Power Systems: Load Flow, Faults, and Stability

Load flow formulation: Ybus and Jacobian

The Newton-Raphson load flow is built on the bus admittance matrix Y_{bus}. An off-diagonal element Y_{ij} equals the negative of the admittance of the branch joining buses i and j. A diagonal element Y_{ii} equals the sum of all branch admittances incident on bus i, plus any shunt admittance and half of the total line charging connected at that bus. Series susceptances are negative, so a positive shunt susceptance such as a capacitor or line charging makes the diagonal less negative than the plain sum of series admittances. If the diagonal equals that sum exactly, no capacitive element is present at that bus or on its lines.

A junction node that carries no load can be removed by Kron reduction, Y^{red}_{ij}=Y_{ij}-Y_{iJ}Y_{Jj}/Y_{JJ}, where J is the eliminated node. The reduced matrix keeps the size of the bus set and stays symmetric.

The Jacobian relates the mismatch vector of real and reactive powers to the corrections in voltage angles and magnitudes. It has four blocks: H (derivatives of P with respect to angle), N (derivatives of P with respect to magnitude), M (derivatives of Q with respect to angle), and R (derivatives of Q with respect to magnitude). The unknowns are the angles at all non-slack buses and the magnitudes at PQ buses, so the real power equations number N-1 and the reactive power equations number N_L. The block M therefore has N_L rows, one for each PQ bus, and N-1 columns, one for each non-slack bus. When P generator buses (the slack included) and Q=N-P load buses are present, the unknown count is (N-1)+Q, so the Jacobian is square with order 2N-1-P.

In a meshed network, total I^2R loss is \sum_k R_k|I_k|^2 over all branches. Opening one branch of a loop makes the network radial, and then each branch current is fixed by the load currents beyond it, independent of the branch impedances. The best branch to open is therefore the one that gives the smallest value of \sum_k R_k|I_k|^2 for the resulting radial network. For example, two branches of resistance 1 and 2 pu carrying currents of 1 and 2 pu give a loss of 1\cdot1^2+2\cdot2^2=9 units.

Economic dispatch incremental cost

Economic dispatch divides a fixed load among generators so that total fuel cost is minimal. The incremental cost of unit i is \lambda_i=dC_i/dP_i, the extra cost per extra MW. With network losses neglected and the total load P_D fixed, the optimum occurs when all units operate at the same incremental cost, \lambda_1=\lambda_2=\cdots=\lambda, subject to \sum_i P_i=P_D. For a quadratic cost C_i=a_i+b_iP_i+c_iP_i^2, the incremental cost is linear, \lambda_i=b_i+2c_iP_i, so the equal-\lambda condition reduces to linear equations in the outputs.

For example, take \lambda_1=10+0.1P_1 and \lambda_2=12+0.05P_2 with P_D=150 MW. Equating the two gives P_2=2P_1-40, and the load condition 3P_1-40=150 gives P_1\approx63.3 MW and P_2\approx86.7 MW, with a common \lambda of about 16.3 per MWh.

Generation limits modify this rule. If the equal-\lambda solution would require a unit beyond its maximum rating, that unit is held at its limit and the remaining load is shared among the other units at equal incremental cost. A unit whose incremental cost at zero output already exceeds the common \lambda stays at zero output, and the rest of the load is shared by the others.

Changing a unit's cost coefficient shifts the optimum. If a unit's cost rises at every output level, the optimum moves load away from that unit toward the cheaper one, and the minimum total cost cannot decrease.

Symmetrical components of unbalanced currents

An unbalanced set of phasors can be resolved into three balanced sets: the zero-sequence set (three equal phasors), the positive-sequence set (phases in the order A, B, C), and the negative-sequence set (phases in the order A, C, B). With a=1\angle120^\circ and a^2=1\angle240^\circ=1\angle-120^\circ, the phase currents are

I_A=I_0+I_1+I_2, I_B=I_0+a^2I_1+aI_2, I_C=I_0+aI_1+a^2I_2.

The zero-sequence component is common to all three phases, so it is removed by subtracting the same value from each phase reading. The inverse relation is I_1=\frac{1}{3}(I_A+aI_B+a^2I_C). For example, a pure positive-sequence current I_1=1\angle0^\circ gives phase B the reading a^2I_1=1\angle-120^\circ and phase C the reading aI_1=1\angle120^\circ.

For a bolted line-to-line fault between phases b and c, I_0=0 and the positive and negative currents are equal and opposite, I_2=-I_1. The fault current in phase b is then I_b=(a^2-a)I_1=-j\sqrt{3}\,I_1, so its magnitude is \sqrt{3}\,|I_1|.

A transformation between phase quantities and two components must be invertible, so its matrix must have a nonzero determinant. A matrix with identical rows, such as \begin{bmatrix}1&1\\1&1\end{bmatrix}, is singular and cannot recover the components. Among invertible forms, the usual orthogonal two-phase decomposition is V_p=V_\alpha+V_\beta and V_q=V_\alpha-V_\beta, or the same with the sign of V_\alpha reversed; both have determinant -2. A matrix that is invertible but gives one component to a phase alone is not this orthogonal decomposition.

The zero-sequence component is the average V_0=\frac{1}{3}(V_A+V_B+V_C). If every phasor is rotated by the same angle \Delta\theta with its magnitude unchanged, the sum is multiplied by e^{j\Delta\theta}. Thus V_0 keeps its magnitude but its phase angle shifts by \Delta\theta.

Per-unit reactance network for stability

For stability studies, each machine is replaced by a constant internal EMF E\angle\delta behind its transient or synchronous reactance, and all reactances are converted to a common base in per unit. Series reactances in the same path add directly, so a generator, a line and a motor in series form one equivalent reactance X=X_g+X_{line}+X_m. For a lossless path between two EMFs, the real power is

P=\frac{E_1E_2}{X}\sin(\delta_1-\delta_2).

For example, EMFs of 1.1 and 1.0 pu with reactances 0.5, 0.2 and 0.3 pu in series give X=1.0 pu and P=1.1\sin(\delta_1-\delta_2). The coefficient of the sine is set by the reduced network, and the transfer limit is that coefficient.

In the equal-area criterion, the power-angle curve changes at a fault and again after clearing. During the fault, the electrical power falls below the mechanical input, so the rotor accelerates and the accelerating area grows from the initial angle \delta_0. After clearing, the post-fault curve P'\sin\delta decelerates the rotor, with stable equilibrium \delta_s where P'\sin\delta_s=P_m and the largest swing limit \delta_{max}=\pi-\delta_s. The critical clearing angle \delta_c is the angle at which the accelerating area equals the decelerating area between \delta_c and \delta_{max}.