1- # BranchEMT Model
1+ # BranchLumpedConstant Model
22
3- ` BranchEMT ` represents a lumped-parameter EMT transmission line. The nominal
4- pi model is obtained by spatially discretizing the telegrapher equations over
3+ ` BranchLumpedConstant ` represents a lumped-parameter EMT transmission line.
4+ The nominal $\pi$- model is obtained by spatially discretizing the telegrapher equations over
55a segment of length $\Delta x$, with a half shunt placed at each terminal.
6- Series current $\mathbf{i}_ s$ is directed from bus 1 to bus 2. The positive
7- flow direction is into buses. All parameters are $3 \times 3$ matrices
8- capturing self and mutual coupling between phases.
6+ Series current $\mathbf{i}$ is directed from bus 1 to bus 2. Bus residual
7+ current injections are positive into buses. All electrical parameter matrices
8+ are $3 \times 3$ and capture self and mutual coupling between phases.
9+
10+ <div align =" center " >
11+ <img align =" center " src =" ../../../../../docs/Figures/EMT/lumped_constant_diagram.svg " >
12+
13+ Figure 1: Lumped constant EMT branch model
14+ </div >
915
1016## Model Parameters
1117
@@ -17,7 +23,7 @@ $\mathbf{G}'$ | [S/m] | Shunt conductance matrix per unit length | $
1723$\mathbf{C}'$ | [ F/m] | Shunt capacitance matrix per unit length | $\mathbb{R}^{3 \times 3}$
1824$\Delta x$ | [ m] | Line segment length | $\mathbb{R}$
1925
20- ### Model Derived Parameters
26+ ## Model Derived Parameters
2127
2228``` math
2329\begin{aligned}
@@ -34,7 +40,7 @@ $\Delta x$ | [m] | Line segment length | $
3440
3541Symbol | Units | Description | Note
3642-----------------|--------|-----------------------|---------------------------------
37- $\mathbf{i}_ s $ | [ A] | Series branch current | $\mathbb{R}^3$ , directed bus 1 to bus 2
43+ $\mathbf{i}$ | [ A] | Series branch current, directed bus 1 to bus 2 | $\mathbf{i} = [ i_a, i_b, i_c ] ^T \in \mathbb{R}^3$
3844
3945#### Algebraic
4046
@@ -51,8 +57,8 @@ of equations.
5157
5258Symbol | Units | Description | Note
5359-----------------|--------|--------------------------|------------------
54- $\mathbf{v}_ 1$ | [ V] | Terminal voltage at bus 1 | $\mathbb{R}^3$, owned by EMT bus
55- $\mathbf{v}_ 2$ | [ V] | Terminal voltage at bus 2 | $\mathbb{R}^3$, owned by EMT bus
60+ $\mathbf{v}_ 1$ | [ V] | Terminal voltage at bus 1, owned by bus 1 | $\mathbf{v} _ 1 = [ v _ {1,a}, v _ {1,b}, v _ {1,c} ] ^T \in \mathbb{R}^3$
61+ $\mathbf{v}_ 2$ | [ V] | Terminal voltage at bus 2, owned by bus 2 | $\mathbf{v} _ 2 = [ v _ {2,a}, v _ {2,b}, v _ {2,c} ] ^T \in \mathbb{R}^3$
5662
5763#### Algebraic
5864
6268
6369### Differential Equations
6470
65- ``` math
66- \dot{\mathbf{i}}_s = \mathbf{L}^{-1}\left((\mathbf{v}_1 - \mathbf{v}_2) - \mathbf{R}\,\mathbf{i}_s\right)
67- ```
68-
69- (or, alternatively)
7071
7172``` math
72- 0 = ( \mathbf{v}_1 - \ mathbf{v}_2) - \mathbf{R}\, \mathbf{i}_s - \mathbf{L}\dot{\ mathbf{i}}_s
73+ 0 = \mathbf{R}\,\ mathbf{i} + \mathbf{L}\dot{ \mathbf{i}} + \mathbf{v}_2 - \ mathbf{v}_1
7374```
7475
7576### Algebraic Equations
@@ -82,11 +83,11 @@ The lumped line contributes to the KCL residual at each terminal bus.
8283Each expression is accumulated into the owning bus residual.
8384
8485``` math
85- \Delta \ mathbf{i}_1 = -\mathbf{i}_s - \dfrac{\mathbf{C }}{2}\,\dot{\ mathbf{v}} _1 - \dfrac{\mathbf{G }}{2}\,\mathbf{v}_1
86+ \mathbf{i}^\text{inj} _1 : = - \dfrac{\mathbf{G }}{2}\,\mathbf{v}_1 - \dfrac{\mathbf{C }}{2}\,\dot{\ mathbf{v}}_1 - \mathbf{i}
8687```
8788
8889``` math
89- \Delta \ mathbf{i}_2 = +\mathbf{i}_s - \dfrac{\mathbf{C }}{2}\,\dot{\ mathbf{v}} _2 - \dfrac{\mathbf{G }}{2}\,\mathbf{v}_2
90+ \mathbf{i}^\text{inj}_2 := - \dfrac{\mathbf{G }}{2}\,\mathbf{v}_2 - \dfrac{\mathbf{C }}{2}\,\dot{\ mathbf{v}}_2 + \mathbf{i}
9091```
9192
9293## Initialization
@@ -95,10 +96,10 @@ The initialization assumes a balanced three-phase system. Given bus
9596voltages $\mathbf{v}_ 1(0)$, $\mathbf{v}_ 2(0)$ and their time
9697derivatives $\dot{\mathbf{v}}_ 1(0)$, $\dot{\mathbf{v}}_ 2(0)$ from
9798the EMT bus, and the power flow phasor series current
98- $I_s = |I_s | \angle \theta$, the initial series current is:
99+ $I = |I | \angle \theta$, the initial series current is:
99100
100101``` math
101- \mathbf{i}_s (0) = \sqrt{2}\,|I_s |
102+ \mathbf{i}(0) = \sqrt{2}\,|I |
102103\begin{bmatrix}
103104 \cos(\theta) \\
104105 \cos(\theta - \tfrac{2\pi}{3}) \\
@@ -110,5 +111,13 @@ The initial derivative is then given by the series branch equation for
110111DAE consistency:
111112
112113``` math
113- \dot{\mathbf{i}}_s (0) = \mathbf{L}^{-1}\left(( \mathbf{v}_1(0) - \mathbf{v}_2(0)) - \mathbf{R}\,\mathbf{i}_s (0)\right)
114+ \dot{\mathbf{i}}(0) = \mathbf{L}^{-1}\left(\mathbf{v}_1(0) - \mathbf{v}_2(0) - \mathbf{R}\,\mathbf{i}(0)\right)
114115```
116+
117+ ## Model Outputs
118+
119+ Candidate monitorable outputs include the series branch current components
120+ $i_a$, $i_b$, and $i_c$.
121+
122+ Terminal current injection expressions are documented above as
123+ $\mathbf{i}^\text{inj}_ 1$ and $\mathbf{i}^\text{inj}_ 2$.
0 commit comments