Complex Imaginary, Exam 1, Problem 18

The questions is as follows:

A power distribution panel is serving a single-phase 3.3kV, 75 kW load, with a 0.90 lagging power factor.  The feeder to the load has an impedance of 0.2 + j0.5 ohms.  What is the actual voltage for this feeder at the panel?

I understand this is a reverse voltage problem. However, the solution they give does not multiple the impedance by 2.  Shouldn't we have to multiple by 2 since this is a single phase circuit?

My understanding is that all the power is dissipated in the load in this question, causing no current in the return feeder.

With no current, no extra voltage drop to add.  The only Vd you take into account is the feeder with current in it.

Take a look at Complex book 2, question 10 for comparison.

Hmm, ok that makes sense actually.  Thank you!

Millerific said:

The questions is as follows:

A power distribution panel is serving a single-phase 3.3kV, 75 kW load, with a 0.90 lagging power factor.  The feeder to the load has an impedance of 0.2 + j0.5 ohms.  What is the actual voltage for this feeder at the panel?

I understand this is a reverse voltage problem. However, the solution they give does not multiple the impedance by 2.  Shouldn't we have to multiple by 2 since this is a single phase circuit?

Click to expand...

Hi

When feeder impedance is given in totality you need not take loop impedance as double of that. The impedance given in this question is meant for both wires (P & N). So If we take the voltage at the terminals of the load as 3300<0 and the current in the load as 75/(.9*3.3)=25.25 A <-cosinverse(.9)=25.25<-25.84

So the voltage at the panel is V= V at Load+ voltage dropped in line

                                                   = 3300<0+25.25<-25.84 * (.2+j.5)=3310<.15 

Is this the answer? Provide me your email for further discussion.

Millerific said:

The questions is as follows:

A power distribution panel is serving a single-phase 3.3kV, 75 kW load, with a 0.90 lagging power factor.  The feeder to the load has an impedance of 0.2 + j0.5 ohms.  What is the actual voltage for this feeder at the panel?

I understand this is a reverse voltage problem. However, the solution they give does not multiple the impedance by 2.  Shouldn't we have to multiple by 2 since this is a single phase circuit?

Click to expand...

Can anyone explain why we're supposed to consider feeder impedance includes Line and neutral impedance values ? I end up multiplying by 2 whenever I see single phase Voltage drop problems.