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G-S one iteration with π-model lines

Short answer

Using nominal-π Ybus with line charging, one G-S iteration gives V2=1.0281∠−4.6106° pu and V3=1.0796∠−7.6322° pu. Slack bus complex power is S1=1.6206−j0.1054 pu.

The single line diagram of a power system network is shown in figure below. The per km line series reactance is 0.001 p.u and shunt susceptance is 0.0016 p.u. [Figure: line 1-2 is 100 km, line 1-3 is 200 km, line 2-3 is 400 km] The power and bus voltages are as follows:

Bus No.PLoadQLoadPgenQgenV
100??1.05∠0°
21.00.600?
30.80.300?

Using Ybus from the nominal pi-model of lines, compute the voltage at bus 2 & 3 using G-S method for one iteration and also the slack bus complex power using the computed and specified bus voltages. Start the analysis assuming unknown bus voltage magnitudes as 1 p.u. & phase angles zero degree.

Each line is modelled as a nominal-π: series reactance = 0.001 × length, total shunt susceptance = 0.0016 × length, half at each end. Bus 1 is the slack bus (1.05∠0∘1.05\angle 0^\circ); buses 2 and 3 are load buses with P2−jQ2=−1.0+j0.6P_2 - jQ_2 = -1.0 + j0.6 and P3−jQ3=−0.8+j0.3P_3 - jQ_3 = -0.8 + j0.3 pu.

Line parameters

LineLength (km)XX (pu)y=1/jXy = 1/jX (pu)B/2B/2 each end (pu)
1-21000.1−j10-j10j0.08j0.08
1-32000.2−j5-j5j0.16j0.16
2-34000.4−j2.5-j2.5j0.32j0.32

Y-bus

Y11=−j10−j5+j0.08+j0.16=−j14.76Y22=−j10−j2.5+j0.08+j0.32=−j12.10Y33=−j5−j2.5+j0.16+j0.32=−j7.02Y12=j10,Y13=j5,Y23=j2.5\begin{aligned} Y_{11} &= -j10 - j5 + j0.08 + j0.16 = -j14.76 \\ Y_{22} &= -j10 - j2.5 + j0.08 + j0.32 = -j12.10 \\ Y_{33} &= -j5 - j2.5 + j0.16 + j0.32 = -j7.02 \\ Y_{12} &= j10, \quad Y_{13} = j5, \quad Y_{23} = j2.5 \end{aligned} Ybus=[−j14.76j10.00j5.00j10.00−j12.10j2.50j5.00j2.50−j7.02] puY_{bus} = \begin{bmatrix} -j14.76 & j10.00 & j5.00 \\ j10.00 & -j12.10 & j2.50 \\ j5.00 & j2.50 & -j7.02 \end{bmatrix}\ \text{pu}

Bus voltages after one G-S iteration

Iteration 1

V2(1)=1Y22[P2−jQ2V2∗−Y21V1−Y23V3]P2−jQ2V2∗=−1.0000+j0.60001.0000=−1.0000+j0.6000∑k≠2Y2kVk=j13.0000V2(1)=(−1.0000+j0.6000)−(j13.0000)−j12.1000=1.0248−j0.0826=1.0281∠−4.6106∘ pu\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{-1.0000 + j0.6000}{1.0000} = -1.0000 + j0.6000 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= j13.0000 \\ V_2^{(1)} &= \frac{(-1.0000 + j0.6000) - (j13.0000)}{-j12.1000} = 1.0248 - j0.0826 \\ &= 1.0281\angle -4.6106^\circ\ \text{pu} \end{aligned} V3(1)=1Y33[P3−jQ3V3∗−Y31V1−Y32V2]P3−jQ3V3∗=−0.8000+j0.30001.0000=−0.8000+j0.3000∑k≠3Y3kVk=0.2066+j7.8120V3(1)=(−0.8000+j0.3000)−(0.2066+j7.8120)−j7.0200=1.0701−j0.1434=1.0796∠−7.6322∘ pu\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.8000 + j0.3000}{1.0000} = -0.8000 + j0.3000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= 0.2066 + j7.8120 \\ V_3^{(1)} &= \frac{(-0.8000 + j0.3000) - (0.2066 + j7.8120)}{-j7.0200} = 1.0701 - j0.1434 \\ &= 1.0796\angle -7.6322^\circ\ \text{pu} \end{aligned}

Slack bus complex power

Using the specified V1V_1 and the computed V2V_2, V3V_3:

P1−jQ1=V1∗(Y11V1+Y12V2+Y13V3)=(1.05)(1.5434+j0.1003)=1.6206+j0.1054S1=P1+jQ1=1.6206−j0.1054 pu\begin{aligned} P_1 - jQ_1 &= V_1^{*}\left(Y_{11}V_1 + Y_{12}V_2 + Y_{13}V_3\right) \\ &= (1.05)(1.5434 + j0.1003) = 1.6206 + j0.1054 \\ S_1 &= P_1 + jQ_1 = 1.6206 - j0.1054\ \text{pu} \end{aligned}

(Bus 1 has no load, so SG1=S1S_{G1} = S_1. The negative Q1Q_1 shows that the slack generator absorbs reactive power, because the long lines produce a lot of charging current at this light loading.)

Answer: V2=1.0281∠−4.6106∘V_2 = 1.0281\angle -4.6106^\circ pu, V3=1.0796∠−7.6322∘V_3 = 1.0796\angle -7.6322^\circ pu, S1=1.6206−j0.1054S_1 = 1.6206 - j0.1054 pu.

✓ Answer checked by Claude Sonnet 5.5

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