Electric Engineering Task

 

 

 

 

 

 

 

Electric Engineering Task

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Given that:

Figure 01: SimulationCalculation for the given question

Figure 02: Value for the SG3

The generator and load are represented in the circuit shown below by their respective symbols.The three bus bars are labeled SG1, SD1, and SG2. The load is connected across all three bars.The generator is connected to one side of each bar, with its other side connected to the ground. The load is also connected to the ground through its two terminals.Inductors represent S13, S23, S31, and S32 that connect the generators and loads. There is an additional inductor between SD1 and SD2 that represents So2. To satisfy the complex power balance and KCL for the three buses, S3 must be zero, and S31+S32 must equal one.The complex power balance at bus 1 holds well since there is no current flow through the inductor and no voltage across it.

So,SG1 = 1+/1

BUS Symbol SD1 = 1 -j1

SG2 = 0.5 + j0.5 2 Load Symbol So2 = 1 + j1 0.4 +j0.2 -0.4+ j0.2 S13

SG3= S31+S32 = 0.4+1.8j+0.7-0.7j

Let’s prove this with another method:

Figure 03:  Values of Sg1, SG2, and SG3

So, the value of SG3= S31+S32

= 0.4+1.8j+0.7-0.7j

SG3 = 0.5+1.1j

If we justify the given diagram, we must first satisfy each node’s KCL equation.The voltage source node is at the center of the transmission line and is, therefore, a current source. The inductor will have an infinite impedance, so no current can flow through it. The capacitor will have a short circuit to ground and therefore has no impedance. The current source node will have no current flowing into or out of it because no other nodes are attached to it. It also satisfies KCL because its voltage equals its current multiplied by its impedance: (V0-1)/(1+j1). The load node will have a short circuit to the ground and, therefore, an infinite impedance. No current flows through this node, which means that all of the power delivered by the generator must be dissipated in this one node.

To simulate the system, we will use Power World. The first step is to choose our base units. We will use apparent power and line voltage as our base units for this simulation.

The simulation results are as follows:

Nodal Voltages: V_{SG3} = [0.5j]

Power Flow: P_SG3 = 0.5+1.1j*100MVA= 50.5kW

Line Voltages: V_{D} = 80kV+0.5j*132.7kV= 132.7kV+0.5j