Determine the load voltage levels in feeder systems

[rob1985]

Homework Statement

Hi guys, I was just wondering if anybody could help me with the below question

Homework Equations

The Attempt at a Solution

Radial

Total load = 3.2 /_-36.87deg + 2.3 /_ 53.13deg
=(2.56 - j1.92) +(1.38 +j1.84)
= (3.94 -j0.08)
= 3.94 /_-1.16 MVA
Line current (I) = s/sqrt3 * VL
= 3.94 * 10^6/sqrt3*11*10^3
= 207 A

Phase voltage is 6351 volts
therefore 6351 - (4.6 * 207) = 5398.8
so volt drop is 952.2 volts
which is 15% regulation per phase

Does this answer the question for the radial circuit and if so how do I complete the parallel and the ring feeder systems

 

[cnh1995]

Line current (I) = s/sqrt3 * VL
= 3.94 * 10^6/sqrt3*11*10^3
= 207 A

That's not correct. You are not considering the j4.6 ohm line reactance. Also, you don't know the power angle. Which equations describe the active and reactive power flow between two adjacent nodes?

 

[rob1985]

That is what the lesson information gives when tackling a similar question, ill be honest I am not sure of the equations, that is something I have not covered during the lessons

 

[cnh1995]

That is what the lesson information gives when tackling a similar question, ill be honest I am not sure of the equations, that is something I have not covered during the lessons

This will help.
https://www.physicsforums.com/threads/ac-power-analysis-part-2-network-analysis-comments.864530/

 

[rob1985]

So the only equation I can think that will help is Q =(Va-Vb)/X Q = (11KV- 3.2MVA)/4.6

 

[cnh1995]

Line current (I) = s/sqrt3 * VL
= 3.94 * 10^6/sqrt3*11*10^3
= 207 A

You have used the sending end voltage here. It should be the receiving end voltage, which is unknown. The apparent power S that you've calculated is associated with the load, hence you need to use the receiving end voltage in the formula, which you have to calculate in the first place.

You have two unknowns in the equation, receiving end voltage V and power angle δ.
What are the equations for real power P and reactive power Q that contain V and δ?

 

[rob1985]

P = ((Vs * Vr) /X)* sinδ
Q = Vr/X * (Vs*cosδ - Vr)

These would be the equations your referring to I hope. How I work that into the question I've got, I don't know

 

[cnh1995]

These would be the equations your referring to I hope.

Yes.
You know P, Q and Vs.
You'll have two equations with two unknowns, Vr and δ.
Solve them simultaneously.

 

[rob1985]

I think that is beyond my capability

 

[cnh1995]

I think that is beyond my capability

It is somewhat lengthy. You need to do some algebraic and trigonometric manipulations. I solved it yesterday, but I can't just post the solution here as it is against the rules.

There should be some other way of doing it. I'll think and post later.

 

[anorlunda]

I make sign errors all the time, so don't rely on my answer.

Obviously, you can choose any reference for zero degrees. I just thought it easiest to make the 11KV @ZERO degrees.

But here's another simple way to look at it. With a reactor in series, if the load is reactive then we have a simple voltage divider circuit and the receiving voltage is necessarily <11. But the net load is negative MVAR, a capacitor, so that receiving voltage might be >11.

 

[cnh1995]

But the net load is negative MVAR, a capacitor, so that receiving voltage might be >11.

For the first load i.e. 3.2 MVA, 0.8 pf lag, isn't the load inductive? The current should be lagging, hence, the MVAR should be leading because S=VI*.
So wouldn't it be S=2.56+j1.92?

Or should we take the MVA as lagging, which means the current is leading?

 

[anorlunda]

There were two loads, one leading one lagging. It is the sum that counts. The OP had the following, is that correct?

Total load = 3.2 /_-36.87deg + 2.3 /_ 53.13deg
=(2.56 - j1.92) +(1.38 +j1.84)
= (3.94 -j0.08)

 

[cnh1995]

is that correct?

No.
For the second load, 1.8MVA @0.6 pf lead, it should be
1.8*0.6+j1.8*0.8= 1.08+j1.44.

Their sum should be 3.64-j.48.

But what is the significance of that lagging/leading label?
Does @0.8 pf lagging mean the load is inductive or the MVAR is negative (which means the load is capacitive)?

 

[anorlunda]

I've always been dyslexic about arbitrary sign conventions, can't remember which is lead and which lag. But one load was lead, the other lag. Only the sum counts. An inductive load draws positive VARs, so if the sum is +VAR, then the receiving voltage is <11, if negative than V>11.

 

[cnh1995]

I've always been dyslexic about arbitrary sign conventions, can't remember which is lead and which lag. But one load was lead, the other lag. Only the sum counts. An inductive load draws positive VARs, so if the sum is +VAR, then the receiving voltage is <11, if negative than V>11.

Ok.
So the sum is negative, means the net load is capacitive, and yes, receiving end voltage is more than 11kV.