Voltage Drop Calculations

[BebeshKing PE]

No, it only appears that way because with the smaller wires, the resistance has more of an impact. As the wire size gets larger, the effective Z becomes closer to XL.

It's because the angle of the load cannot be directly translated to the angle of the "circuit." The phase angle of the circuit is different from the power factor of the load.  The impedance of the conductors will change the power factor of the current supplying the load. 

So if the given is the circuit power factor, we need to use the approximate formula(per NEC)?

And if the given is the load power factor, we need to use the complex number formula ?

If I understand this problem correctly, the 0.75 pf is the load power factor and we need to use the complex number formula and the answer would be 1.13V?

 

[Orchid PE]

So in summary:

Effective Z will give a close approximation of voltage drop based on the current and power factor of the load.

The reason the above is different from doing V = (R + jX)*(I) is because the current and PF given are of the load in ideal conditions (or when it was tested by the mfg). As soon as we connect the load to the conductors, the actual current and PF will change. The only way to know the exact current magnitude and angle are to know the R + jX of the load.

When to use either method comes down to knowing what the question on the exam is asking, based on the information provided. We can't just give a blanket statement that says "every time you come across 'this type' of question use effective Z, and every time you come across 'that type' of problem use actual Z."

 

[Orchid PE]

Let's look at the problem you posted earlier, but I'm not concerned about the whole question. Just read how they specify the motor information.

"The motor full-load current is 550A at a power factor of 0.9 lagging" All this means is that the factory specifications are 550 A @ 0.9 lagging. This does not mean when we install it on the system, the motor will draw exactly that amount of current at that power factor. As soon as it is connected to the system, everything changes. Maybe not by much, but the R + jX of the conductors will affect the angle of current. Therefore, we are given note 2 to calculate a close approximation of voltage drop if we connected this motor to the specified conductors.

(In the problem below, we don't actually have to find effective Z, so just ignore everything aside from how the motor info is specified)

 

[Dude99]

imo

if they give conduit type and reference the NEC use their approximation method

if they give Z in complex form use it

isn't system pf = cos(line ang - load ang)?

 

[Orchid PE]

if they give conduit type and reference the NEC use their approximation method

if they give Z in complex form use it

This isn't the best advice, because NCEES will put in red herrings.

Again, the best solution is read the question and know exactly what is being asked based on the provided information. Q512 is a good example of how more than enough information is provided in order to throw off those that don't completely understand. We saw this when it was asked why effective Z wasn't calculated for that problem.

 

[Dude99]

This isn't the best advice, because NCEES will put in red herrings.

Again, the best solution is read the question and know exactly what is being asked based on the provided information. Q512 is a good example of how more than enough information is provided in order to throw off those that don't completely understand. We saw this when it was asked why effective Z wasn't calculated for that problem.

Who's to say?I've passed PE EE (general, decades ago)  EE power and PE controls both within the last 10 yrs, and am taking the PE ME HVAC.

any of the methods will get you there adjusted for pf

 

[Orchid PE]

I gave the explanation, it's easy to understand.

 

[Dude99]

More than 1 way to skin a cat.  In my opinion the test is looking for minimum competency, comprehension/understanding of concepts, not memorization or math skills.  They are not trying to 'trick' anyone.

 

[Dude99]

I gave the explanation, it's easy to understand.

You also thought 0.046 was 4 awg cmil.

It's easy if they give you LOAD pf. But you don't need to go that far.

 

[Orchid PE]

They are not trying to 'trick' anyone.

Wrong. Many questions on the exam provide more information than is needed to actually solve the question. Those questions have red herrings to make sure individuals know exactly what is needed to solve the problem. This is a known fact.

Again, the best approach is always treat it on a problem by problem basis and to know exactly what the question is asking. 

 

[Dude99]

Wrong. Many questions on the exam provide more information than is needed to actually solve the question. Those questions have red herrings to make sure individuals know exactly what is needed to solve the problem. This is a known fact.

Again, the best approach is always treat it on a problem by problem basis and to know exactly what the question is asking. 

Wrong...again

that is not a trick, that is seeing if you can separate the wheat from the chaff.  Most real world problems require gather all information and eliminating the 'noise'.  Obviously it is important to know what the 'question is asking'. lol The 'best' approach varies by individual and is not the exclusive domain of one.

 

[ME->EE]

7 hours ago, Chattaneer PE said:

Thanks for the video and spending the time responding to all the questions @Chattaneer PE, It's still tough for me to wrap my head around but it seems like the vast majority of questions I've come across are looking for the actual voltage drop so the Note 2 equation isn't used.  I'll try to make more sense of this when I get some time.