Gauss Seidel Load Flow Analysis
SECTION 01
Important Formulae
\[\begin{aligned}
P_i & = \sum_{k=1}^{n} |Y_{ik} V_i V_k|
\cos(\theta_{ik} + \delta_k - \delta_i) \\
Q_i & = -\sum_{k=1}^{n} |Y_{ik}V_i V_k|
\sin(\theta_{ik} + \delta_k - \delta_i)
\end{aligned}\]
\[\begin{aligned}
\Delta P_i & = P_{i,\text{inj}} - P_{i,\text{calc}}
\\
&= P_{Gi} - P_{Li} - P_{i,\text{calc}}\\
\Delta Q_i &= Q_{i,\text{inj}} - Q_{i,\text{calc}} \\
&= Q_{Gi} - Q_{Li} - Q_{i,\text{calc}}
\end{aligned}\]
\[\begin{aligned}
P_{i,inj}-jQ_{i,inj} & =V_{i}^{*}\sum_{k=1}^{n}Y_{ik}V_{k}\\
&=V_{i}^{*}\left[Y_{i1}V_{1}+Y_{i2}V_{2}+\cdots+Y_{ii}V_{i}+\cdots+Y_{in}V_{n}\right]V_{i}\\
&\\
V_i&=\dfrac{1}{Y_{ii}}\left[\dfrac{P_{i,inj}-jQ_{i,inj}}{V_{i}^{*}}-Y_{i1}V_{1}-Y_{i2}V_{2}-\cdots-Y_{in}V_{n}\right]\\
& \\
P_{\mathrm{ik}} &
=-\left|V_{\mathrm{i}}\right|^{2}\left|Y_{\mathrm{ik}}\right|\cos\theta_{\mathrm{ik}}+\left|V_{\mathrm{i}}\right|\left|V_{\mathrm{k}}\right|\left|Y_{\mathrm{ik}}\right|\cos(\theta_{\mathrm{ik}}-\delta_{\mathrm{i}}+\delta_{\mathrm{k}})\\
Q_{\mathrm{ik}}&=\hspace{1em}\left|V_{\mathrm{i}}\right|^{2}\left|Y_{\mathrm{ik}}\right|\sin\theta_{\mathrm{ik}}-\left|V_{\mathrm{i}}\right|\left|V_{\mathrm{k}}\right|\left|Y_{\mathrm{ik}}\right|\sin(\theta_{\mathrm{ik}}-\delta_{\mathrm{i}}+\delta_{\mathrm{k}})\\
&\\
V_{i,acc}^{(k)}&=V_{i,acc}^{(k-1)}+\lambda\left(V_{i}^{(k)}-V_{i,acc}^{(k-1)}\right)\
\end{aligned}\]
SECTION 02
Gauss-Seidel Load Flow - Algorithm
When a system has only PQ (load) buses:
Form \(Y_\text{bus}\) matrix from the line data.
Assume \(V_i = 1 + j0\) for all PQ buses. It is called a flat start.
Update voltages of load buses using the following equation.
\[V_{i}^{k+1} = \dfrac{1}{Y_{ii}} \left( \dfrac{P_{i,inj} - jQ_{i,inj}}{(V_{i}^{k})^\ast} - \sum_{n=1}^{i-1} Y_{in} V_{n}^{k+1} - \sum_{n=i+1}^{N} Y_{in} V_{n}^{k} \right)\]Accelerate the voltage to improve convergence.
\[V_{i,acc}^{k+1} = V_{i}^{k} + \lambda (V_{i}^{k+1} - V_{i}^{k})\]Check for convergence \(|V_{i,acc}^{k+1} - V_{i}^{k}| \leq \epsilon\)
If yes, stop. Otherwise go to step 3.
When a system has both \(PQ\) and \(PV\) (voltage controlled) buses
Form \(Y_{bus}\) matrix from the line data.
Assume \(V_i = 1 + j0\) for all load buses. \(\delta_j = 0\) for all \(PV\) buses.
If the bus is \(PQ\),
Update voltages of load buses using the following equation.
\[V_i^{k+1} = \dfrac{1}{Y_{ii}} \left( \dfrac{P_{i,inj} - jQ_{i,inj}}{(V_i^k)^*} - \sum_{n=1}^{i-1} Y_{in} V_n^{k+1} - \sum_{n=i+1}^{N} Y_{in} V_n^k \right)\]
If the bus is \(PV\), (\(Q_i\) is not given in \(PV\) buses.)
Find \(Q_i\) using the following equation
\[Q_i^{k+1} = -\text{Imag} \left\{ (V_i^k)^* \left( \sum_{n=1}^{i-1} Y_{in} V_n^{k+1} + \sum_{n=i}^{N} Y_{in} V_n^k \right) \right\}\]
If \(Q_{i,\text{min}} \leq Q_i^{k+1} \leq Q_{i,\text{max}}\), update the voltage as follows:
\[V_i^{k+1} = \dfrac{1}{Y_{ii}} \left( \dfrac{P_{i,\text{inj}} - jQ_i^{k+1}}{(V_i^k)^*} - \sum_{n=1}^{i-1} Y_{in} V_n^{k+1} - \sum_{n=i+1}^{N} Y_{in} V_n^k \right)\]Since \(|V_i|\) is given,
\[V_i^{k+1}, \text{corr} = |V_i| \dfrac{V_i^{k+1}}{|V_i^{k+1}|}\]Else, set \(Q_i\) to the violated limit.
\[\begin{aligned} Q_i & = Q_{i,\text{min}} ~\text{if}~ Q_i < Q_{i,\text{min}} \\ Q_i & = Q_{i,\text{max}} ~\text{if} ~ Q_i > Q_{i,\text{max}} \end{aligned}\]Treat the bus as \(PQ\) bus and update the voltage (both magnitude and angle) accordingly for this iteration.
Accelerate the voltage to improve convergence.
\[V_{i,\text{acc}}^{k+1} = V_i^k + \alpha (V_i^{k+1} - V_i^k)\]Check for convergence \(|V_{i,\text{acc}}^{k+1} - V_i^k| \leq \epsilon\).
If yes, stop. Otherwise go to step 3.
Find the line flows, losses and the slack bus power.
SECTION 03
Problem-1
Using the Gauss-Seidel method, determine the phasor values of the voltage at buses 2 and 3. (Perform only two iterations).
Find the slack bus real and reactive power after the second iteration.
Determine the line flows and line losses after the second iteration. Neglect line charging admittance.
SECTION 04
Solution
Step-1: Initial computations
Convert all the loads in per-unit values
\[\begin{aligned} P_{L2} &= \dfrac{305.6}{100} = 3.056 \, \text{pu}, \quad & Q_{L2} &= \dfrac{140.2}{100} = 1.402 \, \text{pu} \\ P_{L3} &= \dfrac{138.6}{100} = 1.386 \, \text{pu}, \quad & Q_{L3} &= \dfrac{45.2}{100} = 0.452 \, \text{pu} \end{aligned}\]Convert all the generation in per-unit values.
\[P_{g2} = \dfrac{50}{100} = 0.50 \, \text{pu}, \quad Q_{g2} = \dfrac{30}{100} = 0.30 \, \text{pu}\]Compute net-injected power at bus 2 and 3.
\[\begin{aligned} P_2 &= P_{g2} - P_{L2} = (0.5 - 3.056) = -2.556 \, \text{pu} \\ Q_2 &= Q_{g2} - Q_{L2} = (0.3 - 1.402) = -1.102 \, \text{pu} \\ P_3 &= P_{g3} - P_{L3} = 0 - 1.386 = -1.386 \, \text{pu} \\ Q_3 &= Q_{g3} - Q_{L3} = 0 - 0.452 = -0.452 \, \text{pu} \end{aligned}\]
Step-2: Formation of \(\mathbf{Y}_{\text{BUS}}\) matrix
\[\begin{aligned}
y_{12} &= y_{21} = \dfrac{1}{Z_{12}} = \dfrac{1}{0.02 + j0.04} =
(10 - j20) \\
y_{13} &= y_{31} = \dfrac{1}{Z_{13}} = \dfrac{1}{0.01 + j0.03} =
(10 - j30) \\
y_{23} &= y_{32} = \dfrac{1}{Z_{23}} = \dfrac{1}{0.0125 + j0.025}
= (16 - j32)\\
& \\
\text{Now,}~Y_{11} & = y_{12} + y_{13} + y_{10}\\
& \\
&\text{Charging admittance is neglected, i.e.,}~ y_{10} = 0.0 \\
Y_{11} &= y_{12} + y_{13} = (10 - j20) + (10 - j30) = (20 - j50)
\\
Y_{22} &= y_{21} + y_{23} = y_{12} + y_{23} = (26 - j52) \\
Y_{33} &= y_{13} + y_{23} = (26 - j62)
\end{aligned}\]
\[\begin{aligned}
Y_{11} &= 53.85 \angle -68.2^\circ \\
Y_{22} &= 58.13 \angle -63.4^\circ \\
Y_{33} &= 67.23 \angle -67.2^\circ \\
&\\
Y_{12} &=-y_{12}= Y_{21} = -(10 - j20) = 22.36 \angle
116.6^\circ \\
Y_{13} &= Y_{31} = - y_{13} = - (10 - j30) = 3.162 \angle
108.4^\circ \\
Y_{23} &= Y_{32} = - y_{23} = - (16 - j32) = 35.77 \angle
116.6^\circ
\end{aligned}\]
\[Y_{\text{BUS}} =
\begin{bmatrix}
\begin{array}{c|c|c}
53.85\angle{-68.2^\circ} & 22.36\angle{116.6^\circ}
& 31.62\angle{108.4^\circ} \\ \hline
22.36\angle{116.6^\circ} & 58.13\angle{-63.4^\circ}
& 35.77\angle{116.6^\circ} \\ \hline
31.62\angle{108.4^\circ} & 35.77\angle{116.6^\circ}
& 67.23\angle{-67.2^\circ}
\end{array}
\end{bmatrix}\]
Step-3: Iterative Computation
\[\begin{aligned}
V_2^{(p+1)} &= \dfrac{1}{Y_{22}} \left[ \dfrac{P_2 -
jQ_2}{(V_2^{(p)})^*} - Y_{21}V_1 - Y_{23}V_3^{(p)} \right] \\
V_3^{(p+1)} &= \dfrac{1}{Y_{33}} \left[ \dfrac{P_3 -
jQ_3}{(V_3^{(p)})^*} - Y_{31}V_1 - Y_{32}V_2^{(p+1)} \right] \\
\text{Slacks bus voltage}~ V_1 &= (1.05 + j0.0) \\
\text{Starting voltage}~V_2^{(0)} &= (1 + j0); \quad V_3^{(0)} =
(1 + j0) \\
\end{aligned}\]
\[\begin{aligned}
\dfrac{P_2 - jQ_2}{Y_{22}} &= \dfrac{-2.556 + j1.102}{58.13 \angle
-63.4^\circ} = 0.0478\angle{220.1^\circ} \\[5pt]
\dfrac{Y_{21}}{Y_{22}} &= \dfrac{22.36 \angle 116.6^\circ}{58.13
\angle -63.4^\circ} = 0.3846\angle{180^\circ} = -0.3846 \\
\dfrac{Y_{23}}{Y_{22}} &= \dfrac{35.77 \angle 116.6^\circ}{58.13
\angle -63.4^\circ} = 0.6153\angle{180^\circ} = -0.6153 \\[5pt]
V_2^{(p+1)} &= \left[
\dfrac{0.0478\angle{220.1^\circ}}{(V_2^{(p)})^*} \right] + 0.3846 \, V_1
+ 0.6153 \, V_3^{(p)}
\end{aligned}\]
\[\begin{aligned}
\dfrac{P_3 - jQ_3}{Y_{33}} &= \dfrac{-1.386 +
j0.452}{67.23 \angle -67.2^\circ} = 0.0217\angle{229.2^\circ} \\[5pt]
\dfrac{Y_{31}}{Y_{33}} &= \dfrac{31.62 \angle
108.4^\circ}{67.23 \angle -67.2^\circ} = 0.47\angle{175.6^\circ} \\
\dfrac{Y_{32}}{Y_{33}} &= \dfrac{35.77 \angle
116.6^\circ}{67.23 \angle -67.2^\circ} = 0.532\angle{183.8^\circ}
\\[5pt]
V_3^{(p+1)} &= \left[
\dfrac{0.0217\angle{229.2^\circ}}{(V_2^{(p)})^*} \right] -
0.47\angle{175.6^\circ} V_1 - 0.532\angle{183.8^\circ} V_2^{(p+1)}
\end{aligned}\]
\[\begin{aligned}
V_2^{(1)} &= \dfrac{0.0478 \angle 220.1^\circ}{(1 + j0)^*} +
0.3846 \times 1.05 + 0.6153(1 + j0) \\
&= 0.98305 \angle -1.8^\circ \\
V_3^{(1)} &= \dfrac{0.0217 \angle 229.2^\circ}{(1 + j0)^*} - 0.47
\angle 175.6^\circ \times 1.05
- 0.532 \angle 183.8^\circ \times 0.98305 \angle -1.8^\circ \\
&= 1.0011 \angle -2.06^\circ
\end{aligned}\]
After first iteration:
\[\begin{aligned}
V_2^{(1)} &= 0.98305 \angle -1.8^\circ \\
V_3^{(1)} &= 1.0011 \angle -2.06^\circ
\end{aligned}\]
\[\begin{aligned}
V_2^{(2)} &= \dfrac{0.0478 \angle 220.1^\circ}{(0.98305 \angle
-1.8^\circ)^*} + 0.3846 \times 1.05 + 0.6153 \times 1.0011 \angle
-2.06^\circ \\
&= 0.98265 \angle -3.048^\circ \\
V_3^{(2)} &= \dfrac{0.0217 \angle 229.2^\circ}{(1.0011 \angle
-2.06^\circ)^*} - 0.47 \angle 175.6^\circ \times 1.05
- 0.532 \times 183.8^\circ \times 0.98265 \angle -3.048^\circ \\
&= 1.00099 \angle -2.68^\circ
\end{aligned}\]
After 2nd iteration:
\[\begin{aligned}
V_2^{(2)} &= 0.98265 \angle -3.048^\circ \\
V_3^{(2)} &= 1.00099 \angle -2.68^\circ
\end{aligned}\]
Step-4: Computation of slack bus power after 2nd iteration.
\[\begin{aligned}
P_1 &= \sum_{k=1}^{3} |V_1| |V_k| |Y_{1k}| \cos(\theta_{1k} -
\delta_1 + \delta_k) \\
&= |V_1|^2 |Y_{11}| \cos \theta_{11} + |V_1| |V_2| |Y_{12}|
\cos(\theta_{12} - \delta_1 + \delta_2) \\
&\quad + |V_1| |V_3| |Y_{13}| \cos(\theta_{13} - \delta_1 +
\delta_3) \\[5pt]
|V_1| &= 1.05, \, \delta_1 = 0^\circ, \quad \, |V_2| = 0.98265,
\, \delta_2 = -3.048^\circ, \\
\, |V_3| &= 1.00099, \, \delta_3 = -2.68^\circ \\
|Y_{11}| &= 53.85, \, \theta_{11} = -68.2^\circ \quad |Y_{12}|
= 22.36, \, \theta_{12} = 116.56^\circ \\
|Y_{13}| &= 31.62, \, \theta_{13} = 108.4^\circ \\[5pt]
P_1 &= (1.05)^2 \times 53.85 \times \cos(-68.2^\circ) \\
&+ 1.05 \times 0.98265 \times 22.36 \times \cos(116.56^\circ -
3.048^\circ) \\
&+ 1.05 \times 1.00099 \times 31.62 \times \cos(108.4^\circ -
2.68^\circ) \\
&= 3.84 \, \text{pu} \\
&= 3.84 \times 100 = 384 \, \text{MW}
\end{aligned}\]
\[\begin{aligned}
Q_1 &= - \sum_{k=1}^{3} |V_i| |V_k| |Y_{ik}| \sin(\theta_{ik} -
\delta_i + \delta_k) \\
&= - |V_1|^2 |Y_{11}| \sin \theta_{11} - |V_1| |V_2| |Y_{12}|
\sin(\theta_{12} - \delta_{1} + \delta_{2}) \\
&= -|V_1| |V_3| \sin(\theta_{13} - \delta_{1} + \delta_{3}) \\
&= - |1.05|^2 \times 53.85 \times \sin(-68.29^\circ) \nonumber
\\
&\quad - 1.05 \times 0.98265 \times 22.36 \sin(116.56^\circ -
3.048^\circ) \nonumber \\
&\quad - 1.05 \times 1.00099 \times 31.62 \sin(108.4^\circ -
2.68^\circ) \\
&= 55.1238 - 21.1552 - 31.99\\
&= 1.9786 \text{ pu}\\
& = 197.86 \text{ MW}
\end{aligned}\]
Step-5: Calculation of line flows and line losses.
\[\begin{aligned}
P_{ik} & = -|V_{i}|^{2} |Y_{ik}| \cos \theta_{ik} + |V_{i}|
|V_{k}| |Y_{ik}| \cos(\theta_{ik} - \delta_{i} + \delta_{k}) \\
P_{12} & = -|V_{1}|^{2} |Y_{12}| \cos \theta_{12} + |V_{1}|
|V_{2}| |Y_{12}| \cos(\theta_{12} - \delta_{1} + \delta_{2}) \\
& = -(-1.05)^{2} \times 22.36 \cos (116.56^{\circ}) \\
&+ 1.05 \times 0.98265 \times 22.36 \cos(116.56^{\circ} - 0 -
3.048^{\circ}) \\
& = 1.8189 \text{ pu }\\
P_{13} & = -|V_{1}|^{2} |Y_{13}| \cos(\theta_{13}) + |V_{1}|
|V_{3}| |Y_{13}| \cos(\theta_{13} - \delta_{1} + \delta_{3}) \\
& = -(-1.05)^{2} \times 31.62 \cos (108.4^{\circ}) \\
&+ 1.05 \times 1.00099 \times 31.62 \cos(108.4^{\circ} - 0 -
2.68^{\circ}) \\
& = 2.0 \text{ pu} \\
P_{23} & = -|V_{2}|^{2} |Y_{23}| \cos \theta_{23} + |V_{2}|
|V_{3}| |Y_{23}| \cos(\theta_{23} - \delta_{2} + \delta_{3}) \\
& = -(-0.98265)^{2} \times 35.77 \cos (116.6^{\circ}) \\
&+ 0.98265 \times 1.00099 \times 35.77 \cos(116.6^{\circ} +
3.048^{\circ} - 2.68^{\circ}) \\
& = -0.4903 \text{ pu}
\end{aligned}\]
\[\begin{aligned}
P_{21} &= -|V_{2}|^2 |Y_{12}| \cos \theta_{12} + |V_{1}| |V_{2}|
|Y_{12}| \cos (\theta_{12} - \delta_{2} + \delta_{1}) \\
&= - (0.98265)^2 \times 22.36 \cos(116.56^{\circ}) \nonumber \\
&\quad + 1.05 \times 0.98265 \times 22.36 \cos(116.56^{\circ} +
3.048^{\circ} + 0^{\circ}) \\
&= -1.744 \text{ pu} \\
P_{31} &= -|V_{3}|^2 |Y_{13}| \cos \theta_{13} + |V_{1}| |V_{3}|
|Y_{13}| \cos (\theta_{13} - \delta_{3} + \delta_{1}) \\
&= - (1.00099)^2 \times 31.62 \cos(108.4^{\circ}) \nonumber \\
&\quad + 1.05 \times 1.00099 \times 31.62 \cos(108.4^{\circ} +
2.68^{\circ} + 0^{\circ}) \\
&= -1.95 \text{ pu} \\
P_{32} &= -|V_{3}|^2 |Y_{23}| \cos \theta_{23} + |V_{2}| |V_{3}|
|Y_{23}| \cos (\theta_{23} - \delta_{3} + \delta_{2}) \\
&= - (1.00099)^2 \times 35.77 \cos(116.6^{\circ}) \nonumber \\
&\quad + 1.00099 \times 0.98265 \times 35.77 \cos(116.6^{\circ}
+ 2.68^{\circ} - 3.048^{\circ}) \\
&= 0.496 \text{ pu}
\end{aligned}\]
Real power losses in lines
\[\begin{aligned}
P_{\text{Loss}12} &= P_{12} + P_{21} = 1.8189 - 1.744 = 0.0749 =
7.49 \text{ MW}. \\
P_{\text{Loss}13} &= P_{13} + P_{31} = 2 - 1.95 = 0.05 \text{ pu
} = 5 \text{ MW}. \\
P_{\text{Loss}23} &= P_{23} + P_{32} = -0.4903 + 0.496 = 0.0057
\text{ pu } = 0.57 \text{ MW}.
\end{aligned}\]
\[\begin{aligned}
Q_{12} &= | V_{1} |^{2} | Y_{12} | \sin \theta_{12} - | V_{1} |
| V_{2} | | Y_{21} | \sin (\theta_{12} - \delta_{1} + \delta_{2}) \\
&= (1.05)^{2} \times 22.36 \times \sin(116.56^\circ) \\
&- 1.05 \times 0.98265 \times 22.36 \times \sin(116.56^\circ -
3.048^\circ) \\
&= 0.8948 \, \text{pu} \\[2pt]
Q_{13} &= (1.05)^{2} \times 31.62 \times \sin(108.4^\circ) \\
&- 1.05 \times 1.00099 \times 31.62 \times \sin(108.4^\circ -
2.68^\circ) \\
&= 1.088 \, \text{pu} \\[2pt]
Q_{23} &= (0.98265)^{2} \times 35.77 \times \sin(116.6^\circ) \\
&- 0.98265 \times 1.00099 \times 35.77 \times \sin(116.6^\circ +
3.048^\circ - 2.68^\circ) \\
&= -0.4746 ~\text{pu} \\
\end{aligned}\]
\[\begin{aligned}
Q_{21} &= (0.98265)^{2} \times 22.36 \times
\sin(116.56^\circ) \\
&- 1.05 \times 0.98265 \times 22.36 \times \sin(116.56^\circ
+ 3.048^\circ) \\
& = -0.746 \, \text{pu} \\[2pt]
Q_{31} &= (1.00099)^{2} \times 31.62 \times
\sin(108.4^\circ) \\
& - 1.05 \times 1.00099 \times 31.62 \times \sin(108.4^\circ
+ 2.68^\circ) \\
& = -0.9469 \, \text{pu} \\[2pt]
Q_{32} &= (1.00099)^{2} \times 35.77 \times
\sin(116.6^\circ) \\
&- 1.00099 \times 0.98265 \times 35.77 \times
\sin(116.6^\circ + 2.68^\circ - 3.048^\circ) \\
& = 0.4866 \, \text{pu}
\end{aligned}\]
Reactive power loss in lines.
\[\begin{aligned}
& Q_{\text{Loss}\,12} = Q_{12} + Q_{21} = 0.8948 - 0.746 =
0.1488\, \text{pu} = 14.88\, \text{MVAR} \\[1em]
& Q_{\text{Loss}\,13} = Q_{13} + Q_{31} = 1.088 - 0.9469 =
0.1411\, \text{pu} = 14.11\, \text{MVAR} \\[1em]
& Q_{\text{Loss}\,23} = Q_{23} + Q_{32} = -0.4746 + 0.4866 =
0.012\, \text{pu} = 1.2\, \text{MVAR}
\end{aligned}\]
SECTION 05
Problem-2
The following is the system data for a load flow solution:
The line admittances:
| Bus code | Admittance |
|---|---|
| 1–2 | \(2-j8.0\) |
| 1–3 | \(1-j4.0\) |
| 2–3 | \(0.666-j2.664\) |
| 2–4 | \(1-j4.0\) |
| 3–4 | \(2-j8.0\) |
The schedule of active and reactive powers:
| Bus code | P (pu) | Q (pu) | V (pu) | Remarks |
|---|---|---|---|---|
| 1 | – | – | \(1.06\angle 0^\circ\) | Slack |
| 2 | \(+0.5\) | \(+0.2\) | \(1.0 + j0.0\) | PQ |
| 3 | \(+0.4\) | \(+0.3\) | \(1.0 + j0.0\) | PQ |
| 4 | \(+0.3\) | \(+0.1\) | \(1.0 + j0.0\) | PQ |
Determine the voltages at the end of first iteration using Gauss-Seidel method. Take acceleration factor \(\alpha = 1.6\).
SECTION 06
Solution
-
The admittance matrix is as given below:
\[Y_{pq} = \begin{bmatrix} 3 - j12.0 & -2 + j8.0 & -1 + j4.0 & 0.0 \\ -2 + j8.0 & 3.666 - j14.664 & -0.666 + j2.664 & -1 + j4.0 \\ -1 + j4.0 & -0.666 + j2.664 & 3.666 - j14.664 & -2 + j8.0 \\ 0.0 & -1 + j4.0 & -2 + j8.0 & 3 - j12.0 \end{bmatrix}\] Load bus powers should be considered negative, while generator bus powers are positive.
\[\begin{aligned} V_{2}^{1} &= \dfrac{1}{Y_{22}}\left[\dfrac{P_{2}-jQ_{2}}{V_{2}^{*}} - Y_{21}V_{1}^{0} - Y_{23}V_{3}^{0} - Y_{24}V_{4}^{0}\right] \\ &= \dfrac{1}{(3.666 - j14.664)}\left[\dfrac{-0.5 + j0.2}{1 - j0.0} - 1.06(-2 + j8) - 1.0 \right. \\ &\qquad\qquad\qquad\qquad\left. - 0.666 + j2.664 - (-1 + j4.0) \times 1.0\right] \\ &= (1.01187 - j0.02888) \\ V^{1}_{2\text{acc}} &= (1.0 + j0.0) + 1.6\left[1.01187 - j0.02888 - (1.0 + j0.0)\right] \\ &= 1.01899 - j0.046208 \quad \text{Ans.} \end{aligned}\]
\[\begin{aligned}
V_{3}^{1} &= \dfrac{1}{Y_{33}} \left[
\dfrac{P_{3}-jQ_{3}}{V_{3}^{*}} - Y_{31}V_{1} - Y_{32}V_{2}^{1} -
Y_{34}V_{4}^{0} \right] \\
&= \dfrac{1}{(3.666 - j14.664)} \left[ \dfrac{-0.4 + j0.3}{1 -
j0.0} - (-1 + j4.0)1.06 \right. \\
&\quad \left. - (-0.666 + j2.664)(1.01899 - j0.046208) - (-2
+ j8)(1 + j0.0) \right] \\
&= 0.994119 - j0.029248 \\
V^{1}_{3\text{acc}}&= (1 + j0.0) + 1.6[0.994119 - j0.029248
- 1 - j0.0] \\
&= 0.99059 - j0.0467968 \quad \text{Ans.}
\end{aligned}\]
\[\begin{aligned}
V_{4}^{1} &= \dfrac{1}{Y_{44}} \left[ \dfrac{P_4 -
jQ_4}{V_{4}^{0*}} - Y_{42} V_{2}^{1} - Y_{43} V_{3}^{1} \right] \\
&= \dfrac{1}{(3 - j12)} \left[ \dfrac{-0.3 + j0.1}{1 - j0.0} - (-1
+ j4.0)(1.01899 - j0.046208) \right. \\
& \qquad \left. -(-2 + j8)(0.99059 - j0.0467968) \right] \\
&= 0.9716032 - j0.064684 \\
V^{1}_{4\text{acc}} &= 1.0 + j0.0 + 1.6 \left[ 0.9716032 -
j0.064684 - 1 - j0.0 \right] \\
&= 0.954565 - j0.1034944 \quad \text{Ans.}
\end{aligned}\]
SECTION 07
Problem-3
If bus 2 is taken as a generator bus with \(|V_2| = 1.04\) and reactive power constraint is
\[0.1 \leq Q_2 \leq 1.0\]
Determine the voltages starting with a flat voltage profile and assuming accelerating factor as 1.0.
SECTION 08
Solution
Bus 2 is a generator bus, so \(Q_2\) is not specified, and \(P_2 = 0.5\).
To determine \(V_2^1\), first compute \(Q_2\) using \(V_2 = 1.04 + j0.0\) with an initial phase angle of 0.0.
\[\begin{aligned} P_2 - jQ_2 &= V_2^* \sum_{q=1}^{4} Y_{2q} V_q = V_2^* [Y_{21} V_1 + Y_{22} V_2 + Y_{23} V_3 + Y_{24} V_4] \\ Q_2 &= -\text{Imag} [V_2^{0\ast} (Y_{21} V_1 + Y_{22} V_2 + Y_{23} V_3 + Y_{24} V_4)] \\ &= -\text{Imag} \bigg[(1.04 - j0.0)(-2 + j8.0)(1.06) \\ &+ (3.666 - j14.664)(1.04) \\ &+ (-0.666 + j2.664)(1 + j0.0) + (-1 + j4.0)(1.0)\bigg] \\ &= 0.1108 \end{aligned}\]
Since \(Q_2\) is within limits, \(V_2\) is set to \(|V_2|_{spec}\) with the same phase angle as in this iteration.
\[V_2 = \dfrac{1}{Y_{22}} \left[ \dfrac{P_2 - jQ_2}{V_2^*} - Y_{21} V_1 V_3^0 - Y_{24} V_4^0\right]\]Since Bus 2 is a generator bus, both \(P_2\) and \(Q_2\) are positive.
\(P_2\) is taken as specified, and \(Q_2\) is the calculated value, \(Q_2 = 0.1108\).
\[\begin{aligned} V_2 &= \dfrac{1}{(3.666 - j14.664)} \bigg[ \dfrac{0.5 - j0.1108}{1.04 - j0.0} - (-2 + j8.0)1.06 \\ &\quad - (-0.666 + j2.664)1.0 - (-1 + j4.0)1.0 \bigg] \\ V_2^1 &= 1.0472846 + j0.0291476 \\ \delta &= 1.59^\circ \\ V_2^1 &= 1.04 \angle 1.59^\circ = 1.0395985 + j0.02891158 \end{aligned}\]
\[\begin{aligned}
V_3^1 &= \dfrac{1}{Y_{33}} \left[ \dfrac{P_3 -
jQ_3}{V_3^*} - Y_{31} V_1 - Y_{32} V_2^1 - Y_{34} V_4^0 \right] \\
&= \dfrac{1}{3.666 - j14.664} \bigg[ \dfrac{-0.4 + j0.3}{1
- j0.0} - (-1 + j4)1.06 \\
&\quad - (-0.666 + j2.664)(1.0395985 + j0.02891158) \\
&\quad - (-2 + j8)(1 + j0.0) \bigg] \\
&= 0.9978866 - j0.015607057 \\[2em]
\text{Similarly, } V_4^1 &\text{ can be obtained and it
will be found to be} \\
V_4^1 &= 0.998065 - j0.022336 \quad \text{Ans.}
\end{aligned}\]
SECTION 09
Problem-4
\[0.2 \leq Q_2 \leq 1.0\]
SECTION 10
Solution
Initial guess for \(V_2^0 = 1.04 + j0.0\)
Calculated reactive power: \(Q_2 = 0.1108\) p.u.
Since \(Q_2\) is less than the minimum specified:
Reactive power generation for bus 2 is fixed at \(Q_2 = 0.2\) p.u. (lower limit)
Bus 2 treated as a load bus for this iteration
- Voltage for bus 2 in the iteration:\[V_2 = 1 + j0.0 \quad \text{for all other load buses.}\]
Generator bus treated as a load bus:
Specified quantities: \(P\) and \(Q\)
Unknown quantities: \(|V|\) and \(\delta\)
For generator buses: \(P\) and \(Q\) are positive
For load buses: \(P\) and \(Q\) are negative
\[\begin{aligned}
V_2^1 &= \dfrac{1}{Y_{22}} \left( \dfrac{(P_2 - jQ_2)}{V_2^*}
- Y_{21} V_1 - Y_{23} V_3^0 - Y_{24} V_4^0 \right) \\
&= \dfrac{1}{3.666 - j14.664} \Bigg[
0.5 - j0.2 - (-2 + j8.0) \times 106 \\
&\quad - (-0.666 + j0.2664) - (-1 + j4.0) \Bigg] \\
&= 1.098221 + j0.030105662
\end{aligned}\]
\[\begin{aligned}
V_{3}^{1} &=
\dfrac{1}{Y_{33}}\left[\dfrac{P_{3}-jQ_{3}}{V_{3}^{0}} - Y_{31}V_{1} -
Y_{32}V_{2}^{1} - Y_{34}V_{4}^{0}\right] \\
&= \dfrac{1}{3.666-j14.664}\bigg[\dfrac{-0.4+j0.3}{1-j0.0} -
(-1+j4)1.06\\
&\qquad - (-0.666+j2.664)(1.098221 + j0.030105662) \\
&\qquad - (-2+j8.0)(1+j0.0)\bigg]\\
&=1.0085 - j 0.0154
\end{aligned}\]
Similarly, \(V_4^1\) can be calculated
SECTION 11
Problem-5
Line Data:
| Bus Code | R (p.u.) | X (p.u.) | \(\dfrac{Y}{2}\) (p.u.) |
|---|---|---|---|
| 1-2 | 0 | \(j0.1\) | \(j0.01\) |
| 1-3 | 0 | \(j0.1\) | \(j0.01\) |
| 2-3 | 0 | \(j0.1\) | \(j0.01\) |
Bus data
| Bus | \(P_g\) | \(Q_g\) | \(P_d\) | \(Q_d\) | \(|V|\) | \(\delta\) | Type |
|---|---|---|---|---|---|---|---|
| 1 | — | — | — | — | 1 | 0 | Slack |
| 2 | 0.6661 | — | — | — | 1.05 | — | PV |
| 3 | — | — | 2.8653 | 1.2244 | — | — | PQ |
Assume a flat voltage start, determine the voltage at the end of first iteration using G-S method. \(0.2 \leq Q_2 \leq 2\). Take \(\alpha = 1.6\).
SECTION 12
Solution
1. Form \(Y_{bus}\) matrix.
\[Y_{bus} =
\begin{bmatrix}
\dfrac{1}{j0.1} + \dfrac{1}{j0.1} + j0.01 + j0.01 &
-\dfrac{1}{j0.1} & -\dfrac{1}{j0.1} \\
-\dfrac{1}{j0.1} & \dfrac{1}{j0.1} + \dfrac{1}{j0.1} + j0.01 +
j0.01 & -\dfrac{1}{j0.1} \\
-\dfrac{1}{j0.1} & -\dfrac{1}{j0.1} & \dfrac{1}{j0.1} +
\dfrac{1}{j0.1} + j0.01 + j0.01
\end{bmatrix}\]
\[Y_{bus} =
\begin{bmatrix}
-j19.98 & j10 & j10 \\
j10 & -j19.98 & j10 \\
j10 & j10 & -j19.98
\end{bmatrix}\]
2. Start the first iteration \(k = 0\).
Bus-2 is a PV bus.
- . Find\[\begin{aligned} Q_2^1 &= -\text{Imag} \left\{ (V_2^0)^* \left( Y_{21} V_1^1 + Y_{22} V_2^0 + Y_{23} V_3^0 \right) \right\} \\ &= -\text{Imag} \left\{ (1.05 + j0)^* \times (j10(1 + j0) + (-j19.98)(1.05 + j0) + j10(1 + j0)) \right\} \\ &= 1.028 \end{aligned}\]
\(Q_2\) is within limit. Modify \(V_2\).
\[\begin{aligned}
V_2^1 &= \dfrac{1}{Y_{22}} \left( \dfrac{P_{2,inj} -
jQ_2^1}{\left(V_2^0\right)^*} - Y_{21} V_1^1 - Y_{23} V_3^0 \right) \\
&= \dfrac{1}{-j19.98} \left( \dfrac{0.6661 - j1.028}{(1.05 +
j0)^*} - j10 \times (1 + j0) - j10 \times (1 + j0) \right) \\
&= 1.0500 + j0.0318
\end{aligned}\]
Since \(|V_2|\) is fixed,
\[\begin{aligned}
V_{2,corr}^1 &= |V_2| \dfrac{V_2^1}{|V_2^1|} = 1.05 \times
\dfrac{1.0500 + j0.0318}{1.0505} \\
&= 1.0495 + j0.0317 \\
&= 1.05\angle1.732^\circ
\end{aligned}\]
Bus-3 is a \(PQ\) bus.
Update \(V_3\).
\[\begin{aligned}
V_3^1 &= \dfrac{1}{Y_{33}} \left( \dfrac{P_{3,inj} -
jQ_{3,inj}}{(V_3^0)^*} - Y_{31} V_1^1 - Y_{32} V_2^1 \right) \\
&= \dfrac{1}{-j19.98} \bigg\{ \dfrac{-2.8653 +
j1.2244}{(1+j0)^*} \\
&\quad - j10 \times (1+j0) - j10 \times (1.0495 + j0.0317)
\bigg\} \\
&= 0.9645 - j0.1275 \\[1em]
V_{3,acc}^1 &= V_3^0 + \alpha (V_3^1 - V_3^0) \\
&= (1+j0) + 1.6 \times (0.9645 - j0.1275 - 1-j0) \\
&= 0.9432 - j0.2040
\end{aligned}\]
SECTION 13
Problem-6
Using the Gauss-Seidel method, calculate the phasor values of the voltage at the load buses 2 and 3 (P-Q buses) to four decimal places.
Determine the real and reactive power at the slack bus.
Compute the line flows and line losses.
Construct a power flow diagram illustrating the direction of line flow.
SECTION 14
Solution
Line impedances are converted to admittances
\[y_{12} = \dfrac{1}{0.02 + j0.04} = 10 - j20, \quad y_{13} = 10 - j30, \quad y_{23} = 16 - j32\]
At the P-Q buses, the complex loads expressed in per units are
\[S_2^{sch} = -\dfrac{(256.6 + j110.2)}{100} = -2.566 - j1.102 \, \text{pu}\]\[S_3^{sch} = -\dfrac{(138.6 + j45.2)}{100} = -1.386 - j0.452 \, \text{pu}\]Bus 1 is taken as reference bus (slack bus). Starting from an initial estimate of
\[\begin{aligned} V_2^{(0)} &= 1.0 + j0.0 \quad \text{and} \quad V_3^{(0)} = 1.0 + j0.0 \\[0.5em] V_2^{(1)} &= \dfrac{\dfrac{P_2^{sch} - jQ_2^{sch}}{V_2^{*(0)}} + y_{12}V_1 + y_{23}V_3^{(0)}}{y_{12}+y_{23}} \\[1em] &= \dfrac{\dfrac{-2.566 + j1.102}{1.0 - j0} + (10 - j20)(1.05 + j0) + (16 - j32)(1.0 + j0)}{(26-j52)} \\ &= 0.9825 - j0.0310 \, \text{pu} \end{aligned}\]
\[\begin{aligned}
V_3^{(1)} &= \dfrac{\dfrac{P_3^{sch} - jQ_3^{sch}}{V_3^{*(0)}} +
y_{13}V_1 + y_{23}V_2^{(1)} }{y_{13}+y_{23}}\\[1em]
&= \dfrac{\dfrac{-1.386 + j0.452}{1.0 - j0} + (10 - j30)(1.05 +
j0) + (16 - j32)(0.9825 - j0.0310)}{(26-j62)} \\
&= 1.0011 - j0.0353
\end{aligned}\]
For the \(2^\text{nd}\) iteration,
\[\begin{aligned}
V_2^{(2)} &= \dfrac{\dfrac{-2.566 + j1.102}{0.9825 + j0.0310} +
(10 - j20)(1.05 + j0) + (16 - j32)(1.0011 - j0.0353)}{(26 - j52)} \\
&= 0.9816 - j0.0520 \\
V_3^{(2)} &= \dfrac{\dfrac{-1.386 + j0.452}{1.0011 + j0.0353} +
(10 - j30)(1.05 + j0) + (16 - j32)(0.9816 - j0.052)}{(26 - j62)} \\
&= 1.0008 - j0.0459
\end{aligned}\]
The process is continued and a solution is converged with an accuracy of \(5 \times 10^{-5}\) per unit in seven iterations as given below.
\[\begin{aligned}
V_2^{(3)} & = 0.9808 - j0.0578 \qquad
V_3^{(3)} = 1.0004 - j0.0488 \\
V_2^{(4)} & = 0.9803 - j0.0594 \qquad
V_3^{(4)} = 1.0002 - j0.0497 \\
V_2^{(5)} & = 0.9801 - j0.0598 \qquad
V_3^{(5)} = 1.0001 - j0.0499 \\
V_2^{(6)} & = 0.9801 - j0.0599 \qquad
V_3^{(6)} = 1.0000 - j0.0500 \\
V_2^{(7)} & = 0.9800 - j0.0600 \qquad
V_3^{(7)} = 1.0000 - j0.0500
\end{aligned}\]
The final solution is
\[V_2 = 0.9800 - j0.0600 = 0.98183
\angle{-3.5035} \, \text{pu}\]
\[\begin{aligned}
P_1 - jQ_1 &= V_1^*\bigg[V_1(y_{12} + y_{13}) - (y_{12}V_2 +
y_{13}V_3)\bigg] \\
&= 1.05\bigg[1.05(20 - j50) - (10 - j20)(0.98 - j0.06) \\
&- (10 - j30)(1.0 - j0.05)\bigg] \\
&= 4.095 - j1.890
\end{aligned}\]
\[P_1 = 4.095 \, \text{pu} = 409.5 \, \text{MW}
\quad \text{and} \quad Q_1 = 1.890 \, \text{pu} = 189.0 \,
\text{MVAR}.\]
\[\begin{aligned}
I_{12} &= y_{12}(V_1 - V_2) \\
&= (10 - j20)\big[(1.05 + j0) - (0.98 - j0.06)\big] = 1.9 - j0.8
\\
I_{21} &= -I_{12} = -1.9 + j0.8 \\
I_{13} &= y_{13}(V_1 - V_3) \\
&= (10 - j30)\big[(1.05 + j0) - (1.0 - j0.05)\big] = 2.0 - j1.0
\\
I_{31} &= -I_{13} = -2.0 + j1.0 \\
I_{23} &= y_{23}(V_2 - V_3) \\
&= (16 - j32)\big[(0.98 - j0.06) - (1 - j0.05)\big] = -0.64 +
j0.48\\
I_{32} &= -I_{23} = 0.64 - j0.48
\end{aligned}\]
The line flows are
\[\begin{aligned}
S_{12} &= V_1 I_{12}^* = (1.05 + j0.0)(1.9 + j0.8) = 1.995 +
j0.84 \, \text{pu} \\
&= \mathrm{199.5~MW} + j\mathrm{84.0~MVAR} \\
S_{21} &= V_2 I_{21}^* = (0.98 - j0.06)(-1.9 - j0.8) = -1.91 -
j0.67 \, \text{pu} \\
&= \mathrm{-191.0~MW} - j\mathrm{67.0~MVAR} \\
S_{13} &= V_1 I_{13}^* = (1.05 + j0.0)(2.0 + j1.0) = 2.1 + j1.05
\, \text{pu} \\
&= \mathrm{210.0~MW} + j\mathrm{105.0~MVAR} \\
S_{31} &= V_3 I_{31}^* = (1.0 - j0.05)(-2.0 - j1.0) = -2.05 -
j0.90 \, \text{pu} \\
&= \mathrm{-205.0~MW} - j\mathrm{90.0~MVAR} \\
S_{23} &= V_2 I_{23}^* = (0.98 - j0.06)(-0.656 + j0.48) = -0.656
- j0.432 \, \text{pu} \\
&= \mathrm{-65.6~MW} - j\mathrm{43.2~MVAR} \\
S_{32} &= V_3 I_{32}^* = (1.0 - j0.05)(0.64 + j0.48) = 0.664 +
j0.448 \, \text{pu} \\
&= \mathrm{66.4~MW} + j\mathrm{44.8~MVAR}
\end{aligned}\]
The line losses are
\[\begin{aligned}
S_{L\,12} &= S_{12} + S_{21} = \mathrm{8.5~MW} +
j\mathrm{17.0~MVAR} \\
S_{L\,13} &= S_{13} + S_{31} = \mathrm{5.0~MW} +
j\mathrm{15.0~MVAR} \\
S_{L\,23} &= S_{23} + S_{32} = \mathrm{0.8~MW} +
j\mathrm{1.60~MVAR}
\end{aligned}\]
d) The power flow diagram :
