calculation of relay settings for differentialFull description
protection
Differential Protection fr Motors and transfrmersFull description
Bus Bar differential protection
Full description
CT connection
Protection scheme required for the protection of power system components against abnormal conditions such as faults etc., and that essentially consists of protective relaying and circuit breaker. Protective relay senses the fault and determines the l
method to calculate relay protection settingFull description
Principios y Aplicaciones_Protección diferencial Digital.
Full description
VERIFICATION OF CURRENT TRANSFORMERS FOR TRANSFORMER DIFFERENTIAL PROTECTION APPLICATIONS
TRANSFORMER GENERAL DATA:
KVA: Z%: Vp (V): Vs (V): Tap min (%): Tap max (%):
30000 10 34500 6300 -5 5
Rated Power Impedance in % Primary Voltage in Volts Secondary Voltage in Volts Minimum Tap in % Maximum Tap in %
TRANSFORMER PRIMARY SIDE: CT Primary Amps / CT Secondary Amps CTp Ratio: 600 5 VAprim: 50 CT Rated Power Class: 5P20 Accuracy %, Type (P,C), Limit Factor (Maximum CT times to satisfy accuracy). Eg. 10P20 RCTp (Ω): 0.648 CT Resistance in ohms LCTp (m): Cross Section (mm2):
800
Cable Length from CT to Protective relay (in meters)
16
Cable cross section (in mm2)
TRANSFORMER SECONDARY SIDE: CT Primary Amps / CT Secondary Amps CTs Ratio: 3000 5 VAsec: 15 CT Rated Power Class: 5P20 Accuracy %, Type (P,C), Limit Factor (Maximum CT times to satisfy accuracy). Eg. 10P20 RCTs (Ω): 1.13 CT Resistance in ohms LCTs (m): Cross Section (mm2):
20
Cable Length from CT to Protective relay (in meters)
6
Cable cross section (in mm2)
PROTECTION RELAY DATA: RELAY: ABB REF-542 Relay model (For information only) RELAY 0.1 Relay Burden in VA BURDEN (VA): MAXIMUM EXTERNAL FAULT. CURRENT THROUGH: Current seen by: Shortcircuit at: HV Side LV Side PRIMARY SIDE (A): 3000 8000 SECONDARY SIDE (A): 800 15000
1) UNIT VALUES Primary Current Full Load Ip = KVA/(√3*Vp/1000) Ip = 502.04 A
Ip_pu = Ip_pu =
Ip/CTp 4.1837
Secondary Current Full Load Is = KVA/(√3*Vs/1000) Is = 2749.29 A
Is_pu = Is_pu =
Is/CTs 4.5821
Primary Current Maximum Fault Current through Ifp = Max [Ifault prim, Ifault sec] Ifp_pu = Ifp = 3000 A Ifp_pu =
Ifp/CTp 25.0000
Secondary Current Maximum Fault Current through Ifs = Max [Ifault prim, Ifault sec] Ifs_pu = Ifs = 15000 A Ifs_pu =
Ifs/CTp 25.0000
2) BURDEN CALCULATIONS Primary Lead Resistance RPkm 1.15 (ohms/Km): Rp (Ω) =
2 * LCTp * RPkm/1000
Secondary Lead Resistance RSkm 3.08 (ohms/Km): Rs (Ω) =
[Obtained from cooper cable data table] Rp (Ω) =
1.84
2 * LCTs * RSkm/1000
Rs (Ω) =
0.1232
Rrelay (Ω) =
0.004
Bprimary (Ω) =
1.844
(ICT_sec2): 1A or 5 A
Total External Resistance Burden Rp + Rrelay
Ip_pu2 * Rp (VA) =
32.206
Rs_VA @ FLC =
Is_pu2 * Rs (VA) =
2.587
[Obtained from cooper cable data table]
Relay Resistance Rrelay (Ω) = VArelay/(ICT_sec2)
Bprimary (Ω) =
Rp_VA @ FLC =
Bsecondary (Ω) = Rs + Rrelay
Bsecondary (Ω) = 0.1272
3) CALCULATION OF CT BURDEN AND DRIVING VOLTAGES 3a) FULL LOAD OPERATION Driving Voltage. Primary side Vsp (V) = (Bprimary + RCTp) * Ip_pu
CT Primary Amps / CT Secondary Amps 5 CT Rated Power Accuracy %, Type (P,C), Limit Factor (Maximum CT times to satisfy accuracy). Eg. 10P20 CT Resistance in ohms
400
Cable Length from CT to Protective relay (in meters)
10
Cable cross section (in mm 2)
PROTECTION RELAY DATA: RELAY: ABB REF-542 Relay model (For information only) RELAY BURDEN 0.1 Relay Burden in VA (VA): MAXIMUM EXTERNAL FAULT. CURRENT THROUGH: External Shortcircuit (A) 8000
Maximum Generator contribution to external shortcircuits
1) UNIT VALUES Generator Full Load Current Igen = KVA/(√3*Vp/1000) Igen = 1574.59 A
IL_pu = IL_pu =
Maximum through Fault Current for external shortcircuits Ies = Ext. Shortcircuit IesL_pu = Ies = 8000 A IesL_pu =
Igen/CTL 3.9365
IN_pu = IN_pu =
Igen/CTN 3.9365
Ies/CTL 20.0000
IesN_pu = IesN_pu =
Ies/CTN 20.0000
2) BURDEN CALCULATIONS Line Lead Resistance RLkm 1.83 (ohms/Km): RL (Ω) =
2 * LCTL * RLkm/1000
Secondary Lead Resistance RNkm 1.83 (ohms/Km): RN (Ω) =
Maximum CT voltage accros terminals to guarantee rated accuracy (error)
RESULTS: BURDEN AND VOLTAGE COMPARISON Line side CT Limit Zb (Ω)
2 Vacr_Line Limit(V) 200
BLine (Ω) 1.468
OK. Line Burden is lower than CT limit
VsfL (V) 45.36
OK. Driving voltage is lower than Vacr limit
Neutral side 2
BNeutral (Ω) 1.468
Vacr_Neutral Limit (V)
VsfN (V)
200
45.36
CT Limit Zb (Ω)
OK. Neutral Burden is lower than CT limit
OK. Driving voltage is lower than Vacr limit
CALCULATED BY: DATE:
LUIS FERNANDEZ FERNANDEZ MAR. 22, 2007
REV. 00
VERIFICATION OF CURRENT TRANSFORMERS FOR MOTOR DIFFERENTIAL PROTECTION APPLICATION
MOTOR GENERAL DATA: KW: 1000 VOLTAGE (V): 11000 FLA (A) 150 LRA [Times FLA] 6.5 MOTOR LINE SIDE: CTL Ratio: VALine: Class: RCTL (Ω):
15 5P20 0.2
LCTL (m): Cross Section (mm2):
Rated Power in KW (Only for information. Not required for calculations) Rated Voltage in Volts (Only for information. Not required for calculations) Full Load current in amperes Locked Rotor curren in multiples of FLA
CT Primary Amps / CT Secondary Amps 200 5 CT Rated Power Accuracy %, Type (P,C), Limit Factor (Maximum CT times to satisfy accuracy). Eg. 10P20 CT Resistance in ohms
300
Cable Length from CT to Protective relay (in meters)
16
Cable cross section (in mm 2)
MOTOR NEUTRAL SIDE: CT Primary Amps / CT Secondary Amps CTN Ratio: 200 5 VANeutral: 15 CT Rated Power Class: 5P20 Accuracy %, Type (P,C), Limit Factor (Maximum CT times to satisfy accuracy). Eg. 10P20 RCTN (Ω): 0.2 CT Resistance in ohms LCTN (m): Cross Section (mm2):
300
Cable Length from CT to Protective relay (in meters)
16
Cable cross section (in mm 2)
PROTECTION RELAY DATA: RELAY: ABB REF-542 Relay model (For information only) RELAY BURDEN 0.1 Relay Burden in VA (VA): STARTING CURRENT. CURRENT THROUGH: External Shortcircuit (A)
Maximum current through
975
1) UNIT VALUES Motor Full Load Current
FLA =
150.00
A
FLAL_pu = IL_pu =
FLA/CTL 3.7500
FLAN_pu = FLA/CTN IN_pu = 3.7500
975
A
LRAL_pu = IesL_pu =
LRA/CTL 24.3750
LRAN_pu = LRA/CTN IesN_pu = 24.3750
Motor Starting current
LRA =
2) BURDEN CALCULATIONS Line Lead Resistance RLkm 1.15 (ohms/Km): RL (Ω) =
2 * LCTL * RLkm/1000
Secondary Lead Resistance RNkm 1.15 (ohms/Km): RN (Ω) =
