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3-bus load flow with Q limits (cut off)

Short answer

Form Ybus and start with a flat voltage profile. For PV bus 2, compute Q2, check Q-limits; if violated fix Q and treat as PQ, else correct angle and reset |V2|. Update PQ bus 3 with the Gauss–Seidel formula. Iterate until convergence, then compute slack bus power and line flows to obtain total losses.

Suppose you are given a 3-bus power system network with one reference bus, one load bus and one generator bus with reactive power limits and asked to perform the [rest of the question is cut off in the scan: ...unknown variables and total losses in the network].

The scanned question is cut off. It is taken here as: describe how to carry out the load flow of such a 3-bus system (Gauss-Seidel method) to find the unknown variables and the total losses. Bus 1 is the reference (slack) bus, bus 2 the generator (PV) bus with Q-limits, and bus 3 the load (PQ) bus.

Known and unknown quantities

BusTypeKnownUnknown
1Reference (slack)∣V1∣\lvert V_1\rvert, δ1=0\delta_1 = 0P1P_1, Q1Q_1
2Generator (PV)P2P_2, ∣V2∣\lvert V_2\rvert, QminQ_{min}, QmaxQ_{max}Q2Q_2, δ2\delta_2
3Load (PQ)P3P_3, Q3Q_3∣V3∣\lvert V_3\rvert, δ3\delta_3

Step-by-step procedure

  1. Form YbusY_{bus} from the line data: YiiY_{ii} = sum of admittances connected to bus ii (including half line charging), Yij=−yijY_{ij} = -y_{ij}.
  2. Net injections: Pi=PGi−PLiP_i = P_{Gi} - P_{Li}, Qi=QGi−QLiQ_i = Q_{Gi} - Q_{Li}.
  3. Flat start: V2(0)=∣V2∣spec∠0∘V_2^{(0)} = |V_2|_{spec}\angle 0^\circ, V3(0)=1∠0∘V_3^{(0)} = 1\angle 0^\circ.
  4. PV bus 2: find the reactive power
Q2(k+1)=−Im{V2(k)∗∑j=13Y2jVj}Q_2^{(k+1)} = -\text{Im}\left\{V_2^{(k)*}\sum_{j=1}^{3}Y_{2j}V_j\right\}
  1. Check Q-limits. If Qmin≤Q2≤QmaxQ_{min} \le Q_2 \le Q_{max}, keep bus 2 as PV. If not, set Q2Q_2 to the violated limit and treat bus 2 as a PQ bus (its voltage magnitude is then not reset).
  2. Update V2V_2:
V2(k+1)=1Y22[P2−jQ2V2(k)∗−Y21V1−Y23V3(k)]V_2^{(k+1)} = \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{(k)*}} - Y_{21}V_1 - Y_{23}V_3^{(k)}\right]

If bus 2 is still PV, keep only the angle and reset the magnitude to ∣V2∣spec|V_2|_{spec}. 7. Update V3V_3 using the newest V2V_2:

V3(k+1)=1Y33[P3−jQ3V3(k)∗−Y31V1−Y32V2(k+1)]V_3^{(k+1)} = \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{(k)*}} - Y_{31}V_1 - Y_{32}V_2^{(k+1)}\right]
  1. Convergence test: stop when ∣Vi(k+1)−Vi(k)∣<ε|V_i^{(k+1)} - V_i^{(k)}| < \varepsilon (e.g. 0.0001) for all buses. Otherwise repeat from step 4. An acceleration factor of about 1.6 may be used on PQ buses.
  2. Slack bus power:
P1−jQ1=V1∗(Y11V1+Y12V2+Y13V3)P_1 - jQ_1 = V_1^{*}\left(Y_{11}V_1 + Y_{12}V_2 + Y_{13}V_3\right)

and PG1=P1+PL1P_{G1} = P_1 + P_{L1}, QG1=Q1+QL1Q_{G1} = Q_1 + Q_{L1}. 10. Line flows: Sij=Vi[(Vi−Vj)yij+Vi ysh,ij]∗S_{ij} = V_i\left[(V_i - V_j)y_{ij} + V_i\,y_{sh,ij}\right]^{*}. 11. Total losses:

Ploss=∑i=13Pi=P1+P2+P3=∑linesRe(Sij+Sji)\begin{aligned} P_{loss} &= \sum_{i=1}^{3}P_i = P_1 + P_2 + P_3 \\ &= \sum_{\text{lines}}\text{Re}(S_{ij} + S_{ji}) \end{aligned}

and similarly Qloss=∑QiQ_{loss} = \sum Q_i.

 Read data -> form Ybus -> flat start
            |
            v
 +--> Q2 from present voltages
 |    check Q-limits (PV or PQ?)
 |    update V2 (reset |V2| if PV)
 |    update V3
 |    converged? --no--+
 |                     |
 +---------------------+
            | yes
            v
 slack P1,Q1 -> line flows -> losses

✓ Answer checked by Claude Sonnet 5.5

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