Traction Motors and Control

SPPU
Class-TE-Electrical
Subject-UEE

1. DEPARTMENT OF ELECTRICAL ENGINEERING JSPMS BHIVARABAI SAWANT INSTITUTE OF TECHNOLOGY AND RESEARCH , WAGHOLI, PUNE A.Y. 2019-20 (SEM-II) Class: T.E. Subject: Utilization of Electrical Energy Prepared by Prof. S. D. Gadekar [email protected] Mob. No-9130827661

2. Unit No 6: Traction Motors and Control Topics Desirable characteristic of traction motor Suitability of DC series, AC Series and Linear Induction motor for traction purpose Speed Control of DC Traction Motor Series Parallel Control Efficiency for series parallel starting of two motors Transition Methods Numerical on Series Parallel Control Self relieving property of DC series motor Numerical on Series Parallel Control and maximum speed Regenerative braking French method of regenerative braking Numerical on regenerative braking

3. Desirable characteristic of traction motor i ) Mechanical features a) Robust and capable to withstand continuous vibrations b) The weight of traction motor should be minimum in order to increase the pay load capacity of the vehicle c) It must be small in overall dimensions specially in its overall diameter . d) It must be totally enclosed type .

4. Desirable characteristic of traction motor ii) Electrical characteristics a) High starting torque b) Simple speed control c) Self relieving property – The speed torque characteristics of the motor should be such that the speed may fall with the increase in load. d) Possibility of dynamic or regenerative braking. e) Capability to withstand voltage fluctuation f) Capability of withstanding temporary interruption of supply. g) Overload capacity. h) Parallel running

5. Suitability of DC series motor for traction purpose i ) The dc series motor develops high torque at start which is an essential requirement of traction service. ii) The series motor is amenable to simple speed control methods. iii) In case of dc series motor , Thus The power drawn from supply mains varies as the square root of the load torque .  

6. Suitability of DC series motor for traction purpose iv) Series motor can be suitable for regenerative braking with some special arrangements. Such as French method of regenerative braking . v) Due to excellent commutation , replacement of brushes is not required frequently. vi) Flux varies as the armature current. Thus, torque only depends on current and independent of line voltage thus unaffected by variations in the supply voltage. vii) Up to point of saturation, Thus, dc series motors are capable of withstanding excessive loads . viii) In parallel operations dc series motor share load almost equally.  

7. Suitability of AC series motor for traction purpose The speed of an ac series motor may be controlled efficiently by taps on a transformer, which is not possible in case of a dc series motor . Torque - Speed characteristic of the single phase series motor is similar to that of the dc series motor i.e. high starting torque and decrease in speed with increase in load making it to have self-relieving property from heavy excessive load, so such a machine is particularly useful for traction services. These motors are used for main line services. Single phase AC series motor have better performance (improved power factor, higher efficiency, improved commutation , better starting and fewer poles for a given output ).

8. Suitability of Linear Induction motor for traction purpose The linear synchronous speed Vs of linear induction motor is given as s = 2 𝞃 f meters / second Where 𝞃 is the pole pitch in meters and f is the supply frequency in Hz . s = slip of an induction motor. Where V is the actual speed of rotor plate. Thrust or force or tractive effort is given as, F = W here is the actual power supplied to the rotor.  

9. Suitability of Linear Induction motor for traction purpose In linear induction motor, tractive effort is a function of slip s. It gives thrust-speed characteristic similar to a conventional rotary induction motor. Tractive effort, F can be controlled by varying both frequency and voltage simultaneously so that induction density remains constant . Transverse edge effect reduces the effective thrust and increases the losses. The end effect do not contribute to useful thrust but only towards motor losses .

10. Speed Control of DC Traction Motor Rheostatic Control Series Parallel Control Field Control Motor Generator Locomotive control Diesel Electric Locomotive Control

11. Series Parallel Control M1 M1 + - Step 1 M1 M1 + - Step 2 V V M1 M1 + - V Step 3 M1 M1 + - V Step 4

12. Efficiency for series parallel starting of two motors- Let I=Current Amperes; V=Line Voltage volts and T=Accelerating Time seconds . At the starting instant, both motors are connected in series along with the starting resistance, the voltage across the motors is zero and across the external resistance is V volts. As the motor speed up, the external resistance is gradually reduced to zero. During this period voltage acting across each motor gradually increases from 0 to volts . The time required is seconds. Energy utilized by each motor = Average voltage acting across each motor * current * time  

13. Efficiency for series parallel starting of two motors- When the motors are changed from series parallel, the external resistance is again inserted in the motor circuit. The voltage across each motor is V/2 volts and the current per motor is I amperes. Total current drawn from the line is 2I Amp. The motor speed up, the external resistance is gradually reduced to zero and voltage across each motor increases gradually from V/2 to V volts . Energy utilized by each motor = Average voltage acting across each motor * current * time  

14. Efficiency for series parallel starting of two motors- = =VIT Watt –Seconds = Overall Starting Efficiency  

15. Transition Methods The methods of changing-over the connections from one grouping to another are known as transition methods. Open-circuit transition Shunt transition Bridge Transition

16. Transition Methods Open-Circuit T ransition

17. 2. Shunt Transition

18. 2. Bridge Transition

19. Line voltage, V = 600 volts Current per motor, I = 400 amperes Starting period, Ts = 20 seconds Motor resistance, R = 0.1 Ohm Example-1 Two 600 volts motors are started by series parallel control. Each motor takes 400 A during starting time of 20 sec. and has 0.1 ohm resistance. Calculate, Energy lost in controller and motor , ii ) Motor output, iii) Total energy input from line , iv) Efficiency at starting .

20. Back emf of each motor in full series position , 400*0.1=260 Volts Back emf of each motor in full parallel position , since motors take 20 seconds to build up 560 volts, therefore, time taken to build up 260 volts emf .   Example-1 Two 600 volts motors are started by series parallel control. Each motor takes 400 A during starting time of 20 sec. and has 0.1 ohm resistance. Calculate, Energy lost in controller and motor, Motor output, Total energy input from line, Efficiency at starting.

21. Voltage drop in the starting rheostat in series combination at the starting instant. = V – 2IR = 600 – 2 * 400 * 0.1 = 520 volts ……..Which reduces to zero in full series position. Voltage drop across the starting resistance in first parallel position is equal to i.e. 300 volts which gradually reduces to zero . Energy utilized by each motor = Average voltage acting across each motor * current * time So energy dissipated in starting resistance series and parallel for two motor is given by, =0.268 kwh+0.357 kwh=0.625 Energy dissipated in starting resistance during series and parallel combination per motor, =0.312 kwh   Example-1 Two 600 volts motors are started by series parallel control. Each motor takes 400 A during starting time of 20 sec. and has 0.1 ohm resistance. Calculate, Energy lost in controller and motor, Motor output, Total energy input from line, Efficiency at starting.

22. =0.089 kwh Energy output of motor for one motor= =0.623 kwh Total Energy input per motor=Output +Loss=0.623+0.089+0.312 =1.024 kwh Total Energy input for two motor=1.024*2=2.048 kwh Efficiency of starting=   Example-1 Two 600 volts motors are started by series parallel control. Each motor takes 400 A during starting time of 20 sec. and has 0.1 ohm resistance. Calculate, Energy lost in controller and motor, Motor output, Total energy input from line, Efficiency at starting .

23. Self relieving property of DC series motor The speed-torque characteristic of the motor should be such that the speed may fall with the increase in load . The motors having such speed-torque characteristics are self protective against excessive overloading as power output of a motor is proportional to the product of torque and speed .  

24. Line voltage, V = 600 volts Current per motor, I = 300 amperes Starting period, Ts = 15 seconds Motor resistance, R = 0.1 Ohm   Example-2 Two 600 volts motors are started by series parallel control. Each motor takes 300 A during starting time of 15 sec. and has 0.1 ohm resistance. The train speed at the end of this period is 29 kmph . Calculate, Rheostatic losses in kwh during a) The series and b) The parallel combination of motors , iv ) The train speed at which transition from series to parallel must be made.

25. Back emf of each motor in full series position , *0.1=270 Volts Back emf of each motor in full parallel position , since motors take 20 seconds to build up 570 volts, therefore, time taken to build up 270 volts emf .   Example-2 Two 600 volts motors are started by series parallel control. Each motor takes 300 A during starting time of 15 sec. and has 0.1 ohm resistance. The train speed at the end of this period is 29 kmph . Calculate, Rheostatic losses in kwh during a) The series and b) The parallel combination of motors , iv) The train speed at which transition from series to parallel must be made.

26. Voltage drop in the starting rheostat in series combination at the starting instant. = V – 2IR = 600 – 2 *300 * 0.1 = 540 volts ……..Which reduces to zero in full series position. Voltage drop across the starting resistance in first parallel position is equal to i.e. 300 volts which gradually reduces to zero . Energy utilized by each motor = Average voltage acting across each motor * current * time So energy dissipated in starting resistance series and parallel for two motor is given by, =0.16 kwh+0.198 kwh   Example-2 Two 600 volts motors are started by series parallel control. Each motor takes 300 A during starting time of 15 sec. and has 0.1 ohm resistance. The train speed at the end of this period is 29 kmph . Calculate, Rheostatic losses in kwh during a) The series and b) The parallel combination of motors , iv) The train speed at which transition from series to parallel must be made.

27. Acceleration, =1.933 Kmphps Speed at the end of series period= * =1.933*7.1 =13.7 Kmph   Example-2 Two 600 volts motors are started by series parallel control. Each motor takes 300 A during starting time of 15 sec. and has 0.1 ohm resistance. The train speed at the end of this period is 29 kmph . Calculate, Rheostatic losses in kwh during a) The series and b) The parallel combination of motors , iv) The train speed at which transition from series to parallel must be made.

28. R egenerative braking in traction In regenerative braking the motors remain connected to the supply and return the braking energy to the supply. For regenerative braking it is necessary that traction motors must generate power at a voltage higher than the supply voltage and at a reasonable constant voltage. Regenerative braking is an inherent characteristic of dc shunt wound motors and does not require any change of connections. The dc series motors cannot be used readily for regenerative braking and some modification of shunt winding or separate excitation of the series field at low voltage is necessary to be employed to enable the motor to act as a generator. One method of obtaining regenerative braking with series motors is French method. Regenerative braking is an inherent characteristic of an induction motor since an induction motor operates as an induction generator when run at speeds above synchronous speed and it feeds power back to the supply line. The induction motor is not self exciting as a generator and is required to be connected to a system supplied from synchronous generator.

29. Regenerative braking in traction Advantages : i . The wear of the brake shoes and wheel tyres is reduced to considerable extent, therefore, their life is increased and replacement cost is reduced. ii. Higher value of braking retardation is obtained so that the vehicle can be brought to rest quickly and running time is considerably reduced. iii. Higher speeds are possible while going down the gradients because the high braking retardation can be obtained with regenerative braking . Disadvantages : i. Additional equipment is required for control of regeneration and for protection of equipment and machines, hence initial as well as maintenance cost is increased. ii. For regenerative energy the operation of the substations becomes complicated and difficult. iii. Regenerative braking is employed down to a speed of 16 kmph , then rheostatic braking to about 6.5 kmph and then mechanical braking is required to bring locomotive to rest.

30. French method of regenerative braking If there is a single series motor as incase of a trolley buses, tramways, it is provided with a main series field winding connected in parallel with auxiliary field winding. During regeneration (braking period) the auxiliary field windings are put in series with each other and are switched across the supply . Motoring Generating The machine acts as a compound generator slightly differentially compounded .

31. If there are several motors, we do not require any auxiliary winding . During normal running the motors are connected in parallel with field winding connected in series with their respective armatures. But, during regeneration the motors are connected such that all armatures are connected in parallel and series field windings of all motors are connected in series and placed across the supply. French method of regenerative braking

32. Example-3 A train weighing 300 tonne has speed reduced from 80 kmph to 30 kmph while going down an incline of 1 in 100 through a distance of 3 km by employing regenerative braking. Calculate the electrical energy teturned to the line assuming overall efficiency of 75 % Tractive resistance is 4 kg per tonne and allow rotational inertia of 8 %. Given- Accelerating weight, we = 1.1 W = 1.1 * 300 = 330 tonnes Distance travelled during period, s = 3 km Gradient, G = 1 % Efficiency of conversion, = 75 % = 0.75 Tractive resistance=4 kg per tonne =4*9.81 Newton per tonne   Energy available due to reduction in speed =0.01072*330*( =19456.8 watt-hours=19.45 kwh   Tractive effort required while going down the gradient , =(300*4*9.81)-(98.1*300*1) =– 17658 Newton That is tractive effort of 17658 Newton available.  

33. Example-3 A train weighing 300 tonne has speed reduced from 80 kmph to 30 kmph while going down an incline of 1 in 100 through a distance of 3 km by employing regenerative braking. Calculate the electrical energy teturned to the line assuming overall efficiency of 75 % Tractive resistance is 4 kg per tonne and allow rotational inertia of 8 %. Energy available on account of moving down the gradient a distance of 3 km =14.715 kwh Total energy available = 19.4568 + 14.715 = 34.1718 kwh Energy returned to the line considering 75 % efficiency = 0.75 * 34.1718 = 25.6288 kwh.  

34. Example-4 A 2340 tonne train including loco proceeds down a gradient of 1 in 80 for 5 minutes during which its speed gets reduced from 60 kmph to 36 kmph by application of regenerative breaking. Find the energy returned to the lines if the tractive resistance is 5 kg/ tonne , rotational inertia 10 % and overall efficiency of the motor during regeneration is 70 %. Given- Accelerating weight, we = 1.1 W = 1.1 *2340 = 2574 tonnes Regenerative times, t=5 minutes=5*60=300 Seconds Gradient, G = 1 in 80, Efficiency of conversion, = 70 % = 0.70 Tractive resistance=5 kg per tonne =5*9.81 Newton per tonne   Energy available due to reduction in speed =0.01072*2574*( =63575 watt-hours=63.57 kwh   Tractive effort required while going down the gradient , =(2340*5*9.81)-(98.1*2340*1.25) =– 172165.5 Newton That is tractive effort of 172165.5 Newton available.  

35. Example-4 A 2340 tonne train including loco proceeds down a gradient of 1 in 80 for 5 minutes during which its speed gets reduced from 60 kmph to 36 kmph by application of regenerative breaking. Find the energy returned to the lines if the tractive resistance is 5 kg/ tonne , rotational inertia 10 % and overall efficiency of the motor during regeneration is 70 %. Distance moved during regeneration is given by calculating the area under the speed-time curve for regeneration period, D=4 Km Energy available on account of moving down the gradient a distance of 3 km =191.3 kwh Total energy available = 63.575+191.3= 254.875 kwh  

36. Example-4 A 2340 tonne train including loco proceeds down a gradient of 1 in 80 for 5 minutes during which its speed gets reduced from 60 kmph to 36 kmph by application of regenerative breaking. Find the energy returned to the lines if the tractive resistance is 5 kg/ tonne , rotational inertia 10 % and overall efficiency of the motor during regeneration is 70 %. Total energy available = 63.575+191.3= 254.875 kwh Energy returned to the line considering 75 % efficiency = 0.70 *254.875 = 178.4 kwh .