# Voltage Drop Calculations



## ME->EE (Mar 12, 2020)

When calculating the effective impedance when we would we use Chapter 9 Table 9 Note 2: Ze = R x PF + XLsin[arccos(PF)] rather than just using the R and XL values from the table in the following formula: Z = (distance) * (R+jXL)

Most of my practice problems don't use the Note 2 equation at all so I'm not seeing when I should use that.

Thanks!


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## Orchid PE (Mar 13, 2020)

Let me know if either of these help.


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## Orchid PE (Mar 13, 2020)

@ME-&gt;EE Also this:


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## ME->EE (Mar 13, 2020)

Thanks @Chattaneer PE, so if I'm understanding this correctly you never use the Note 2 equation for finding conductor impedance but you will use the Note 2 equation when you need to find the voltage drop and you are only given the circuit PF (not voltage or current PF separately)?  Otherwise when finding the voltage drop you only use Z = (distance) * (R+jXL)?


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## Orchid PE (Mar 13, 2020)

ME-&gt;EE said:


> Thanks @Chattaneer PE, so if I'm understanding this correctly you never use the Note 2 equation for finding conductor impedance but you will use the Note 2 equation when you need to find the voltage drop and you are only given the circuit PF (not voltage or current PF separately)?  Otherwise when finding the voltage drop you only use Z = (distance) * (R+jXL)?


Essentially.

If you know the current drawn by a load, and the load properties, you can calculate voltage drop.

For example, if a 3-ph 10kVA load is draws 20 amps @ 0.9 PF, what is the voltage drop across an uncoated copper 4 AWG wire that is 500ft long in PVC conduit?

R + X for 4 AWG per 1000 ft: 0.31 + 0.048

For 500ft: 0.155 + 0.024

Effective Z @ 0.9: 0.155*PF + 0.024*sin(acos(PF)) = 0.13175 + 0.012643 = *0.148\4*

Voltage drop: 20 * 0.144 = 2.89

----

This is useful when calculating voltage drop for building electrical systems, because industrial equipment will generally give you the amps and PF in the specification sheets.


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## ME->EE (Mar 13, 2020)

Wouldn't X = 0.048 since it's in a PVC conduit?  I think the 0.06 comes from using a steel conduit.

R + X for 4 AWG per 1000 ft: 0.31 + 0.048

For 500ft: 0.155 + 0.024

theta = arccos(0.9) = 25.84 degrees

V = (20A)( @ 25.84 degrees)*(0.155 + 0.024)

V = 3.58 (@ 25.84 degrees)


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## Orchid PE (Mar 13, 2020)

I could be wrong, but this is what I'm seeing:







But this is just a table I found online since I don't have the NEC in front of me. But regardless of the actual values, the principle remains the same.


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## Dude99 (Mar 13, 2020)

Z = 0.9 x 0.155 x sin(acos0.9) x 0.024 = 0.150

0.150 x 20 = 3 V

4 awg ~42,000 cmil


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## BebeshKing PE (Mar 13, 2020)

Chattaneer PE said:


> Essentially.
> 
> If you know the current drawn by a load, and the load properties, you can calculate voltage drop.
> 
> ...


@Chattaneer PE,  what situation we can use the voltage drop equation in complex number?   Vdrop =(R+jXL)(I&lt;theta) ?


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## Orchid PE (Mar 13, 2020)

BebeshKing said:


> @Chattaneer PE,  what situation we can use the voltage drop equation in complex number?   Vdrop =(R+jXL)(I&lt;theta) ?


Whenever you're given the values to do so. _Actual_ voltage drop is defined exactly as you stated: Vdrop = (R + jX) * (I&lt;theta). If you are given a current with an angle, go for it.

This effective Z method is used because most industrial equipment will generally give you the amps and PF in the specification. An angle is _generally _not given in equipment specifications, just the power factor. The effective Z method just gives an approximation of the voltage drop.

For example, this is from a datasheet:




Notice they give full load current and power factor. They do not give an angle for current. The effective Z method would be used to approximate Vdrop.


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## Dude99 (Mar 13, 2020)

imho the NEC note is confusing

they give you R and X hence the line ph ang = atan x/r

assume 500' #4 R = 0.155 and X = 0.024

Z = R x cos(atan x/r) + j X sin(atan  /r) = 0.153/8.8 deg

using the NEC method of 0.9 x R + X x sin(acos 0.9) = 0.150

moot on 100 A load: 15.3 vs 15.0 drop


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## BebeshKing PE (Mar 13, 2020)

Chattaneer PE said:


> Whenever you're given the values to do so. _Actual_ voltage drop is defined exactly as you stated: Vdrop = (R + jX) * (I&lt;theta). If you are given a current with an angle, go for it.
> 
> This effective Z method is used because most industrial equipment will generally give you the amps and PF in the specification. An angle is _generally _not given in equipment specifications, just the power factor. The effective Z method just gives an approximation of the voltage drop.


@Chattaneer PE, I asked this because I encountered a problem that I used both formulas (complex number vs approximation formula) to compare each other but the answer is way far from each other. Here's the problem:

An 90 foot circuit provides power to a single phase load connected to a 120V panel. The conductor size is 1AWG uncoated copper ran in PVC conduit. Calculate the voltage drop in volts if the amps drawn by the load is 40 with lagging power factor at 0.75.

Here's what I did:

*Using complex number:*

Vdrop= 2(R+jXL)(I&lt;theta) = 2(0.15+j0.046)(90/1000)(40&lt;-41.4)=*1.13*&lt;-24.4 volts

*Using approximate formula:*

Vdrop=2(Rcos(theta)+XL(sin(theta))(I)=2(0.15(0.75)+0.046(0.66))(90/1000)(40)=*1.03* volts

 what's your thought about this?


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## Dude99 (Mar 13, 2020)

The way I do it

X = 0.024, R = 0.155, Z Line phase = 8.8 deg, old pf = 0.85 31.79 deg, new 0.9 25.84 deg

Z at 0.85 = 0.145 from table at 0.85 pf

Z at 0.9 = Z x cos(Z line - Z load old/(cosZ line - Z load new)

plugging Z = 0.1506 

any method will get you 'close enough)

if a motor load ph ang = acos(motor pf)


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## Dude99 (Mar 13, 2020)

The real part of 1.13/-24.4 = 1.03

exactly 1.0291, same as the NEC 1.0291


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## BebeshKing PE (Mar 13, 2020)

Dude99 said:


> The real part of 1.13/-24.4 = 1.03
> 
> exactly 1.0291, same as the NEC 1.0291


@Dude99, So you were saying if we encounter this type of problem and we use the complex number formula, the answer would be the Real part of the Volt drop?


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## Dude99 (Mar 13, 2020)

BebeshKing said:


> @Chattaneer PE, I asked this because I encountered a problem that I used both formulas (complex number vs approximation formula) to compare each other but the answer is way far from each other. Here's the problem:
> 
> An 90 foot circuit provides power to a single phase load connected to a 120V panel. The conductor size is 1AWG uncoated copper ran in PVC conduit. Calculate the voltage drop in volts if the amps drawn by the load is 40 with lagging power factor at 0.75.
> 
> ...


What were the answer choices?


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## BebeshKing PE (Mar 13, 2020)

Dude99 said:


> What were the answer choices?


This was not a multiple choice question, though. In any way, I just want to know the concept for this.

Also,    In the NCEES #129 solution , they used the complex number formula but did not consider the real part of the voltage drop to get the voltage at the load side. 

Little confuse.


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## Orchid PE (Mar 13, 2020)

BebeshKing said:


> @Chattaneer PE, I asked this because I encountered a problem that I used both formulas (complex number vs approximation formula) to compare each other but the answer is way far from each other. Here's the problem:
> 
> An 90 foot circuit provides power to a single phase load connected to a 120V panel. The conductor size is 1AWG uncoated copper ran in PVC conduit. Calculate the voltage drop in volts if the amps drawn by the load is 40 with lagging power factor at 0.75.
> 
> ...


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.


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## Orchid PE (Mar 13, 2020)




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## Orchid PE (Mar 13, 2020)

BebeshKing said:


> So you were saying if we encounter this type of problem and we use the complex number formula, the answer would be the Real part of the Volt drop?


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.


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## BebeshKing PE (Mar 13, 2020)

Chattaneer PE said:


> 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.






Chattaneer PE said:


> 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?


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## Orchid PE (Mar 13, 2020)

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."


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## Orchid PE (Mar 13, 2020)

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)


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## Dude99 (Mar 13, 2020)

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)?


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## Orchid PE (Mar 13, 2020)

Dude99 said:


> 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.


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## Dude99 (Mar 13, 2020)

Chattaneer PE said:


> 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


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## Orchid PE (Mar 13, 2020)

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


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## Dude99 (Mar 13, 2020)

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.


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## Dude99 (Mar 13, 2020)

Chattaneer PE said:


> 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.


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## Orchid PE (Mar 13, 2020)

Dude99 said:


> 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.


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## Dude99 (Mar 13, 2020)

Chattaneer PE said:


> 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.


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## ME->EE (Mar 13, 2020)

> 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.


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