Determine the load voltage levels in feeder systems

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Discussion Overview

The discussion revolves around determining load voltage levels in feeder systems, specifically focusing on radial, parallel, and ring feeder configurations. Participants explore calculations related to apparent power, line current, voltage drops, and the implications of reactive power in these systems.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a calculation for total load and line current for a radial circuit, but questions how to extend this to parallel and ring systems.
  • Another participant challenges the initial calculation, noting the omission of line reactance and the unknown power angle, prompting a discussion on the equations governing active and reactive power flow.
  • Several participants express uncertainty regarding the relevant equations and how to apply them to the problem at hand.
  • Participants discuss the implications of leading and lagging power factors, with one questioning whether the current should be considered leading or lagging based on the load characteristics.
  • There is a debate about the significance of the net load being capacitive or inductive and its effect on the receiving voltage in relation to the sending voltage.
  • One participant mentions difficulties with algebraic manipulations required to solve the equations, indicating a perceived complexity in the problem.
  • Another participant reflects on their confusion regarding sign conventions for leading and lagging loads, emphasizing that only the sum of the loads matters for determining the overall effect on voltage.

Areas of Agreement / Disagreement

Participants express varying levels of agreement on specific calculations and interpretations of leading and lagging loads. There is no consensus on the correct approach to solving the problem, with multiple competing views and uncertainties remaining throughout the discussion.

Contextual Notes

Participants acknowledge missing assumptions and unresolved mathematical steps, particularly regarding the power angle and the correct application of equations for real and reactive power.

rob1985
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Homework Statement



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

upload_2017-7-19_11-7-2.png
upload_2017-7-19_11-7-27.png


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
 
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rob1985 said:
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?
 
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
 
So the only equation I can think that will help is Q =(Va-Vb)/X Q = (11KV- 3.2MVA)/4.6
 
rob1985 said:
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?
 
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
 
rob1985 said:
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.
 
I think that is beyond my capability
 
  • #10
rob1985 said:
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.
 
  • #11
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
anorlunda said:
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?
 
  • #13
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
anorlunda said:
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)?
 
  • #15
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
anorlunda said:
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.
 

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