N-R two iterations, 2-bus system

Newton-Raphson (polar form) is applied to the two-bus system; the line has $z = 0.12 + j0.16$ pu, and the load of 100 MW + j50 MVAr becomes $1 + j0.5$ pu on the 100 MVA base.

Bus data and Y-bus

Base 100 MVA. Bus 1: slack, $V_1 = 1\angle 0^\circ$. Bus 2: load bus with $P_2 = -1.00$ pu and $Q_2 = -0.50$ pu (negative, as power is taken out). Unknowns: $\delta_2$, $|V_2|$. Flat start: $|V_2|^{(0)} = 1$, $\delta_2^{(0)} = 0$.

$$\begin{aligned} y_{12} &= \frac{1}{0.12 + j0.16} = \frac{0.12 - j0.16}{0.04} = 3 - j4 \\ Y_{11} &= Y_{22} = y_{12} = 3.0000 - j4.0000 = 5.0000\angle -53.13^\circ \\ Y_{12} &= Y_{21} = -y_{12} = -3.0000 + j4.0000 = 5.0000\angle 126.87^\circ \end{aligned}$$

Equations used

With $Y_{ij} = |Y_{ij}|\angle\theta_{ij}$, $V_i = |V_i|\angle\delta_i$ and $\phi_{ij} = \theta_{ij} - \delta_i + \delta_j$:

$$\begin{aligned} P_i &= \sum_{j=1}^{n} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ Q_i &= -\sum_{j=1}^{n} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \end{aligned}$$

Jacobian elements (polar form, $j \ne i$):

$$\begin{aligned} \frac{\partial P_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial |V_i|} &= 2|V_i||Y_{ii}|\cos\theta_{ii} + \sum_{j\ne i} |V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial P_i}{\partial |V_j|} &= |V_i||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{j\ne i} |V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_j|} &= -|V_i||Y_{ij}|\sin\phi_{ij} \end{aligned}$$

For bus 2, $[\Delta P_2;\ \Delta Q_2] = J\,[\Delta\delta_2;\ \Delta|V_2|]$.

Newton-Raphson iterations

Iteration 1

Calculated powers and mismatches with the values from iteration 0:

$$\begin{aligned} P_2^{(0)} &= 0.0000, \quad \Delta P_2 = -1.0000 - (0.0000) = -1.0000 \\ Q_2^{(0)} &= 0.0000, \quad \Delta Q_2 = -0.5000 - (0.0000) = -0.5000 \end{aligned}$$

Jacobian elements:

$$\begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0000)(1.0000)(5.0000)\sin(126.87^\circ) = 4.0000 \\ \frac{\partial P_2}{\partial |V_2|} &= 2(1.0000)(5.0000)\cos(-53.13^\circ) + (1.0000)(5.0000)\cos(126.87^\circ) = 3.0000 \\ \frac{\partial Q_2}{\partial \delta_2} &= (1.0000)(1.0000)(5.0000)\cos(126.87^\circ) = -3.0000 \\ \frac{\partial Q_2}{\partial |V_2|} &= -2(1.0000)(5.0000)\sin(-53.13^\circ) - (1.0000)(5.0000)\sin(126.87^\circ) = 4.0000 \end{aligned}$$ $$\begin{bmatrix} -1.0000 \\ -0.5000 \end{bmatrix} = \begin{bmatrix} 4.0000 & 3.0000 \\ -3.0000 & 4.0000 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix}$$ $$\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} = \begin{bmatrix} -0.1000 \\ -0.2000 \end{bmatrix}$$ $$\begin{aligned} \delta_2^{(1)} &= 0.0000 + (-0.1000) = -0.1000\ \text{rad} = -5.7296^\circ \\ |V_2|^{(1)} &= 1.0000 + (-0.2000) = 0.8000\ \text{pu} \end{aligned}$$

Iteration 2

Calculated powers and mismatches with the values from iteration 1:

$$\begin{aligned} P_2^{(1)} &= -0.7875, \quad \Delta P_2 = -1.0000 - (-0.7875) = -0.2125 \\ Q_2^{(1)} &= -0.3844, \quad \Delta Q_2 = -0.5000 - (-0.3844) = -0.1156 \end{aligned}$$

Jacobian elements:

$$\begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (0.8000)(1.0000)(5.0000)\sin(132.60^\circ) = 2.9444 \\ \frac{\partial P_2}{\partial |V_2|} &= 2(0.8000)(5.0000)\cos(-53.13^\circ) + (1.0000)(5.0000)\cos(132.60^\circ) = 1.4157 \\ \frac{\partial Q_2}{\partial \delta_2} &= (0.8000)(1.0000)(5.0000)\cos(132.60^\circ) = -2.7075 \\ \frac{\partial Q_2}{\partial |V_2|} &= -2(0.8000)(5.0000)\sin(-53.13^\circ) - (1.0000)(5.0000)\sin(132.60^\circ) = 2.7195 \end{aligned}$$ $$\begin{bmatrix} -0.2125 \\ -0.1156 \end{bmatrix} = \begin{bmatrix} 2.9444 & 1.4157 \\ -2.7075 & 2.7195 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix}$$ $$\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} = \begin{bmatrix} -0.0350 \\ -0.0773 \end{bmatrix}$$ $$\begin{aligned} \delta_2^{(2)} &= -0.1000 + (-0.0350) = -0.1350\ \text{rad} = -7.7345^\circ \\ |V_2|^{(2)} &= 0.8000 + (-0.0773) = 0.7227\ \text{pu} \end{aligned}$$

i) Bus 2 voltage after the 2nd iteration

$|V_2| = 0.7227$ pu, $\delta_2 = -7.735^\circ$, i.e. $V_2 = 0.7227\angle -7.735^\circ$ pu.

ii) Slack bus power and network losses

$$\begin{aligned} S_1 = P_1 + jQ_1 &= V_1\left[Y_{11}V_1 + Y_{12}V_2\right]^{*} = 1.2408 + j0.8439\ \text{pu} \\ &= 124.08\ \text{MW} + j84.39\ \text{MVAr} \end{aligned}$$ $$\begin{aligned} I_{12} &= (V_1 - V_2)y_{12} = 1.2408 - j0.8439 = 1.5006\angle -34.2208^\circ\ \text{pu} \\ S_{loss} &= S_{12} + S_{21} = V_1I_{12}^{*} - V_2I_{12}^{*} = 0.2702 + j0.3603\ \text{pu} \\ P_{loss} &= 27.02\ \text{MW}, \quad Q_{loss} = 36.03\ \text{MVAr} \end{aligned}$$

(The loss is found from the line current with the 2nd-iteration voltages. If it is instead taken as $S_1 - S_{load}$, it gives $0.2408 + j0.3439$ pu; the two differ because after two iterations the calculated bus 2 power has not yet reached the specified load (the solution is not fully converged). The converged answer is $V_2 = 0.7071\angle -8.130^\circ$ pu.)

Answer: $V_2 = 0.7227\angle -7.735^\circ$ pu; slack bus supplies $P_1 = 124.08$ MW and $Q_1 = 84.39$ MVAr; losses $\approx 27.02$ MW and $36.03$ MVAr.