Bus Way Impedance

[Humble2000]

My question is that how the bus duct resistance and impedance can change with load Pf?

http://www.sea.siemens.com/busway/product/BD-Busway-Guide.pdf

page 19
According to Siemens publication, there are different values for R and X for different load pf. I also do not get the same voltage drop as it’s published for Al or Copper?
Thank you,

 

[stevenal]

Where does it say load pf? The value I see is bus pf= Rleg/Zleg.

 

[rbulsara]

That appears to be funky way of saying X/R ratio. Those pf values appear to be cosine of the phase angle based on X and R. I am not sure why.

 

[davidbeach]

I agree, it is not load pf, it is the bus pf. Not sure why they would indicate it that way, but it is the X/R of the bus as indicated above. Load pf enters the picture when calculating voltage drop as the relationship of current angle and line angle does make a difference.

 

[Humble2000]

So...
When it says "Resistance, Reactance and Impedance Values",these are not bus values?

The table at the top shows "Voltage Drop by % Load Power Factor—Volts"

How they got those voltage drop numbers?

 

[dpc]

The power factor of the load will have some effect on the voltage drop (on a per-amp basis), but that's not quite the question you asked orginally.

The load power factor does not change the busway impedance - it changes the angle between the load current and the voltage. This can change the voltage drop.

 

[Humble2000]

Yes, I understand that the voltage drop changes by load pf.
The question is different impedances shown for different load pf which I don't understand.

 

[rbulsara]

It is not the load pf. They are just mentioning the phase angle of the "Busway".

The pf mentioned there is found thus, I believe.

Cos (tan^-1 (X/R))

I do not see any usefulness of that info as you already know X and R. Do not confuse that with load pf.

 

[waross]

Some of us may at first have missed the table at the top of page 19. This table is load power factor. A table near the bottom of page 19 gives the bus power factor. (Bus resistance divided by bus impedance.)

Humble2000;
Try resolving a sample current at a chosen power factor into the real or in-phase component and the reactive component.
Next, use the in-phase current and the resistance to calculate the in phase voltage drop.
Use the quadrature or reactive current and the reactance to calculate the reactive voltage drop.
Finally, add the in-phase voltage drop and the reactive voltage drop vectorily.
Please let us know if this method will reproduce the values given in the table.
respectfully

 

[Humble2000]

I am not sure what I am doing wrong?

Pf=.8 from the top table…
Pf=.8=cos(a)
Sin(a)=.6
Vd=I*sqrt(3)*(Rcos(a)+Xsin(a))

R=46.15x10^(-6) per foot from the table in page 19-bottom
X=18.2x10^(-6) per foot

Vd=225x1.732x(46.15x10^(-6)x100x.8+ 18.2x10^(-6)x100x.6)=1.864

The voltage drop is 3.22

 

[rbulsara]

The voltage drop you calculated is "Per Phase" value. For line voltage drop value you need to multiply it by sqrt of 3.

1.864*1.732 = 3.22

 

[rbulsara]

The writer of the cutsheet refers to Per Phase as Per Leg. Not an egineering term, but that is what it seems to mean.

 

[rbulsara]

My last two posts were not to validate or invalidate the equation itself.

See the link below, it has a good explanation.

http://www.gilbertelectrical.com/library/technical.discussions/economics/economics.htm

 

[Humble2000]

In my calculations, I had multiplied the equation by 1.732.
Does this mean that:

Single phase voltage drop:Vd=I*sqrt(3)*(Rcos(a)+Xsin(a))
Three phase volage drop: Vd=3*I*(Rcos(a)+Xsin(a))

This formula is in the same page too.