Analysis of unsymmetrical faults using bus impedence matrix
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1. The document analyzes unsymmetrical faults using the bus impedance matrix method. It describes connecting sequence networks for positive, negative, and zero sequences for line-to-ground faults. 2. Current is injected into the faulted bus in each sequence network. Voltages are then calculated based on the impedances in the bus impedance matrices. 3. Post-fault voltages and currents can be determined for all buses to analyze the effects of unsymmetrical faults on the system.
Analysis of unsymmetrical faults using bus impedence matrix
- 1. ANALYSIS OF UNSYMMETRICAL FAULTS USING BUS IMPEDENCE MATRIX
- 2. INTRODUCTION OF UNSYMMETRICAL FAULTS Most Of the faults that occur on power systems are unsymmetrical faults, which may consist of unsymmetrical short circuits, unsymmetrical faults through impedances, or open conductors. Unsymmetrical Faults occur as single line--‐to--‐ground faults, line--‐to--‐line faults, or double line--‐to--‐ground faults. The Path of the fault current from line to line or line to ground may or may not contain impedance. One Or two open conductors result in unsymmetrical faults, through either the breaking of one or two conductors or the achon of fuses and other devices that may not open the three phases simultaneously
- 3. Connection of sequence network for LG fault 𝑖 𝑡ℎ bus Consider that there are n number of buses in a System and a LG fault occurs 𝑖 𝑡ℎ bus of this system The positive sequence network is replaced by its Thevenin’s equivalent i.e. the prefault voltage. 𝑉1−𝑖 0 Of bus I in series with the passive positive sequence Network. As there is no prefault negative and zero sequence Voltage ,both are passive network only
- 4. for passive positive sequence network 𝑉1−𝑏𝑢𝑠 = Z1−𝑏𝑢𝑠 . I1−𝑏𝑢𝑠 Where 𝑉1−𝑏𝑢𝑠 = 𝑉𝑖−1 𝑉𝑖−2 .. 𝑉𝑖−𝑛 = 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑏𝑢𝑠 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 Z1−𝑏𝑢𝑠 = 𝑍𝑖−11 ⋯ 𝑍𝑖−1𝑛 ⋮ ⋱ ⋮ 𝑍𝑖−𝑛1 ⋯ 𝑍𝑖−𝑛𝑛 = 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑏𝑢𝑠 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑚𝑎𝑡𝑟𝑖𝑥 I1−𝑏𝑢𝑠 = 𝐼𝑖−1 𝐼𝑖−2 . . 𝐼𝑖−𝑛 = 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑏𝑢𝑠 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟
- 5. According to sequence network connection , current −𝐼1−𝑖 𝑓 is injected only at the faulted 𝑖 𝑡ℎ bus of the positive sequence network. Hence, I1−𝑏𝑢𝑠 = 0 0.. . −𝐼1−𝑖 𝑓 . . 0 Hence the positive sequence voltage at the 𝑖 𝑡ℎ bus of the passive positive sequence network is 𝑉1−𝑖 = −Z1−𝑖𝑖 . I1−𝑖 𝑓 Thus, the passive positive sequence network present an impedance Z1−𝑖𝑖 to the positive sequence current I1−𝑖 𝑓
- 6. For a negative sequence network. 𝑉2−𝑏𝑢𝑠 = Z2−𝑏𝑢𝑠 . I2−𝑏𝑢𝑠 The negative sequence network is injected with current I1−𝑖 𝑓 at the 𝑖 𝑡ℎ bus only.hence, I2−𝑏𝑢𝑠 = 0 0.. . −𝐼2−𝑖 𝑓 . . 0 ∴𝑉2−𝑖 = −Z2−𝑖𝑖 . I2−𝑖 𝑓 Thus negative sequence network offers an impedance Z2−𝑖𝑖 to the negative sequence current 𝐼2−𝑖 𝑓
- 7. For zero sequence network 𝑉0−𝑏𝑢𝑠 = Z0−𝑏𝑢𝑠 . I0−𝑏𝑢𝑠 I0−𝑏𝑢𝑠 = 0 0.. . −𝐼0−𝑖 𝑓 . . 0 𝑉0−𝑖 = −Z0−𝑖𝑖 . I0−𝑖 𝑓 Thus zero sequence network offers an impedance Z0−𝑖𝑖 to the zero sequence current 𝐼0−𝑖 𝑓
- 8. From sequence network connection, we can write, 𝐼1−𝑖 𝑓 = 𝐼2−𝑖 𝑓 = 𝐼0−𝑖 𝑓 = 𝑉1−𝑖 0 𝑍1−𝑖𝑖 + 𝑍2−𝑖𝑖 + 𝑍0−𝑖𝑖 + 3𝑍 𝑓 Similarly we can compute sequence current for LL & LLG fault. For passive positive sequence network, the voltage developed at bus k due to injection of current −𝐼1−𝑖 𝑓 at bus I is , 𝑉1−𝑘 = −Z1−𝑖𝑘 . I1−𝑖 𝑓 Hence the postfault positive sequence voltage at bus K is, 𝑉1−𝑘 𝑓 = 𝑉1−𝑘 0 −Z1−𝑖𝑘 . I1−𝑖 𝑓 ; k=1,2,…n. Where, 𝑉1−𝑘 0 =prefault positive sequence voltage at bus K Z1−𝑖𝑘 = 𝑖𝑘 𝑡ℎ component of 𝑉1−𝑏𝑢𝑠
- 9. As the prefault negative sequence bus voltage is zero, the postfault negative sequence bus voltage is, 𝑉2−𝑘 𝑓 = 0 + V2−𝑘 𝑉2−𝑘 𝑓 = −Z2−𝑖𝑘 . I2−𝑖 𝑓 Where Z2−𝑖𝑘=𝑖𝑘 𝑡ℎ component of 𝑍2−𝑏𝑢𝑠 The postfault zero sequence bus voltage is 𝑉0−𝑘 𝑓 = 𝑉0−𝑘 0 −Z0−𝑖𝑘 . I1−𝑖 𝑓 ;k=1,2,…n. Where Z0−𝑖𝑘=𝑖𝑘 𝑡ℎ component of 𝑍0−𝑏𝑢𝑠
- 10. After computing postfault sequence voltage , the sequence current in the line can be computed as , for line pg. having sequence admittance 𝑌1−𝑝𝑔, 𝑌2−𝑝𝑔and 𝑌0−𝑝𝑔 𝐼1−𝑝𝑔 𝑓 = 𝑌1−𝑝𝑔(𝑉1−𝑝 𝑓 − 𝑉1−𝑔 𝑓 ) 𝐼2−𝑝𝑔 𝑓 = 𝑌2−𝑝𝑔(𝑉2−𝑝 𝑓 − 𝑉2−𝑔 𝑓 ) 𝐼0−𝑝𝑔 𝑓 = 𝑌0−𝑝𝑔(𝑉2−𝑝 𝑓 − 𝑉2−𝑔 𝑓 ) After computing sequence voltage and current, phase volage and current can be easily computed as , 𝑉𝑝 = 𝐴 𝑣𝑠 𝐼 𝑝 = 𝐴𝑖𝑠
- 11. As this method requires computation of bus impedance matrices of all the three sequence network, it seems to be more tedious and time consuming . But once the bus impedance matrices have been formed, fault analysis can be easily can be easily carried out for all the buses which is the aim of fault analysis. Also for any changes in power network bus impedance matrix can be easily modified