Chapter 9 T9 Note 2

[anthonysolino]

Can any PE's chime in on how to apply the formula in relation to voltage drop for a different power factor, lets say a motor with a 30%PF I am attempting to do a calculation on a fire pump motor and I understand the K value should be replaced with the the correct power factor of the load being the table is based on .85% PF.

example

1.732x6.91(K-value PF%)X404A(locked rotor current)X150FT/Cmils

my question to you is how does the engineer come up with the K value based on the given power phase angle?

I was attempting to solve the R cos Theta+ X sin Theta
then it goes on to provide Z=R X PF+ X(L) sin[arccos (PF)]
essentially I am trying to solve Z=R X PF+ X(L) sin[arccos (PF)] I cant seem to understand what [arccos (PF)] values need to be based on the table.

 

[oldsparky52]

You are over my head, but I just can't help comment about you posting this stuff on a Sunday morning at 2 AM. You really love this stuff, huh?

Much better than trying to sleep off too much fun.

 

[anthonysolino]

lol I enjoy the challenge it brings ill never size the conductor with out an engineers stamp ,but I thought it would be cool to see if there is any PE's in the threads that might know!

 

[tortuga]

Well the 2017 handbook has a pretty good example. P1118:

A 270-A continuous load is present on a feeder. The circuit consists of a
single 4-in. PVC conduit with three 600-kcmil XHHW/USE aluminum
conductors fed from a 480-V, 3-phase, 3-wire source. The conductors are
operating at their maximum rated temperature of 75°C. If the power fac-
tor is 0.7 and the circuit length is 250 ft, is the voltage drop excessive?
Solution
Step 1. Using the Table 9 column “XL (Reactance) for All Wires,” select
PVC conduit and the row for size 600 kcmil. A value of 0.039 ohm per
1000 ft is given as this XL. Next, using the column “Alternating-Current
Resistance for Aluminum Wires,” select PVC conduit and the row for size
600 kcmil. A value of 0.036 ohm per 1000 ft is given as this R.
Step 2. Find the angle representing a power factor of 0.7. Using a cal-
culator with trigonometric functions or a trigonometric function
table, find the arccosine (cos –1) θ of 0.7, which is 45.57 degrees. For
this example, call this angle θ.

I'll add for clarity θ = .7953 = 45.57 degrees

Then it goes on:

Step 3. Find the impedance (Z) corrected to 0.7 power factor (Zc):
Zc = (R × cos θ) + (XL × sin θ)

So they are saying
cos .7953 = .7
sin .7953 = .7141

= (0.036 × 0.7) + (0.039 × 0.7141)
= 0.0252 + 0.0279
= 0.0531 ohm to neutral

Is that what your looking for?
Then the steps go on to show how to calculate voltage drop.

 

[ramsy]

Well the 2017 handbook has a pretty good example. P1118:

...The conductors are operating at their maximum rated temperature of 75°C. If the power factor is 0.7 and the circuit length is 250 ft ..
....
= (0.036 × 0.7) + (0.039 × 0.7141)
= 0.0252 + 0.0279
= 0.0531 ohm to neutral

Excellent example

0.0531 = Ohms per 1000 ft @ 75c

Adjusting for 250ft
0.0531 / 1000 * 250ft = 0.0133 Ohms @ 75c

Solve for actual Temperature if below 75c

0.036 (1+IF(EXACT(Wire,"Copper"),0.00323,0.0033)*(Tmp.Rise-75))
=0.0129 Ohms @ 59.5c

Table 9 values assume 75c
R2 adjust, NEC Tbl. 8, Note 2:
R2 = R1 [1 + α (T2-75)]
T2 = (Ambient + Temp.Rise)
T2 = T1 + (TR-T1) * Load^2 / Imax^2

Tmp.Rise =Ambient+(90-Ambient)*270^2/(CHOOSE(Table,310.16 #600@75c)*VLOOKUP(Table 310-15(b)(2)(a),3))^2

Where: Load = Circuit Amps
T1 = Ambient Temperature
T2 = Estimated Conductor Temperature
TR = Conductor Max Temperature Rating
Imax = Max Amps at TR, derated for ccc's
In conduit with > 3 ccc's. NEC table values must be adjusted

 

[anthonysolino]

Well the 2017 handbook has a pretty good example. P1118:


I'll add for clarity θ = .7953 = 45.57 degrees

Then it goes on:


So they are saying
cos .7953 = .7
sin .7953 = .7141


Is that what your looking for?
Then the steps go on to show how to calculate voltage drop.

Clarify:
(This is the formula that appears at the end of note 2)
Ze=Effectieve impedance
R= Resistant of uncoated wires
PF=Desired power factor
xL= Reactance
what I am trying to find is what values are designed to be inserted to= [arccos(PF)]

And also for the first formula in note 2
are cos and sin constants? and if so what do they represent? phase angles?

I wish I had the handbook seems like theres a great example in there

 

[anthonysolino]

Excellent example

0.0531 = Ohms per 1000 ft @ 75c

Adjusting for 250ft
0.0531 / 1000 * 250ft = 0.0133 Ohms @ 75c

Solve for actual Temperature if below 75c

0.036 (1+IF(EXACT(Wire,"Copper"),0.00323,0.0033)*(Tmp.Rise-75))
=0.0129 Ohms @ 59.5c

Table 9 values assume 75c
R2 adjust, NEC Tbl. 8, Note 2:
R2 = R1 [1 + α (T2-75)]
T2 = (Ambient + Temp.Rise)
T2 = T1 + (TR-T1) * Load^2 / Imax^2

Tmp.Rise =Ambient+(90-Ambient)*270^2/(CHOOSE(Table,310.16 #600@75c)*VLOOKUP(Table 310-15(b)(2)(a),3))^2

Where: Load = Circuit Amps
T1 = Ambient Temperature
T2 = Estimated Conductor Temperature
TR = Conductor Max Temperature Rating
Imax = Max Amps at TR, derated for ccc's
In conduit with > 3 ccc's. NEC table values must be adjusted

this one I figured out I believe your response was related to note 2 on table 8 for DC my question was for table 9

 

[ramsy]

..what I am trying to find is what values are designed to be inserted to= [arccos(PF)]

PF is the only variable that can be inserted.
If load Power Factor is unknown, I use the table default 0.85

 

[tortuga]

Clarify:
(This is the formula that appears at the end of note 2)
Ze=Effectieve impedance
R= Resistant of uncoated wires
PF=Desired power factor
xL= Reactance
what I am trying to find is what values are designed to be inserted to= [arccos(PF)]

Ze=Effective impedance
R= Resistant of uncoated wires
PF=Power Factor
xL= Reactance
We need to create a angle theta from PF:
θ = arccos(PF)

Ze = R × PF + XL sin[arccos(PF)]
Ze = (R × cos θ) + (XL × sin θ)

And also for the first formula in note 2
are cos and sin constants? and if so what do they represent? phase angles?

I wish I had the handbook seems like theres a great example in there

They are all relationships to power factor.

 

[synchro]

And also for the first formula in note 2
are cos and sin constants? and if so what do they represent? phase angles?

See figure 17 on page 30 of the following document. Fyi, they use Φ instead of θ and I'll use their nomenclature where I can.
The current "I" through a load with P.F. = cosΦ is at an angle Φ from the voltage at the load (receiver voltage).
I*Z is the voltage developed between the source (sending end) and the load (receiving end) of a conductor with a complex impedance Z = R + jX. So I*R is the voltage across the resistance of the conductor, and I*X is the voltage across the reactance of the conductor.

In answer to your question, I*R cosΦ represents the component of the voltage developed across the conductor resistance R that is aligned (I.e., in-phase) with the load (receiver) voltage. I*X sinΦ represents the component of the voltage developed across the conductor reactance X that is aligned with the load voltage.

At the bottom right of figure 17 you can see that these two components are added together to come up with an "estimated" voltage drop I*R cosΦ + I*X sinΦ which is the expression that you were referring to. This simplified expression is more than accurate enough for most purposes. See the equations on page 35 and 36 if you need even more accuracy. This document also provides a number of examples and other info.

https://pdhonline.com/courses/e426/e426content.pdf

 

[anthonysolino]

PF is the only variable that can be inserted.
If load Power Factor is unknown, I use the table default 0.85

how ever I am trying to do the calculation with a power factor of .30. what does "Arccos" mean in relation to mathematics?

See figure 17 on page 30 of the following document. Fyi, they use Φ instead of θ and I'll use their nomenclature where I can.
The current "I" through a load with P.F. = cosΦ is at an angle Φ from the voltage at the load (receiver voltage).
I*Z is the voltage developed between the source (sending end) and the load (receiving end) of a conductor with a complex impedance Z = R + jX. So I*R is the voltage across the resistance of the conductor, and I*X is the voltage across the reactance of the conductor.

In answer to your question, I*R cosΦ represents the component of the voltage developed across the conductor resistance R that is aligned (I.e., in-phase) with the load (receiver) voltage. I*X sinΦ represents the component of the voltage developed across the conductor reactance X that is aligned with the load voltage.

At the bottom right of figure 17 you can see that these two components are added together to come up with an "estimated" voltage drop I*R cosΦ + I*X sinΦ which is the expression that you were referring to. This simplified expression is more than accurate enough for most purposes. See the equations on page 35 and 36 if you need even more accuracy. This document also provides a number of examples and other info.

https://pdhonline.com/courses/e426/e426content.pdf

this probably the best article ive seen about the topic I will be reading it tonight it looks pretty detailed and explains it in detail thank you sir for the input ill attempt to solve the problem, ill let you know how I do!

 

[steve66]

sin, cos, arcsin, and arcsin are normally found using the corresponding buttons on a calculator. They are all based on right triangles, and how the lengths of sides change as the angles change. arcsin is also sometimes noted sin-1 (I cant do a superscript here - but the -1 is normally printed above sin). Same with arccos and cos-1.

The power factor is basically designed as cos theta, which means that if you know the power factor, then theta = cos-1 PF.

So for a PF of 0.85, theta = 31.78 degrees (found by cos -1 of .85). And a PF of 0.3 would have a theta of 72.5. You have to make sure you have your calculator set for degrees, and not radians (I hope mine is set right.) And on many calculators, you enter the number first, and then hit cos-1.

So if the PF= 0.85, and theta = 31.78, you would find sin(arccos(theta) by taking the sin of 31.78 degrees = 0.527.

Notice that .85 squared + .527 squared = 1.000. Taking the sine and the cosine of the same angle, squaring and adding the results will always equal 1. If I remember correctly (and that's a big if), that's because sin and cos are complementary functions. As one gets smaller, the other gets larger, and vice versa.

 

[PaulMmn]

sin, cos, arcsin, and arcsin are normally found using the corresponding buttons on a calculator. They are all based on right triangles, and how the lengths of sides change as the angles change. arcsin is also sometimes noted sin-1 (I cant do a superscript here - but the -1 is normally printed above sin). Same with arccos and cos-1.

The power factor is basically designed as cos theta, which means that if you know the power factor, then theta = cos-1 PF.

So for a PF of 0.85, theta = 31.78 degrees (found by cos -1 of .85). And a PF of 0.3 would have a theta of 72.5. You have to make sure you have your calculator set for degrees, and not radians (I hope mine is set right.) And on many calculators, you enter the number first, and then hit cos-1.

So if the PF= 0.85, and theta = 31.78, you would find sin(arccos(theta) by taking the sin of 31.78 degrees = 0.527.

Notice that .85 squared + .527 squared = 1.000. Taking the sine and the cosine of the same angle, squaring and adding the results will always equal 1. If I remember correctly (and that's a big if), that's because sin and cos are complementary functions. As one gets smaller, the other gets larger, and vice versa.

I've always understood the 'arc' functions as the non-arc functions in reverse:
"arcsin - the angle whose sin is .85"
"arccos - the angle whose cos is xxx"

 

[anthonysolino]

sin, cos, arcsin, and arcsin are normally found using the corresponding buttons on a calculator. They are all based on right triangles, and how the lengths of sides change as the angles change. arcsin is also sometimes noted sin-1 (I cant do a superscript here - but the -1 is normally printed above sin). Same with arccos and cos-1.

The power factor is basically designed as cos theta, which means that if you know the power factor, then theta = cos-1 PF.

So for a PF of 0.85, theta = 31.78 degrees (found by cos -1 of .85). And a PF of 0.3 would have a theta of 72.5. You have to make sure you have your calculator set for degrees, and not radians (I hope mine is set right.) And on many calculators, you enter the number first, and then hit cos-1.

So if the PF= 0.85, and theta = 31.78, you would find sin(arccos(theta) by taking the sin of 31.78 degrees = 0.527.

Notice that .85 squared + .527 squared = 1.000. Taking the sine and the cosine of the same angle, squaring and adding the results will always equal 1. If I remember correctly (and that's a big if), that's because sin and cos are complementary functions. As one gets smaller, the other gets larger, and vice versa.

this is actually really interesting, I enjoy the engineering aspect of the trade I wish I understood it more in greater detail as to phase angles and how that all is relative to every thing I think its awesome stuff
so when I hit cos1 on my calculator with a power factor of .30 I get 72.5 SO after this to find the sin(arccostheta I then take the sin1 of 72.5?

 

[anthonysolino]

this is actually really interesting, I enjoy the engineering aspect of the trade I wish I understood it more in greater detail as to phase angles and how that all is relative to every thing I think its awesome stuff
so when I hit cos1 on my calculator with a power factor of .30 I get 72.5 SO after this to find the sin(arccostheta I then take the sin1 of 72.5?

lol let me talk this through as to what I am calculating

First of I am selecting number three copper from the table correct me if I am wrong please,
(Lets say to make it easy the distance is 1000 feet)

R= 0.25
PF= .30 (In my example)
xL=0.047 (For PVC ohms to neutral per 1000 feet)
to solve for Sin[arccos(PF)
1Cos(.30) = .72
1Sin(.72) = 46.05
there for: sin[arccos(PF) = sin[arccos(46.05)??

then would the formula look something like this?
0.25 X .30 + 0.047 X 46.05?

+ = ADDITION
X=BEING MULTIPLICATION

 

[tortuga]

Nice explanations; (The handbook used θ instead of Φ so thats why I use it. )
Your close, I figure the angle theta representing a power factor first:


Ze = R × PF + XL sin[arccos(PF)]
As the others said arccosine and cosine cancel eachother out
so the formula in the note omits that step.
I really think of it as
Where:
θ = arccosine (PF)
Ze = (R × cos θ) + (XL × sin θ)


accos( .3) = 1.266
cos(1.266) = .3

Ze = (.25 X .3) + (0.047X X sin(1.266))
Ze = .075 + (0.047X.9539)
Ze = .075 + .0448

 

[charlie b]

It seems to me that you did not have the pleasure of taking trigonometry in high school. Or if you did take it, you did not find the experience pleasurable. Let me offer a tidbit or two.

Take pencil and paper, and draw a horizontal line, perhaps 2 or 3 inches long. Label the left hand point “A” and the right hand point “B.” Starting at point B, draw a vertical line a few inches long. Label the point at the top of that line “C.” Draw a line between points A and C. You now have a right triangle. Label the three angles as “a,” “b,” and “c,” with angle “a” being located at point “A,” and so forth.


The terms “sin,” which is shorthand for “sine,” and “cos,” which is shorthand for “cosine,” and the others (i.e., tangent, arcsine, etc.) are never used in isolation. We never say just “sine.” We always say “the sine of that particular angle,” using the shorthand notation of “sin (a).” The various trig functions are defined, using the figure I asked you to draw, and looking at angle “a,” as follows:

Sin(a) = length of side BC divided by length of side AC.
Cos(a) = length of side AB divided by length of side AC.
Tan(a) = length of side BC divided by length of side AB.

If you had drawn the figure such that angle “a” was exactly 30 degrees, then the sine of that angle, or “sin(30),” would be 0.5. And it would not matter how long you had drawn each of the three lines that formed the triangle. As long as you managed to make angle “a” 30 degrees, then the ratio of side BC to side AC is going to be 1:2.

Now if you told me that you knew that the sine of an as-yet unknown angle was 0.5, and wanted to know what that angle would be, you would find it by taking the arcsine of the 0.5. That is, arcsin(0.5) is 30 degrees.

If you are using a calculator, you need to be aware of whether it is expecting angles to be expressed in units of “degrees” or “radians.” There are 360 degrees in a circle. There are about 6.28 radians in a circle (specifically 2 times the value of Pi, which is about 3.14). So that 30 degree angle I mentioned would equate to about .523 radians.

I’ll end this discussion here, and return to your original question in my next post.

 

[charlie b]

sin, cos, arcsin, and arcsin are normally found using the corresponding buttons on a calculator.

I would call that the second thing you do. You need to start by knowing what they mean, before you use a calculator to find any numbers.

 

[charlie b]

You lost me here:

Can any PE's chime in on how to apply the formula in relation to voltage drop for a different power factor. . . I understand the K value should be replaced with the the correct power factor of the load being the table is based on .85% PF.

What has led you to believe that the K value changes with power factor? It does not. There are different values for copper versus aluminum wires. The K factor is based on the properties of the materials and varies with the assumed ambient temperature.

The influence that power factor has on voltage drop comes into play with the value of current you use in the formula. If you started by only knowing the "real power" value, the "kW" value, then you are not ready to calculate VD. You need to also know the PF, in order to calculate the "apparent power," or "kVA." From that value, and also knowing the voltage and whether this is a single phase or three phase circuit, you calculate the current. Then and only then do you apply the formula you cited in your original post.

 

[PaulMmn]

It seems to me that you did not have the pleasure of taking trigonometry in high school. Or if you did take it, you did not find the experience pleasurable.

I have some tidbits as well...

First, a diagram showing the relationships between the lengths of the sides of a right triangle and the sin / cos / tan functions:

Next, an animation showing sin - cos in relation to a circle...


Finally, a diagram showing the relation between a helix (a circle moving through time) and the sin / cos functions: