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.