THE PRINCIPLES OF RELAY PROTECTION IN PROTECTIVE METHOD

Contents CHAPTER 1 : SELECT THE CURRENT TRANSFORMER 1.1 SELECT BI FOR LINE PROTECTION (BI7) .5 1.2 SELECT BI FOR TRANSFORMER PROTECTION .6 1.2.1 Select BI1 (BI4) 1.2.2 Select BI2 (BI5) 1.2.3 Select BI3, BI6 .6 CHAPTER 2: METHOD OF PROTECTION 2.1 METHOD OF PROTECTION FOR TRANSFORMER 2.1.1 The type of faults and abnormal working mode .7 2.1.2 Method of protection 2.2 METHOD PROTECTION FOR LINE CHAPTER 3: THE PRINCIPLES OF RELAY PROTECTION IN PROTECTIVE METHOD 3.1 RESTRAINT-DIFFERENT RELAY PROTECTION 3.1.1 The principle of restraint-different relay protection 3.1.2 Restraint differential relay protection .10 3.2 BUCHHOLZ RELAY PROTECTION 12 3.3 TIME OVERCURRENT RELAY 13 3.4 OVERCURRENT ZERO SEQUENCE PROTECTION .14 3.5 OVERLOAD TRANSFORMER PROTECTION 14 CHAPTER 4: CALCULATION OF SHORT CIRCUIT 16 4.1 CALCULATION OF REACTANCE VALUE .16 4.1.1 Calculation of reactance value of element in maximum power system mode .16 4.1.2 Calculate the resistances in the minimum system capacity mode 17 4.2 CACULATE SHORT-CIRCUIT CURRENT .18 4.2.1 Calculate short-circuit current in maximum system power mode 18 4.2.2 Calculate short-circuit current in minimum system power mode 24 CHAPTER 6: CALCULATE PROTECTION FOR LINE .27 6.1 SHORT-CIRCUIT CURRENT IN SOME CASES ON LINE L 27 6.1.1 Instantaneous overcurrent relay protection 27 6.2 MAXIMUM OVERCURRENT RELAY 29 6.2.1 Maximum power system mode 29 6.2.2 Minimum power system mode .30 6.3 CALCULATE ZERO PROTECTION (TTK) .30 CHAPTER 7: CHECK THE WORK OF PROTECTION FOR THE SUBJECTS .32 7.1 CHECK THE WORK OF THE TRANSFORMER PROTECTION 32 7.1.1 Check the work of the transformer protections compare offset current with braking .32 7.1.2 Check the work of over fast cutting current 34 7.1.3 Zero sequence in earth fault ( TTK) 34 7.2 CHECK THE WORK OF THE PROTECTIONS IN THE LINE 35 7.2.1 The instantaneous overcurrent .35 7.2.2 Maximum current protection 35 7.2.3 Zero sequence in earth fault (TTK) .35 RELAY PROTECTION PROJECT Code: 43554B I Grid diagram II Initial data Content System Transformer Line Initial data SNmax (MVA) 2300 SNmin (MVA) 1700 X0HT/X1HT 1,15 SBđm (MVA) 40 U1/U2: 115/24 kV, tổ đấu dây YN/yn- 12 UN (%) 11,5 Chiều dài Loại dây XLPE-240 Z1(Ω/km) 0,11+j0,33 Z0(Ω/km) 0,32+j0,94 X0L/X1L 2,8 Pmax (MW) 11 Load 0,9 0,5s Impedance: Line type XLPE-240 : 0,119+j0,154 (Ω/km)1 13,5 xTMS I Time characteristics of time overcurrent relay: t = r −1 , with TMS is constant of times set, with time delay t= 0.5s III Performance requirements Select BI for protection Select protection for transformers B1, B2 and line L Tra bảng Sổ tay tra cứu thiết bị điện từ 0,4-500kV, Ngô Hồng Quang The principles of relays protection Calculate the short circuit Calculate protection parameters of current overcurrent, overcurrent, overcurrent and earth fault protection for transformers Calculation of overcurrent protection, fast current overcurrent, overcurrent of current not set for line L Check the work of the guard for the above objects CHAPTER 1 : SELECT THE CURRENT TRANSFORMER 110 kV MC1 22 kV B1 BI1 BI2 MC2 BI3 L MC7 HTÐ MC4 BI4 B2 BI6 P1 BI7 BI5 MC5 Figure 1.1 Diagram of current transformer From the protection requirement of the project, we need to select BI to serve the protection of the transformer and line L In this, BI5 is used for line protection and BI1, BI2, BI3 and BI4 for transformer protection Conditions to select BI:  The rated current:  The rated voltage:  The secondary load:  Dynamic stability current coefficient:  Allowed force on top of BI's porcelain:  Multiples thermal stability: With: - Dynamic stability current coefficient,, which is determined by the manufacturer - a : is the distance between phases - distance from the current transformer to the nearest porcelain support Multiples thermal stability, which is determined by the manufacturer 1.1 Select BI for line protection (BI7) The maximum current working through the line L : I L lvmax=I pt max= P1 max √ U đ m cos = 11 103 =¿ 320,75 (A) √ 22.0,9  Select BI has the rated current 350A and the rated second current 5A, the rate voltage 22 kV The conversion ratio is n BI 7= 350 1.2 Select BI for transformer protection 1.2.1 Select BI1 (BI4) - Current transformer BI1 BI4 are select with the same the conversion ratio The maximum current flows through BI1 : IBI1max = kqt x S dmB √3 U Cdm = 1,4 40 103 = 293,92 (A) √3 110  Select BI has the rated primary current 300A and the rated second current 5A, the rate voltage 110 kV The conversion ratio is n BI 1= - Similar, the conversion ratio of BI4 isn BI 4= 300 300 1.2.2 Select BI2 (BI5) - Current transformer BI2 BI5 are select with the same the conversion ratio Considering the overload condition of the MBA, the maximun current flows through BI1 is: IBI1max=kqt x SdmB √3 U Hdm = 1,4 40 103 = 1469,62 (A) √3 22 Select BI has the rated primary current 1500 A and the rated second current 5A, the rate voltage 22 kV The conversion ratio is: n BI 2= - Similar, the conversion ratio of BI5 is : n BI 5= 1500 1500 1.2.3 Select BI3, BI6 We choose BI3, BI6 same BI1 so the ratio is n BI 3=n BI 6= 300 CHAPTER 2: METHOD OF PROTECTION 2.1 Method of protection for transformer 2.1.1 The type of faults and abnormal working mode Fault types can be divided into two groups: internal faults and external faults: + Internal faults: - Teminals fault - Short-circuit winding fault - Touch ground (short circuit) and short circuit ground - Breakdown of voltage divider - Oil tankers (oil leak) + External faults: - Short circuit on the system - One-phase short circuit in the system + Abnormal working mode: - Overload - Saturated magnetic circuits 2.1.2 Method of protection  Main protection: Restraint-different relay protection and Buchholz relay - Function : main protection for transformers - Protection area: against all types of faults inside the transformer + Restraint differential protection : remove short circuit Single phase or Multiphases inside transformer + Buchholz relay: remove winding faults and oil faults  Back-up protection : time overcurrent protection and instantaneous overcurrent protection - Function : + Back-up protection for transformers + Remove short-circuit faults occuring inside and outside transformer - Protection area : inside in transformers and a part outside - Note : + The impact time of back-up protections must be after the impact time of the main protections + Coordinating time with neighboring protections + If the transformer receives power from multiple sources, then put the power orientation at the connection to the source having smaller impact time + If there is a two-winding transformer then just put the overcurrent protection at one end of the transformer, near the source (because if one coil is overloaded then the remaining coil of the transformer is overloaded too) If there is a transformer with multiple windings, each side must have one set  Neutral overcurrent protection - Function : Ground cover (inner shell) inside the transformer - Protection area :  Overload protection : Overcurrent or Thermal relay - Function : Remove overload fault - Protection area : Figure 2.1 Diagram of protection mode for transformers 2.2 Method protection for line Line L is a medium voltage line, so we need to put protection against short circuits, touching on the line We use time overcurrent relay (50) as the main protection (Inverses-Time Overcurrent), instantaneous overcurrent relay (51) as redundancy To detect and prevent ground faults on line L, use zero overcurrent relay (50N, 51N) Figure 2.2 Diagram of protection mode for line CHAPTER 3: THE PRINCIPLES OF RELAY PROTECTION IN PROTECTIVE METHOD 3.1 Restraint-different relay protection 3.1.1 The principle of restraint-different relay protection - Directly compare the amplitude of the current at the two ends of the protected element - Active when the current deviation between two protected elements exceeds a given value (threshold current): The protective zone of the differential protection is limited by the position of the two current transformers at the begin and the end of the protected element, from which the current signal is received for comparison Figure 3.1.1 Differential relayprotection a) Diagram of the principle; b) Vector graph of current when short circuit outside the zone and in normal mode; c) short circuit in the area In theory =0 However, reality may be different by the effect of unbalanced currents by some of the following reasons: + Due to the error of the current transformer has a small value + Due to magnetic circuit saturation of BI: occurs when the transformer is no loaded or short circuit outside has large value in the short time + Due to voltage regulation The threshold current is defined as follows: with: =0,1 : homogeneous coefficient BI the same select : noncyclical coefficients, =1 =1,8 : the largest external short circuit 3.1.2 Restraint differential relay protection Restraint differential relay protection is differential relay has been added restraint element to increase the sensitivity and reliability of the protection (With conventional dimension from the busbar to the line) Figure 3.1.2 Restraint differential relay protection Threshold current of restraint differential relay has changes when the current in branch of the protection circuit The comparator part of the relay compares the absolute value of the two currents Differential current: I ˙SL = I˙ = I ˙T 1− I ˙T Restraint current: I˙H = I ˙T + I ˙T Relay impacts when ISL ≥ IH - When the short circuit is outside the protected area: The two secondary current vectors IT1 and IT2 have a small deviation angle so the restraint current is larger than the differential current, the relay does not work 10 - Short circuit N(1): XN2∆= XN22∑ + XN20∑ = 0,083+0,084 = 0,167 Current of phase A in positive: (1 ) I Na1 = EHT = =4 X N 21+ X N ∆ 0,083+ 0,167 Short circuit current at N2: (1) (1) (1) I N =m I Na 2=3.4=12 Short circuit current in zero: (1) (1) (1) I N =3 I Na 0=3 I Na1=12 - Short circuit N(1,1): X N 2∆= X N 22 Σ X N 20 Σ 0,083.0,084 = =0,042 X N 22 Σ+ X N 20 Σ 0,083+0,084 Current of phase A in positive: (1,1) I Na1 = E HT = =8,02 X N 21 + X N ∆ 0,083+0,042 Short circuit current at N2: (1,1) (1,1) I N =m √ I Na1 =√ 1− (1,1) √ ¿ √ 1− X N 22 Σ X N 20 Σ (1,1) ( X ¿ ¿ N 22 Σ+ X N 20 Σ ) I Na1 ¿ 0,083.0,084 8,02=12,02 ( 0,083+0,084)2 Three-phase shortcircuit current in zero: (1,1) ( 1,1 ) I (1,1) N =3 I Na =−3 I Na1 ¿−3.8,02 X N 22 Σ X N 22 Σ + X N 20 Σ 0,083 =−11,95 0,083+0,084 Similar calculate, we have: Table of short circuit of max mode with transformer working N1 N2 N3 N4 N5 N6 0,011 X1 0,083 0,093 0,103 0,112 0,122 X2 0,011 0,083 0,093 0,103 0,112 0,122 X0 0,012 0,084 0,111 0,139 0,166 0,193 XΔ(1,1) 0,0057 0,042 0,05 0,06 0,07 0,07 m(1,1) 1,500 1,50 1,502 1,506 1,509 1,513 I (3) 90,91 12,05 10,78 9,76 8,92 8,2 INa1(1) 29,41 4,00 3,37 2,91 2,56 2,29 IN(1) 88,23 12,00 10,11 8,73 7,69 6,86 I0(1) 88,23 12,00 10,11 8,73 7,69 6,86 23 INa1(1,1) 59,88 8,02 6,98 6,2 5,58 5,08 IN(1,1) 89,85 12,02 10,48 9,33 8,42 7,69 I0(1,1) 85,91 11,95 9,52 7,91 6,76 5,91 4.2.2 Calculate short-circuit current in minimum system power mode The type of short circuit: N(2)- Double phase short circuit N(1) – Line to ground short circuit N(1,1)- Double phase short circuit with ground connection a) transformers working parallel  Short circuit current at N1 negative, zero when the short circuit at the point N1  The total resistance positive, ' ' ' ' ' ' X N 11 Σ =X 11=0,015 X N 12 Σ =X N 11 Σ=0,015 X N 10 Σ =X 10=0,017 - Short circuit N(2) ' ' X N Δ=X N 12 Σ=0,015 Short-circuit current at N1: I N =m I Na 1=√ (2) - (2) E (2) X HT ' N 11 Σ +X Δ =√ =57,74 0,015+0,015 Short circuit N(1): XN’1∆= XN’12∑ + XN’10 = 0,015+0,017 = 0,032 Short-circuit current at N1: I N 1(1)=m (1) I (1) Na = E HT =3 =63,83 X N 11+ X N ∆ 0,015+ 0,032 Short cirtcuit current in zero: (1) I (1) N =I N 1=63,83 - Short circuit N(1,1): X N 1∆= X N 12 Σ X N 10 Σ 0,015.0,017 = =0,008 X N 12 Σ+ X N 10 Σ 0,015+0,017 Current of phase A in positive: (1,1) I Na1 = E HT = =43,54 X N 11 + X N ∆ 0,015+0,008 Short-circuit current at N1: 24 (1,1) (1,1) I N =m √ I Na1 =√ 1− (1,1) √ ¿ √ 1− X N 12 Σ X N 10 Σ (1,1) ( X ¿ ¿ N 12 Σ+ X N 10 Σ ) I Na1 ¿ 0,015.0,017 43,54=65,35 (0,015+0,017)2 Three-phase shortcircuit current in zero: (1,1) ( 1,1 ) (1,1) I N =3 I Na =−3 I Na1 ¿−3.43,54 X N 12 Σ X N 12 Σ + X N 10 Σ 0,015 =−61,22 0,015+ 0,017 Similar calculate, we have: Table of short circuit of mode with transformer working in parallel N1 N2 N3 N4 N5 N6 X1 0,015 0,051 0,061 0,071 0,08 0,09 X2 0,015 0,051 0,061 0,071 0,08 0,09 X0 0,027 0,053 0,08 0,108 0,135 0,162 (1,1) XΔ 0,008 0,026 0,035 0,043 0,05 0,058 (1,1) m 1,501 1,500 1,505 1,511 1,516 1,52 (2) I 57,74 16,98 14,26 12,28 10,79 9,62 (1) INa1 21,28 6,45 4,96 4,02 3,39 2,92 (1) IN 63,83 19,35 14,87 12,07 10,16 8,77 (1) I0 63,83 19,35 14,87 12,07 10,16 8,77 (1,1) INa1 43,54 12,99 10,49 8,84 7,66 6,76 (1,1) IN 63,65 19,48 15,79 13,36 11,61 10,28 (1,1) I0 61,22 19,11 13,56 10,51 8,58 7,25 b) transformer works independently  Short cirtcuit current at N1, N1’ The same in transformer mode works in parallel  Short cirtcuit current at N2, N2’ The total resistance positive, negative, zero when the short circuit at the point N2 : XN21 = X’11 + X ' 21 = 0,015+ 0,072 = 0,087 X’N22 = X’N21 = 0,087 X’N20 = X’10 + X ' 21 = 0,017 + 0,072 = 0,089   - Short circuit N(2) ' ' X N Δ=X N 22 Σ =0,087 Short circuit current at N2: I N =m I Na2= √3 (2) (2) (2) E HT =√3 =9,95 X ' N 21+ X ' N ∆ 0,087+0,087 25 - Short circuit N(1): XN’2∆= XN’22∑ + XN’20∑ = 0,087+0,089 = 0,176 - Short circuit current at N2: (1) (1) (1) I N =m I Na1=3 E HT =3 =11,41 X ' N 21+ X ' N ∆ 0,087+ 0,176 Short circuit current in zero: 11,41 - Short circuit N(1,1): X N 2∆= X ' N 22 Σ X ' N 20 Σ 0,087.0,089 = =0,044 X ' N 22 Σ + X ' N 20 Σ 0,087+0,089 Current of phase A in positive: (1,1) I Na1 = E HT = =3,8 X N ' 21+ X N '2 ∆ 0,087+ 0,176 Short circuit current at N2: (1,1) (1,1) I N =m √ I Na1 =√ 1− (1,1) √ ¿ √ 1− X N 22 Σ X N 20 Σ (1,1) ( X ¿ ¿ N 22 Σ+ X N 20 Σ ) I Na1 ¿ 0,087.0,089 3,8=11,45 (0,087+0,089)2 Three-phase shortcircuit current in zero: (1,1) X N 22 Σ X N 22 Σ + X N 20 Σ 0,087 ¿−3.3,8 =−11,32 0,087+0,089 ( 1,1 ) (1,1) I N =3 I Na =−3 I Na1 Similar calculate, we have: Table of short circuit of mode with transformer working N1 N2 N3 N4 N5 N6 X1 0,015 0,087 0,097 0,107 0,116 0,126 X2 0,015 0,087 0,097 0,107 0,116 0,126 X0 0,027 0,089 0,116 0,144 0,171 0,198 X 0,008 0,044 0,053 0,061 0,069 0,077 1,501 1,5 1,502 1,505 1,509 1,512 57,74 9,95 8,95 8,13 7,45 6,87 21,28 3,8 3,23 2,81 2,48 2,22 63,83 11,41 9,69 8,42 7,44 6,67 (1,1) Δ m (1,1) I (2) INa1 I (1) (1) N 26 I0(1) 63,83 11,41 9,69 8,42 7,44 6,67 INa1(1,1) 43,54 7,63 6,69 5,97 5,39 4,98 IN(1,1) 63,65 11,45 10,04 8,98 8,14 7,45 I0(1,1) 61,22 11,32 9,11 7,62 6,55 5,75 27 CHAPTER 6: CALCULATE PROTECTION FOR LINE 6.1 Short-circuit current in some cases on line L The resistance of transformer when two transformers work in parallel in half when a transformer works independently Therefore, the largest short-circuit current on the line when a short circuit occurs will be the max system power mode and two transformers work in parallel The smallest short-circuit current on the line when a short-circuit occurs on the line will be the system power mode and transformer works independently Table 6.1 Short-circuit currents in max system power mode, transformers: N2 N3 N4 N5 N6 IN(3) 21,28 12,23 10,93 9,88 9,01 IN(1) 21,12 11,29 9,60 8,35 7,39 I0(1) 21,12 11,29 9,60 8,35 7,39 IN(1,1) 21,12 11,82 10,39 9,29 8,41 I0(1,1) 20,09 10,48 8,56 7,23 6,26 Table 6.2 Short-circuit currents in system power mode, transformer: N2 N3 N4 N5 N6 I (2) 9,95 8,95 8,13 7,45 6,87 IN(1) 11,41 9,69 8,42 7,44 6,67 I0(1) 11,41 9,69 8,42 7,44 6,67 IN(1,1) 11,45 10,04 8,98 8,14 7,45 I0(1,1) 11,32 9,11 7,62 6,55 5,75 We calculte instantaneous overcurrent relay, maximum overcurrent relay, zero overcurrent relay prrotection for line L 6.1.1 Instantaneous overcurrent relay protection The threshold current of protection The threshold current : with: kat – safety factor, kat = 1,2 I Nngmax – The maximum external short-circuit current is the largest shortcircuit current that is usually equal to the short-circuit current value of the bar on the end of line 28 - The external short-circuit current of line L: The maximum external short-circuit current: 9,01 The minimum external short-circuit current: 7,39 - The threshold current for instantaneous overcurrent relay protection of line L : = 1,2 9,01 =10,81 Defining protection area: Protect area of instantaneous overcurrent relay 30 25 20 Max shortcircuit current Min shortcircuit current Threshold current 15 10 N2 N3 N4 N5 N6 The protected area is considered the length of the protected line, calculated from the beginning of the line (from the protection point) to the point where the short-circuit current is equal to the thresold current of the protection - The largest protection area: (Lmax) Trong đó: The largest protection area: 29 [ ] 0,072 23 −(0,011+ ) = = 6,25(km) 10,81 25.0,154 6,25 100%= 89,29% => Protect for 89,29% of line 6.2 Maximum overcurrent relay 6.2.1 Maximum power system mode The threshold current : with: kat –Safety factor, kat = 1,2 kmm – Open factor, kmm = kv - Return coefficient selected for digital relay: kv = 0,95 Ilvmax- The largest working current, Ictlvmax = 238,57 A Convert maximum working current to relative unit system: 23=0,38  The threshold current  0,38 = 0,96 - Construction work time characteristics: Overcurrent protection definite time : t= 13,5 x TMS I ¿ −1 with with : TMS: constant time set of relay (s) I ¿: Short-circuit current through relay 30 - Short circuit point N6: IN6 max = 9,01 9,01 0,96 = 9,39 Impact time characteristics at N6 t  = 13,5 x TMS I ¿ −1 13,5 xTMS 1= 9,39−1  TMS = 0,62 Short circuit point N5: IN5 max = 9,88 I N max I ¿= 9,88 = 10,29 I kdI >¿ = ¿ 0,96 t  = 13,5 x TMS I ¿ −1 13,5 x 0,62= 0,9(s) t5= 10,29−1 Similar calculations for short circuits on the line we have Table 6.3 Impact time of the max system   IN max 21,28 12,23 10,93 9,88 9,01 Ikđ 0,96 I* 22,17 12,74 11,39 10,29 9,39 TMS 0,62 t 0,40 0,71 0,81 0,90 1,00 6.2.2 Minimum power system mode Table 6.4 Impact time of the system   IN max 9,95 8,95 8,13 7,45 6,87 Ikđ 0,96 I* 10,36 9,32 8,47 7,76 7,16 TMS 0,62 t 0,89 1,01 1,12 1,24 1,36 31 CHARACTERISTICS OF THE TIME OF OVERCURRENT PROTECTION t max mode t mode 003 002 001 001 001 001 001 001 001 001 001 001 000 000 6.3 Calculate zero protection (TTK) -Threshold current of protection with: k0 - adjustment coefficient, k = 0,3 IdmBI - Rated current of the current transformer set for the line The working time of overcurrent protection does not have time to select according to independent characteristics: 1.2 0.8 1 1 0.6 0.4 0.2 N2 N3 N4 N5 N6 Time characteristics impact of overcurrent protection TTK 32 CHAPTER 7: CHECK THE WORK OF PROTECTION FOR THE SUBJECTS 7.1 Check the work of the transformer protection 7.1.1 Check the work of the transformer protections compare offset current with braking a Checking safety factor in the brake when short circuit in external over current To checksafety factor in the brake when short circuit in external over current, we check when short circuit in enternal over current is the largest - Consider the largest current through short circuit at N2 The largest short circuit current go through protection in every MBA is N (3) short circuit current in the max power system, transformer work independently (3) (3) I N BVmax=I N 2=¿ 12,05 The largest short circuit current go through protection in every transformer is N (1,1) in the max power system, transformer work independently (1,1) (1.1) I N BVmax =I N =12,02 Conclusion, short circuit in external protection by: INngmax = 12,05 I SL=I kcbtt max=f imax K dn K kck I N ngmax ¿ 0,1.1 1,8.12,05=2,17 I H =2 I N ngmax =2.12,05=24,1 Safety factor in the brake is defined by fomula: With IHtt is calculate restrain current Straight line ISL = 1,34 cut characteristic (c) so: K atH = I H 24,1 = = 4,65 > I Htt 5,18  Protecting brake stady, not affected when short circuit outside the protection zone b Short-circuit sensitivity in the protection zone When short-circuited in the protection zone, because of the supply, the I SL difference current is always equal to the I H damping current, so the theoretical relays always work Check the working of the relay we check the sensitivity : 33 With: Ikd – starting current in protection To check the sensitivity of the protection, we consider the smallest short-circuit current when a short circuit occurs in the protection zone (at N1’ and N2’)  When short circuit at N1’: Following to calculate in chapter 4, The smallest short-circuit current that passes through short circuit protection at N1 'is a two-phase short-circuit when the system power is minimum INmin= 57,74 We have ISL = IH = INmin = 57,74 Straight line IH= 57,74 cut characteristic line (d)  Ikd =  Sensivity of protection : = 57,74 =7,22 >2  So, protection ensure to cut safely when having short circuit at N1’  When short circuit at N2’ Following to calculate in chapter 4, the smallest short-circuit current passing through short circuit protection at N2 'is a two-phase short-circuit when the minimum system power, two transformer parallel operation INmin= 16,98 We have ISL = IH = INmin = 16,98 Straight line IH= 16,98 cut starting characteristic current line at characteristic segment (c)  Ikd = -1,25+0,5IH = 4,56  Sensivity of protection: K N = I SL 16,98 = =3,72>2 I SLtt 4,56  So, protection ensure to cut safely when having short circuit at N2’ 34 The impact characteristics of the braking differential protection applied to the MBA 7.1.2 Check the work of over fast cutting current  Check sensivity of protection Sensivity of protection is defined by formula: k N= I N I kd with condition Sensivity of protection : (2) I Nmin I N 9,95 kN= = = =2,81>1,5 I kd I kd 3,54  So protection get requirement for sensitive level 7.1.3 Zero sequence in earth fault ( TTK)  Check sensivity of protection Sensivity of protection is defined by formula: k N= I N I kd with condition Sensivity of protection : 35 I Nmin I (1,1) 11,45 N2 kN= = = =38,17> 1,5 I kd I kdTTK 0,3  So protection get requirement for sensitive level 7.2 Check the work of the protections in the line 7.2.1 The instantaneous overcurrent  Check sensivity of protection kN= I Nmin I (1) 6,67 N6 = = =6,95> 1,5 I kd I kd 0,96  So protection get requirement for sensitive level  Check sensivity of protection The smallest short circuit currenr TTK go through protecting is short circuit N(1,1)current at N6 in the minimum power system 5,75 Sensivity of protection: kN= I Nmin 5,75 = =49,9> 1,5 I kd 0,12 => So protection get requirement for sensitive level 7.2.2 Maximum current protection  Check sensivity of protection Sensivity of protection is defined by formula: k N= I N I kd with condition Sensivity of protection : I Nmin I (2) 6,87 N6 kN= = = =7,16> 1.5 I kd I kd 0,96 => So protection get requirement for sensitive level 7.2.3 Zero sequence in earth fault (TTK)  Check sensivity of protection Sensivity of protection is defined by formula: k N= I N I kd with condition The sensitivity of protection : 36 kN= I Nmin 5,75 = =49,9> 1.5 I kd 0,12  So protection meets the sensitivity requirement Tài liệu tham khảo Bảo vệ hệ thống điện GS Trần Đình Long NXB khoa học kỹ thuật Ngắn mạch đứt dây hệ thống điện PGS.TS Phạm Văn Hòa NXB khoa học kỹ thuật 37 ... considered the length of the protected line, calculated from the beginning of the line (from the protection point) to the point where the short-circuit current is equal to the thresold current of the protection. .. RELAY PROTECTION IN PROTECTIVE METHOD 3.1 Restraint-different relay protection 3.1.1 The principle of restraint-different relay protection - Directly compare the amplitude of the current at the. .. accumulate in the lid of the Buchholz relay When the amount of gas accumulated is large enough, it will sink the float and close the contact, the relay sends a level warning - When the incident