voltage drop

[Mike01]

when evaluating voltage drop and the forumla calls for (R cos theta + X sin theta) where do you get the values for the R and the X??

 

[charlie b]

I haven't seen a voltage drop formula that uses those terms. But I think you can get what you need from Table 9. The column showing "Effective Z at 0.85 pf" can be used to calculate R and X.

 

[Smart $]

I haven't seen a voltage drop formula that uses those terms.

The former IEEE formula used those terms. The terms provided are also equivalent to the formula at the very end of the Notes to Table 9, where...

Ze = R ? PF + XL sin[arccos (PF)]

...and...

PF = cos θ
sin[arccos (PF)] = sin θ

...therefore...

Ze = R cos θ + XL sin θ.​

The IEEE changed the formula a few years ago to the "exact" formula...

...leaving adjustments to R for operating conditions as an excercise for the user :grin:

But I think you can get what you need from Table 9. The column showing "Effective Z at 0.85 pf" can be used to calculate R and X.

The values for R and XL are in the preceding columns (i.e. if one reads left to right :wink: )

 

[Mike01]

is this right??

is this right??

In a square d catalog for busway systems it gives an example for 1000A 50% pf bus duct as follows:

Sqrt(3) * I (R cos? + X sin?)

Sqrt(3) * 1000 (.00163 *.5 + 0.00093 * .866)

What I do not understand is where are the numbers in bold derived from?

If I am right the R cos? is equal to the R value given 1.63/1000 ???
Than the Xsin? is equal to the X value (at 60hz) /1000 ???
but where does the 0.866 come from?? The sin of 60 is .866 is that for the frequency??

 

[Smart $]

In a square d catalog for busway systems it gives an example for 1000A 50% pf bus duct as follows:

Sqrt(3) * I (R cos? + X sin?)

Sqrt(3) * 1000 (.00163 *.5 + 0.00093 * .866)

What I do not understand is where are the numbers in bold derived from?...

Umm... I don't see any numbers in bold type

The √3 operand should be obvious. The 1000(A) is given, as is the 50% power factor (which is the .5 in the equation, because cos θ = PF, as noted in my previous post). The .00163 is resistance in ohms per distance unit(s).?generally that is 1000 ft. working in the imperial system of measures. ...And the 0.00093 is inductive reactance, in the same ohms per distance measure.

...Than the Xsin? is equal to the X value (at 60hz) /1000 ???
but where does the 0.866 come from?? The sin of 60 is .866 is that for the frequency??

No... 60 is theta(θ) in degrees (i.e. 60?). X, more appropraiately X sub L is the inductive reactance portion of the impedance in ohms per 1000 ft. (?). That value has already taken 60Hz operation into consideration. What multiplying X by sin θ does is it adjusts X for the current's phase shift from the applied voltage...

The voltage drop formula, to eliminate the depicted error, was changed to the so-called "exact" formula I posted earlier.

 

[bob]

In a square d catalog for busway systems it gives an example for 1000A 50% pf bus duct as follows:

Sqrt(3) * I (R cos? + X sin?)

Sqrt(3) * 1000 (.00163 *.5 + 0.00093 * .866)

What I do not understand is where are the numbers in bold derived from?

If I am right the R cos? is equal to the R value given 1.63/1000 ???
Than the Xsin? is equal to the X value (at 60hz) /1000 ???
but where does the 0.866 come from?? The sin of 60 is .866 is that for the frequency??

Cos? and Sin? come from the power factor. PF = 0.50
In your example Cos? =.50 and ? = 60 degrees. Sin 60 = 0.866.