Electric Trains

8/6/2019 Electric Trains

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GROUP # 6: CP BEM UNITEN

1. WAN ABDUL BARIE BIN WAN MD MARZUKI (CP086367)

2. WEE KIM YONG (CP087169)

3. ZAKARIA BIN ABDUL RAHMAN (CP087180)

4. ZULIHANIS BT ISMAIL (CP086368)

5. WAN MAHRIMI BIN WAN AMBAK (CP087201)

6. AHMAD ARIPIN BIN SAIMAN (CP087200)

7. SHEFIAN BIN MOHD DOM (CP087205)

ELECTRIC TRAIN

EEEB283

ELECTRICAL MACHINES AND DRIVES


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TABLE OF CONTENTS PAGES

Abstract 2

1.0

1.1

1.2

1.3

1.4

1.5

Literature Survey On Electric Vehicle

Electric Traction Power

Harmonic From Electric Mass Transport

Traction Transformer

Induction And Permanent Magnet Motors/ Motor Design

Reduction In Carbon Foot Print

3

2.02.1

2.2

Impact Of Electric Mass TransportThe Various Electric Motor Design And Traction Transformer For

Various Types Of Electric Trains

The Various Energy Storage Technologies Employed In The Existing

Electric Trains

5

3.0 Case Study 8

5.0 Bibliography 11

6.0 Appendix A 12

7.0 Appendix B 21

8.0 Appendix C 29


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1.0 LITERATURE SURVEY ON ELECTRIC VEHICLE

1.1 ELECTRIC TRACTION POWER

An electric vehicle (EV), also referred to as an electric drive vehicle, uses one or more

electric motors or traction motors for propulsion. For mass transportation, electric train (Figure

1) are used which is powered by electricity from overhead lines (Figure 2), a third rail (Figure 3)

or an on-board energy storage device (Figure 4) (such as a chemical battery or fuel cell) such as

trams, light rail, rapid transit, people movers, commuter rail and monorail suspension railways.

The distinguishing design features of electric train are the type of electrical power used,

either alternating current or direct current, the method for store (batteries, ultra capacitors) or

collecting (transmission) electrical power and the means used to mechanically couple the tractionmotors to the driving wheels (drivers).

1.2 HARMONIC FROM ELECTRIC MASS TRANSPORT

An AC electric trains having thyristors or pulsewidth-modulation (PWM)-controlled

converters inject harmonic currents into the feeding overhead lines. Harmonic currents in the

electric train are one of the biggest concerns. The current harmonics injected from an ac electric

train propagate through power-feeding circuits. Being a distributed RLC circuit, the feedingcircuit can experience parallel resonance at a specific frequency. The harmonic current is

amplified by the resonance, and the amplified harmonic current usually induces various

problems, including interference in adjacent communication lines and the railway signaling

system, overheating, and vibration at the power capacitors, and erroneous operation at the

protective devices. Therefore, the harmonic current flow must be assessed exactly in the

designing and planning stage of the electric traction system.

1.3 TRACTION TRANSFORMER

A traction transformer is a single transfer point for energy between catenary and motors,

and must therefore meet high reliability levels. Traction transformer is a special voltage power

transformer, which should be able to endure the acute changes of traction load and the external

frequent short. Besides high demands on reliability and performance, traction transformers must


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also be compact and lightweight and display high efficiency. Generally speaking, traction

transformers have the capability to 100% overload.

1.4 INDUCTION AND PERMANENT MAGNET MOTORS (MOTOR DESIGN)

Various motor is used to move the train, permanent magnet and induction motor mostly

used. The biggest drawback of the induction machine is the always lagging power factor,

because the machine is magnetized from the stator, in other words, there is a magnetizing current

flowing in the stator winding even at no-load conditions. This means that less torque is available

with a given current than for example with a PMSM, or alternatively, more current is required to

produce an equal torque, which leads to an inverter with a higher current rating. More detail is in

Table 1. T

There are 2 types of motor design, AC and DC motor. By the early 1980s, power

electronics had progressed to the stage where the 3-phase AC motor (Figure 5) became a serious

and more efficient alternative to the DC motor. Details are in Table 2. From a transmission point

of view, AC is better than DC because it can be distributed at high voltages over a small size

conductor wire, whereas DC needs a large, heavy wire or, on many DC railways, an extra rail.

For simply put, simply as AC for long distance and DC for short distance railway.

1.5 REDUCTION IN CARBON FOOT PRINT

Acarbon footprint is a measure of the impact our activities have on the environment, and

in particular climate change. It relates to the amount of greenhouse gases produced in our day-to-

day lives through burning fossil fuels for electricity, heating and transportation etc. Electric

Trains have always had no carbon emissions because they are run entirely by internal electric

motors. However, the means of generating the electricity used to power these motors was

predominately by burning fossil fuels or coal, both of which produce a large amount of carbon

emissions. With the emergence of 'clean energy' generation, electrical trains actually run with

very low environmental impact. For example, the proposal for the high speed rail line between

San Francisco and Los Angeles in California has the potential for zero greenhouse gas emissions,

with the 3,350 GWh each year being generated by California's extensive infrastructure of

renewable energy sources.


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2.0 IMPACT OF ELECTRIC MASS TRANSPORT

Mass transportation enables economic activity by connecting people, businesses and

resources. Although transportation contributes to economic productivity it also imposes

significant economic costs. Electrical mass transportation cost includes civil (infrastructure),

equipment, projects and financing cost. It is varied with design requirement (safety requirement

e.g. anti earth-quake, immunity typhoon and high humidity) and geographical location. Table 3

shows some sample new railway project costs as published in the railway trade press.

Although we often hear people talking of mass transportation making profit, what they

are actually referring to is that the railway takes in more fare revenue than it spends on operating

costs. But there are systems that work well and able to provide positive fare box ratio one over

100% such as Seoul Metro (Korea) (Refer Table 4). These systems are successful because of two

features - high patronage and management efficiency.

Governments, of any variety, may opt to subsidize public transport, for social,

environmental or economic reasons. Key motivations are the need to provide transport to people

those who cannot afford or are physically or legally incapable of using an automobile, and to

reduce congestion, land use and emissions of local air pollution and greenhouse gases. Other

motives may be related to promote business and economic growth, or urban renewal in formerly

deprived areas of the city. This transportation services may be commercial, but receive greater

benefits from the government compared to a normal company.

Gases emitted by automobiles have been cited as major contributors to the issues

addressed in green initiatives. A single person, 20-mile round trip by car can be replaced using

public transportation and result in a net CO2 emissions reduction of 4,800 lbs/year. Using public

transportation saves CO2 emissions in more ways than simply travel as public transportation can

help to alleviate traffic congestion as well as promote more efficient land use.

In order to understand the electric vehicles carbon footprint, two phases must be

distinguished: energy production and vehicle use. In the use phase, the electric vehicle does not

emit any C02, but it does consume a certain quantity of energy. When using an electric vehicle,


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neither carbon gases nor pollutants are emitted. In the energy production phase means source of

the electricity used to charge them. An electric vehicle that is charged with energy from a clean

source, like hydroelectric power, will produce very little pollution, unlike unclean source (coal).

Carbon footprint for vehicles is measured by average number of C02 grams emitted per

kWh. In France, this C02 content is in the region of 85 g/kWh, which is much less than the

European average (of around 400 g/kWh). And for good reason: the electricity produced in

France comes predominantly from nuclear power (80%), with the balance coming from water-

powered stations (15%) and from gas or coal (5%), according to the IEA (the International

Energy Agency). The situation is different elsewhere in the world where nuclear power and

renewable energy are rarer. It hits 850 g/kWh in China and 875 g/kWh as it is produced in the

majority by coal-powered stations.

Table 5 shows comparison of the pollutants emitted by internal combustion engine

vehicles and electric vehicles charged by various energy sources. Notice that carbon monoxide

and hydrocarbon emissions are negligible when using electric vehicles, but that the emission of

sulfur oxides increases. Also note the increasing emissions of carbon dioxide, nitrogen oxides,

and sulfur oxides that occur when using greater proportions of coal and oil derived electricity.

2.1 THE VARIOUS ELECTRIC MOTOR DESIGN AND TRACTION

TRANSFORMER FOR VARIOUS TYPES OF ELECTRIC TRAINS

Various type of motor design mainly separated to 2 types, which are AC and DC motor.

There are however a number of DC motor choices where series-wound machine is the best

machine for use in traction applications primarily; it is due to its torque/speed characteristics

being well matched to the demands required for traction application. More details are in Table 6.

There are 2 types of AC traction motors, synchronous and induction motor

(asynchronous). Synchronous AC motors are also occasionally used, as in the French TGV high

speed train. Three-phase cage induction (asynchronous) motors have been utilized for

traction over the years, and it was again the advent of power electronic devices within

variable-frequency converters that made their use feasible. These motors are mechanically


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the simplest of the traction machine types with no electrical connections necessary for the

rotating field winding. This motor widely used nowadays such as commuter, intercity etc.

A new form of traction which has appeared in recent years is the linear motor. The

conventional DC motor consists of a fixed part (the stator) and a moving part (the rotor) (Figure

6). Both parts are contained in a case on the train and the rotor is connected to the axle by a

pinion/gear arrangement. When the armature turns, the wheel turns. The efficiency of the linear

motor is about 60% of the conventional motor but it has the advantage of less moving parts and it

does not have the reliance on adhesion of the conventional motor. Many countries had used

linear motor in electric such as in Figure 7.

Traction Transformers for trains transform the overhead contact line voltage, which

ranges mainly from 15 kV or 25 kV to voltages suitable for traction converters (0.7 kV and 1.5

kV). Traction Transformers have ratedpower from 500 kVA to 10 MVA where EMUs mostly

used 1 to 3MVA traction transformers and 3 to 5.5MVA for high sped train.

2.2 THE VARIOUS ENERGY STORAGE TECHNOLOGIES EMPLOYED IN THE

EXISTING ELECTRIC TRAINS

A regenerative brake is an energy recovery mechanism which slows a vehicle by

converting its kinetic energy into another form, which can be either used immediately or stored

until needed. Traction motors provide resistance to the wheels turning, slowing them down. By

capturing the energy in these traction motors, engineers were able to build regenerative braking

into existing train systems. Figure 8 show energy flow of braking and accelerating electric train. .

Basically, there are 2 type of energy storage applied in rolling stocks, stationary (Figure

9) and mobile (Figure 10). For stationary energy storage, electric double-layer capacitor is used.

This storage helps stabilization of line voltage due to increase of the minimum line voltage. The

energy storage system reduces primary energy consumption without affecting transport capacity

and punctuality. For mobile energy storage, NiMH-Traction battery can also be used instead of

electric double-layer capacitor. This mobile energy storage can help higher utilization of power

supply due to less voltage drop along the overhead contact line.


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3.0 CASE STUDY

Case information

Given a 460V, 25hp, 60Hz, 4 poles, Y-connected wound rotor induction motor has the following

impedances in ohms per phase referred to the stator circuit

R1=0.641 , R2=0.332 , X1=1.106 , X2=0.464 , XM=26.3

Discussion and Observation

Problem 2 (c) (i) : Formulate a wound rotor induction model to plot a base torque-speed

curve for slip speed from 1 to 0.

Equivalent circuit for respective induction motor depicted in Figure A,

Figure A

Investigating the torque-speed curve in Figure B using MATLAB software, it is apparent that the

rotor circuit resistance has significant impact on speed at which maximum torque occurs. The plots

above illustrate five cases, range from low rotor resistance R2 gradually up to the high rotor

resistance 5R2. By increasing the rotor resistance, the breakdown torque peak is shifted left to zerospeed. Note that this torque peak is much higher than the starting torque available with base rotor

resistance, R2. For further reference, the details calculation, curve torque-speed, programming and

comparison table was attached in Appendix B.

0.641 j1.106 j0.464 0.332

j26.3


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Figure B: Torque-Speed Curve with varying of R2 for Problem 2(c) (i)

Problem 2 (c) (ii) : Determine the rotor resistances of this induction machine to get the

starting torque that half and double the based starting torque.

Table A: Manual calculation analysis for Problem 2(c) (ii)

Table B: MATLAB curve analysis for Problem 2(c) (ii)

Rotor

Resistance, R2SlipMax, Smax

Mechanical

speed, mTorque Start, start

Torque Max,

max

R2 = 0.332 0.1979 1444 103.81 229

R2 = 0.151 0.0900 1638 51.91 229

R2 = 1.005 0.5992 721 207.62 229

RotorResistance, R2

Slip Max,Smax

Mechanicalspeed, m

Torque Start, startTorque Max,

maxR2 = 0.332 0.2 1440 106.6 230.8

R2 = 0.151 0.1 1620 53.54 230.2R2 = 1.005 0.6 720 211.4 230.8


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Figure C: Torque-Speed Curve with varying of torque start for Problem 2(c) (ii)

The above plots show the change in slip is attained by changing the value of rotor

resistances. As the value of rotor resistance increase, the slope of the motor curve will decrease,

shifting the stable operating point for the given load curve to a point with higher slip. Thus the

speed control is achieved in the rotor resistance control. As we know, slip is proportional to rotor

resistance and pullout torque is proportional to slip. Thus, high torque is produced while starting.

In induction motor, starting torque must be in non-linear part of torque-speed curve while

normal operating torque must be at linear part. By refer to the curve where starting torque occurs

at double base torque in Figure C, it shows that the curve have high starting torque at the

beginning of non-linear part but end up with high slip at linear part. Since the slip produced is

high, then it will contribute to low efficiency of induction motor. On the other hand, when the

curve starting torque occurs at half base torque in same figure, it shows that the curve will begin

with low starting torque and produce low slip value at the end of curve thus giving higher

efficiency to respective induction motor. For further reference, the details calculation, curve

torque-speed, programming and comparison table was attached in Appendix C.


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4.0 BIBLIOGRAPHY

4.1Mom, G., 2004. The Electric Vehicle: Technology and Expectations in the Automobile Age. JohnsHopkins University Press, pg. 31.

4.2History of the Electric Car. Energy Efficiency and Renewable Energy, Department of Energy. Lastupdated November 22, 2005.

4.3Hu Yu, Yuan Yue, Chen Zhe, Chen Zhifei, Tao Ye, Research on the Selection of Railway TractionTransformer, IPEC, 2010 Conference Proceedings.

4.4Yuen, KH; Pong, MH; Lo, EWC; Ye, ZM, Modeling of electric railway vehicle for harmonicanalysis of traction power-supply system using spline interpolationin frequency domain, The 14th

Annual Applied Power Electronics Conference and Exposition, Dallas.TX, 14-18 March 1999, v. 1,

p.458-4634.5Changmu Lee; Jang, G.; Sae-hyuk Kwon, Harmonic analysis of the Korean high-speed railway using

the eight-port representation model, IEEE TRANSACTIONS ON POWER DELIVERY, VOL. 21,

NO. 2, APRIL 2006

4.6Jussi Puranen, Induction Motor Versus Permanent Magnet Synchronous Motor in Motion ControlApplications: A Comparative Study, Lappeenranta 2006 147 p. , Acta Universitatis

Lappeenrantaensis 249 , Diss. Lappeenranta University of Technology

4.7Nicholson, T.J. , DC and AC Traction Motor, Electric Traction Systems, 2008 IET ProfessionalDevelopment Course,

4.8ABB is contributing to high speed trains, ABB Review 2 2010,4.9 http://www.railway-technical.com/index.shtml4.10 http://www.uitp.org/Public-Transport/sustainabledevelopment/4.11 http://www.physics.ohio-state.edu/~wilkins/writing/Samples/policy/voytishlong.html4.12 http://www.energy.siemens.com/co/en/power-transmission/transformers/traction-transformers/4.13 http://www.futurecars.com/technology/how-regenerative-braking-works4.13.1 http://en.wikipedia.org/wiki/KTM_Komuter


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5.0 Appendix A

Figure 1

Figure 2


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Figure 3

Figure 4


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Figure 5

Permanent Magnet Induction motor

Smooth torque possible Excellent dynamics with proper control

High efficiency High speed operation possible

High torque/volume Low price and simple construction

High pull-out torque possible Durable

Good heat Dissipation-good overloading

Capability

Several suppliers available

Expensive Complicated control

Danger of demagnetization of the magnets Always lagging power factor

Poor field weakening Lower efficiency with lighter loads

Table 1


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AC motor advantage against DC motor:

They are simpler to construct, they require no mechanical contacts to work(such as brushes) and they are lighter than DC motors for equivalent power.

Modern electronics allow AC motors to be controlled effectively to improveboth adhesion and traction.

AC motors can be microprocessor controlled to a fine degree and canregenerate current down to almost a stop whereas DC regeneration fades

quickly at low speeds.

They are more robust and easier to maintain than DC motors.Table 2

Railway Date Type of SystemCost per

km(US$)Distance Notes

Australia - Brisbane

Airport Link1998 Airport Line $16.2 million 8.5 kms Surface

Norway -Oslo/Gardemoen

1998Express AirportLine

$11.3 million 66 kms 21 % tunnel

West Rail - Hong Kong 1999 Heavy Metro $220 million 30.5 kms 38% tunnel

Taiwan High Speed 1998High Speed

Passenger$49 million 345 kms Mostly surface

Singapore North East

Line1998 Heavy Metro $150 million 20 kms 100% tunnel

Caracas, Venezuela 1998 Suburban $31.6 million 9.3 kmsNo intermediate

stations

Meteor, Paris 1998 Metro $130 million 8.5 kms 100% tunnel

Hamburg -Wurzburg 1999High Speed

Passenger$47.5 million Mostly surface

TGV Est Phase 1,

France1999

High Speed

Passenger$ 11 million 310 kms Surface


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BART SFO Extension 1999 Heavy Metro $112 million 14 kms 70% tunnel

Shanghai China 1999 Heavy Metro $91 million 16.5 kms 100% tunnel

Kuala Lumpur,

Malaysia1999

Airport/suburban

link$14 million 57 kms 100% surface

Manila Line 3

Extension1999 Light Metro $50 million 5.2 kms Elevated

Porto Portugal 1999 Light Rail $13 million 70 kms10% tunnel. Part

existing.

Kaoshiung, Taiwan 1999 Heavy Metro $140 million 43 kms 85% tunnel

Salt Lake City, Utah,

US1998 Light Rail $13 million 24 kms At grade

Hudson-Bergen NJ,

USA 1999 Light Rail $72 million 15.3 kms

Part elevated, incl.

15 yearconcession.

Bangkok, Thailand 1999 Metro $73.6 million 23.1 kms 100% elevated

Bangkok, Thailand 1999 Metro $139 million 20 kms 100% tunnel

Jubilee Line, London,

UK1999 Heavy Metro $336 million 16 kms 100% tunnel

Lewisham Extension,

DLR, London, UK1999 Light Rail $76 million 4.2 kms

25% tunnel, 75%

elevated

Hanover, Germany

Line D Extn. 1999 Light Rail $20.6 million 9.6 kms At grade

Cali, Colombia 2000 Light Rail $31.3 million 18.8 kms

Seoul-Pusan, Korea 1999High Speed

Passenger$37.3 million 412 kms

46% tunnel, 26%

viaducts

San Juan, Puerto Rico 1998 Metro $31.6 million 17.2 kms15% tunnel, 50%

elevated

Tripoli-Ras Jedir,

Lybia2000 Main line $2.5 million 191 kms Surface

Haikou-Sanya, China 2010 High Speed line $10million 308 kms

Table 3


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RailwayRatio of Revenue to

Operating Costs

Kuala Lumpur PUTRA System - Malaysia 40%

RATP Paris 50%

Sao Paulo Metro Brazil 70%

BNSF Aurora (Chicago, Il. USA) 75%

Kuala Lumpur STAR Elevated - Malaysia 90%

London Underground 125%

Seoul Metro Korea 140%

Mass Rapid Transit Singapore 150%

Santiago Chile 160%

Manila Line 1 Philippines 170%

Manchester Metrolink UK 190%

Mass Transit Railway - Hong Kong 220%

Table 4

Region Studied LA Basin Germany IdealInternal

Combustion

Amount of Electricity

from Coal or Oil21% 49% 100% (for comparison)

Carbon Monoxide 0.007 0 0 1

Carbon Dioxide 0.34 1 2.5 1

Hydrocarbons 0.01 0 0 1

Nitrogen Oxides 0.27 1 3.3 1

Sulfur Oxides 1.72 10 25 1

Table 5


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Series-Wound Advantage

The torque from a series-wound motor is determined by the current andbroadly independent of the supply voltage, unlike the shunt-wound motor

where the torque drops as the voltage drops.

The field winding only has a few fairly large-section copper with a lowvoltage difference between turns, giving a low insulation to copper ratio

allowing a small and robust design of coil.

The series-wound motor inherent characteristic where the torque isinversely proportional to the speed, i.e. as the speed increase the torque

decreases, this prevents overloading of the motor.

Series-wound motors when connected in series will all do the same thing,as for a given current the voltage effectively fixes the motor speed.

Table 6

Figure 6


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Figure 7

Figure 8


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Figure 9

Figure 10


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6.0 Appendix B

Rotor Resistance, R2Slip Max,

Smax

Mechanical speed,

m

Torque Start,

start

Torque Max,

max

1*R2 = 0.332 0.1979 1444 103.81 2292*R2 = 0.664 0.3959 1087 170.47 229

3*R2 = 0.996 0.5938 731 207.29 229

4*R2 = 1.328 0.7918 375 224.05 229

5*R2 = 1.66 0.9897 19 228.58 229

Table 7: Manual calculation analysis for Problem 2(c) (i)

Table 8: MATLAB curve analysis for Problem 2(c) (i)

Figure 11: Torque-Speed curve with varying of R2 for Problem 2(c) (i)

Rotor Resistance, R2Slip Max,

Smax

Mechanical speed,

m

Torque Start,

start

Torque Max,

max

1*R2 = 0.332 0.2 1440 106.6 230.8

2*R2 = 0.664 0.4 1080 174.1 230.83*R2 = 0.996 0.6 720 210.7 230.8

4*R2 = 1.328 0.8 360 226.9 230.8

5*R2 = 1.66 1 0 230.8 230.8


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%CP BEM UNITEN

%EEEB283 Electrical Machine and Drives%Group 6 (Semester 1 2011/2012)%Dr Ungku Anisa Bte Ungku Amirulddin

%Programming for Problem 2.(c).(i)%M-File: problem2ci.m

%M-File to create plot of Torque Induced(Nm) vs Speed(rpm)%Varying the rotor resistance(R2)on wound rotor induction motor

%Initial valuer1=0.641;x1=1.106;r2=0.332;x2=0.464;xm=26.3;

v_phase=460/sqrt (3);n_sync=1800;w_sync=188.5;

% Calculate thevenin voltage and impedancev_th=v_phase*(xm/sqrt(r1^2+(x1+xm)^2));z_th=((j*xm)*(r1+j*x1))/(r1+j*(x1+xm));r_th=real(z_th);x_th=imag(z_th);

%Calculate slips=(0:1:50)/50;s(1)=0.001;nm=(1-s)*n_sync;

%Calculate torque with R2 valuefor ii=1:51

t_ind1(ii)=(3*v_th^2*r2/s(ii))/...(w_sync*((r_th+r2/s(ii))^2+(x_th+x2)^2));

end

%Calculate torque with 2R2 valuefor ii=1:51

t_ind2(ii)=(3*v_th^2*(2*r2)/s(ii))/...(w_sync*((r_th+(2*r2)/s(ii))^2+(x_th+x2)^2));

end



end

%Calculate torque with 4R2 value


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for ii=1:51t_ind4(ii)=(3*v_th^2*(4*r2)/s(ii))/...

(w_sync*((r_th+(4*r2)/s(ii))^2+(x_th+x2)^2));end



end

%Plot torque versus speed curveplot(nm,t_ind1,'color','r','linewidth',2.0);hold on;plot(nm,t_ind2,'color','b','linewidth',2.0);plot(nm,t_ind3,'color','g','linewidth',2.0);plot(nm,t_ind4,'color','k','linewidth',2.0);plot(nm,t_ind5,'color','m','linewidth',2.0);

xlabel('\itn_{m}','fontweight','bold');ylabel('\tau_{ind}','fontweight','bold');title('Induction Motor Torque-Speed Curve','fontweight','bold');legend('R_{2}','2R_{2}','3R_{2}','4R_{2}','5R_{2}');grid on;hold off;


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Manual calculation for Problem 2(c)(i)

:: Case 1 at R base = 0.332 = 1R2 = 0.332 ::

The Thevenin voltage is,

VTH = V X m R1 + (X 1 + X m)

= (266)(26.3) (0.641) + (1.106+ 26.3)

= 255.2 Volt

The Thevenin Resistance is,

RTH = R1 Xm

X1 + Xm

= (0.641) 26.31.106 + 26.3

= 0.590

The Thevenin Reactance is,

XTH = X1 = 1.106

To find slip at which maximum torque occurswhen Rbase = 1xRbase = 0.332 ,

S max = R2 (RTH) + (XTH + X2)

S max = 0.332 (0.590) + (1.106 + 0.464)

= 0.3321.677

= 0.1979

To find correspond to a mechanical speed of,

Nm = (1 S ) N sync

= ( 1- 0.1979) (1800)

= 1443.78 rpm

The torque at this speed is,

max = 3 V TH2 W sync (RTH + ( RTH + X2))

= 3(255.2)2(188.5)(0.59 + ((0.590)+(1.106+ 0.464))

= 229 N.m

To find starting torque of this motor is found by settingS = 1,

start = 3 VTH R2W sync ((RTH + R2) + (XTH + X2)

= 3(255.2)(0.332)(188.5) x ((0.590+0.332) + (1.106+0.464))

= 64866.53184624.87373

= 104 N.m


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:: Case 1 at R base = 0.332 = 2R2 = 0.664 ::


VTH = V X m

R1 + (X 1 + X m)

= (266)(26.3) (0.641) + (1.106+ 26.3)

= 255.2 Volt


RTH = R1 Xm

X1 + Xm

= (0.641) 26.3

1.106 + 26.3

= 0.590


XTH = X1 = 1.106

To find slip at which maximum torque occurs

when Rbase = 2Rbase = 0.664 ,


S max = 0.664 (0.590) + (1.106 + 0.464)

= 0.3959


Nm = (1 S ) N sync

= ( 1- 0.3959) (1800)

= 1087 rpm


max = 3 V TH

2 Wsync ((RTH + ( RTH + X2))

= 3(255.2)2(188.5)(0.59 + ((0.590)+(1.106+ 0.464))

= 229 N.m


start = 3 VTH R2W sync ((RTH + R2) + (XTH + X2))

= 3(255.2)(0.664)(188.5) x ((0.590+0.664) + (1.106+0.464))

= 170.47 N.m


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:: Case 1 at R base = 0.332 = 3R2 = 0.996 ::


VTH = V X m

R1 + (X 1 + X m)

= (266)(26.3) (0.641) + (1.106+ 26.3)

= 255.2 Volt


RTH = R1 Xm

X1 + Xm

= (0.641) 26.3

1.106 + 26.3

= 0.590


XTH = X1 = 1.106




S max = 0.996 (0.590) + (1.106 + 0.464)

= 0.5938


Nm = (1 S ) N sync

= ( 1- 0.5938) (1800)

= 731 rpm


max = 3 V TH

2 Wsync (RTH + ( RTH + X2))

= 3(255.2)2(188.5)(0.59 + ((0.590)+(1.106+ 0.464))

= 229 N.m



= 3(255.2)(0.996)(188.5) x ((0.590+0.996) + (1.106+0.464))

= 207.29 N.m


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:: Case 1 at R base = 0.332 = 4R2 = 1.328 ::


VTH = V X m

R1 + (X 1 + X m)

= (266)(26.3) (0.641) + (1.106+ 26.3)

= 255.2 Volt


RTH = R1 Xm

X1 + Xm

= (0.641) 26.3

1.106 + 26.3

= 0.590


XTH = X1 = 1.106




S max = 1.328 (0.590) + (1.106 + 0.464)

= 0.7918


Nm = (1 S ) N sync

= ( 1- 0.7918) (1800)

= 375 rpm


max = 3 V TH


= 3(255.2)2(188.5)(0.59 + ((0.590)+(1.106+ 0.464))

= 229 N.m



= 3(255.2)(1.328)(188.5) x ((0.590+1.328) + (1.106+0.464))

= 224.05 N.m


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:: Case 1 at R base = 0.332 = 5R2 = 1.66 ::


VTH = V X m

R1 + (X 1 + X m)

= (266)(26.3) (0.641) + (1.106+ 26.3)

= 255.2 Volt


RTH = R1 Xm

X1 + Xm

= (0.641) 26.3

1.106 + 26.3

= 0.590


XTH = X1 = 1.106




S max = 1.66 (0.590) + (1.106 + 0.464)

= 0.9897


Nm = (1 S ) N sync

= ( 1- 0.9897) (1800)

= 19 rpm


max = 3 V TH


= 3(255.2)2(188.5)(0.59 + ((0.590)+(1.106+ 0.464))

= 229 N.m



= 3(255.2)(1.66)(188.5) x ((0.590+1.66) + (1.106+0.464))

= 228.58 N.m


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6.0 Appendix C

Table 9: Manual calculation analysis for Problem 2(c) (ii)

Table 10: MATLAB curve analysis for Problem 2(c) (ii)

Figure 12: Torque-Speed curve with varying of R2 for Problem 2(c) (ii)

Rotor

Resistance, R2SlipMax, Smax

Mechanical

speed, mTorque Start, start

Torque Max,

max

R2 = 0.332 0.1979 1444 103.81 229

R2 = 0.151 0.0900 1638 51.91 229

R2 = 1.005 0.5992 721 207.62 229

RotorResistance, R2

Slip Max,Smax

Mechanicalspeed, m

Torque Start, startTorque Max,

maxR2 = 0.332 0.2 1440 106.6 230.8

R2 = 0.151 0.1 1620 53.54 230.2

R2 = 1.005 0.6 720 211.4 230.8


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Choosing the realistic R2 from the quadratic equation

Figure D Figure E

When torque required was given in Problem 2(c) (ii), the manipulation of torque equation to

determine R2 will lead to quadratic solution (Ax + Bx + C) which will produce two different R2

values. For case to produce half of base torque on Figure D, the value R2=0.151 (Torque Half

A curve) was chosen as it produce low slip and higher efficiency compare R2 with value of

R2=18.602 (Torque Half B curve). Same approach will be used in double of base torque onFigure E where the R2=1.005 (Torque Double A curve) was chosen instead of R2=2.798.

(Torque Double B curve).


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%CP BEM UNITEN

%EEEB283 Electrical Machine and Drives%Group 6 (Semester 1 2011/2012)%Dr Ungku Anisa Bte Ungku Amirulddin

%Programming for Problem 2.(c).(ii)%M-File: problem2cii.m

%M-File to create plot of Torque Induced(Nm) vs Speed(rpm)%Varying the starting torque to half or double of base starting torque

%Initial valuer1=0.641;x1=1.106;r2_tsbase=0.332;r2_tshalf=0.15122;r2_tsdouble=1.00545;x2=0.464;xm=26.3;

v_phase=460/sqrt (3);n_sync=1800;w_sync=188.5;

% Calculate thevenin voltage and impedancev_th=v_phase*(xm/sqrt(r1^2+(x1+xm)^2));z_th=((j*xm)*(r1+j*x1))/(r1+j*(x1+xm));r_th=real(z_th);x_th=imag(z_th);

%Calculate slips=(0:1:50)/50;s(1)=0.001;nm=(1-s)*n_sync;

%Calculate base starting torque valuefor ii=1:51

t_ind1(ii)=(3*v_th^2*r2_tsbase/s(ii))/...(w_sync*((r_th+r2_tsbase/s(ii))^2+(x_th+x2)^2));

end

%Calculate half of base starting torque valuefor ii=1:51

t_ind2(ii)=(3*v_th^2*r2_tshalf/s(ii))/...(w_sync*((r_th+r2_tshalf/s(ii))^2+(x_th+x2)^2));

end

%Calculate double of base starting torque valuefor ii=1:51

t_ind3(ii)=(3*v_th^2*r2_tsdouble/s(ii))/...(w_sync*((r_th+r2_tsdouble/s(ii))^2+(x_th+x2)^2));

end


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%Plot torque versus speed curveplot(nm,t_ind1,'color','r','linewidth',2.0);hold on;plot(nm,t_ind2,'color','b','linewidth',2.0);plot(nm,t_ind3,'color','g','linewidth',2.0);

xlabel('\itn_{m} (rpm)','fontweight','bold');ylabel('\tau_{ind} (Nm)','fontweight','bold');

title('Induction Motor Torque-Speed Curve','fontweight','bold');legend('Torque Base','Torque Half','Torque Double');grid on;hold off;


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When start = 2 x 104 Nm = 208 Nm,

start = 3 VTH R2W sync (RTH + R2) + (XTH + X2)

208 Nm = 195381.12 R2530.2505 + 222.43R2 + 188.5R2

110292.104+46265.44R2+39208R2 = 195381.12 R2

390208R2-149115.68 R2+110292.104 = 0

R2A = 2.79774, R2B = 1.00545 (chosen)


when R2 = 1.00545,

S max = R2

(RTH) + (XTH + X2)

S max = 1.1005

(0.590) + (1.106 + 0.464)

= 0.5992


Nm = (1 S ) n sync = ( 1- 0.5992)(1800)

= 721 rpm


max = 3 V TH2 W sync (RTH + ( RTH + X2))

= 3(255.2)

2(188.5)(0.590 + ((0.590)+(1.106+ 0.464))

= 229 N.m

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