chapter 1 - fundamental concept

8/7/2019 Chapter 1 - Fundamental Concept

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33--phase Motorsphase Motors

Fundamental ConceptFundamental Concept

What is motor?What is motor?

Motor is defined as an electromagnetic

device that convert electrical energy

into mechanical energy.

Electrical energy is measure in WATT

Mechanical energy is measure in HOUSEPOWER.


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33--phase Motorsphase MotorsFundamental ConceptFundamental Concept-- Power (unit: Watt)Power (unit: Watt)

Motor output power =

Motor input power X motor efficiency

Motor input Power Motor Motor Output power

(Electrical power) (Mechanical power)

(Main Supply)

Power loss in Heat

Tenaga Elektrik

Tenaga Mekanik

Tenaga Haba


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33--phase Motorsphase MotorsFundamental ConceptFundamental Concept

-- Horsepower (hp)Horsepower (hp)

Power is defined as the rate of doing work.

Work is defined in term of weight in pounds multipliedby distance in feet

Work = Pound x Distance (unit: foot-pounds)

Power = Work/Time (unit: foot-pounds/second)

1 Horsepower = 550 foot-pounds/second

1 foot-pound/second = (1/500) Horsepower ----------(1)


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33--phase Motorsphase MotorsFundamental ConceptFundamental Concept-- Relationship between Watt and HorsepowerRelationship between Watt and Horsepower

The Watt is the measure of electrical power and is defined in term of voltage andcurrent or ampere,

Watt = Voltage X AmpereVoltage is defined as 1 joule per coulomb of electron (J/C)Ampere is defined as the flow of 1 coulomb of electron per

second (C/T)

So, Watt = J/C X C/T = J/TSince, 1 Joule = 0.737 foot-poundSo, Watt = 0.737 foot-pound/T = 0.737 foot-pound/second

1 foot-pound/second = Watt/0.737 ------------(2)(1) = (2),

(1/550) Horsepower = Watt/0.737Horsepower = Watt X 550 / 0.737 = 746 Watt

1 HP = 746W


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33--phase Motorsphase MotorsFundamental ConceptFundamental Concept-- Relationship between Watt and HorsepowerRelationship between Watt and Horsepower

Either Horsepower or Kilowatt (KW)(1KW = 1000W)can be used to indicate the capacity rating of themotor.

For example, we can call 1 horsepower motor as0.746KW motor or 0.75KW motor.

Please note that the motor power (KW)orhorsepower(HP) that indicated on the motor nameplateor catalog is refer to the motor output power.

DONT BE CONFUSED


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-- Motor Input PowerMotor Input Power

How to calculate the motor input power???

For AC motor, the Voltage and Current are not in phase wherecurrent will lag voltage for lagging angle from 0deg to 90deg.

So, Watt = Voltage (rms) X Ampere (rms) must be changed to

Watt = Voltage (rms) X Ampere (rms) X Cosine ----1 phase

Watt = 3 (Voltage (rms) X Ampere (rms) X Cosine ) --3phase= 3 ( (Voltage (p-p) / 3) X Amp (rms) X Cosine )= 3 (Voltage (p-p) X Amp (rms) X Cosine )= 3 (Voltage (p-p) X Amp (rms) X Power factor)Where, Cosine = power factor

Watt = 3 (Voltage (p-p) X Amp (rms) X Power factor)


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33--phase Motorsphase MotorsFundamental ConceptFundamental Concept-- Motor Input PowerMotor Input Power

For example:

What is the motor input power for the motor with the following performance data?

Motor output power = 0.75KWRated current (ampere) = 1.8A (ac current)Rated voltage = 415VacPower factor (Cosine )= 0.8Efficiency = 80% or 0.80

(All this data are indicated on the motor nameplate)

Motor input power, Watt = 3 (415V X 1.8 X 0.8) --------Method 1

= 1035W= 1KW

Motor Input power = Motor output power / motor efficiency ------Method 2= 0.75KW / 0.80= 0.94KW= 1KW


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Question 1.1:

What is the motor input power for the motor with thefollowing performance data?

Motor output power = 22KWRated current (ampere) = 39A (ac current)

Rated voltage = 415VacPower factor (Cosine ) = 0.87Efficiency = 90.7% or 0.907


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Answer:

Motor input power, Watt = 3 (415V X 39 X 0.87)

= 24388.9W= 24.4KW = 24KW

Motor input power = Motor output power / motor efficiency= 22KW / 0.907= 24.25KW= 24KW


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33--phase Motorsphase MotorsFundamental ConceptFundamental Concept-- Motor SpeedMotor Speed

Motor may be single speed, dual speed,multiple speed and variable speed. The

later can be adjusted to any value withinthe range of the design.

The speed of the motor is given inrevolution per minute (RPM).


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Speed type:

a) Motor Synchronous speed (RPM)=a) Motor Synchronous speed (RPM)=Motor rated speed(RPM) + Motor slip(RPM)Motor rated speed(RPM) + Motor slip(RPM)

b) Motor base speed = Motor actual speed =b) Motor base speed = Motor actual speed =Motor rated speedMotor rated speed


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What is motor slip?What is motor slip?The rotating of motor rotor is due to the running magnetic fieldThe rotating of motor rotor is due to the running magnetic fieldthat generated by the induction of the magnetic field from motorthat generated by the induction of the magnetic field from motorstator winding upon the rotor. If the motor rotor turn at the samestator winding upon the rotor. If the motor rotor turn at the sameRPM with this magnetic field then relative motion betweenRPM with this magnetic field then relative motion betweenRotor and the magnetic field. Therefore no current would beRotor and the magnetic field. Therefore no current would beInduced into the rotor and no magnetic field would be created toInduced into the rotor and no magnetic field would be created toCause rotor to turn.Cause rotor to turn.

Also,Also,

Synchronous speed (RPM)= 120 X Frequency (Hz)No. of pole


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For example:For example:

Motor rated frequency = 50Hz,Motor rated frequency = 50Hz,No. of pole = 4No. of pole = 4

Motor Synchronous speed (RPM)= 120 X 50 / 4Motor Synchronous speed (RPM)= 120 X 50 / 4= 1500RPM.= 1500RPM.


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No. of pole Rated Frequency Synchronousspeed(RPM)

2 50 3000 4 50 1500 6 50 1000 8 50 750 10 50 600


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Following is the typical motor base speed (actual speed)

No. of pole Rated Frequency Base speed(RPM)2 50 2840

4 50 14006 50 9358 50 700

NOTE1 : Actual speed must be lower than Synchronous speed

NOTE2: Higher the pole, lower the speed.

Base speed (RPM) = Synchronous speed(RPM) slip(RPM)


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-- Motor SpeedMotor Speed

For Example:For Example:For 0.75KW, 4pole motor, Base speed = 1400RPMFor 0.75KW, 4pole motor, Base speed = 1400RPM

a) Slip(RPM) = 1500a) Slip(RPM) = 1500 14001400

= 100RPM= 100RPM

b) Slip (%) =b) Slip (%) = 15001500 14001400 X 100%X 100%15001500

= 6.7%= 6.7%

Motor slip (RPM) = Synchronous speedMotor slip (RPM) = Synchronous speed base speedbase speed

Motor Slip (%) =Motor Slip (%) = (Synchronous speed(Synchronous speed base speed) X 100%base speed) X 100%Synchronous speedSynchronous speed


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Question 1.2:

a) What is the synchronous speed of

14pole, 50Hz motor?

b) What is the actual speed of this motor if

the motor slip is 7%?


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Answers:

a)a) Synchronous speed of 14 pole motor =Synchronous speed of 14 pole motor =

(120 x 50)/14 = 429RPM(120 x 50)/14 = 429RPM

b)b) Actual speed = 429 ((7 x 429)/100)

= 398.97 RPM


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-- Motor Driven end (DE) and Non Driven End (NDEMotor Driven end (DE) and Non Driven End (NDE))DEDE NDENDE


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-- Direction of RotationDirection of Rotation

Motor may rotor in only one direction, or they may bereversible.

Standard direction of rotation isclockwise when view from nondriven end (NDE).

For this motor the motorcooling system become lessefficiency if run in clockwisewhen view from NDE.


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-- Motor EfficiencyMotor Efficiency

MotorOutput Input power=1KWpower= 0.75KW

Power loss in heat = 0.25KW which is 25% of input power


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-- Motor EfficiencyMotor Efficiency

Efficiency (%) =100% - ( Motor input power Motor output power X100%)

Motor input power

Example: For 0.75KW motor,Motor input power =1KWMotor output power = 0.75KWMotor efficiency = 100% -(1KW 0.75KW) X 100%

1KW

= 100% - 25%= 75%

1KW

0.75KW (75%) Putaran Angker

0.25KW (25%) - Haba


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-- Power Factor (Power Factor (Faktor Kuasa)Faktor Kuasa)

Example: 0.745KW motor, 4pole, 415Vac, 1.8A,Efficiency = 74.5%, Motor output power = 745W,Motor input power = 745W = 1000W

0.745Power factor = 1000 .

415 X 1.8 X 1.732= 0.772= 0.8

Power factor = Watt ----------------1 phaseVoltage x Ampere

Power factor = Watt -------3 phaseVoltage x Ampere X1.732


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-- Power Factor (Power Factor (Faktor Kuasa)Faktor Kuasa)

Bilangan motor yang tinggi akanmerendahkan faktor kuasa bekalanelektrik.

Keadaan ini akan merendahkan kegunaanbekalan elektrik dan akan mendapatidenda daripada TNB


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-- Motor TorqueMotor Torque (Kilas)(Kilas)

Torque (T) is defined as a turning force (F).Torque (T) is defined as a turning force (F).

Torque = Force X distance

Torque = kBI

Where, k = constantB = flux densityI = rotor current

The unit of the torque is kgm^2 or Nm

-------F

omula (1)

--------------------------------Fomula (2)


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-- Electromagnetic Force (F)Electromagnetic Force (F) (Daya Elektromagnetik)(Daya Elektromagnetik)

Magnetic circuit is consisted of the magneticflux.

The source of flux is come from either coil orpermanent magnet.

Following show thefield produced by acircular solenoid(coil). Right hand ruleis applied.


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-- Magnetomotive force (MMF)Magnetomotive force (MMF)

How to increase the flux density of the circularsolenoid?

The answer is to increase the current in the coilor increase the number of turn.

If we double the current or double the numberof turn then we double the total flux, therebydouble the flux density as well.


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-- Magnetomotive force (MMF)Magnetomotive force (MMF) (Daya gerak magnet)(Daya gerak magnet)

We quantify the ability of the coil to produce flux

in term of the magnetomotive force (MMF)

The SI unit of MMF is ampere-turns

MMF is proportional to Flux density.

MMF = number of turn (N) X current (I)MMF = number of turn (N) X current (I)


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-- Reluctance (Reluctance (00))

What is Reluctance???

The reluctance gives a measure of howdifficult it is for the magnetic flux tocomplete its circuit.


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Analogy between Magnetic circuit and Electric circuit

MMF is analog to Voltage

Flux is analog to current

Reluctance is analog to resistance

So

is analog toFlux (Flux (JJ)) == MMF (MMF (NINI)) ..

Reluctance (Reluctance (00))

Currrent (I) =Currrent (I) = Voltage (V) .Voltage (V) .Resistance (R)Resistance (R)


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FromFrom Flux = MMF/reluctanceFlux = MMF/reluctance we can see thatwe can see thatwe can increase the flux by decreasing thewe can increase the flux by decreasing thereluctance of the magnetic circuit.reluctance of the magnetic circuit.

Air is the poor magneticAir is the poor magneticmaterial (high reluctance).material (high reluctance).

So, air should be replaced bySo, air should be replaced by

good magnetic material suchgood magnetic material suchas iron as shown below.as iron as shown below.

Note: The reluctance of the airNote: The reluctance of the airis typically 1000 times greateris typically 1000 times greaterthan the reluctance of the iron.than the reluctance of the iron.


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Almost all the flux is confined within the iron, ratherthan spreading out into the surrounding air.Therefore the iron can be sharp in order to guide

the flux to wherever it is needed. Also that inside the iron the

flux density remain uniformover the whole cross-

section.

The flux density in the airgap has the same highvalue as it does inside the

iron


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Since the air gap is very small as air gap in themotor (air gap between stator and rotor) so the fluxjump across the air gap without any tendency to

balloon out into the surrounding air. With most ofthe flux lines going straight across the air gap.

If air gap is big then the fluxwill leak into the air. This iscalled Flux leakage.


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-- Air Gap Flux DensityAir Gap Flux Density

MMF also defined as the force that requiredmoving the magnetic flux to complete the

magnetic circuit.

Since iron has a low reluctance so the MMFthat required to move the flux in iron is very

small (can be neglected) so most of the MMFis used to move the flux in the air.


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This equation show that,This equation show that,-- if g is increase thenif g is increase then 00 alsoalsoincrease andincrease and

-- if A is increase thenif A is increase then 00 willwill

decrease.decrease.

How to calculate the reluctance in the air gap?How to calculate the reluctance in the air gap?

The reluctance of the air with crossThe reluctance of the air with cross--sectional area Asectional area A

and length g is given byand length g is given by

00 == gg ..QQ AA

Where,Where, QQ is the magnetic space constantis the magnetic space constantor permeability of the free space.or permeability of the free space.It is equal to 4It is equal to 4TT X10X10--7H/m7H/m


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Magnetic flux at air gap,Magnetic flux at air gap,

JJ == NINI == NIANIAQQ00 gg


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Flux density in the air gap,Flux density in the air gap,

So, the flux density is proportional to MMF and invert proportionalSo, the flux density is proportional to MMF and invert proportionalto g.to g.

E

xample:E

xample: suppose the magnetizing coil have 250 turns, thesuppose the magnetizing coil have 250 turns, thecurrent is 2 A, and the gap is 1mm. The flux density iscurrent is 2 A, and the gap is 1mm. The flux density isgiven bygiven by

B =B = 250 X 2 X 4250 X 2 X 4TT X10X10--77 = 0.63 Tesla= 0.63 Tesla1X101X10--33

B =B = JJ == NIANIAQQ == NINIQQ

A gA gA gA g


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If the cross section area of the iron were constant at allpoints, the flux density would be 0.63 T everywhere.

However if the crosssection area of the iron isreduced to half of the airgap as shown below then

the flux is compressed inthe narrower section andthe flux density is doubledwhich is 2 X 0.63 = 1.26 T.


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-- SaturationSaturation

If we try to work the iron at higher flux densities, itIf we try to work the iron at higher flux densities, itbegin to exhibit significant reluctance.begin to exhibit significant reluctance.

This sketch showing how the effective reluctance ofThis sketch showing how the effective reluctance ofiron increase rapidly as the flux density approachesiron increase rapidly as the flux density approachessaturationsaturation..

chapter 1 - fundamental concept

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