Calculations

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Bjenks

Senior Member
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East Coast of FL
Smart $ said:
If anyone reading that thread wants to download my Excel neutral calculator, which allows for entry of and accounts for power factor, go to the following post:

http://forums.mikeholt.com/showpost.php?p=1065087&postcount=11
Click to expand...

Nice spreadsheet. That is a great learning tool for anyone wanting to better understand the vectors of power flow. It looks like you could easily update this spreadsheet to take into consideration the voltage not being in phase either.
 
Besoeker

Besoeker

Senior Member
Location
UK
hillbilly said:
When I first read the OP, and did a quick calculation in my head, my mind said....worst case, 50A on the neutral.

Try as I might, I cannot repeat the calculation.:roll:

I had either a flash of brilliance, or a moment of delusion.:D

steve
Click to expand...
It could be worse, I suppose, but maybe not very likely.
To get the figures, I modified a spreadsheet I previously made for harmonic analysis. It was easier than doing it longhand, and has the merit of producing a chart of the various currents.
 
Smart $

Smart $

Esteemed Member
Location
Ohio
Besoeker said:
The loads also add vectorially in the neutral. That's how you end up with zero in the neutral if you have a balanced load. Suppose you have 1A at zero, 1A at 2Pi/3, and 1A at 4Pi/3.

Then:
Ia = 1
Ib = -0.5 - jsqrt(3)/2
Ic = -0.5 + jsqrt(3)/2

In = Ia + Ib + Ic
= 0

QED
;)
Click to expand...
Thanks for the refresher, on someone else's behalf... 'cause I already know this. Would have been hard to make my Excel calculator without knowing it :D


Besoeker said:
But my point was that masterinbama could only have arrived at the 40A for Ia by arithmetic addition of the currents he gave (20+10+10)
Click to expand...
Ahh!... I see what you are getting at... masterinbama's attempt to substitute equivalent loading... and the three loads he presented are not equivalent to the OP's.
 
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Smart $

Smart $

Esteemed Member
Location
Ohio
Besoeker said:
It shouldn't.
But given that appears as an object in a Word document.........
Click to expand...

Open the word doc...
Right-click on the excel object and choose Worksheet Object>Open...
The object opens in Excel (at least this is how it supposed to work) where you can Save As to your local directory and modify to your heart's content ;)
 
Smart $

Smart $

Esteemed Member
Location
Ohio
Besoeker said:
Then maybe you would like to review the content of post #15.
Click to expand...
Why? ... what I wrote in post #15 holds true.

masterinbama did not state the total current per line, he only stated the neutral current, which is correct at 10A for the loads he listed for only one of the loads was line to neutral.
 
Besoeker

Besoeker

Senior Member
Location
UK
Smart $ said:
Why? ... what I wrote in post #15 holds true.
Click to expand...

Your post #15
There isn't any adding required for the neutral conductor. The loads only add, vectorially, at the line connection.

My post #17
The loads also add vectorially in the neutral. That's how you end up with zero in the neutral if you have a balanced load. Suppose you have 1A at zero, 1A at 2Pi/3, and 1A at 4Pi/3.

Your post #22
Thanks for the refresher, on someone else's behalf... 'cause I already know this.

Do you see now?
 
Besoeker

Besoeker

Senior Member
Location
UK
Smart $ said:
Open the word doc...
Right-click on the excel object and choose Worksheet Object>Open...
The object opens in Excel (at least this is how it supposed to work) where you can Save As to your local directory and modify to your heart's content ;)
Click to expand...
Nice work.
How do you deal wih leading power factor?
 
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