When computing V3 in iteration 1, which value of V2 is used?

The newly computed V2 of the same iteration

The value from the previous run

Why is the network real power loss zero, even though P1 = 2.432 pu is not exactly 2.7 pu after two iterations?

Because lines have no resistance; the gap is only due to non-convergence after two iterations

Because the loads draw no real power

Because the slack bus absorbs the loss

Because the shunt admittance is negligible

Asked 2/24
2079 Bhadra · 10 marks
2070 Chaitra · 10 marks

G-S load flow, j0.12 lines, losses

Short answer

Using G-S with Ybus elements j8.3333 pu (off-diagonal) and −j16.6667 (diagonal), two iterations gave V2 = 0.9576∠−8.87° pu and V3 = 0.9443∠−8.78° pu. Slack bus power: P1 = 2.432 pu, Q1 = 1.005 pu. Since line resistance is zero, real power loss = 0 pu; reactive power loss = 0.418 pu (not fully converged after two iterations).

For a network given below, series impedance of each line is 0 + j0.12 p.u. Shunt admittance of the line is negligible. Calculate: (i) Voltage at bus-2 and bus-3 by using G-S method (upto 2 iteration). (ii) Slack bus real and reactive power. (iii) Network real and reactive power losses. [Figure: three buses with lines 1-2, 1-3 and 2-3; bus 1 is the slack bus with generator, V1 = 1.0∠0°; load at bus 2 = 1.5 + j0.4 p.u; load at bus 3 = 1.2 + j0.5 p.u]

Data and Ybus

Each line: $z = j0.12$ , $y = 1/j0.12 = -j8.3333$ pu. Every bus has two lines:

Y_{bus} = \begin{bmatrix} -j16.6667 & j8.3333 & j8.3333 \\ j8.3333 & -j16.6667 & j8.3333 \\ j8.3333 & j8.3333 & -j16.6667 \end{bmatrix}

Bus 1: slack, $V_1 = 1.0\angle 0^\circ$ . Buses 2 and 3 are PQ buses with net injections $S_2 = -1.5 - j0.4$ and $S_3 = -1.2 - j0.5$ pu. Flat start: $V_2^{(0)} = V_3^{(0)} = 1.0\angle 0^\circ$ .

G-S equation (latest values used immediately):

V_i^{(r+1)} = \frac{1}{Y_{ii}}\left[\frac{P_i - jQ_i}{V_i^{*(r)}} - \sum_{k \ne i}Y_{ik}V_k\right]

(i) Iteration 1

\begin{aligned} V_2^{(1)} &= \frac{1}{-j16.6667}\left[\frac{-1.5 + j0.4}{1.0} - j8.3333(1.0) - j8.3333(1.0)\right] \\ &= \frac{-1.5 - j16.2667}{-j16.6667} = 0.9760 - j0.0900 = 0.9801\angle -5.2685^\circ \\ V_3^{(1)} &= \frac{1}{-j16.6667}\left[\frac{-1.2 + j0.5}{1.0} - j8.3333(1.0) - j8.3333(0.9760 - j0.0900)\right] \\ &= \frac{-1.95 - j15.9667}{-j16.6667} = 0.9580 - j0.1170 = 0.9651\angle -6.9630^\circ \end{aligned}

Iteration 2

\begin{aligned} \frac{-1.5 + j0.4}{V_2^{*(1)}} &= -1.4865 + j0.5469 \\ V_2^{(2)} &= \frac{(-1.4865 + j0.5469) - (0.9750 + j16.3167)}{-j16.6667} \\ &= 0.9462 - j0.1477 = 0.9576\angle -8.8715^\circ \\ \frac{-1.2 + j0.5}{V_3^{*(1)}} &= -1.1714 + j0.6650 \\ V_3^{(2)} &= \frac{(-1.1714 + j0.6650) - (1.2307 + j16.2182)}{-j16.6667} \\ &= 0.9332 - j0.1441 = 0.9443\angle -8.7797^\circ \end{aligned}

Here $\sum_{k \ne 2}Y_{2k}V_k = j8.3333(V_1 + V_3^{(1)}) = 0.9750 + j16.3167$ and $\sum_{k \ne 3}Y_{3k}V_k = j8.3333(V_1 + V_2^{(2)}) = 1.2307 + j16.2182$ .

Iteration	$V_2$ (pu)	$V_3$ (pu)
0	$1.0\angle 0^\circ$	$1.0\angle 0^\circ$
1	$0.9801\angle -5.27^\circ$	$0.9651\angle -6.96^\circ$
2	$0.9576\angle -8.87^\circ$	$0.9443\angle -8.78^\circ$

(ii) Slack bus power (with iteration-2 voltages)

\begin{aligned} I_1 &= Y_{11}V_1 + Y_{12}V_2 + Y_{13}V_3 = 2.4318 - j1.0052 \\ S_1 &= V_1I_1^* = 1.0(2.4318 + j1.0052) \end{aligned}

So $P_1 = 2.4318$ pu and $Q_1 = 1.0052$ pu.

(iii) Network losses

Line flows $S_{ik} = V_i[y_{ik}(V_i - V_k)]^*$ :

Line	$S_{ik}$ (pu)	$S_{ki}$ (pu)	Loss $S_{ik} + S_{ki}$
1-2	$1.2307 + j0.4485$	$-1.2307 - j0.2426$	$j0.2059$
1-3	$1.2011 + j0.5567$	$-1.2011 - j0.3464$	$j0.2103$
2-3	$-0.0121 + j0.1068$	$0.0121 - j0.1053$	$j0.0015$

Answer:

$V_2 = 0.9576\angle -8.87^\circ$ pu, $V_3 = 0.9443\angle -8.78^\circ$ pu after 2 iterations.
Slack bus: $P_1 = 2.432$ pu, $Q_1 = 1.005$ pu.
Losses: real power loss $= 0$ (lines have no resistance); reactive power loss $= 0.418$ pu.

Note: after only two iterations the solution has not fully converged, so $S_1$ minus the loads ( $2.432 - 2.7 = -0.27$ ) does not yet equal the losses; with more iterations $P_1$ tends to 2.7 pu.

✓ Answer checked by Claude Sonnet 5.5

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