Section 4.5, AS3008 - Voltage DropBased on Load Power Factor

Yes, the load power factor does affect voltage drop, but the formulas are used for 2 cases: 3 phases then Vd=sqrt(3)*I*Z, one phase then Vd=2*I*Z

Why it is related to load power factor, it is because Z=R*cos(phi)+X*sin(phi), cos(phi) is the power factor so that is why.

One last thing, there is no definition of cable power factor. You just need to care about the power factor of the load it connects to.

Regards,

P.

How come in section 4.3.1 and I quote:

"Voltage drop in a circuit represents the vectorial difference in voltage between the origin or supply end and the load end. For the purpose of determining the maximum voltage drop value in Clause 4.2, the voltage drop (Vd) has been related to the impedance of the cables forming the circuit when the power factor of the cable is equal to the power factor of the load, in which case-

Vd = IZc

Zc = impedance of cable, in ohms

= √(R2c + X2c)

"

?

Ah, it is beyond my knowledge, sorry. It is my mistake about Z=R*cos(phi)+X*sin(phi)

The formulas are stated that : Vd=sqrt(3)*I*(R*cos(phi)+X*sin(phi)) or Vd=2*I*(R*cos(phi)+X*sin(phi)), so I jump to conclusion that Z=R*cos(phi)+X*sin(phi)

It is true that Z=R*cos(beta)+X*sin(beta) and cos(beta) is not usually equal to cos(phi). (I checked that before)

but to call that power factor of cable is quite new to me.

To my understanding, voltage drop is applied when the current (of the load) run through the cable (with impedance) and loses an amount of power (sqrt(3)*U*I) and the voltage reach the end will be lower or higher (depends on L or C of cable), usually lower and calculated as above formulas.

The words quoted may say it in an academical way. :)

Sorry if no helping for you.

To maximize power transfer, output and input impedances must be balanced at all times. To maintain this balance in impedance and power factors both for output (load) and input (source), the load must always be maintained balanced for all 3-phases. Once some loads were taken off line, a corresponding load imbalance is created which will be reflected with a corresponding voltage changes on the supply line as this is also is reflected as a change in the original power factor of your load, etc.

Hope this helps..

When you have imbalanced phases, this does not change the load and supply power factors. It merely means either your phase voltages change. Current and voltage at the supply side will be of the same angle difference at the load side afaik.

Don't you effectively change your load impedance when a load imbalance is created?

Someone has to know the answer!!