N-R two iterations, y12 = 10∠−73.74°

Newton-Raphson (polar form) is applied to the two-bus system. The line admittance is $y_{12} = 10\angle -73.74^\circ = 2.8 - j9.6$ pu, and the load 150 MW + j50 MVAr is $1.5 + j0.5$ 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 $P_2 = -1.50$ pu and $Q_2 = -0.50$ pu (negative, as power is taken out). Unknowns: $\delta_2$, $|V_2|$. Flat start: $|V_2|^{(0)} = 1$, $\delta_2^{(0)} = 0$.

$$\begin{aligned} y_{12} &= 10\angle -73.74^\circ = 2.8 - j9.6 \\ Y_{11} &= Y_{22} = y_{12} = 2.8000 - j9.6000 = 10.0000\angle -73.74^\circ \\ Y_{12} &= Y_{21} = -y_{12} = -2.8000 + j9.6000 = 10.0000\angle 106.26^\circ \end{aligned}$$

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:

$$\begin{aligned} P_2^{(0)} &= 0.0000, \quad \Delta P_2 = -1.5000 - (0.0000) = -1.5000 \\ Q_2^{(0)} &= 0.0000, \quad \Delta Q_2 = -0.5000 - (0.0000) = -0.5000 \end{aligned}$$

Jacobian elements:

$$\begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0000)(1.0000)(10.0000)\sin(106.26^\circ) = 9.6000 \\ \frac{\partial P_2}{\partial |V_2|} &= 2(1.0000)(10.0000)\cos(-73.74^\circ) + (1.0000)(10.0000)\cos(106.26^\circ) = 2.8000 \\ \frac{\partial Q_2}{\partial \delta_2} &= (1.0000)(1.0000)(10.0000)\cos(106.26^\circ) = -2.8000 \\ \frac{\partial Q_2}{\partial |V_2|} &= -2(1.0000)(10.0000)\sin(-73.74^\circ) - (1.0000)(10.0000)\sin(106.26^\circ) = 9.6000 \end{aligned}$$ $$\begin{bmatrix} -1.5000 \\ -0.5000 \end{bmatrix} = \begin{bmatrix} 9.6000 & 2.8000 \\ -2.8000 & 9.6000 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix}$$ $$\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} = \begin{bmatrix} -0.1300 \\ -0.0900 \end{bmatrix}$$ $$\begin{aligned} \delta_2^{(1)} &= 0.0000 + (-0.1300) = -0.1300\ \text{rad} = -7.4485^\circ \\ |V_2|^{(1)} &= 1.0000 + (-0.0900) = 0.9100\ \text{pu} \end{aligned}$$

Iteration 2

Calculated powers and mismatches with the values from iteration 1:

$$\begin{aligned} P_2^{(1)} &= -1.3403, \quad \Delta P_2 = -1.5000 - (-1.3403) = -0.1597 \\ Q_2^{(1)} &= -0.3822, \quad \Delta Q_2 = -0.5000 - (-0.3822) = -0.1178 \end{aligned}$$

Jacobian elements:

$$\begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (0.9100)(1.0000)(10.0000)\sin(113.71^\circ) = 8.3320 \\ \frac{\partial P_2}{\partial |V_2|} &= 2(0.9100)(10.0000)\cos(-73.74^\circ) + (1.0000)(10.0000)\cos(113.71^\circ) = 1.0751 \\ \frac{\partial Q_2}{\partial \delta_2} &= (0.9100)(1.0000)(10.0000)\cos(113.71^\circ) = -3.6590 \\ \frac{\partial Q_2}{\partial |V_2|} &= -2(0.9100)(10.0000)\sin(-73.74^\circ) - (1.0000)(10.0000)\sin(113.71^\circ) = 8.3160 \end{aligned}$$ $$\begin{bmatrix} -0.1597 \\ -0.1178 \end{bmatrix} = \begin{bmatrix} 8.3320 & 1.0751 \\ -3.6590 & 8.3160 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix}$$ $$\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} = \begin{bmatrix} -0.0164 \\ -0.0214 \end{bmatrix}$$ $$\begin{aligned} \delta_2^{(2)} &= -0.1300 + (-0.0164) = -0.1464\ \text{rad} = -8.3885^\circ \\ |V_2|^{(2)} &= 0.9100 + (-0.0214) = 0.8886\ \text{pu} \end{aligned}$$

i) Bus 2 voltage after the 2nd iteration

$|V_2| = 0.8886$ pu, $\delta_2 = -8.389^\circ$, i.e. $V_2 = 0.8886\angle -8.389^\circ$ pu.

ii) Slack bus power and network losses

$$\begin{aligned} S_1 = P_1 + jQ_1 &= V_1\left[Y_{11}V_1 + Y_{12}V_2\right]^{*} = 1.5830 + j0.7976\ \text{pu} \\ &= 158.30\ \text{MW} + j79.76\ \text{MVAr} \end{aligned}$$ $$\begin{aligned} I_{12} &= (V_1 - V_2)y_{12} = 1.5830 - j0.7976 = 1.7726\angle -26.7403^\circ\ \text{pu} \\ S_{loss} &= S_{12} + S_{21} = V_1I_{12}^{*} - V_2I_{12}^{*} = 0.0880 + j0.3016\ \text{pu} \\ P_{loss} &= 8.80\ \text{MW}, \quad Q_{loss} = 30.16\ \text{MVAr} \end{aligned}$$

(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 $0.0830 + j0.2976$ 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 $V_2 = 0.8879\angle -8.420^\circ$ pu.)

Answer: $V_2 = 0.8886\angle -8.389^\circ$ pu; slack bus supplies $P_1 = 158.30$ MW and $Q_1 = 79.76$ MVAr; losses $\approx 8.80$ MW and $30.16$ MVAr.