A number of forum threads discuss how the step-up transformer connecting an Alternator to the Grid is able to control voltage-alignment using a Tap-Changer. Also discussed is the question, "What happens when tap-position is altered?" Thus far, in my opinion, there has not been a viable conclusion.
Therefore it is the intent of this multi-part thread to explain how tap-change impacts reactive-current, hence kVAr flow. I will use the concept of a tap-changer as a Turns-Ratio-Controller.
Caveat:
This presentation/participation does requires a rudimentary knowledge of Vector Algebra. A Vector is a quantity having both magnitude and direction, such as Voltage, Current, and Impedance. Let’s start with the very simple network consisting of just three components:
o Gen’r connected to a bus having a Voltage magnitude, Vs, and direction θsº.
o Load connected to a bus having a Voltage magnitude, Vr, and direction θrº.
o Cable whose reactance, Xc, connects the two busses.
Concept of Load-Flow (or Power-Transfer)
Personally, I prefer the former, because 'Load', to me, denotes having two vector-components:
o Real, having as its magnitude, W or Ps, and direction, 0º.
o Reactive, having as its magnitude, VAr or Qs, and direction, 90º.
An Aside
One vector, being at 0º, and the other 90º means, mathematically, they are in ”quadrature”!
Basic Equations
There are just two, but quite crucial, equations representing Ps and Qs. They are,<pre>
o Ps = The Real-component, W, is transferred from the Gen-Bus,
Vs∠θsº, to the Load-Bus, Vr∠θrº,
= (Vs*Vr/Xc)*Sin(θrº-θsº)
o Qs = The Reactive-component, VAr, is transferred from Gen-Bus,
Vs∠θsº, to the Load-Bus, Vr∠θrº,
= (Vs/Xc)*[Vs–Vr*Cos(θrº-θsº)]</pre>
Before continuing with Part 2, please calculate Ps and Qs for the following three Exercises:<pre>
.... | Vs | |∠θsº| | Vr | |∠θrº| | X |
Ex1 | |1.00| |0.00| |1.00| |0.00| |0.10|
Ex2 | |1.05| |0.00| |1.00| |1.00| |0.10|
Ex3 | |1.05| |0.00| |1.00| |22.0| |0.10|</pre>
I think you ll will be pleasantly surprised at the results. And, it be a major step for understanding Part 2.
Please post your answers to Control.com or send them to me at: [email protected].
Ps: If you see the symbol '□’ substitute the angle 'theta'
Therefore it is the intent of this multi-part thread to explain how tap-change impacts reactive-current, hence kVAr flow. I will use the concept of a tap-changer as a Turns-Ratio-Controller.
Caveat:
This presentation/participation does requires a rudimentary knowledge of Vector Algebra. A Vector is a quantity having both magnitude and direction, such as Voltage, Current, and Impedance. Let’s start with the very simple network consisting of just three components:
o Gen’r connected to a bus having a Voltage magnitude, Vs, and direction θsº.
o Load connected to a bus having a Voltage magnitude, Vr, and direction θrº.
o Cable whose reactance, Xc, connects the two busses.
Concept of Load-Flow (or Power-Transfer)
Personally, I prefer the former, because 'Load', to me, denotes having two vector-components:
o Real, having as its magnitude, W or Ps, and direction, 0º.
o Reactive, having as its magnitude, VAr or Qs, and direction, 90º.
An Aside
One vector, being at 0º, and the other 90º means, mathematically, they are in ”quadrature”!
Basic Equations
There are just two, but quite crucial, equations representing Ps and Qs. They are,<pre>
o Ps = The Real-component, W, is transferred from the Gen-Bus,
Vs∠θsº, to the Load-Bus, Vr∠θrº,
= (Vs*Vr/Xc)*Sin(θrº-θsº)
o Qs = The Reactive-component, VAr, is transferred from Gen-Bus,
Vs∠θsº, to the Load-Bus, Vr∠θrº,
= (Vs/Xc)*[Vs–Vr*Cos(θrº-θsº)]</pre>
Before continuing with Part 2, please calculate Ps and Qs for the following three Exercises:<pre>
.... | Vs | |∠θsº| | Vr | |∠θrº| | X |
Ex1 | |1.00| |0.00| |1.00| |0.00| |0.10|
Ex2 | |1.05| |0.00| |1.00| |1.00| |0.10|
Ex3 | |1.05| |0.00| |1.00| |22.0| |0.10|</pre>
I think you ll will be pleasantly surprised at the results. And, it be a major step for understanding Part 2.
Please post your answers to Control.com or send them to me at: [email protected].
Ps: If you see the symbol '□’ substitute the angle 'theta'