# Homework Help: Determine the load voltage levels in feeder systems

1. Jul 19, 2017

### rob1985

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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

Last edited by a moderator: Jul 19, 2017
2. Jul 19, 2017

### cnh1995

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?

3. Jul 19, 2017

### rob1985

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

4. Jul 19, 2017

### cnh1995

5. Jul 20, 2017

### rob1985

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

6. Jul 20, 2017

### cnh1995

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?

7. Jul 20, 2017

### 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 ive got, I don't know

8. Jul 20, 2017

### cnh1995

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

9. Jul 20, 2017

### rob1985

I think that is beyond my capability

10. Jul 20, 2017

### cnh1995

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.

11. Jul 22, 2017

### Staff: Mentor

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.

12. Jul 22, 2017

### cnh1995

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?

13. Jul 22, 2017

### Staff: Mentor

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)

14. Jul 22, 2017

### cnh1995

No.
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)?

15. Jul 22, 2017

### Staff: Mentor

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.

16. Jul 22, 2017

### cnh1995

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