Q3Power System 2

Question

Q.3. Figure shows the single-line diagram of a simple three-bus power system with generators at buses 1 and 3. The magnitude of voltage at bus-1 is V1 = 1.025<0 degrees per unit. Voltage magnitude at bus-3 is fixed at 1.03 per unit with a real power generation of 300 MW. A load consisting of 400 MW and 200 Mvar is taken from bus-2. Line impedances are marked in per unit on a 100 MVA base, and the line charging susceptances are neglected. Line reactances: bus1-bus3 = j0.05 pu, bus1-bus2 = j0.025 pu, bus2-bus3 = j0.025 pu.

(i) Form Y-Bus
(ii) Obtain the power flow solution up to two-iterations by Gauss-Seidel method.

Answer 1

(i) Formation of Y-Bus

Given line reactances (per unit, 100 MVA base): bus1-bus2: X12=0.025 pu; bus2-bus3: X23=0.025 pu; bus1-bus3: X13=0.05 pu (line charging neglected, lines purely reactive/lossless).

The line admittances (reciprocal of the series impedances, purely susceptive since resistance is neglected) are:

y_{12} = \frac{1}{j 0.025} = - j 40 pu, y_{23} = \frac{1}{j 0.025} = - j 40 pu, y_{13} = \frac{1}{j 0.05} = - j 20 pu

The Y-bus matrix elements are formed using the standard rule: each diagonal element Yii equals the sum of all admittances connected to bus i, and each off-diagonal element Yij equals the negative of the admittance directly connecting bus i and bus j:

Y_{11} = y_{12} + y_{13} = - j 60 pu, Y_{22} = y_{12} + y_{23} = - j 80 pu, Y_{33} = y_{13} + y_{23} = - j 60 pu

Y_{12} = Y_{21} = - y_{12} = j 40, Y_{13} = Y_{31} = - y_{13} = j 20, Y_{23} = Y_{32} = - y_{23} = j 40 pu

giving the complete Y-bus matrix (in pu, all values purely imaginary since the network is lossless):

Y_{b u s} = - j 60 j 40 j 20 j 40 - j 80 j 40 j 20 j 40 - j 60 pu

(ii) Gauss-Seidel Load Flow Solution - Two Iterations

Bus data: Bus 1 (slack): V1=1.025<0 degrees pu (fixed). Bus 2 (PQ/load bus): load of 400 MW+j200 Mvar, so scheduled injected power S2 = -(4.0+j2.0) pu (on 100 MVA base; negative since it is a load, consuming rather than injecting power). Bus 3 (PV/generator bus): |V3|=1.03 pu fixed, with real power generation P3=300 MW=3.0 pu (Q3 to be calculated each iteration). Flat start: V2^(0)=1.0+j0, V3^(0)=1.03+j0 (angle initialized to zero).

Iteration 1

Update V2 (PQ bus formula):

V_{2}^{(1)} = \frac{1}{Y _{22}} [\frac{P _{2} - j Q _{2}}{V _{2}^{(0) *}} - Y_{21} V_{1} - Y_{23} V_{3}^{(0)}]

Substituting numerical values (P2-jQ2 = -4.0+j2.0, since S2=-4-j2 so conj(S2)=-4+j2) and evaluating gives:

V_{2}^{(1)} = 1.0025 - j 0.05 pu (∣ V_{2}^{(1)} ∣ = 1.00375, angle = - 2.85 5^{\circ})

Calculate Q3 at the PV bus, using the updated V2 and current V3:

Q_{3}^{(1)} = - Im [V_{3}^{(0) *} (Y_{31} V_{1} + Y_{32} V_{2}^{(1)} + Y_{33} V_{3}^{(0)})] \approx - 1.236 pu

Update V3 (PV bus formula, then rescale magnitude to the fixed value 1.03):

V_{3}^{(1)} = \frac{1}{Y _{33}} [\frac{P _{3} - j Q _{3}^{(1)}}{V _{3}^{(0) *}} - Y_{31} V_{1} - Y_{32} V_{2}^{(1)}], then rescale ∣ V_{3}^{(1)} ∣ \to 1.03

V_{3}^{(1)} \approx 1.0299 + j 0.0158 pu (angle \approx 0.8 8^{\circ}, after rescaling to ∣ V_{3} ∣ = 1.03)

Iteration 2

Repeating the same three steps using the iteration-1 values of V2 and V3:

V_{2}^{(2)} \approx 1.00008 - j 0.04060 pu (∣ V_{2}^{(2)} ∣ \approx 1.00091, angle \approx - 2.32 5^{\circ})

Q_{3}^{(2)} \approx - 1.369 pu

V_{3}^{(2)} \approx 1.0298 + j 0.0221 pu (angle \approx 1.2 3^{\circ}, after rescaling to ∣ V_{3} ∣ = 1.03)

Summary after two iterations: V2 has progressed from the flat-start value of 1.0+j0 pu toward approximately 1.00008-j0.0406 pu (magnitude approximately 1.0009 pu, angle approximately -2.33 degrees), reflecting the voltage drop expected at the loaded bus 2, while V3's angle has progressed from 0 degrees toward approximately +1.23 degrees (with magnitude held fixed at the specified 1.03 pu throughout, consistent with its role as a voltage-controlled PV bus), and the bus-3 reactive power generation required to maintain this fixed voltage has been calculated as approximately -1.37 pu (i.e., approximately 137 Mvar of reactive power absorption, indicating that generator 3 must absorb, rather than supply, reactive power to hold its terminal voltage at 1.03 pu given the relatively light net reactive loading of this small three-bus system after accounting for the reactive power supplied by the lines' own effect on the voltage profile). Further iterations would be required to reach full numerical convergence (typically indicated by successive voltage changes falling below a small tolerance, such as 0.0001 pu), consistent with the generally slow, linear convergence characteristic of the Gauss-Seidel method discussed in the corresponding comparison answer elsewhere in this paper.

It is useful to note that although the given problem specifies bus 3 as a voltage-controlled (PV) bus, the reactive power required at bus 3 to maintain its scheduled voltage magnitude is not fixed in advance but must instead be calculated at each iteration from the current network voltage estimates, exactly as carried out above — this is precisely the additional computational step that distinguishes handling of a PV bus from the more straightforward direct voltage update formula used for an ordinary PQ (load) bus such as bus 2. In a complete Gauss-Seidel solution, this same three-step iterative procedure (update V2, recompute Q3, update and rescale V3) would be repeated for further iterations until successive changes in all bus voltages fall below a specified small convergence tolerance, after which the converged bus voltages could be used to compute the final line power flows and losses throughout the network using the standard branch power flow equations.

Q3Power System 2

Question

Answer

(i) Formation of Y-Bus

(ii) Gauss-Seidel Load Flow Solution - Two Iterations

Iteration 1

Iteration 2