Question-5 [75 Points] Two time-overcurrent relays protect adjacent sections of a radial system. Bus 3 is at the end of the radial line. Approximately 7000 amps of fault current will flow for a close-in fault at point A and approximately 5000 amps of fault current will flow for a line-end fault at point B. Load current at bus 2 is 100A and load current at bus 3 is 350A. Both load currents are at unity power factor. Model this situation using ASPEN One Liner software and simulate the aforementioned faults. Observe the relay curves and the trip times. Investigate the phase and sequence voltages and currents for each bus and line after simulating both faults. Grid Procedure: ● Bus 1 ● ● 600/5 51 CO-9 100 A Name Length = Bus 2 ● Right-click in the workspace and choose New Bus. The Bus Info pop-up menu should appear. Fill in the Name as Bus 1, the Bus no. as 1, the Nom. KV as 41.6. The buses can be moved by dragging them around the workspace. Do this again for buses 2 and 3. Click on Bus 1, hold down the shift key, and right-click on Bus 2. Choose New Line. The Transmission Line Data menu should appear. Fill in the following spaces: = In the file menu, choose New to start a new project. Choose 100 MVA as the system base. Line 1 = 10 mi X = 0.10025 Xo = 0.1 400/5 CT ratio 51 CO-9 = ID=Relay 1 120 Leave everything else the same. Repeat this for the line between buses 2 and 3, which will be called line 2. However, use X=0.0774. Everything else should be the same as line 1. Right-click on line 1 near Bus 1 and choose New Relay Group. On the menu that appears, click on the Add OC Phase Relay button. Fill the spaces as follows: Time dial = 4 Tap unit = GC2 Bus 3 350 A Pickup = 4 Curve CO-9 Make sure the directional option is NOT selected. Click OK, then Done. Do the same for line 2 near Bus 2, except for the following. ● Simulation: ● ID=Relay 2 CT ratio=80 Time dial = 0.5 Tap unit = GC2 Make sure the directional option is NOT selected. Right-click on Bus 1. Choose new, then generator. The parameter menu for the generator will automatically appear. Use R=0 and X=0.1 for each impedance, including neutral impedance. The unit rating should be 100 MVA and the scheduled generation should be 100 MW. Leave the rest the same. Click OK, then Done. • Right-click on relay 2 and choose Specify Fault. Choose Close-in-Fault, no outage. Choose the 3LG phase connections. Deselect all other phase selections, since here we are only interested in a balanced 3-phase fault. Click Simulate. On the toolbar, click on the clock icon to see the relay operating time and the fault current. Analysis: ● Pickup = 5 Curve CO-9 Right Click on Bus 2. Choose New, then Load. The parameter menu for the load will automatically appear. Enter 0 MW for constant power, 7.2 MW for constant current, and 0 MW for constant Impedance. Repeat for Bus 3, except use 25.22 MW of constant current. • Repeat this for a line end fault with no outage. Make sure to de-select close-in- fault. ● By holding down the control key and double-clicking on a line or bus, a graphical representation of the voltages and currents can be seen. To see a text file of simulation results, click on the TTY icon on the toolbar. After simulating a fault, right-click on the relay near Bus 1. Choose View Relay Curves and click OK. A graph of the time-overcurrent curve for relay 1 should appear. On the curve graph, click on the Add menu and select Relay Curves in Vicinity. Select Relay 2 from the window that appears and hit OK. The overcurrent curve for both relays should now be on the same graph. To see when the relays tripped, go into the Show menu and select Relay Operations for All Faults. Solution should consist of the relay coordination curves, a text file of simulation results, graphical representation of voltages and currents, screenshot depicting the relay operations and screenshot of the oneline diagram.

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Question-5 [75 Points] Two time-overcurrent relays protect adjacent sections of a radial system. Bus 3 is at the end of the radial line. Approximately 7000 amps of fault current will flow for a close-in fault at point A and approximately 5000 amps of fault current will flow for a line-end fault at point B. Load curren