What happens at a PV bus if the computed Q exceeds its limit?

Q is fixed at the limit and the bus is treated as PQ

Why must a slack bus exist?

Losses are unknown before the solution, so one bus supplies the balance of P and Q

To make all voltages equal 1 pu

2076 Asoj · 8 marks

Bus classification and 4-bus Ybus

Short answer

Buses are slack (known |V|, δ; unknown P, Q), PV (known P, |V|; unknown Q, δ), or PQ (known P, Q; unknown |V|, δ). For a four-bus system, Ybus diagonal elements are sum of admittances at each bus, off-diagonals are negative mutual admittances, and zero if no direct connection.

Explain bus classification in power flow analysis with their known and unknown quantities and hence formulate the bus admittance matrix for four bus system.

In load-flow analysis each bus has four quantities: $P$ , $Q$ , $|V|$ and $\delta$ . Two are specified and two are found by the solution. Buses are classified according to which two are known.

Bus classification

Bus type	Known	Unknown	Example
Slack / swing / reference bus	$\lvert V\rvert$ , $\delta$ (usually $\delta = 0$ )	$P$ , $Q$	Largest generating station
PV / generator / voltage-controlled bus	$P$ , $\lvert V\rvert$ (with $Q_{min}$ , $Q_{max}$ )	$Q$ , $\delta$	Generator buses, buses with SVC
PQ / load bus	$P$ , $Q$	$\lvert V\rvert$ , $\delta$	Load buses (about 80% of buses)

Slack bus: only one in the system. Its P and Q make up the difference between scheduled generation and load + losses (losses are unknown before the solution). Its voltage angle serves as reference.
PV bus: real power output is set by the turbine and voltage magnitude by the AVR. If the computed Q exceeds its limits, Q is fixed at the limit and the bus becomes a PQ bus.
PQ bus: net injection $P = P_G - P_D$ , $Q = Q_G - Q_D$ is known; voltage is found.

Ybus for a four-bus system

Take a four-bus network with lines 1-2, 1-3, 2-3, 2-4 and 3-4. Each line has series admittance $y_{ik}$ and shunt $y'_{ik}/2$ at each end; total shunt at bus $i$ is $y_{i0}$ .

Applying KCL:

\begin{aligned} I_1 &= y_{10}V_1 + y_{12}(V_1 - V_2) + y_{13}(V_1 - V_3) \\ I_2 &= y_{20}V_2 + y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) + y_{24}(V_2 - V_4) \\ I_3 &= y_{30}V_3 + y_{13}(V_3 - V_1) + y_{23}(V_3 - V_2) + y_{34}(V_3 - V_4) \\ I_4 &= y_{40}V_4 + y_{24}(V_4 - V_2) + y_{34}(V_4 - V_3) \end{aligned}

\mathbf{Y}_{bus} = \begin{bmatrix} y_{10} + y_{12} + y_{13} & -y_{12} & -y_{13} & 0 \\ -y_{12} & y_{20} + y_{12} + y_{23} + y_{24} & -y_{23} & -y_{24} \\ -y_{13} & -y_{23} & y_{30} + y_{13} + y_{23} + y_{34} & -y_{34} \\ 0 & -y_{24} & -y_{34} & y_{40} + y_{24} + y_{34} \end{bmatrix}

General rule: $Y_{ii} = y_{i0} + \sum_{k\ne i} y_{ik}$ and $Y_{ik} = -y_{ik}$ ( $= 0$ if no line, e.g. $Y_{14}$ ).

These elements are then used in the power-flow equations

P_i - jQ_i = V_i^*\sum_{k=1}^{4} Y_{ik}V_k

which are solved for the unknowns listed in the table above.

✓ Answer checked by Claude Sonnet 5.5

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