QUESTION 1 1.1 The bus impedance matrix of Figure 1 is given as j0.2 5 J0.125...
The single-line diagram of a four-bus system and its bus impedance matrix are shown below BUS 2 0.25 j0.2 0.125 0.25 0.4 BUS 3 BUS 1 BUS 4 j0.1 0.1 j0.2 j0.2 ground is the reference node) 0.25 0.2 0.16 0.14 0.2 0.23 0.15 0.151 ZBUs =기0.16 0.15 0.196 0.1 0.14 0.151 0. 0.195 A solid three-phase fault occurs at bus 2 of the network. (a) Calculate the initial symmetrical RMS current in the fault. (b) Determine the voltages during...
Consider the impedance diagram of a simple energy distribution network seen in fig. 1. The transmission lines that connect the buses have the line impedances as shown in the figure. The generators connected to buses 1 and 2 are known to be E1=1.0 pu (per unit) and E2=0.5 pu, respectively, on a 1 MV base. a) (05) Draw the admittance diagram for the system shown in fig. 1. b) (10) Obtain the bus admittance matrix Ybus for the system. c)...
Questions from protective Relaying course Need help ... Thank you In Figure 1.1, a three-phase short circuit at bus f can be represented c) (6) as connection through a transition impedance Zf to the ground. The faulted network can be decomposed as two parts. If the elements of impedance matrix of the network are given, G is the set of buses connected with current sources, what is the voltage of any bus i before the fault happens? d) With the...
ystem Design December 17, 2019 2. (20 pts.) Using the "DC" power flow method find the real power flows in MW between the buses and the power generated in MWW by the generator on bus 1. The impedance values shown for the lines in the figure are in per unit with a base of 100 MVA and 230 kV. PG2 = 250 MW P2 = 50 MW 120 1 3 j0.25 j0.2 j0.4 P3 = 300 MW
The one-line diagram of a three-bus power system is shown in Figure 4. All impedances are expressed in per unit on a common MVA base. All resistances and shunt capacitances are neglected. Information on each component in this system is given below: • Each generator is represented by an emf behind the sub-transient reactance of j0.045 and their neutrals are connected to the ground. • Line 1-2 has reactance of j0.88 • Line 2-3 has reactance of j0.65 • Line...
3- The bus impedance matrix of a four-bus network with values in per unit is Z??? = [?0.15 ?0.08 ?0.04 ?0.07 ?0.08 ?0.15 ?0.06 ?0.09 ?0.04 ?0.06 ?0.13 ?0.05 ?0.07 ?0.09 ?0.05 ?0.12 ] Generators connected to busses 1 and 2 have their sub-transient reactances included in Zbus. Neglect pre-fault currents and find sub-transient current in pu in the fault for a three-phase fault on bus 4. Assume the voltage at fault is 1∠0 before the fault occurs. Also, find...
Qustion 2.120 marks) a) Figure I below shows a system with plant-I and plant-2 connected to bus-1 and bus-2 respectively. There are two loads and 3 branches. The bus-I is the reference bus with 1.040° pu voltage. Base MVA is 100 If the branch currents and branch impedance values are as follows -2.5-j0.50pu -2-j040 pu I 1.5-j0.3 pu Calculate: 7-0.05+j0.20 pu -0.02+j0.08 pu Z 0.025+j0.10 pu i. Current distribution factors (4 marks) ii. Power factor angles at bus 1 and...
Answer part D The single line diagram of a power network is shown in the figure. Bus#1 is a slack bus. The scheduled powers for bus#2 and bus#3 are given. The impedances shown in the figure are all in per-unit considering a power base of 100 MVA. 30 400 MW 320 MVAr Slack V-140 j0.0125 jo.0s 00 MW 270 MVAr A. Use the Gauss Seidel technique to determine voltages at bus#2 & bus#3. (Start with an initial guess 140 for...
Consider the single line diagram of a 3-bus power system shown in Figure 2. Slack bus 3 Figure 2. The data for this system are given in Tables 1 and 2. Bus Table 1 Generation Load Assumed PG QGPLQL bus voltage (MW) (MVar) (MW) (MVar) 1.05 +10.0 - - 1.0 + 0.0 50 30 305.6 140.2 1.0 +0.0 0.0 0.0 138.6 45.2 slack bus) Table 2 Bus-to-bus Impedance 0.2 + j0.04 .01 +0.03 2.3 0.0125 + j0.025 (0) Convert all...
You are given the four-bus problem as follows (Figure 6.34): The data for this problem are as follows: x12 = 0.2;x14 = 0.4;x13 = 0.25;x24 = 0.5;x34 = 0.1; Form the 4 × 4 Bx matrix for this network. Use of Matlab is strongly recommended. The first row will be Bx = (1/ x12+1/ x13+1/ x14)−1/ x12−1/ x13−1/ x14 You cannot invert the Bx matrix, so you should replace all elements in the fourth column by zeros, and replace...