeepe said:

Another attempt following up on Louis's help: Table 5 says THHN #1 is 100.8 mm2, so 6x100.8 = 604.8 mm2. Table 4 for IMC over 2 wires (last column) next higher to 600 is 937 mm2. This is Trade size 2. Am I correct?

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your answer is the book answer for the conduit size. but I think what I am confused about is if I use the last column should I used 40% of the total? in another word should the value be 604.8 * 40%? or I should use the total 604.8 to determine the conduit size?

GAZOO said:

your answer is the book answer for the conduit size. but I think what I am confused about is if I use the last column should I used 40% of the total? in another word should the value be 604.8 * 40%? or I should use the total 604.8 to determine the conduit size?

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No. 40% is for conduit area and not for current. You take the current as it is. Table has the 40% multiplication for area done for you. Column 4 from left in Table 4 has 100% which is 2341. And 2341*0.4=937 is what you read from the 40% column.

In fact, it is even easier when you used Table C.4 in Annex C. All you do is find THHN #1 row and go right looking for 6 conductors. You go past 5 then hit 9 which is the right size. It is Metric 53 and Trade size 2 (top of the column). You didn't have to do any current calculation.

But I am not an expert in this area. So let's see if the experts approve my approach.