A single-line diagram of a planned network is shown in Figure B1, and component parameters given in Table B1. Motor M1 takes an electrical input power of 2 MW from a supply voltage of 11 kV. The fault level capacity of motor M1 is 10 MVA. Calculate: (i) (ii) The corresponding reactances using a common base of 5 MVA and draw a per-unit reactance diagram, showing the location of the fault. The fault level and fault current which would flow into a three-phase symmetrical short circuit fault at the 11 kV busbar. Component Generator G1 Generator G2 Transformer T1 Rating or Length 12.9 MW, 0.86 pf 8.5 MW, 0.85 pf Voltage 6.6 kV Reactance XG1 = 0.1 pu 6.6 kV XG2 = 0.16 pu 25 MVA 6.6/66 kV XT1 = 0.08 pu Transformer T2 Transformer T3 25 MVA 12.5 MVA 66/11 kV XT2 = 0.2 pu 11/0.4 kV XT3 = 0.2 pu Motor M1 2 MW, 0.89 pf, eff=83% 11 kV XM1 = 0.25 pu Line L1 12 km 66 kV 0.9 Q/phase/km Table B1

A single-line diagram of a planned network is shown in Figure B1, and component parameters given in Table B1. Motor M1 takes an electrical input power of 2 MW from a supply voltage of 11 kV. The fault level capacity of motor M1 is 10 MVA. Calculate: (i) (ii) The corresponding reactances using a common base of 5 MVA and draw a per-unit reactance diagram, showing the location of the fault. The fault level and fault current which would flow into a three-phase symmetrical short circuit fault at the 11 kV busbar. Component Generator G1 Generator G2 Transformer T1 Rating or Length 12.9 MW, 0.86 pf 8.5 MW, 0.85 pf Voltage 6.6 kV Reactance XG1 = 0.1 pu 6.6 kV XG2 = 0.16 pu 25 MVA 6.6/66 kV XT1 = 0.08 pu Transformer T2 Transformer T3 25 MVA 12.5 MVA 66/11 kV XT2 = 0.2 pu 11/0.4 kV XT3 = 0.2 pu Motor M1 2 MW, 0.89 pf, eff=83% 11 kV XM1 = 0.25 pu Line L1 12 km 66 kV 0.9 Q/phase/km Table B1

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter7: Symmetrical Faults

Section: Chapter Questions

Problem 7.2MCQ: Even though the fault current is not symmetrical and not strictly periodic, the rms asymmetrical...

See similar textbooks

Similar questions

Q.3 When a line-to-ground fault occurs, the current in faulted phase 'a' is 100A. The zero-sequence current in phase 'c' is r
In a short circuit test on 132 kV, 3-phase system, the breaker gives the following results: Power factor of the fault = 0.45. Recovery voltage 0.9 time of full line voltage. The breaking current is symmetrical. The restriking transient has a natural frequency of 15 kHz. Calculate the rate of rise of restriking voltage (RRRV) in the following types of faults: (1) Grounded fault (2) Undergrounded fault.
A three phase fault is applied at the point P as shown in the figure. Find the criticalclearing angle for clearing the fault with simultaneous openings of the breakers 1 and 2. Thegenerator is delivering 1.0 pu at the instant preceding the fault
Q.3 When a line-to-ground fault occurs, the current in faulted phase 'a' is 100A. The zero-sequence current in phase 'c' is
The sequence components of the fault current are as follows: Ipositive = j2 p.u., Inegative = - j0.5 p.u. and Izero = -j 1.5 p.u. The type of fault in the system is
Figure 2 shows the circuit of a simple power system. The ratings of the generators, transformers and the motors are as shown in the figure. The reactance of lines 1 and 2 are 40 ohms and 50 ohms respectively. Assume that system is unloaded, and the generators are generating their rated voltage. Select the generator G₁ ratings as base values. Choose the fault location in bus 1 And give the below answers For a symmetrical three phase fault on the bus (a).Determine the Thevenin's equivalent circuit which can be used to calculate the fault current. (b) Calculate the fault current contribution from the generator connected to the bus on which you have selected for fault calculation
Two alternators rated 15 MVA and 20 MVA have per unit reactance of 0.20 and 0.60 respectively are connected to a common 12-kV bus bar. Solve the fault current if a symmetrical 3-phase fault occurs at the bus bar
complet The equivalent reactance of the double circuit line under post fault condition is................than fault condition.
answer both i will rate.... a. In sequence networks of power system, why does voltage sources appear only in the positive sequence? b. Describe the main differences in fault protection objectives when dealing with a temporary fault vs a permanent fault.
1. The subscript d in the generator subtransient reactance refers to: 1. Generator impedance 2. If the available fault current slightly exceeds the breaker published 2. Generator 3. Direct axis 4. 1 and 2 reactance interrupting rating, then it is safe to use the breaker. 1. True | 2. False 3. Maybe 3. The rms symmetrical fault current times an asymmetry factor K, is equal to the ac fault current. | 2. False 4. The most common fault on a 3-phase power system is: 1. True 3. Maybe | 2. DLG 3. L-L 1. SLG 5. All rotating and non-rotating load impedances are usually included in a power system fault study 1. True 2. False 3. Мaybe
Q2. The single-line diagram of a simple three-bus power system is shown in Figure-2. Each generator is represented by an emf behind the sub-transient reactance. All impedances are expressed in per unit on a common MVA base. All resistances and shunt capacitances are neglected. The generators are operating on no load at their rated voltage with their emfs in phase. A three-phase fault occurs at bus 3 through a fault impedance of Zf = j0.19 per unit. (i) Using Th'evenin's theorem, obtain the impedance to the point of fault and the fault current in (ii) Determine the bus voltages per unit. ) j0.05 j0.075 j0.75 2 j0.30 j0.45 Figure-2: Single line diagram of the power system network for Q2 3
Data obtained from a short circuit test on a 132 kV 3-phase circuit breaker with earthed neutral are as follows: -the power factor fault was 0.3. -the recovery voltage was 0.75 of the full line value. -the breaking current was symmetrical. -the restriking transient had a natural frequency of 1600 Hz. Estimate the rate rise of the recovery voltage. Assume that the fault is grounded.