Bus Way Impedance
Bus Way Impedance
(OP)
My question is that how the bus duct resistance and impedance can change with load Pf?
h ttp://www. sea.siemen s.com/busw ay/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,
h
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,






RE: Bus Way Impedance
RE: Bus Way Impedance
RE: Bus Way Impedance
RE: Bus Way Impedance
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?
RE: Bus Way Impedance
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.
RE: Bus Way Impedance
The question is different impedances shown for different load pf which I don't understand.
RE: Bus Way Impedance
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.
RE: Bus Way 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
RE: Bus Way Impedance
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
RE: Bus Way Impedance
1.864*1.732 = 3.22
RE: Bus Way Impedance
RE: Bus Way Impedance
See the link below, it has a good explanation.
http:/
RE: Bus Way Impedance
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.