Calculating voltage drops in cables with vectors?

  • Thread starter EVriderDK
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  • #1
EVriderDK
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Hey.

If i Have a 3x400VAC system, and I want to calculate the voltage drop over the cable "vectorially" - or what you call it :) - how do I do this?
U.nominal=400V

I have tried this method:
[itex]\Delta[/itex]U.conductor=I.1*Z.cable1+I.2*Z.cable2+etc. and then I will have to find the new voltage ->
[itex]\Delta[/itex]U.phase=U.conductor*[itex]\sqrt{3}[/itex]
U.new=U.nominal-[itex]\Delta[/itex]U.phase

When I'm calculating this with angles, I actually get a higher voltage, after the voltage drop :D
The voltage drop is cause by a electrical motor starting.

In school we learn it this way:
[itex]\varphi[/itex].motor=80 degrees

U.new=U.nominal-[itex]\sqrt{3}[/itex]*I1*Length.cable1*(R.cable1*cos[itex]\varphi[/itex]+X.cable1*sin[itex]\varphi[/itex])+[itex]\sqrt{3}[/itex]*I2*Length.cable2(R.cable2*cos[itex]\varphi[/itex]+X.cable2*sin[itex]\varphi[/itex])+etc.
But that is not with vectors.

Hope you understand?

Thanks in advance.
 
Last edited:

Answers and Replies

  • #2
meBigGuy
Gold Member
2,324
406
I'm pretty weak with respect to power factor stuff, but sin and cos are intimately tied to the vector quantities.

Acosβ represents the X component and Asinβ the Y component of a vector of magnitude A at angle β (usually called real and imaginary parts).

If you want to add two vectors you can add their respective X parts and Y parts to determine the new X and Y parts of the resulting vector.

You can probably go through the equations you posted and see where the vector math is happening.
 

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