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Q1: 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.See Answer
Q2:Question-3 [30 Points] (a) A wye-connected generator has a nameplate rating of 200 MVA, 20 kV, and its subtransient reactance (X) is 1.2 pu. Determine its reactance in ohms. (b) The generator is now connected to a power system where the base is specified as 100 MVA, 13.8 kV. What is the generator reactance in per unit on this system base? (c) Convert the per-unit answer calculated in part (b) to ohms. Does this match the value determined in part a?See Answer
Q3: ECE 473/5561 Intro to Protection Assignment#1 Protection Philosophy and Per Unit Representation Submission Deadline: 02/05/2024 Answer all the questions and upload the solution on Canvas by 11:59 PM 02/05/2024. Question-1 [75 Points] For the power system arrangement shown in Figure. 1, complete the given table, taking into account the operation of the circuit breakers as shown for each fault case. Please note that, as in the example discussed in the class, some of the circuit breakers that operated may have done so unnecessarily. Table 1: Relay/breaker operations Case F1 F2 F3 Breaker that operated 2,3,4,5 21,22,23,24,27 10,11,17,19 Name :- S.No - Breakers that maloperated Tripped by primary protection 2,5 Tripped by back up protection Bajo Anchicayá Tabor 10x 15 *11 Pailon 115 kV 115 kV 115 kV Termoyumbo XI 115 kV Chipichape 115 kV 181 201 Chipichape 34.5 kV 24 Diesel 34.5 kV Figure. 1: Power System Oneline Diagram San Luis 115 kV 8 -TE: *271 San Luis 34.5 kVSee Answer
Q4:Question-2 [40 Points] For the system shown in Figure. 2, directional overcurrent relays are used at breakers B12, B21, B23, B32, B34, and B43. Overcurrent relays alone are used at B1 and B4. (a) For a fault at P1, which breakers do not operate? Which breakers should be coordinated? Repeat (a) for a fault at (b) P2, (c) P3. (d) Explain how the system is protected against bus faults i.e. mention the breakers that operate for the bus faults. B1 B12 Load L1 P₂ B21 0 2 B23 0 Load L2 P₂ 3 B32 834 0 P₁ Load L3 Figure. 2: Oneline Diagram of a Radial System B43 B4 口 Load L4See Answer
Q5:Question-5 T5 Paints Two time-overcament relays protect adjacent sections of a radial system. Has 3 in at the end of the radial line. Appimately 7000 amps of fault current will flow for a coes-in fault at point A and approximately 5000 amps of fault curt will flow fo line-end fault at point II. Load current at bus 2 is 100A and load current at 3504 Hoth load carts at unity power fastor. Modell this situation using ASPEN Che Liner dware and simulate the aforementioned faults Observe the relay curves and the triptis. Investigate the phase and sequence voltage and currents for ach bus and lize after slmosting both faulte, (23) I • In the file menu, choose New to start a new project. Choose 100 MVA as the • Eight-click in the workspace and choose Now Blue. The Hus Info pop-up men should appear. Fill in the Name as Has 1, the lus a as 1, the No. KV 41.6. The buses can be moved by dragging them around the workspace. Do this again 2 and 3. Click on 1, hold down the shift key, and right-click on las 2. Choose New Line. The Transion Line Data mou should appear. Fill in the following Name = Line |_ Longth = 10 mi Laame srverythling olee the same. Repeat this for the line between buses 2 and 3, which will be called line 2. However, X-0.0774. Everything should be the same as line 1. • Eight-click on line 1 near This 1 and choose New Holay Group. On the mens that appears, click on the Add OC Phase Relay button. Fill the space 11-Relay ||GT ratio = 12 Timedial Tap unit Pickup-4 Dure=1002 Fer Make the dictional option is NOT selected. Click OK, then Ins. Do the same for line 2 near las 2, exopt for the following. ||II-Relay 2 CT ratio=0 Time del=0.8 Tap unit Pickup = 5 Curve Ch Make sure the directional option is NOT selected. Eight-click on us 1. Choose now, then getor. The parameter for the generator will automatically appear. Use E-0 and X-0.1 for each impedanc including neutral impedance. The unit rating should be 100 MVA and the scheduled gestion should be 100 MW. Leave the most the same. Click OK for the • Right click on us 2. Choose New, then Load. The parameter load will automatically appear. Kate MW for constant power, 7.2 MW for constant current, and 0 MW far constant Impedance. Hopeat for H3, cept use 25.22 MW of constant current • Eight-click on olay 2 and choose Specity Faulk. Choose Close-in-Vault, no outage Choose the phone connections. Doselect all other phase selections, since here we are only intested in a balanced 3-phase fault. Click Simulat . On the to, click on the clock icon to see the olay operating time and the fault current • Repeat this for a line end fault with no outage. Make sure to do-select close-is- . By holding down the control key and double-clicking on a line or bus, a graphical rotation of the voltages and currents can be see To sea text file of simation sults, click on the TTY icon on the toolbar. After ting fault, right-click on the relay near Blus 1. Chaos View Holay Curves and alak Di & gph of the tliese cooooumeut curve four relay 1 should appear. On the curve graph, click on the Add me and select Hlay Cri Vivinity. • Select Relay 2 from the window that appears and hit OK. The curat curve for both relays should now be on the same graph. To see when the relays tripped, go into the Show me 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 events, screenshot depicting the relay operations and screenshot of theSee Answer
Q6:Problem #3 In the power system network shown in Figure 1. bus 1 is a slack bus with V₁1.0/0° per unit and bus 2 is a load bus with S₂ 280 MW + j60 Mvar. The line impedance on a base of 100 MVA is Z = 0.02 +0.04 per unit. (a) Using Gauss-Seidel method, determine V₂. Use an initial estimate of V1.0+ 0.0 and perform four iterations. (b) If after several iterations voltage at bus 2 converges to V₂=0.90-j0.10, determine S, and the real and reactive power loss in the line. Z12=0.02 +0.04 2 FIGURE 1 One-line diagram for Problem #3 S₂ = 280 MW +60 MvarSee Answer
Q7:Problem #2 - Find Ybus matrix by inspection. j0.25 j0.25 0000 j0.1 j1.15 81.250 j0.125 j0.4 momo j0.2 reeeeteeee( j0.1 jl.15 0.85/-45°See Answer
Q8:Problem #1: Two generators rated 10 MVA, 13.2 KV and 15 MVA, 13.2 KV are connected in parallel to a bus bar. They feed supply to 2 motors of inputs 8 MVA and 12 MVA respectively. The operating voltage of motors is 12.5 KV. Assuming the base quantities as 50 MVA, 13.8 KV, draw the per unit reactance diagram. The percentage reactance for generators is 15% and that for motors is 20%.See Answer
Q9:Question 5 If a single-phase VT is connected to a resistive burden taking 200VA at 110V through wiring having a loop resistance of 1.50, what errors, additional to those of the VT itself, would result?See Answer
Q10:Question 3 a b In the example on restricted earth-fault protection in Section 4.3.6, (lead loop resistance = 10) decide the minimum relay operating voltage if the CT secondary winding resistance is 30 and the standard relay operating voltages are 50, 75, 100, 125, 150 and 175V. Using the curve (a) shown below, calculate the primary current to operate the relay if the mean core diameter is 20cms, the knee-point Bm = 1.2 Telsa and the relay operating current is 10mA. (Assume CT knee-point voltage = 2 x relay setting). 2.0 Flux Density - B max. (Tesla) 1.5 in 1.0 0.5 0.5 (b) 1.0 (c) 1.5 Exciting Forss-(AT/smr.ma) (a) 2.0 50 Hz magnetisation curves (a) Hot-rolled non-oriented silicon steel (b) Cold-rolled oriented silicon steel (c) Nickel-iron (80% nickel)See Answer
Q11:Question 2 a b с Calculate the VA output required from a CT of 5A rated secondary current when the burden consists of relays requiring 10VA at 5A plus a loop lead resistance of 0.80. What output VA would be required from a CT of 1A rated secondary current for a burden of 10VA at 1A plus lead resistance as in a? An earth fault relay has a burden of 1.0VA at setting and is set at 0.2A. What output voltage must a CT produce to ensure linear operation of the relay at 20 times setting? Assume that CT plus lead resistance is 1.50 and that the relay coil impedance saturates to its unsaturated value at 20 times setting.See Answer
Q12:Question 1 a b C Calculate the net core section of ring core CT of 1000/1 turns ration having a knee-point emf of 1500V at 50Hz. (Assume knee-point Bm = 1.4 Tesla) If the mean diameter of the core is 25cm and using curve (b) of Figure 4.2.2A, calculate the value of exciting current at an emf of 1000V. With 20kA primary current, and assuming the total impedance of the secondary circuit (including the winding) is 4002, calculate the percentage exciting current.See Answer
Q13: QUESTION: Problem P2 (20 points): A synchronous 60 Hz, 18 kV, 360 MVA generator is connected to an infinite bus (ideal voltage source- zero internal impedance) through a step-up transformer and a series capacitor compensated transmission line as in Figure P2. The generator's transient reactance (positive sequence) is 0.20 per unit on the generator ratings. The negative and zero sequence impedances of the generator are j0.21 and j0.11 pu on the generator ratings. The transformer is 360 MVA, 18 kV/230 kV, delta/wye connected (wye on the 230 kV side) with a leakage impedance of j0.08 pu on the transformer ratings. The transmission line positive sequence impedance is j0.62 ohms per mile, the negative sequence impedance is also j0.62 ohms per mile and the zero-sequence impedance is j1.85 ohms per mile. The length of the line is 165 miles long. The absolute value of the capacitor impedance is 45% of the absolute value of the transmission line positive sequence impedance. For initial conditions (prior to any faults) you can consider one of the two options below: Option 1: The internal voltage (generated) by the generator is 1.0 pu and the voltage of the infinite source is 1.0 pu. Determine the phase angles of these voltages so that prior to the fault there is zero power flow through the high side voltage of the transformer. Option 2: The generator delivers rated power at rated voltage at unity power factor at the terminals of the generator. (a) Compute the transmission line impedance in per-unit on the same base as for the 230 kV side of the transformer. (b) Compute the series capacitor impedance in per-unit on the same base as for the 230 kV side of the transformer. (c) Compute the three phase fault current for a three phase fault on the high side bus of the transformer in per-unit and in Amperes. (d) Compute the single line to ground fault current for a single line to ground fault on the high side bus of the transformer in per-unit and in Amperes. Figure P2 Hint: A similar problem is provided in Chapter 4, problem P4.3. NEED DETAILED HANDWRITTEN SOLUTIONS:/nTH 360 MVA 18/230 kV 8% G 360 MVA 18 kV Z₁ =j 0.20 230 kV Line Figure P2 Series Capacitor Infinite BusSee Answer
Q14:Question 3 A proposed 132kV/33kV substation is to be equipped with two 90MVA Star/Delta transformers which will operate in parallel at the 33kV busbar and will be supplied as transformer feeders from a 400/132kV Grid supply point, as shown in the diagram below. The connection will be by 20km of 175mm² ACSR double circuit overhead steel tower line; the line has positive sequence reactance of 0.225% per km and a zero sequence reactance of 0.587% per km. The positive sequence impedance of the transformers is expected to be 25% and the zero sequence impedance 21.25%. All impedances are quoted to a base of 100MVA. The negative sequence impedance equals the positive sequence impedance for all plant. The star point of each transformer will be connected directly to Earth. The fault level at the grid supply point is 2500MVA and the earth fault current is 13122A (equivalent to 3000MVA). Calculate for a line to ground short circuit on the red phase 132kV bushing of one transformer the current in amperes that would flow in: (a) each phase of the high voltage windings of each transformer; (b) each transformer star point, and (c) the red phase of the supply circuit. Calculate the zero sequence reactance in ohms of zigzag connected earthing transformers connected to the 33kV bushings of the main transformers which would limit the earth fault current on the 33kV busbar to 3150A. HELPING ADVICE: • The zero sequence impedance of the source is derived from the Earth Fault current and the fact that the impedance for an earth fault is the sum of the positive, negative and zero sequence impedances. • Use of the relationship (a+a)= -1 where a is the 120 degree operator should simplify your calculations. 132kV 33kVSee Answer
Q15:Question 1 The power system shown in Figure 1 develops a single-phase-to-earth fault at the point F on the 275kV terminals of the transformer at Station C. Ignoring the resistive components of the overhead-line impedances, determine the value of the earth-fault current at F and, in addition, the value of the current in the transformer neutral at Station C. All three generators have a rating of 275MW, 324MVA, with positive and negative sequence impedance values of j21.5% and j18.0% respectively. The generator step-up transformers have an impedance of j17.0% on their rating of 324MVA and the transformer at Station C has an impedance of j15.0% on its rating of 120MVA. The 275 kV over-head-line has a positive-sequence impedance of 0.034 + j0.325 ohms per km and a zero-sequence impedance of 0.15 + j0.813 ohms per km, the distances from Stations A, B and C to the tee point being 96km, 128km and 32km respectively. It can be assumed that the pre-fault phase-to-phase voltage at F is equal to the rated value of 275 kV, and that no loads are connected at Busbars A, B and C. Figure 1. 2 D 275kV iço Ⓒ questions_1708304124673.pdfSee Answer
Q16: A3. Transmission line equivalent circuit Due Tuesday, February 22 via Canvas Consider again the three phase transmission line from A2, Problem 1. The correct values (hopefully you got that from A2) for the series impedance and the shunt admittance per unit of length are: z = 0.13562+ 0.84869i y = 4.94335ix 106 Problem 1 1.1) 1.2) 1.3) 1.4) ohm / mi Compute the propagation constant and the characteristic impedance of the transmission line. Build the transmission matrix as a function of the transmission line length Assume the line is feeding a 161kV, 70MW/10MVAr load. Plot the efficiency of the line (in %) as function of the length of the line. What happens to the line efficiency as the length of the line increases? Problem 2 siemens / mi Assume now that z is purely imaginary. Repeat 1.3). What's the source of the losses on the transmission line? Problem 3 Assume now the transmission line is connected to a constant voltage source of 161 kV at the sending end and it is open (nothing connected to it) at the receiving end. Using the transmission matrix, plot the magnitude of the receiving end voltage as a function of the length of the line. What's going on here? (Hint: google Ferranti effect) Consider now the exact II equivalent circuit for the transmission line and assume the line is 100 mi long. 3.1) Compute Zs and Y. 3.2) Assume |V₁|=|V2|=161kV and 012-8°. Compute from the equivalent circuit the complex power at the sending end, receiving end and the dissipation of complex in each element. Check the conservation of complex power theorem. Problem 4 Consider the transmission line shown in the figure below. As can be seen, the series impedance is purely imaginary ( Z =jX) and the shunt impedance has been neglected. V₁/0■ P, Q V₁=1 per unit V₂=1 per unit X= 0.6 per unit 0=5⁰ jX 4.1) Develop an analytical expression for P and Q as a function of V₁, V2, 0 and X. 4.2) Typical values (in per unit) for V₁, V2, 0 and X are the following: V₂/0 Problem 1: 20 points Problem 2: 20 points Problem 3: 20 points Problem 4: 20 points Professionalism: 20 points ӘРӘ ӘР aQ Compute for the values specified above. What can you say about 9 9 2 ᎧᎾ Ꮎ ᎧᏙ , ' Ꮩ the sensitivity of P and Q with respect to 0 and V₁?See Answer
Q17:Probiam P2 (23 pointa]: Conside the 75MMA, 2007/13.88Y, Xrj0.138 pu thuse-pls da- we coned variable cap sep-up mandarmer indicated in Figure P2. This mandamer ispred with a dally. The deside is the 138 kV side. The candomer radiosae and forth 200kV and 128 k oddes, angesctionly. 24 R O R R 0 Patlas R R Figure:P2: 1. Computt wavormuariou 1 of the chave power mansforms of Figure 2. 2. Compute theding camets copecaiog oil of the illaial nelaryngeably (எ 3-phase balanced operating conditions. The specting car should be provided as a function of efer the 138 IV e eedaan the 230 kV e cuteid of pare A ou the 1KV side of the saetome the couve und the 2. Select the cacear ration forum the puovided lie baeilo tharailwale the operating arE untde tml operaring conditialone a poob Foryourtion of the CThis ttthenaincopa kal aporer facree tar the lone vielge ie afthe randomthetpse 4. ratio (up) has roceed a 24112. Compவo the opticumen (oTevitial) in the differentially under these conditions. You may use the computer program WhalGS to perform S. For your selection of the CTsiname that the come upecates at norainal power, al and porern facsoe at the love vulge ice afthe randform l ratio (ap) tax rmoned o 204114.4. Compuசை the operating curmen (clatleyrctial) in the differentially under these conditions. You may use the computer program WhalGS to perform Table P2: Current Transformer Standard Ratios 1200:5 2000:5 3000:54000-55000-5 100:5 300:5 2005 00:5 300:5 500:5 கை Secondary 500:5 600:5 2005 15005 2000:5 300:5 500:5 500-5 500:5 1000 1000 2005 1500:5 15005 2000:5 20005 2500:5 2500:5 900:5 1000:5 11005 1200:5 13005 5005 2000:5 3000:5 2500:5 3500:5 900:5 1600:5 3200:5; 4000-5 40005 10005 2000:5 2500:5 50005 12005 3000:5 GOOD S 600:5 1200 1900:5 24005 3000-5 3000-5 4000-5 4000-5 6000:5/nSee Answer
Q18:Problem 5 (10 points) Two generating units rated at 300 MW and 400 MW have governor speed regulation (droop) of 6.0 and 4.0 percent, respectively. The initial load of the system is 200 MW that is shared equally between these two units. (I.e. 100 MW each). If the load is changed from 200 MW to 300 MW, what is the output of each unit?See Answer
Q19:Problem 4 (10 points) The fuel cost functions in $/h for three thermal plants are given by C1 = 100+ 7.0 P1 +0.004 (P1)² C2 = 200+ 8.0 P2 +0.002 (P2)² C3 = 300+ 6.0 P2 +0.003 (P3)²2 a) Find an economic dispatch for PD = 1000 MW. What are P1, P2, and P3? b) What is the system incremental cost for above dispatch? c) What is the generation cost in $/h for each generator?See Answer
Q20:Problem 3 (10 points) A transmission system has a bus fault at one of its 138 kV buses. The positive, negative, and zero impedance of the bus are given as, Z1 = j0.10 pu, Z2 = j 0.15 pu, and 20 =j 0.2 pu respectively. a) (2 points) Calculate the magnitude of a 3-phase fault current in amperes. b) (3 points) Calculate the magnitude of a single line to ground fault current in ampereSee Answer

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