LT/HT MOTORS PROTECTION TESTING AND MAINTENANCE

Why the winding of motor fails ?

Temperature rise

Heat

Losses in the form of Copper, friction and windage

Temperature

Heat

Losses

Classification of Insulation

ClassMax.Working temp.Y90A105E120B130F155H180CABOVE180

Typically the following abnormal conditions may be observed in an induction motors:Over loads : Sustained Overload or Momentary OverloadSupply Failures: Single phasingMotor Internal faults , i.e., Phase to Phase faults, Phase to Earth faults or Open circuit faultsStarting Failures: Prolonged Starting of the Motor Rotor Failures : Blocked Rotor

Low Rating LV Motors: Up to 100 KW motors, Contactors shall be provided with the following protections:

Bi-metallic thermal relay for overload protection Fuses for Short Circuit protection

Bimetal relayPower fusesPower ContactorMotorT/F 415/110Remote stopLocal stopStartCC-1RGC-2C-3415V SupplySwitch4A2A2AO/LLink

Bimetal relayFeNi

Bimetal Overload relayThese are popular in LT Motor applications. Bending of one or more bimetal strips causes movement of a common lever, which in turn operates the trip contact in the case of overloads. These relays can be provided directly in series with the load current. For the larger rating motors, these bimetallic relays can be connected through CTs.These relays are also either self reset type or hand reset type. Before selecting the bimetallic relay, the following shall be monitored:

Bimetal Overload relaySelect the relay rating nearer to the motor FL current rating.

Set the relay current at the normal operating current.

FUSEIn the case of the short circuits, bimetal overload relays will operate in delay and may worsen further the situation. Hence, for quick isolation is required in case of short circuits. HRC ( High Rupturing Capacity) fuses shall be used for this purpose.The rating of this fuses shall be depends on the starting currents of the motors.

Low Voltage Motors beyond 100 kW: Air Break Circuit Breakers shall be provided with protective relays with the following functions:

Overload Protection by IDMT RelaysShort Circuit Protection with High set relaysLocked rotor Protection with Definite time relays

MOTOR WITH CIRCUIT BREAKER

Power ContactorOne Coil for starting / StoppingMotor stops when supply fails and will not start when the supply restores, as the control supply is derived from power supply.CBClose coil for closingTrip coil for trippingMotor stops when supply fails and will start when the supply restores, if the trip command was not given

Power Contactor & CB

Inverse Definite Time Over Current Relay

This Relay operates when ever the current exceeds the Set value. The relay operation time is inversely proportional to the magnitude of CurrentApplication: E/F prot., LT Motor Prot.Example : CDG Relay of EE make

CDG Relay

Instantaneous Over Current RelayThis Relay operates when ever the current exceeds the Set value (No time delay)Set slightly more than the starting current (in general it is 6 to 7 times)

Application : Generator Inter turn, GT Restricted E/F prot, Instantaneous prot LT motors.

Example : CAG Relay of EE make

CAG Relay

CAG Relay

Definite Time Over Current Relay

This Relay operates when ever the current exceeds the Set value and that current shall persist for defined time

Application : Blocked Rotor Protection, Over Load Alarm etc..

Example : CTU Relay of EE make&Operation

Current

Time

High Voltage Motors: These motors shall be generally controlled by Circuit Breakers and these shall have a typical Motor Protection relay (MPR) with the following functions:

Thermal Overload Protection with alarm and TripShort Circuit protectionUnbalanced ProtectionLocked Rotor ProtectionEarth Fault Protection by Residual Connection or with CBCTLimitation of Number of startsBus Under Voltage Protection

ABB MPR

Front panelUser configurableLEDsFixed function LEDsLocal communicationsDownload / Monitor portMenu browserAlarm viewerLithium batteryFacility for lead seal

Thermal Overload ProtectionThermal overload protection relays gives protection against heating of the rotor due to negative sequence currents caused by an unbalance system. Multiple of square of Negative sequence current is added to the thermal element. The equivalent current Ieq is calculated as:Ieq = ( I1 2 + Ke I2 2)

I1 = Positive Sequence current of the motorI2 = negative sequence current of the motorKe = Negative sequence current recognition factor (adjustable) ( For CTM Relays, Ke = 6)

Unbalance Protection Unbalance protection is generally provided in the following ways:

By measuring Negative phase sequence currentsBy Monitoring the difference between Max and Minimum phase currents

EARTH FAULT Protection If the source is from 3 wire supply derived form the Transformer having neutral solidly earthed or low resistance earthed, earth fault protection shall be extended to trip the motor. If the similar supply is derived form the Delta Transformer or the transformer with resistance earthing, earth fault protection need not be given for tripping of the motor and can be connected to an alarm.

EARTH Fault Protection

EARTH FAULT Protection Earth fault current generally set at 30% of the full load current.

Negative Sequence Protection Negative sequence currents in the motor are due to following reasons: Non closure of one or two poles of the Circuit BreakerFailure of Power fuses in one or two fusesSustained Faults in the network

Negative Sequence ProtectionNegative sequence currents results in rotating magnetic field revolves in opposite direction. This field induces double frequency induced currents in the rotor body and conductors, gives rise to heat due to copper losses. The rotor gets heated and the temperature of the motor winding increases beyond the limits. Negative sequence protection can be either definite time or with IDMT characteristic.

Differential Protection The motors with rating more than 1500 kW, shall be protected additionally with Differential protection. This protection covers the switchgear, Motor and the power cables.

Differential Protection

Differential ProtectionFor better stability of the protection, Stabilizing resistors shall be connected in series with the relays. To safeguard the entire protection system in the case of open CTs, Non-linear resistor ( Metrocils) shall be connected parallel to the Relay and resistor circuits.

TESTING when protection acted Pre-commissioning

When motor tripped on ProtectionCheck the flag relays and note down on what protection the motor tripped.If the motor tripped on motor protectionwhat protection acted ?Inform to maintenance wing with details.

When motor tripped on protectionWhether any damage at the switchgear?Whether any smoke observed at the TB ?Do not go for further starting of motor until, getting clearance from maintenance wing.

When motor tripped on protectionAsk the shift operator whether earth alarm came on busCheck the IR value of motor it should be > 10M-ohmsif it is Zero :Open the power cable at the TB and then take the IR value of both the motor and the cable.

When motor tripped on protectionIf the motor tripped on short circuit protection:check the winding resistance and inductance of the winding : --- if it is DELTA connected Open the Links and measure each Phase, it must be equal --- if it is STAR connected internally, measure Phase to Phase, it must be equal

When motor tripped on protectionIf the motor tripped on Over load protection:-- Rotate the drive rotor for One or Two rotations --- if it is not rotating freely, then disconnect the coupling and then check whether problem is with motor or pump. --- Plan to replace bearings accordingly

Pre-Commissioning testsNote down the Name plate details.Check IR valueMeasure winding resistance & inductance and note down in the history record for future comparison.Check the differential CT connections (if any)Note down the No Load currents of all phasesCheck for proper alignment.

IP PROTECTION Degree of Protection of EnclosuresThe degree of protection provided by an enclosure is indicated by the IP code in the following way -Code Letters IP 3 1 (International Protection)First Numeral(Resistance to solid objects 0-6)Second Numeral(Resistance to water 0-8)

IP PROTECTION FOR FIRST LETTER

IPTESTSHORT DESCRIPTION0NO IP PROTECTION1 50 mmA large surface of the body, such as a hand (but no protection against deliberate access) solid objects exceeding 50mm in Diameter2 12.5 mmFingers or similar objects not exceeding 80mm in length; solid objects exceeding 12.5 mm in diameter3 2,5 mmTools, wires, etc...., of diameter or thickness greater than 2.5mm; solid objects exceeding 2.5 mm in diameter

IP PROTECTION FOR FIRST LETTER

IPTESTSHORT DESCRIPTION4 1 mmWires or strips of thickness greater than 1.0mm; solid objects exceeding 1.0 mm in diameter5dustIngress of dust is not totally prevented but dust does not enter in sufficient quantity to interfere with satisfactory operation of the equipment6Dust-tightNo ingress of dust

IP PROTECTION FOR SECOND LETTER

IPTESTSHORT DESCRIPTION0NO IP PROTECTION1DrippingDripping water (vertically falling drops) shall have no harmful effect215oVertically dripping water shall have no harmful effect when the enclosure is tilted at any angle up to 15 from its normal position Protected against spraying water Water falling as a spray3SprayingWater falling as a spray at an angle up to 60 from the vertical shallhave no harmful effect4splashingWater splashed against the enclosure from any direction shall have no harmful effect

IP PROTECTION FOR SECOND LETTER

IPTESTSHORT DESCRIPTION5water jetsWater projected by a nozzle against the enclosure from any direction shall have no harmful effect6heavy seasWater from heavy seas or water projected in powerful jets shall notenter the enclosure in harmful quantities7Immersion 15 cmIngress of water in a harmful quantity shall not be possible when the enclosure is immersed in water under defined conditions of pressure and time8Submersion (meter)The equipment is suitable for continuous submersion in water under conditions which shall be specified by the Manufacturer

Preventive MaintenanceInspectTouchFeel

Preventive MaintenanceInspect: Hear the sound and observe the surroundings. Touch: Sense the temperature of the motor & cable and vibrationsFeel: Whether it is same as before.

Preventive MaintenanceOpen the terminal box and check for healthiness quarterly According to running hours, fill the grease and replace bearings (in general it is 1000 hrs)

Preventive MaintenanceTake the vibrations and compare with previous readings Observe the sound by rotating the rotor with hand during Overhaul period (All drives).

Preventive Maintenance

Replace the bearings for important drives during Overhaul (Ex: AH main drives) Replace the lugs if they are in damaged condition

Precautions while giving PTW

For Modules:Conform that the motor is in off condition.Switch off the supply and remove only Control fuses

Precautions while giving PTWFor Circuit Breakers:Conform that the motor is in off condition.Open the mechanical inter lock Keep the breaker in TEST position

IEC DUTY CYCLES

S1Continuous dutyThe motor works at a constant load for enough time to reach temperature equilibrium.S2Short-time dutyThe motor works at a constant load, but not long enough to reach temperature equilibrium. The rest periods are long enough for the motor to reach ambient temperature.S3Intermittent periodic dutySequential, identical run and rest cycles with constant load. Temperature equilibrium is never reached. Starting current has little effect on temperature rise.S4Intermittent periodic duty with startingSequential, identical start, run and rest cycles with constant load. Temperature equilibrium is not reached, but starting current affects temperature rise.S5Intermittent periodic duty with electric brakingSequential, identical cycles of starting, running at constant load and running with no load. No rest periods.S6Continuous operation with intermittent loadSequential, identical cycles of running with constant load and running with no load. No rest periods.S7Continuous operation with electric brakingSequential identical cycles of starting, running at constant load and electric braking. No rest periods.S8Continuous operation with periodic changes in load and speedSequential, identical duty cycles run at constant load and given speed, then run at other constant loads and speeds. No rest periods.

DETERMINATION OF POWER RATING Continuous duty and constant load: P= TN KW 975T is the load torque in kg-mN speed in rpm is the product of the efficiency of the driven equipment and that of transmitting equipment.

DETERMINATION OF POWER RATING in case of hoist mechanismsP = Fv KW 2x102 due to counter weight is always one-half of the useful load is always presentsV vary from 0.5 to 1.5m/s for passenger lifts

DETERMINATION OF POWER RATING in case of linear motionP = Fv KW 102F is the Force caused by load in kgv is the velocity of motion of load in m/sec is the product of the efficiency of the driven equipment and that of transmitting equipment.

DETERMINATION OF POWER RATING in case of pumpsP = HQ KW 102 is the density of liquid pumped in kg/m3H is the gross head in metersQ is the delivery of the pump in m3/sec varies from 0.8 to 0.9 for reciprocating.0.4 to 0.8 for centrifugal

DETERMINATION OF POWER RATING for fan motorsP = Qh KW 102 h is the pressure in mm of waterQ is the volume of air/gas in m3/sec may be taken 0.6 for small fans 0.8 for large fans

DETERMINATION OF POWER RATING 2. Continuous duty variable load: n P2i ti Peq= i=1 n ti i=1 P is the load power This is called method of Average power

DETERMINATION OF POWER RATING OF CONSTANT LOAD Ex: primary water pump using in stage-IQ= 11 liter/sec, H= 45 m.Find Power rating: HQ KW 102 = 1000 kg/m3 as the liquid is waterTake = 0.6 as it is the centrifugal pump P: 1000X45x11 = 8.08Kw (9.3kw nearest) 102X0.6x1000

DETERMINATION OF POWER RATING OF VARIABLE LOAD3274273266273227P15403051482001270t

DETERMINATION OF POWER RATING OF VARIABLE LOADEX:Part of Time Required HP (HP)2tCycle sec (t)1 15 32 15,3602 40 74 219,0403 30 27 21,8704 5 32 5,1205 148 66 644,6906 200 27 145,8007 12 32 12,2908 70 27 51,030 520 1,115,200

DETERMINATION OF POWER RATING OF VARIABLE LOAD n P2i ti Peq= i=1 n ti i=1 Peq= 1,115,200520= 2145 = 46.3 HPAdd 10% tolerance: 46.3x1.1=50.9

Why equipment fails Even though we adopt all the protections?

Thank you

V.SRINIVASULUADE/MRT-I/STAGE-IDr.NTTPS

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