# Computer Aided Power System Design and Analysis

Topics: Electrical engineering, Transformer, Per-unit system Pages: 10 (2419 words) Published: December 11, 2011
1. Introductions
In this assignment, a software package named ‘ERACS’ is used for diagram building of a power system, Load flow analysis, Balanced and Unbalanced Short Circuit analysis. Calculate the short circuit level (in MVA) in balanced, the current fed into the fault and the voltages of the healthy at bus 4 of the unbalanced circuit. Analyze and compare these two results which got by theory and analytical procedures with the system performance.

I. Using ERACS software package to construct the network shown below. Figure 1.1 example network

The network diagram constructed by ERACS:
Figure 1.2 constructed network
Setting notice:
The system positive, negative and zero sequence reactances (X1, X2 and X0, respectively) are given in p.u. on 100 MVA base as follows: G1: X1 = 0.3, X2 = 0.2 and X0 = 0.05
G2: X1 = 0.2, X2 = 0.12 and X0 = 0.03
T1: X1 = X2 = X0 = 0.15 T2: X1 = X2 = X0 = 0.1
L1 and L2: X1 = X2 = 0.15 and X0 = 0.3
Generator G1 is modeled as a slack generator with 1∠0o p.u. voltage. Generator G2, PQ type with active power of 50 MW and reactive power of 40 MVAr. For generators and transformers, within zero neutral earthing resistance and reactance in “Neutral Earthing Data” option. Also ignore the effects of resistance in the system.

A. System analysis and problem solving
Choose the data and results options：
Busbar: Voltage (p.u.), Voltage (kV), Voltage angle ( o ) and Three Phase Fault level (MVA). Generators, Transformers, Lines and Loads: Real power (MW) and Reactive power (MVAr).

Perform the load flow analysis (shown as figure 2.1)

Give a table 2.1 for Results of Generators Transformers Lines and Loads： ComponentsP.Q | Generator1| Transformer1| Line1| Line2| Transformer2 | Generator2| Load1 | Load 2| Load3| active power input| | 50Mw| 20Mw| 20Mw| 50 Mw| | 30Mw| 40Mw| 30Mw| active power output| 50Mw| 50Mv| 20Mw| 20 Mw| 50Mw| 50Mw| | | | reactive power input| | 57.674MVAr| 4.303MVAr| 5.039MVAr| 40MVAr| | 22.5MVAr| 30MVAr| 30.606MVAr| reactive power output| 57.674MVAr| 48.934MVAr| 3.566MVAr| 4.303MVAr| 35.646MVAr| 40MVAr| | | | Table 2.1 Results of Generators Transformers Lines and Loads

Comments on active and reactive power:
Observe the entire current flow direction shown in figure 2.3, and combining the results in table 2.1: The active power flows into Load1+Load2 +Load3=100MW
P= 3 * I2 * R * cosѲ
For throughout this work the resistance is ignored, thus R=0, so the active power (P) do not loss in the parts of the system which means the rated active power of Loads (1, 2, 3) equals to that generated by G1&G2. As G1 is a slack generator with 1∠0o p.u. voltage which needed in the network to balance the powers to feed the loads, so active power output by G1 is 50MW. Q= 3 * I2 * X * cosѲ

Due to the existence of reactance, the reactive power (Q) reduced. So in theory: Qgenerators > Qloads.
First to calculate the rated reactive power of Load1, Load2, Load3: Equation of P, Q, p.f.: Q= P*tan(cos-1p.f.)……according to figure 2.4 Load1: P=30MW, pf=0.8:
Q=30*tan(cos-10.8)=22.5MVAr S Q Load2: P=40MW, pf=0.8: Ѳ
Q=40*tan(cos-10.8)=30MVAr P
Load3: P=30MW, pf=0.7: Figure 2.3 relationship analysis chart of P Q S pf Q=30*tan(cos-10.7)≅30.606MVAr
∴The rated reactive power of Load1+Load2+ Load3≅83.106MVAr Reactive power of Generator1 is variable with the need of Loads. The reactive power of G1&G2 =57.674 + 40 =97.674MVAr > 83.106MVAr Considering the reactive power absorb in each part, QG1 is only enough to feed QL3, after flows past L1, the rest goes by lines and busbars with a certain loss to Load1&Load2,QG1 gives enough for Load1&Load2 also lost some in each part. So the performance by...