Question 12.3: Consider a transmission system of two 138 kV lines, three bu......

First, form a Y matrix as follows:

The active and reactive power at bus l (swing bus) can be written from Equations 12.38 and 12.39 as
P_{s}=V_{s} \sum\limits_{r=1}^{r=n}{V_{r} [ (G_{sr} \cos (\theta _{s}-\theta _{r} )+B_{sr} \sin (\theta _{s}-\theta _{r} )]}      (12.38)
Q_{s}=V_{s} \sum\limits_{r=1}^{r=n}{V_{r} [ (G_{sr} \sin (\theta _{s}-\theta _{r} )-B_{sr} \cos (\theta _{s}-\theta _{r} )]}       (12.39)
P_{1}= 1.02 × 1.02[0.474 \cos(0.0 – 0.0) + (-2.428) \sin(0.0 – 0.0)]
+ 1.02V_{2}[(-0.474) \cos(0.0 – θ_{2}) + 2.45 \sin(0.0 – θ_{2})
+ 1.02 × 1.02[0.0 \cos(0.0 – θ_{3}) + 0.0 \sin(0.0 – θ_{3})]
Q_{1}= 1.02 × 1.02[0.474 \sin(0.0 – 0.0) – (-2.428) \cos(0.0 – 0.0)]
+ 1.02V_{2}[(-0.474) \sin(0.0 – θ_{2}) – 2.45 \cos(0.0 – θ_{2})
+ 1.02 × 1.02[0.0 \sin(0.0 – θ_{3}) – 0.0 \cos(0.0 – θ_{3})]
These equations for the swing bus are not immediately required for load flow, but can be used to calculate the power flow from this bus, once the system voltages are calculated to the required tolerance.

Similarly, the active and reactive powers at other buses are written
P_{2} = V_{2} × 1.02[(-0.474) \cos(θ_{2} – 0) + 2.45 \sin(θ_{2} – 0)] + V_{2}
× V_{2}[1.142 \cos(θ_{2} – θ_{2}) + ( – 4.70) \sin(θ_{2} – θ_{2})] + V_{2}
× 1.02[(-0.668) \cos(θ_{2} – θ_{3}) + 2.297 \sin(θ_{2} – θ_{3})]
Substituting the initial values (V_{2} = 1, θ_{2} = 0°), P_{2} = -0.0228.
Q_{2} = V_{2} × 1.02[(-0.474) \sin(θ_{2} – 0.0) – 2.45 \cos (θ_{2} – 0.0)] + V_{2}
× V_{2}[1.142 \sin(θ_{2} – θ_{2}) – ( – 4.70) \cos(θ_{2} – θ_{2})] + V_{2}
× 1.02[(-0.668) \sin(θ_{2} – θ_{3}) + 2.297 \cos(θ_{2} – θ_{3})]
Substituting the numerical values, Q_{2} = -0.142.

P_{3} = 1.02 × 1.02[0.0 \cos(θ_{3} – 0.0) + 0.0 \sin(θ_{3} – 0.0)] + 1.02
× V_{2}[(-0.668) \cos(θ_{3} – θ_{2}) + 2.297 \sin(θ_{3} – θ_{2})] + 1.02
× 1.02[0.668 \cos(θ_{3} – θ_{3}) + (-2.047) sin(θ_{3} – θ_{3})]
Substituting the values, P_{3} = 0.0136.
Q_{3} = 1.02 × 1.02[0.0 \sin(θ_{3} – 0.0) – 0.0 \cos(θ_{3} – 0.0)] + 1.02
× V_{2}[(-0.668) \sin(θ_{3} – θ_{2}) – 2.297 \cos(θ_{3} – θ_{2}) + 1.02
× 1.02[0.668 \sin(θ_{3} – θ_{3}) – (-2.272) \cos(θ_{3} – θ_{3})]
Substituting initial values, Q_{3} = -0.213.
The Jacobian matrix is

The partial differentials are found by differentiating the equations for P_{2}, Q_{2}, P_{3}, etc.
\frac{\partial P_{2}}{\partial \theta _{2}} = 1.02[V_{2}(0.474) \sin θ_{2} + 2.45 \cos θ_{2}] + 1.02[V_{2}(0.668) \sin(θ_{2} – θ_{3}) + V_{2}(2.297) \cos(θ_{2} – θ_{3})]
= 4.842
\frac{\partial P_{2}}{\partial \theta _{3}} = V_{2}(1.02)[(-0.668) \sin(θ_{2} – θ_{3}) – 2.297 \cos(θ_{2} – θ_{3})]
= -2.343
\frac{\partial P_{2}}{\partial V_{2}} = 1.02[(-0.474) \cos θ_{2} + 2.45 \sin θ_{2}]
+ 2V_{2}(1.142) + 1.02[(-0.608) \cos(θ_{2} – θ_{3}) + 2.297 \sin(θ_{2} – θ_{3})]
= 1.119
\frac{\partial Q_{2}}{\partial \theta _{2}} = 1.02[-V_{2}(0.474) \cos θ_{2} + 2.45 \sin θ_{2}] + 1.02[V_{2}(-0.668) \cos(θ_{2} – θ_{3}) + 2.297 \sin(θ_{2} – θ_{3})]
= -1.1648
\frac{\partial Q_{2}}{\partial V _{2}} = 1.02[(-0.474) \sin θ_{2} – 2.45 \cos θ_{2}] + 2V_{2}(4.28)
+ 1.02[(-0.668) \sin(θ_{2} – θ_{3}) – 2.297 \cos(θ_{2} – θ_{3})]
= 4.56
\frac{\partial Q_{2}}{\partial θ_{3}} = 1.02V_{2}[0.668 \cos(θ_{2} – θ_{3}) – 2.297 \sin(θ_{2} – θ_{3})]
= 0.681
\frac{\partial P_{3}}{\partial θ_{2}} = 1.02V_{2}[(0.668) \sin(θ_{3} – θ_{2}) – 2.297 \cos(θ_{3} – θ_{2})]
= 2.343
\frac{\partial P_{3}}{\partial V_{2}} = 1.02[(-0.668) \cos(θ_{3} – θ_{2}) + 2.297 \sin(θ_{3} – θ_{2})]
= 0.681
\frac{\partial P_{3}}{\partial θ_{3}} = 1.02[0.668 \sin(θ_{3} – θ_{2}) + 2.297 \cos(θ_{3} – θ_{2})]
= 2.343
Therefore, the Jacobian is

The system equations are

Inverting the Jacobian gives

The new values of voltages and phase angles are

This completes one iteration. Using the new bus voltages and phase angles the power flow is recalculated. Thus, at every iteration, the Jacobian matrix changes and has to be inverted.
In the first iteration, we see that the bus 2 voltage is 8.4% higher than the rated voltage; the angles are in radians. The first iteration is no indication of the final results. The hand calculations, even for a simple three-bus system, become unwieldy. The converged load flow is shown in Figure 12.4b. A reactive power injection of 48.10 Mvar is required at bus 3 to maintain a voltage of 1.02 per unit and supply the required active power of 0.5 per unit from the source. There is a reactive power loss of 48.68 Mvar in the transmission lines themselves, and the active power loss is 11.25 MW. The bus phase angles are high with respect to the swing bus, and the bus 2 operating voltage is 0.927 per unit, that is, a voltage drop of 7.3% under load. Thus, even with voltages at the swing bus and bus 3 maintained above rated voltage, the power demand at bus 2 cannot be met and the voltage at this bus dips. The load demand on the system is too high, it is lossy, and requires augmentation or reduction of load. A reactive power injection at bus 2 will give an entirely different result.
If bus 3 is treated as a load bus, the Jacobian is modified by adding a fourth equation of the reactive power at bus 3. In this case the bus 3 voltage dips down to 0.78 per unit, that is, a voltage drop of 22%; the converged load flow is shown in Figure 12.4c. At this lower voltage of 0.78 per unit, bus 3 can support an active load of only 0.3 per unit. This is not an example of a practical system, but it illustrates the importance of reactive power injection, load modeling, and its effect on the bus voltages.
The load demand reduces proportionally with reduction in bus voltages. This is because we have considered a constant impedance type of load, that is, the load current varies directly with the voltage as the load impedance is held constant. The load types are discussed further.