Is the Southwire Voltage drop calculator wrong?

[wwhitney]

My guess is a bug in the calculator.

Pretty sure I've figured out the bug--in calculating Z effective, the formula should be Zeff = R * P + X * sqrt(1-P²), where P is the power factor. [Note 2 to Chapter 9 Table 9 uses sin(arcos(P)) as the coefficient of X, but sin(arcos(P)) = sqrt(1-P²) for 0 <= P <= 1.]

However, instead the calculator appears to be using Zeff = R * P + X * sqrt(1-0.9²). I.e. changing the power factor in the web from the default of 0.9 changes the coefficient of R, but not the coefficient of X.

Will figure out how to report this to Southwire.

Cheers, Wayne

 

[dkarst]

Nice job figuring out the issue... always a problem trying to backtrack unseen code. This show again the dangers of blindly applying a web/app solution without knowing what is under the covers. Please post back on Southwire's response. Also remember even the redbook simplification IR cos theata + IX sin theta is an approximation (although usually a very good one)

 

[electrofelon]

The app may have added for Q

I don't think so. Q was an omnipotent being so he would have had the ability to make for 0 voltage drop. But....he was rather mischievous though so there is a chance that he could have added some extra voltage drop in to confuse and make a conundrum* for the starship Enterprise crew

*Note: there was an episode entitled "conundrum" in the 5th season and this situation was not in the plot of that episode.

 

[wwhitney]

Please post back on Southwire's response.

I got a response today. It said in part:

"Thank you for bringing this to our attention. We plan to release a new revision to this calculator later this year and I will review the calculation regarding the power factor. It seems that users rarely ever change the power factor and I was not aware of an issue with it."

Cheers, Wayne

 

[Julius Right]

In my opinion, if the conductor is aluminium 75 [7075] the volume resistivity it is 51.5% from IACS [10.371/0.51.5=20.137 cmils*ohms/ft]
Then #2 aluminium resistance=20.137*1000/66360=0.3035 ohm/kft [d.c. 20oC].
In NEC the resistivity it is according to Neher & McGrath 61% then resistivity=10.371/0.61=17 and the resistance of #2 aluminium =17*1000/66360=0.256 ohm/kft

 

[Frank DuVal]

Actually, Q appeared in The Next Generation, not The Original Series.

Right, I used an odd but correct wording of "post (i.e. after) the original series".