Geometric Mean Distance of 4 Conduits

If there were 4 conduits (A,B,C,D) spaced as:

A <---- x ----> B<---- y ----> C <----- z ------> D

What is the Geometric Mean Distance of the 4 conduits?

Answer 1:

((x)(x+y)(x+y+z)(y+z)(z)) ^ (1/4)

or

Answer 2:

((x)(z)) ^(1/3)

I think the answer is the one described in answer #1, what does others think?

EEVA said:

If there were 4 conduits (A,B,C,D) spaced as:
A <---- x ----> B<---- y ----> C <----- z ------> D

What is the Geometric Mean Distance of the 4 conduits?

Answer 1:

((x)(x+y)(x+y+z)(y+z)(z)) ^ (1/4)

or

Answer 2:

((x)(z)) ^(1/3)

I think the answer is the one described in answer #1, what does others think?

Click to expand...

I agree, I think it's answer #1. I haven't seen any books describe how to calculate geometric mean distance for >3 conductors (since the neutral conductor is not normally current carrying), but this answer is in line with the solution to NCEES #117. I'd say I have 70% confidence in this.

ElecPwrPEOct11 said:

EEVA said:

If there were 4 conduits (A,B,C,D) spaced as:
A <---- x ----> B<---- y ----> C <----- z ------> D

What is the Geometric Mean Distance of the 4 conduits?

Answer 1:

((x)(x+y)(x+y+z)(y+z)(z)) ^ (1/4)

or

Answer 2:

((x)(z)) ^(1/3)

I think the answer is the one described in answer #1, what does others think?

Click to expand...

I agree, I think it's answer #1. I haven't seen any books describe how to calculate geometric mean distance for >3 conductors (since the neutral conductor is not normally current carrying), but this answer is in line with the solution to NCEES #117. I'd say I have 70% confidence in this.

Click to expand...

Careful with Complex Imag Exam #2, prob 6 then.

Isn't the formula for geometric mean G = (d1*d2*d3*.......*dx)^(1/x)? If so, with six distances shown you would need to take the sixth root of all distances multiplied.

vdubEE said:

Isn't the formula for geometric mean G = (d1*d2*d3*.......*dx)^(1/x)? If so, with six distances shown you would need to take the sixth root of all distances multiplied.

Click to expand...

good catch

vdubEE said:

Isn't the formula for geometric mean G = (d1*d2*d3*.......*dx)^(1/x)? If so, with six distances shown you would need to take the sixth root of all distances multiplied.

Click to expand...

@vdubEE- Yea that's the formula for geometric mean. But it's not clear how to apply it to distances & conductors. Per the solution to NCEES #117, it is not as simple as taking the cube root of all the distances from a fixed point.

Ex: NCEES #117 has 3 conduits with conduit B 3' from A and conduit C 4' from B (all in a line in the same direction). A 3' B 4' C

Using the equation you cite, for #117 you'd do (3*7)^(1/2)? The solution instead uses the distance between each of the three conductors.

EEVA- I don't have my practice tests yet but will keep my eyes out for a similar problem. Does the ComplexImaginary solution contradict the NCEES solution? We might have to call out Josh from ComplexImaginary on this one for a more detailed explanation.

ElecPwrPEOct11 said:

@vdubEE- Yea that's the formula for geometric mean. But it's not clear how to apply it to distances & conductors. Per the solution to NCEES #117, it is not as simple as taking the cube root of all the distances from a fixed point.
Ex: NCEES #117 has 3 conduits with conduit B 3' from A and conduit C 4' from B (all in a line in the same direction). A 3' B 4' C

Using the equation you cite, for #117 you'd do (3*7)^(1/2)? The solution instead uses the distance between each of the three conductors.

Click to expand...

The equation I cited does not mean you just use the two distances given, multiply and take the square root to get your answer. You have to know how to apply the equation first before you can just plug and chug to get your answer. The distances in the equation are all the distances between all the points. You cannot assume it is just the given distances only.

On questions like this one, it is nice to have lots of references available. I have access to the Power Reference Manual by Camara which clearly explains how to apply the equation to three conductors. I have also found this reference to have lots of valuable equations in one reference that I didn't think I would need. If you are taking the PE, it might be worthwhile to pick it up as it will probably turn out to be pretty valuable. Not sure I would necessarily buy the practice exam since it has some questions that seem to be very different than the NCEES.

I will say I was not sure how to work the problem at first but found the equation and explanation in the Power Reference Manual. I was able to plug/chug to the answer quickly. When I ran into the four conduit example problem in Complex Imaginary books, I remembered that it was all distances between all conductors/conduits. Figured out all six distances, multiplied them all together, and took the sixth root. If you don't have a reference that talked about it, now is the time to do some internet searching to find the information since it is more about learning stuff before taking the exam. I would rather be scratching my head at home trying to figure it out, searching the internet, and digging through references to learn stuff than sitting in the room taking the exam with a time constraint assuming I know how to apply a random equation correctly.

EDIT: The complex imaginary solutions do not contradict the NCEES solution as they both use all the distances between all the conductors/conduits.

vdubEE said:

The equation I cited does not mean you just use the two distances given, multiply and take the square root to get your answer. You have to know how to apply the equation first before you can just plug and chug to get your answer. The distances in the equation are all the distances between all the points. You cannot assume it is just the given distances only.
On questions like this one, it is nice to have lots of references available. I have access to the Power Reference Manual by Camara which clearly explains how to apply the equation to three conductors. I have also found this reference to have lots of valuable equations in one reference that I didn't think I would need. If you are taking the PE, it might be worthwhile to pick it up as it will probably turn out to be pretty valuable. Not sure I would necessarily buy the practice exam since it has some questions that seem to be very different than the NCEES.

I will say I was not sure how to work the problem at first but found the equation and explanation in the Power Reference Manual. I was able to plug/chug to the answer quickly. When I ran into the four conduit example problem in Complex Imaginary books, I remembered that it was all distances between all conductors/conduits. Figured out all six distances, multiplied them all together, and took the sixth root. If you don't have a reference that talked about it, now is the time to do some internet searching to find the information since it is more about learning stuff before taking the exam. I would rather be scratching my head at home trying to figure it out, searching the internet, and digging through references to learn stuff than sitting in the room taking the exam with a time constraint assuming I know how to apply a random equation correctly.

EDIT: The complex imaginary solutions do not contradict the NCEES solution as they both use all the distances between all the conductors/conduits.

Click to expand...

That's the exact reference that I am relying on most. I wasn't able to find it clearly explained at all. I found 1 sentence on page 12-10 on the definition of the geometric mean (in the context of probability). He talks about it in the power distribution chapter too but doesn't explain it. Can you give me a page number or section that you found it? I certainly agree that NOW is the time to figure all of this out. While Camara has lots of equations I have found his explanations a bit lacking. And you're right, without knowing how to use the equations, you might as well start guessing. Thanks for clarifying this! I bet it'll show up on the test.

EEVA said:

Careful with Complex Imag Exam #2, prob 6 then.

Click to expand...

Huh? My CI Exam2 Problem6 is about a Variable Frequency Drive.

For example, the geometric mean of six values: 34, 27, 45, 55, 22, 34 is:

ElecPwrPEOct11 said:

That's the exact reference that I am relying on most. I wasn't able to find it clearly explained at all. I found 1 sentence on page 12-10 on the definition of the geometric mean (in the context of probability). He talks about it in the power distribution chapter too but doesn't explain it. Can you give me a page number or section that you found it? I certainly agree that NOW is the time to figure all of this out. While Camara has lots of equations I have found his explanations a bit lacking. And you're right, without knowing how to use the equations, you might as well start guessing. Thanks for clarifying this! I bet it'll show up on the test.

Click to expand...

On page 38-7, equation 38.23, it talks about the equivalent distance for conductors that are not symmetrically arranged in a three-phase transmission system. The equation shows De = cube root(Dab * Dbc * Dca) for three conductors that are not symmetrically arranged. So by looking at the equation and seeing that for three conductors, there is a distance between each conductor to the others (1-2, 2-3, 3-1). Using that, now having four conductors, and examining the distances at each conductor, you have the following:

#1: 1-2, 1-3, 1-4

#2: 2-1, 2-3, 2-4

#3: 3-1, 3-2, 3-4

#4: 4-1, 4-2, 4-3

By eliminating the duplicates, you are left with six: 1-2, 1-3, 1-4, 2-3, 2-4, and 3-4. So for six distances, you take the sixth root. At least that is how I flowed through the solution process.

If anyone has a copy of the old Stevenson "Elements of power systems analysis" I think it has a reasonable explanation.

vdubEE said:

On page 38-7, equation 38.23, it talks about the equivalent distance for conductors that are not symmetrically arranged in a three-phase transmission system. The equation shows De = cube root(Dab * Dbc * Dca) for three conductors that are not symmetrically arranged. So by looking at the equation and seeing that for three conductors, there is a distance between each conductor to the others (1-2, 2-3, 3-1). Using that, now having four conductors, and examining the distances at each conductor, you have the following:
#1: 1-2, 1-3, 1-4

#2: 2-1, 2-3, 2-4

#3: 3-1, 3-2, 3-4

#4: 4-1, 4-2, 4-3

By eliminating the duplicates, you are left with six: 1-2, 1-3, 1-4, 2-3, 2-4, and 3-4. So for six distances, you take the sixth root. At least that is how I flowed through the solution process.

Click to expand...

Your flow through the problem is very good and spot on. I was disappointed to see that nowhere in the book does Camara go through the straightforward couple-sentence explanation you just did. He presents a formula for one case (3 conductors), and moves on to the next subject. Without a description of the formula it is not very helpful for the test-takers.