Motor Protection
Motor Protection
Motor Protection
Niwat Sriklam
Introduction
Many differential application
Differential motor characteristics
Difficult to standardise protection
Protection applied range from
FUSE ÎÎ RELAYS
Introduction
SIZE OF MOTOR,
TYPE & IMPORTANCE
OF THE LOAD
Motor Protection
~
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
Induction Motor
Station field f
fr
(f - fr) = sf
Motor Currents
Induction motor
R2
X2 = 2¶fL
kVs (Stand Still)
Rotor Equivalent Circuit
Standstill :
R2
kVs X2 = 2¶fL = rotor reactance at (Stand Still)
Running :
X2 >> R2
Therefore R2 >> X2
When S is small
Start
Time
Heating
Insulation
Deterioration
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
TRIP
Tmax
Time
t1 t2
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
Start
Time + -
Thermal Hot O/C
Curve TD
Start TD
TD
Time 86
Full load Current
Current
TRIP
Time
+ -
TACHO O/C
TD
Start
TD
Time 86
TD
TRIP
Time
+ -
Cold Stall MSD
TD1+TD2 TD1
TD2
86
TD2
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
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
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
Fuse
50/51/49
MPR
Ts
M 50/51G
Is Icon
Ts > Tfuse at Icont.
High Impedance Winding
Differential Protection
R Stabilizing
87 87 87 Relay
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
I
I’
Stator Current on
Loss Synchronism