N-R Ybus, mismatches and Jacobian with Q limits

Bus 1 is the slack bus, bus 2 a PV bus ($P_2 = 1.5$ pu, $|V_2| = 1.03$ pu, $0 \le Q_{G2} \le 0.8$ pu) and bus 3 a PQ bus ($P_3 = -1.2$, $Q_3 = -0.5$ pu). The shunt admittance j0.04 pu is taken as the total line charging of each line, with $j0.02$ pu at each end. Flat start: $V_2^{(0)} = 1.03\angle 0^\circ$, $V_3^{(0)} = 1\angle 0^\circ$.

(i) Y-bus

$$\begin{aligned} y_{series} &= \frac{1}{0.026 + j0.110} = 2.0351 - j8.6099\ \text{pu} \\ Y_{ii} &= 2y_{series} + 2(j0.02) = 4.0701 - j17.1798 = 17.6553\angle -76.6716^\circ \\ Y_{ij} &= -y_{series} = -2.0351 + j8.6099 = 8.8471\angle 103.2986^\circ \end{aligned}$$ $$Y_{bus} = \begin{bmatrix} 4.0701 - j17.1798 & -2.0351 + j8.6099 & -2.0351 + j8.6099 \\ -2.0351 + j8.6099 & 4.0701 - j17.1798 & -2.0351 + j8.6099 \\ -2.0351 + j8.6099 & -2.0351 + j8.6099 & 4.0701 - j17.1798 \end{bmatrix}\ \text{pu}$$

(ii) Calculated powers at the flat start

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, $j \ne i$):

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

Substituting $|V_1| = |V_2| = 1.03$, $|V_3| = 1.0$ and all angles zero (equivalently $P_i - jQ_i = V_i^{*}\sum_k Y_{ik}V_k$):

$$\begin{aligned} P_2^{0} &= 0.0629\ \text{pu}, \quad Q_2^{0} = 0.2236\ \text{pu} \\ P_3^{0} &= -0.1221\ \text{pu}, \quad Q_3^{0} = -0.5566\ \text{pu} \end{aligned}$$

Q-limit check: $Q_{G2} = Q_2^{0} + Q_{L2} = 0.2236$ pu, which lies between 0 and 0.8 pu. So bus 2 stays a PV bus and only $\Delta P_2$ is written for it.

(iii) Jacobian matrix

Unknowns: $\delta_2$, $\delta_3$, $|V_3|$. Equations: $\Delta P_2$, $\Delta P_3$, $\Delta Q_3$.

$$\begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0300)(1.0300)(8.8471)\sin(103.30^\circ) + (1.0300)(1.0000)(8.8471)\sin(103.30^\circ) = 18.0024 \\ \frac{\partial P_2}{\partial \delta_3} &= -(1.0300)(1.0000)(8.8471)\sin(103.30^\circ) = -8.8682 \\ \frac{\partial P_2}{\partial |V_3|} &= (1.0300)(8.8471)\cos(103.30^\circ) = -2.0961 \\ \frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0300)(8.8471)\sin(103.30^\circ) = -8.8682 \\ \frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0300)(8.8471)\sin(103.30^\circ) + (1.0000)(1.0300)(8.8471)\sin(103.30^\circ) = 17.7364 \\ \frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(17.6553)\cos(-76.67^\circ) + (1.0300)(8.8471)\cos(103.30^\circ) + (1.0300)(8.8471)\cos(103.30^\circ) = 3.9480 \\ \frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0300)(8.8471)\cos(103.30^\circ) = 2.0961 \\ \frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0300)(8.8471)\cos(103.30^\circ) + (1.0000)(1.0300)(8.8471)\cos(103.30^\circ) = -4.1922 \\ \frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(17.6553)\sin(-76.67^\circ) - (1.0300)(8.8471)\sin(103.30^\circ) - (1.0300)(8.8471)\sin(103.30^\circ) = 16.6232 \end{aligned}$$ $$J^{0} = \begin{bmatrix} 18.0024 & -8.8682 & -2.0961 \\ -8.8682 & 17.7364 & 3.9480 \\ 2.0961 & -4.1922 & 16.6232 \end{bmatrix}$$

(iv) Equations for the first N-R iteration

Mismatches:

$$\begin{aligned} \Delta P_2 &= P_2^{sp} - P_2^{0} = 1.5 - 0.0629 = 1.4371 \\ \Delta P_3 &= P_3^{sp} - P_3^{0} = -1.2 - (-0.1221) = -1.0779 \\ \Delta Q_3 &= Q_3^{sp} - Q_3^{0} = -0.5 - (-0.5566) = 0.0566 \end{aligned}$$

General equations:

$$\begin{bmatrix} \Delta P_2 \\ \Delta P_3 \\ \Delta Q_3 \end{bmatrix} = \begin{bmatrix} \frac{\partial P_2}{\partial \delta_2} & \frac{\partial P_2}{\partial \delta_3} & \frac{\partial P_2}{\partial |V_3|} \\ \frac{\partial P_3}{\partial \delta_2} & \frac{\partial P_3}{\partial \delta_3} & \frac{\partial P_3}{\partial |V_3|} \\ \frac{\partial Q_3}{\partial \delta_2} & \frac{\partial Q_3}{\partial \delta_3} & \frac{\partial Q_3}{\partial |V_3|} \end{bmatrix} \begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix}$$

With numbers:

$$\begin{bmatrix} 1.4371 \\ -1.0779 \\ 0.0566 \end{bmatrix} = \begin{bmatrix} 18.0024 & -8.8682 & -2.0961 \\ -8.8682 & 17.7364 & 3.9480 \\ 2.0961 & -4.1922 & 16.6232 \end{bmatrix} \begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix}$$

Solving, $\Delta\delta_2 = 0.0661$ rad, $\Delta\delta_3 = -0.0252$ rad, $\Delta|V_3| = -0.0113$ pu, so after the first iteration

$$\begin{aligned} \delta_2^{1} &= 0 + 0.0661 = 0.0661\ \text{rad} = 3.787^\circ \\ \delta_3^{1} &= 0 + (-0.0252) = -0.0252\ \text{rad} = -1.445^\circ \\ |V_3|^{1} &= 1 + (-0.0113) = 0.9887\ \text{pu} \end{aligned}$$

Answer: $P_2^0 = 0.0629$, $Q_2^0 = 0.2236$, $P_3^0 = -0.1221$, $Q_3^0 = -0.5566$ pu; bus 2 stays PV; $J^0$ as above; first-iteration result $V_2 = 1.0300\angle 3.787^\circ$ pu, $V_3 = 0.9887\angle -1.445^\circ$ pu.