idmt relay grading
Post on 02-Jan-2016
126 Views
Preview:
DESCRIPTION
TRANSCRIPT
at Voltage_1 33000=
Relay 2 Current I21 500:= at Voltage_2 33000=
Required margin Delta_T 0.4:=
Assume Relay 2 TMS of 1.00Determine Relay 2 operating time at this TMSLinearly interpolate to determine the TMS necessary to achieve the required operating time
Relay 1 T1 Relay_1 I12( ):= T1 0.856 seconds=
Relay 2 at TMS = 1.00 T2 SI Pickup_2 1.0, Inst_2, I21,( ):= T2 5.744 seconds=
Ideal TMS TMS_2T1 Delta_T+
T2:=
Ideal Time multiplier TMS_2 0.219=
Select Time multiplier TMS_2 0.22:=
Define Relay 2 Equ'n Relay_2 Current( ) SI Pickup_2 TMS_2, Inst_2, Current,( ):=
IDMT Relay Curve Grading and Log Linear Plot
Define IEC Standard Inverse relay curve
SI Pickup TMS, Inst, Current,( ) if Current 1.001 Pickup⋅< 1000, if Current Inst> 0.02,0.14 TMS⋅
CurrentPickup
⎛⎜⎝
⎞⎠
0.021−
,⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
,⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
:=
Relay 1System voltage Voltage_1 33000:=
Pickup setting current Pickup_1 100:=
Inst setting current Inst_1 1000:=
Time multiplier TMS_1 0.2:=
Define Relay 1 Equ'n Relay_1 Current( ) SI Pickup_1 TMS_1, Inst_1, Current,( ):=
Relay 2System voltage Voltage_2 33000:=
Pickup setting current Pickup_2 150:=
Inst setting current Inst_2 2000:=
Determine Relay 2 TMS to achieve grading at specified currents
Relay 1 Current I12 500:=
Select Time multiplier TMS_3 0.28:=
Define Relay 3 Equ'n Relay_3 Current( ) SI Pickup_3 TMS_3, Inst_3, Current,( ):=
Check Grading Margins
Relay 1 I12 500= at Voltage_1 33000= Relay_1 I12( ) 0.856 seconds=
Relay 2 I21 500= at Voltage_2 33000= Relay_2 I21( ) 1.264 seconds=
Margin Relay_2 I21( ) Relay_1 I12( )−:= Margin 0.408 seconds=
Relay 2 I23 1000= at Voltage_2 33000= Relay_2 I23( ) 0.796 seconds=
Relay 3 I32 500= at Voltage_3 66000= Relay_3 I32( ) 1.198 seconds=
Margin Relay_3 I32( ) Relay_2 I23( )−:= Margin 0.402 seconds=
Relay 3System voltage Voltage_3 66000:=
Pickup setting current Pickup_3 100:=
Inst setting current Inst_3 2000:=
Determine Relay 3 TMS to achieve grading at specified currents
Relay 2 Current I23 1000:= at Voltage_2 33000=
Relay 3 Current I32 500:= at Voltage_3 66000=
Required margin Delta_T 0.4:=
Assume Relay 3 TMS of 1.00Determine Relay 3 operating time at this TMSLinearly interpolate to determine the TMS necessary to achieve the required operating time
Relay 2 T2 Relay_2 I23( ):= T2 0.796 seconds=
Relay 3 at TMS = 1.00 T3 SI Pickup_3 1.0, Inst_3, I32,( ):= T3 4.28 seconds=
Ideal TMS TMS_3T2 Delta_T+
T3:=
Ideal Time multiplier TMS_3 0.280=
100 1 .103 1 .104 1 .105
0.5
1
1.5
2
2.5
3
Relay 1Relay 2Relay 3
IDMT Relay Curves
Current at Relay 1 Base Voltage
Seco
nds
Inst_3 2000=TMS_3 0.28=Pickup_3 100=Voltage_3 66000=Relay 3
Inst_2 2000=TMS_2 0.22=Pickup_2 150=Voltage_2 33000=Relay 2
Inst_1 1000=TMS_1 0.2=Pickup_1 100=Voltage_1 33000=Relay 1
I n( ) Pickup_1 10n⋅:=n 0 0.001, 2.3..:=
Plot curves over a current range based on Relay 1 settings, from 1 to 200 times Relay 1 pickup current setting. Since curves are plotted on a log scale, increment data points logarithmically too.
top related