Motor Protection

Niwat Sriklam
Introduction
„ Many differential application
„ Differential motor characteristics
„ Difficult to standardise protection
„ Protection applied range from
„ FUSE ÎÎ RELAYS
Introduction

COST & EXTENT = POTENTIAL

OF PROTECTION HAZARDS

SIZE OF MOTOR,
TYPE & IMPORTANCE
OF THE LOAD
 Motor Protection

System Motor Circuit Load

-Voltage dips -Insulation failure -Overload
-Voltage Unbalance -Open circuits -Locked rotor
-Loss of Supply -Short circuit -Coupling faults
-Overheating -Bearing faults

~
Motor Protection Application
Switching
Voltage Rating Protection
Device
< 600V < 11kW Contractor i) Fuses
ii) Fuses + Direct acting
Thermal O/L + U/V
releases
< 600V 11-300kW Contractor Fuses + Electronic O/L +
Time Delay Earth Fault
3.3 kV 100kW-1.5MW Contractor Options :- Stalling Under
Current
6.6 kV 1MW-3MW Contractor

6.6 kV >1MW Circuit Breaker As above

+ Instantaneous O/C
11 kV >1MW Circuit Breaker + Differential
Introduction
„ Protection must be able to :-
„ Operate for abnormal conditions
„ Protection must not :-
„ Affect normal motor operation
„ Consideration :-
„ Starting current
„ Starting time
„ Full load current
„ Stall withstand time (hot & cold)
„ Thermal withstand
Motor Currents

Induction Motor
Station field f

fr

„ Define slip, S, as the PER UNIT difference in speed between

the Stator and Rotor field
Slip ‘S’ = (f - fr)/f

„ Speed of stator field relative to rotor

(f - fr) = sf
Motor Currents
„ Induction motor

„ Magnitude of induced voltage : Proportional to sf

„ Frequency of Induction rotor current : Equal to sf

R2
X2 = 2¶fL
kVs (Stand Still)
 Rotor Equivalent Circuit
Standstill :

R2
kVs X2 = 2¶fL = rotor reactance at (Stand Still)

Running :

R2 Rotor Current = skVs

skVs 2¶fL = sX2
[ R22 + S2X22]½

Rotor Current = kVs

[ R22 + X22]½
S2
R2/s
kVs X2
 Motor Starting Characteristic
Rotor Current = skVs time
[ R22 + S2X22]½

Rotor Current = kVs

[ R22 + X22]½
S2

X2 >> R2
Therefore R2 >> X2
When S is small

Start
Time

Full load Current

Current
Mechanical Overload
Overload

Heating

Insulation
Deterioration

Over Load Protection

FUSES Thermal Replica

Motor Heating
HEAT STORED
Î INCREASE
THE MOTOR TEMPERATURE

HEAT DEVELOPED AT
A CONSTANT RATED
DUE TO CURRENT FLOW

HEAT DISPLATED AT
A RATED PROPOTIONAL
TO MOTOR TEMPERATURE
 Motor Heating
Motor Temperature
T = Tmax (1-e-t/ζ)
Tmax

Time
Rate of rise depend on motor thermal time constant ζ
or as temp rise α (Current)2
T = KI2max (1-e-t/ζ)
Motor Heating
I2
I22 T2

I12 T1

IR2 Tmax

Time
t2 t1
 Motor Heating
Time

t1
Thermal Withstand
t2

Current
IR I1 I2
 Motor Heating
Current2
Ieq2

Iθ2

I m2

Time
tTRIP
Iθ2- Im2= ( Ieq2- Im2)(1- e-t/ζ) Or alternatively
Rearrange this express in term of Time t = ζ ℓn {(K2 - a2)/(K2 - l2)}
t = ζ ℓn {(Ieq2 - Im2)/(Ieq2 - Iθ2)}
 Motor Cooling
Cooling Equation :
Current2
Im I2m’ = I2me-t/ζr

Im’
Time
t

After time ‘t’ equivalent motor current is reduced from Im to Im’

 Motor Heating
Temp

TRIP
Tmax

Time
t1 t2

t1 = Motor restart not possible

t2 = Motor restart possible
Cooling Time Constant ζr
Start/Stall Protection

Niwat Sriklam
Stalling Protection
„ Required for :-
„Starting on start-up (Lock Rotor cold stall)
„ Stall during running (Hot stall)
„ With normal 3Ø supply :-
„ Istall = I locked Rotor ~ Istart

„ Cannot disthinguish between ‘STALL’ and ‘START’ by

current alone
„ Most case : tstart < t stall withstand
„ Sometimes : tstart > t stall withstand

„ Stall during Running Condition

„ Increase Load
„ Voltage drop or Dip
Lock Rotor Protection
„ Start Time < Stall Withstand Time

„ Where Staring Time is than Stall Withstand Time :

„ Use Thermal Protection Characteristic
„ Use Dedicated Locked rotor Protection
 Stall Protection
T start < t stall : Use of Thermal Characteristic
Time
Insufficient margin between T start and T stall
Thermal Cold
Curve Cold Use of definite time Overcurrent relay :
Stall
Withstand Time

Start
Time + -
Thermal Hot O/C
Curve TD

Start TD
TD
Time 86
Full load Current
Current

TRIP

Full load I O/C Current

Current
 Stall Protection
T start > t stall : Use of Tachoswitch and definite overcurrent relay time

Time

+ -
TACHO O/C
TD

Start
TD
Time 86
TD

TRIP

Full load I O/C Current

Current
 Stall Protection
T start < t stall : Use of motor start contact and 2 stage definite overcurrent relay

Time

+ -
Cold Stall MSD
TD1+TD2 TD1

Start TD1 Hot Stall TD1 O/C

Time TD2

TD2
86
TD2

Full load I O/C Current

Current TRIP

MSD=Motor switching device

Motor Currents

Negative Sequence Current

Station field f

fr

„ Relative Frequency of stator field = f + fr

But fr = (1 - s)f

„ Therefore

(f + fr) = (2-s) f
Motor Positive and Negative
Sequence Impedances
Positive ½
(R1+ R’2 )2+j(X 1+X’2)2
R1 R’2
S j(X1+X’2) S ½
(R1+ R’2)2+j(X1+X’2)2
At standstill

Negative ½
(R1+ R’2 )2+j(X 1+X’2)2
R1 R’2
2-S j(X1+X’2) 2-S ½
2
(R1+ R’2) +j(X1+X’2)2

2 At normal
running
speed
Operation On Supply Unbalance
„ At normal running speed

Positive SEQ IMP ~ Starting Current .

Negative SEQ IMP Normal Running Current

„ Negative Sequence Impedance is much less than Positive

Sequence Impedance
„ Small unbalance = Relatively Large Negative Sequence current

„ Heating effect of Negative Sequence is greater than equivalent

Positive Sequence Current because they are Higher Frequency
Equivalent Motor Current
„ Heating from Negative Sequence Current Greater than Positive
„ Take this into account in thermal calculation

„ Ieq = (I12 + n I22)½

„ Where : n = Pos Seq Imp : Neg Seq Imp, [ 6 ]

„ Small amount of I2 gives large increase in
Ieq and hence calculated motor Thermal
state
Loss of 1 Phase while Starting
„ Star „ Normal Starting Current
IA = VAN/ Z
„ With 1 Phase open
I’A = VAB/ 2Z = √3 VAN/ 2Z
A = 0.866 x IA

„ SYMMETRICAL COMPONENT
Z I1 = ⅓( I’A + aI’B)
I1 = ⅓( 1 - a) I’B
Z |I1 | = ½ IA

I2 = ⅓( I’A + a2I’B)
Z
B I2 = ⅓( 1 - a2) I’A
C |I2 | = ½ IA
Loss of 1 Phase while Starting
„ Delta „ Normal
IA = √3 VAB/ Z
„ 1 Phase open
I’A = VAB 3/2Z
A = 0.866 x Normal

Z Z

Z
C
B

1 Winding carries twice the current in the other 2

Single Phase Stalling Protection
„ Loss of phase on starting motor remains Stationary
„ Start current = 0.866 normal start I
„ Negative seq component = 0.5 normal start I
„ Clear condition using Negative Sequence Element.
„ Typical Setting = ⅓ I2

„ i.e. 1/6 Normal Starting current

Reverse Phase Sequence Starting
„ Protection required for LIFT motors , Conveyers
„ Instantaneous I2 unit
„ Time delayed thermal trip
„ Separate phase sequence detector for low load current
machines
Undervoltage Protection
„ Cause Low output torque
„ Machine cannot reach rated speed draws high stator current
„ Use time delayed undervoltage protection
Undervoltage Consideration
„ Reduced Torque
„ Increase Starting current
„ Reduced Speed
„ Failure To run-up
Form Of Undervoltage Condition
„ Slight but Prolonged (regulation)
„ Large transient Dip (Fault Clearance)
Undervoltage Protection
„ Disconnects motor from failure supply
„ Disconnects motor after Dip long enough to prevent successful
re-acceleration
Undervoltage Tripping
„ Mean of undervoltage tripping
„AC holding coil for Fuse Contractor
„ Undervoltage release
„ Undervoltage relay for shunt trip
„ DEFINITE TIMP

„ INVERSE TIME

„ Consideration
„ U/V tripping should be delayed for essential motors so that
they may be given a chance to re-accelerate following a
short voltage dip (<0.5 s)
„ Delayed drop-out of fused contractor could be arranged by
using a capacitor in parallel with AC holding coil
Insulation Failure
„ Result of Prolonged or Cyclic Overheating
„ Instantaneous Earth Fault Protection
„ Instantaneous Overcurrent Protection
„ Differential Protection on some Large Machines
 Stator Earth Fault Protection

R-stability

50
50

MOTOR MOTOR

a) Residually Connected CT’s b) Core Balance (Toroidal) CT

Short Circuit Protection
„ Due to the machine construction internal phase-phase faults are
almost impossible
„ Most phase-phase faults occur at the machine terminals or
occasionally in the cabling
„ Ideally the S/C protection should be set just above the max Istart
(I>> 1.25 Istart), however, there is an initial start current of up to
2.5 Istart which rapidly reduces over 3 cycles
„ Increase I>> or delay t>> in small increments according to
start conditions
„ Use special I>> characteristic
Instantaneous Earth Fault or Negative Seq.
Tripping is not permitted with Contactors

Fuse

50/51/49

MPR = Motor Protection

MPR
Ts

M 50/51G
Is Icon
Ts > Tfuse at Icont.
High Impedance Winding
Differential Protection

R Stabilizing

87 87 87 Relay

High Impedance Scheme

Note : Protection must be stable with starting current

Self Balance Winding Differential
Protection
A

87

87

87
Bearing Failure
„ Electrical Interferance
„Induce voltage
„ Results in circulating currents
„ May fuse the bearing
„ Remember to take precaution – earting
„ Mechanical failure
„ Increase Friction
„ Loss of Lubrication
„ heating
Use of RTD
„ RTD sensors as known Stator Hotspotes
„ Absolute temperature measurements to bias the relay thermal
characteristics
„ Monitoring of Motor / Load baring temperatures
„ Ambient air temperature measurement
Synchronous Motors
„ Out of step protection
„Inadequate field or excessive load can cause the machine to
fall out of step. This subjects the machine to overcurrent and
pulsating torque leading to stalling
„ Field current method

ƒ Detect AC current induced in field circuit.

„ Power factor method

ƒ Detect heavy current at low power factor

„ Loss of supply
„ On loss of supply motor should be disconnected if supply
could be restored automatically
„ Overvoltage & Under-frequency

„ Under-power & Reverse power

Power Factor Method

I
I’

Stator Current on
Loss Synchronism