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  1. 1. READING MATERIAL FOR B.E STUDENTS OF RGPV AFFILIATED ENGINEERING COLLEGES PERSUING IN ELECTRICAL AND ELECTRONICS BRANCH Professor MD Dutt Addl General Manager (Retd) BHARAT HEAVY ELECTRICALS LIMITED Professor(retd) in EX Department Bansal Institute of Science and Technology KOKTA ANANAD NAGAR BHOPAL Presently Head of The Department ( EX) Shri Ram College Of Technology Thuakheda BHOPAL Sub Code Ex503 Subject Electrical Machine II UNIT II DC Machines II
  2. 2. RGPV Syllabus EX 503 ELECTRICAL MACHINES II UNIT I I DC MACHINES II Basic operation of DC motor, Torque equation, Operating characteristics of DC motors, Starting of DC motors 2 Point, 3 Point and 4 point starters. Speed control of DC motors. Losses and efficiency of DC machines, testing of DC machines , direct testing, Swinburne test, Hopkinson’s test, Application of DC machines. INDEX Topic Page No 1 Basic operation of DC motor 3 2 Torque equation 4,5 3 Operating characteristics of DC motors, 6,7,8,9 4 Starting of DC motors 2 Point 10,11 5 3 Point starters 12,11 6 4 point starters 13 7 Speed control of DC motors 14,15,16,17 8 Losses and efficiency of DC machines 17,18,19,20 9 Testing of DC machines 21,22 10 Direct testing 23 11 Swinburne test 23,24,25 13 Hopkinson’s test 26,27,28 14 Application of DC machines. 28 15 Formulas 29
  3. 3. Q1 Basic principle of operation of a D.C. motor? When a conductor carrying current is put in a magnetic field, a force is produced in it. Let us consider one such conductor placed in a slot of armature and it is acted upon the magnetic field from the north pole of the motor. By applying L.H.S it is found that the conductor has tendency to move to the L.H.S. Since the conductor is placed on the slot at circumference of rotor , The force acts in a tangential direction of rotor, Thus a torque is developed on the rotor Similar torques are produced on all the rotor conductors .Since the rotor is free to move, It starts rotating. BACK EMF When the motor of armature rotates, its conductors cut the magnetic flux, Therefore the e.m.f. rotation Er is induced in them. In case of a motor, the e.m.f is known as BACK E.M.F or counter e.m.f. The back e.m.f opposes the applied voltage , Since the back E.M.F is induced due to generating action it magnitude is given Eb = PNZ/60A TORQUE OF A DC MACHINE
  4. 4. When the machine is operating as a motor the torque is transferred to the shaft of the rotor and drives the mechanical load, The expression for the torque is same for the generator or motor. The Voltage equation of DC machine is V= E+IaRa Multiplying with Ia We get V Ia = EIa +Ia²Ra VIa = Electrical input to the armature Ia²Ra = copper loss in the armature We also know that Input = output + losses EIa = Electrical equivalent of gross mechanical power developed by armature τav = average electromagnetic torque developed by armature in Newton Mtrs (Nm) Mechanical power developed by the armature Pm = ώτav = 2Пņτav Pm = EIa = ώτav = 2Пņτav But E = nP ΦZ/A Therefore
  5. 5. (nP ΦZ/A) Ia =2Пņτav τav = (PZ/2П A ) X ΦIa This is called Torque equation For agiven machine the value of PZ A are constant Therefore PZ/2П A is also constant = k τav = k ΦIa EQUIVALENT DIAGRAM aaat fff RIEV RIV  
  6. 6. OPERATING CHARACTERISTICS OF DC MOTORS In both cases of shunt or separately excited D.C moter’s . the field is supplied from a constant voltage so that the field current is constant. It is graph between two dependent quantities. SPEED ARMATURE CURRENT CHARECTERISTICS:- In a shunt motor, Ish =V÷Rsh, if V is constant Rsh will also be constant, Hence the flux is constant but at no load the flux decreases slightly due to armature reaction. If the effect of armature Reaction is neglected the flux will remain constant. The motor speed is given by N= V-Ia RA Φ If Φ Is constant N is proportional to V-Ia RA This is the equation of straight line with a negative slope. That is the speed N of motor decreases linearly with increase of current. Since the IaRa at full load is very small compare to V the drop in speed from noload to full load is very small. For all practical purposes the shunt motor is taken a constant speed motor. TORQUE / ARMATURE CURRENT CHARECTERISTICS:- IF the flux Φ is constant in a shunt motor, the torque would increase linearly with armature current Ia, However for larger Ia , the net flux decreases due to the demagnetizing effect of armature reaction. In view of this , the torque current characteristics deviates from straight line as illustrated here below
  7. 7. τav = k ΦIa SPEED TORQUE CHARECTERISTICS::- The speed torque characteristics is also called the mechanical characteristics and under steady state conditions it can be obtained as follows ωm = Vt – Ia Ra KaΦ It is seen from the characteristics that with increase in torque the speed drops.
  8. 8. DC SERIES MOTOR SPEED CURRENT CHARECTERISTICS::- If the armature reaction and saturation is neglected, the main flux is directly proportional to the armature current Ia , therefore it can be written as Φ K Ia where K is constant ωm o = Vt/ Ka Φ = Vt/Ka K Ia Since ωm o is inversely proportional to Ia, the no load speed of series motor becomes dangerously due to small no load current . In view of this, the series motor must always be started and operate under load mechanically coupled with it. TORQUE CURRENT CHARECTERISTICS::- We know that Φ = KIa which shows that the torque is proportional to the Ia and therefore Torque current characteristics is PARABOLA SPEED TORQUE CHARECTERISTICS::- The speed verses Torque curve is HYPERBOLA. It is seen from the curve that the increase in Torque τe does not change the speed drop drastically
  9. 9. DC COMPOUND MOTORS:- SPEED CURRENT CHARECTERISTICS::- With increase in Ia and Φse increases. With increase in Ia the speed drops at faster rate in a cumulative compound motor. TORQUE CURRENT CHARECTERISTICS At no load the value of Ia = 0, Φse=0 and τe=0. As the armature current rises with load the shunt field Φsh remains almost constant but series field Φse rises, as a result motor torque τe SPEED TORQUE CHARECTERISTICS With the increase in motor torque τe , armature current rises, with this field flux Φse rises, Consequently speed drops in cumulatively compound DC motor.
  10. 10. STARTING OF DC MOTORS:- At the time of starting the motor speed is zero. Therefore back emf is also zero. Vt = Ea +Ia Ra Ea = 0 Vt = 0 + IaRa for shunt motor Vt = 0+ Ia (Ra +Rs) for series motor and compound motor With the rated voltage is applied the starting armature current is = Vt/Ra for shunt motor , = Vt/ Ra+Rs for series motor and compound motor Since the value of Ra and Rs is very small the motor draws large starting armature current from supply means. A 10Kw ,250V shunt motor having Ώ armature resistance 0.2 Ώ wiil draw 250/0.2 =1250A where as the rate current is 40A. Such heavy in rush of currents may result the following 1. Detrimental sparking at commutator 2. Damage of armature winding and detoriation of insulation due to overheating 3. High starting torque and quick acceleration which may damage the rotating parts of the rotor 4. Large dips in supply voltage. In view of this , the armature current must be limited to a value that can be commutated safely, by inserting a suitable external resistance in the armature circuit. As the motor accelerates the back EMF is generated in the armature and reduces the armature current to a small value. The external resistance inserted must be gradually reduced as the rotor accelerates. SHUNT AND COMPOUND STARTER’s A Primary function of starter is to limit starting current in the armature circuit during starting or accelerating time. The simplest type of starter is however modified to include few protective devices such as over current release, no volt release etc.
  11. 11. THREE POINT STARTERS FOR DC SHUNT MOTOR The handle H is kept at OFF position by spring S . For starting the motor , the handle H is moved manually and when makes contact with the resistance stud 1 it is in the START position. In position the field winding receives the full supply voltage., but the armature current is limited by the graded resistance R = ( R1+R2+R3+R4). The starter handle is then gradually moved from stud to stud , allowing the speed of the motor to build up until it reaches the RUN position. In this position (a) the motor attains the full speed , (b) The supply is directly across the both winding of motor(c) The resistance R is completely cut out. The handle H is held in RUN position by an electromagnet energized by a no volt Coil NVC. The no no volt trip coil is connected in series with field winding of the motor. In the event of switching off, or when the supply voltage falls below a predetermined value, or complete failure of supply while the motor is running. NVC is deenergized, This results in release of the handle. . which is then pulled back to the OFF position. By the action of spring. The current to the motor is cut off, and the motor is not started without resistance R in the armature circuit.. The NVC also provides protection against an open circuit in the field winding. The NVC is called no-volt or under voltage protection of motor. Without this protection, the supply voltage might be restored with the handle in the RUN position, Consequently full line voltage may be applied directly to the armature resulting in a very large current. The other protective device incorporated in the starter is overload protection. Overload protection is provide by overload trip coil OLC and the NVC. The overload coil is a small electromagnet. It carries the armature and for normal values of armature current the magnetic pull of OLC is insufficient to attract the strip P When the armature current exceeds the normal rated value ( that the motor is overloaded) P is attracted by the electromagnet of OLC and closes the
  12. 12. contact aa, Thus NVC is short circuited , This results in release of handle H which returns to OFF position and the motor supply is cut off. DRAW BACK OF THREE POINT STARTER The motor with large speed variation with armature voltage control suffers . To increase the speed of the motor the field is to be decreased, therefore the current through the shunt field is reduced. The field current may become very low. The very low field current may not develop sufficient electromagnetic hold to over come the force of the spring.
  13. 13. FOUR POINT STARTER The schematic connection diagram is given herewith. The basic difference in the circuit of Four Point starter compare to 3 point starter is that , The holding coil is removed from the shunt field circuit and is connected directly across the line with a current limiting resistance r in series. Such arrangements form three parallel circuits. 1) Armature, starting resistance and overload release. 2) A variable resistance and shunt field winding 3) Holding coil and current limiting resistance. With this arrangement, a change in field current for variation of speed of the motor, does not affect the current through the holding coil, because the two circuits are independent of each other. Now a days automatic push button starters are used. In such starters the ON push button is preset at the time of starting limiting the armature current. The resistors are gradually disconnected by an automatic controlling arrangement until full line voltage is available to the armature circuit. By the pressing of OFF button, the circuit is disconnected.
  14. 14. SPEED CONTROL OF DC MOTORS The speed of DC motor depends on the following equation:- N = ( V-IaRa)/ Φ From the above equation it is clear that the speed is dependent on supply voltage V armature resistance and field flux Φ which is produced by field current. These three factors are used for controlling speed of a DC motor. 1 Variation of armature resistance in the armature circuit. 2 Variation of field flux Φ 3 Variation of armature voltage. ARMATURE RESISTANCE CONTROL :-In this method a variable series resistor Re is connected in the armature circuit. In the case of shunt motor the field is directly connected to the supply and therefore flux Φ Is not affected by this. In the case of series motor the flux gets affected due to change in Ia and RA since Re resistor carries full load armature current it must be designed keeping into this consideration Disadvantages:- i) A large amount of power is wasted in external resistance Re ii) Control is limited to give speeds below the normal rated speed. iii) For a given value of Re the speed reduction is not constant but varies with motor load.
  15. 15. VARIATION OF FIELD FLUX Φ In a shunt motor this is done by connecting a variable resister Re in series with field winding. The resister Re is called the field regulator.. the Re reduces the field current by virtue of this the Φ reduces, decrease in flux causes increase in speed. Increase in flux causes decrease in speed. By this method the speed can be attained above normal rated speed. The variation in field current is done by two methods i) By connecting a variable resistor parallel to the field winding. This is called divertor method ii) The second method is tapped field control.
  16. 16. ADVANTAGES I) This is very easy and convenient method II) Since the field current Ish is very small , the power loss in shunt field is also very small ARMATURE VOLTAGE CONTROL Ward Leonard system of speed control is based on tis method. In this system M is the motor whose speed is to be controlled and G is the separately excite DC generator, The generator is driven by 3 Phase induction motor or synchronous motor. The combination of ac motor and DC generator is called MG Set. By changing generator field current the generated voltage is changed this voltage increases the speed of the motor. With the armature voltage control constant torque variable speed is obtained. ADVANTAGES OF WARD LEONARD DERIVES 1) Smooth control of DC motor on both directions is possible 2) It has inherent regenerative breaking capacity 3) When load is intermittent like rolling mills, IM is used with flywheel to take care of intermittent loading, in case of WARD LEONARD this is possible without flywheel.
  17. 17. 4) By using over excited Synchronous motor a drive the system power factor can be improved DISADVANTAGES OF WARD LEONARD DERIVES 1) Higher initial cost due to MG set 2) Larger size and weight, require more floor area 3) Frequent maintenance and produces more noise. 4) Lower efficiency and higher total losses. SOLID STATE CONTROL Rotating MG sets are now a days replaced with solid state convertors to control the speed of DC motors, The convertors are controlled rectifiers or choppers. In case of AC supply, controlled rectifiers are used to convert fixed ac supply voltage into variable D voltage. When the supply is D.C Choppers are used to obtain variable DC voltage from the fixed DC voltage supply.
  18. 18. LOSSES IN A DC MACHINES and efficiency Following are the losses in the DC machines 1) Electrical or copper loss ( I²R Losses) 2) Core losses or Iron losses 3) Brush Losses 4) Mechanical losses 5) Stray load losses ELECTRICAL LOSSES:- Windings having resistance consumes certain losses, these are termed as copper losses because mostly windings are made of copper. i) Armature copper loss Ia²Ra ( Ia is armature current) ii) Shunt field copper loss Ish²Rsh iii) Copper loss in the series field Ise²Rse iv) Copper loss in the interpole winding which are in series with armature Ia²Ri v) Compound machines both series field and shunt field copper losses are also there vi) Copper losses are there in the compensating winding CORE LOSES:- The core losses are the hysteresis losses and Eddy current losses. Since the machine usually operates at constant flux density and speed, these losses are almost constant. These losses are about 20% of Full load losses. BRUSH LOSES:- There is a power loss at the brush contact with commutator and the carbon brushes. This loss can me measured by the voltage drop at the brush contact and armature current. Pbd = Vbd Ia The voltage drop is more or less remains constant over a wide range of Ia and it is assumed 2V ( approx) MECHANICAL LOSSES:- The losses associated with mechanical effect are called mechanical losses. These consists of friction losses at bearing
  19. 19. and windage losses ( fan losses) . the fans are used to take away the heat produced due to I²R losses and iron losses inside the machine. STRAY LOAD LOSSES:- These are miscellaneous losses which are due to the following reasons:- 1) Distortion of flux due to armature reaction 2) Short circuit currents in the coils due to commutation. These losses are difficult to find out, However they are taken as 1 % of full load power output. EFFICIENCY InputPower Losses InputPower LossesInputPower InputPower OutputPower     1
  20. 20. Let us assume that the R = Total resistance I = Output current Ish = Current through the shunt field Ia armature current I +Ish V is the terminal voltage Power loss in the shunt field = V Ish Mechanical Losses = Friction losses at bearings+ friction losses at commutator + windage losses Stray losses Core losses mechanical losses and shunt field copper losses are considered as combined fixed losses. ή = Output / Input = VI / ( VI + Ia²Rat +Pk =VbdIa Ia = I + Ish Since Ish compare to I is very small we can consider Ia ≡ I ή = VI/ ( VI+ I²Rat +Vbd I +Pk)
  21. 21. LOAD for Maximum efficiency Ifl = Full load current at maximum efficiency Im = current at maximum efficiency Im ² Rat = Pk Im ² = Pk/ Rat Current at Maximum ή = F.L Current X( Pk/F.L Copper loss) ² TESTING OF DC MACHINES:- Testing of Dc machines can be done by following three methods:- 1) Direct testing 2) Swinburne’s Test 3) Hopkinson’s Test DIRECT TESTING :- This method is suitable for small machines. In direct testing method DC machine is subjected to rated load and the entire out put is wasted. The ration of out put power to the input power gives the efficiency. For DC generators the out put power is wasted in resistors, the out put voltage and current gives the out put power. For Dc motors break test is carried out . the belt tightening hand wheels H1 and H2 help in So adjusting the load on the pulley. S1 S2 are the spring balance. These spring balances are calibrated in Kg. The motor out put is given by =ω(S1 –S2)r X 9081 Watts
  22. 22. S1 S2 are the spring balance tight side and slack side readings, r is the effective radius of the pulley in meters = (½ outside pulley diameter + ½ belt thickness) ω is the ( 2Пņ) is the motor speed in rad/sec If Vt is the motor terminal voltage and Il is the line current, then power input to the motor VtIl watts The efficiency ή % = {ω ( S1-S2) 9.81 X100 }/ VtIl For series motor the break should be tight enough before motor is switched on Disadvantages 1) The spring balance are not steady 2) The friction torque , at particular setting of hand wheel H1 and H2 does not remain constant.
  23. 23. SWINBERN’s TEST The machine is run as a motor at rated voltage and speed. Connection diagram is as follows Let V = supply voltage I0 = No load current Ish = Shunt Field current Therefore No load armature current Ia0 = Io- Ish NO load input VIo No load power input supplies the following losses
  24. 24. i) Iron loss in the core ii) Friction losses at bearing and commutator iii) Windage losses iv) Armature cupper loss at no load When the machine is loaded the resistance of the armature and field resistance changes due to temperature rise and due to I²R losses. Let the resistance at the room temperature tOc is made by passing current through armature and field from a low voltage DC supply. Let the rise be 50 degree the hot resistance Rt1 = R0 + α0t1 R(t1=50) = R0{1 + α0 (t1+5050)} α0 = temperature coefficient of resistance at zero degree temperature Stray losses = iron loss +friction loss + windage loss = input at no load =VI0 – Pf –Pa0 = Ps Pc = no load input – no load armature copper loss Pc = Ps +Pf By knowing the constant losses of the machine its efficiency can be calculated at any other load. Let I be the load current Motor input VI Armature cupper loss = Ia²Ra +Pc ή = {VI – (Ia²Ra +Pc)}/ VI Efficiency when running as generator Ia = I + Ish
  25. 25. Out put = VI ή gen = VI/ { VI +( I+Ish) ²Ra +Pc} ADVANTAGES i) It is convenient and economical, since power requirement is very less ii) The efficiency at any load can be determined as the constant losses are known DISADVANTAGES i) This cannot be done on series machines. ii) No account is taken for iron loss from no load to full load. At full load due to the armature reaction iron losses increases.
  26. 26. HPKINSON’S TEST:- This test is also called 1) Regenerative test 2) Back to back test 3) Heat run test For conducting this test two identical machines are required, which are coupled mechanically and also connected electrically parallel. One of them runs as motor and other one runs as generator. When the machines run on full load the supply gives the losses of the both the machines. The switch S is kept open. The field current of motor is so adjusted by Rm to enable to run the system at rated speed. Now the Rg the field rheostat is so adjusted that the voltage across the generator armature is slightly higher than the bus bar voltage by 1 or 2V. When this is achieved the switch S is closed. Under this condition the generator is said to be floating. Any load can be thrown on by changing excitation of machines. Let V = Supply Voltage Il = Line current Im = Motor input current Ig = Output current of generator Iam = Motor armature current Iag = Generator armature current Ishm = Motor shunt field current Ishg = Generator Shunt field current Ra = Armature resistance Rshm = Motor shunt field resistance Rshg = Generator shunt field resistance Eg = Generated Induce voltage Em = Motor induced voltage ( Back EMF) Eg = V + IshgRa Em = V –IshmRa Therefore Eg> em Eg α ΦgN Em α ΦmN
  27. 27. Φg > Φm or Ishg > Ishm Thus the excitation current of generator shall always be greater than motor. Power input = V Il Total losses of both machines Armature copper loss of motor = Iam² Ra Armature copper loss of generator = Iag ²Ra Field copper loss of motor = Ishm² Rshm Field copper loss of generator = Ishg ² Rshg For constant losses Pc ( iron , windage are assumed equal Pc = Power drawn – armature and shunt copper losses of both machine. Since both machines are identical Total constant loss per machine =1⁄2 Pc The efficiency of generator Out put = V Iag Constant losses = 1⁄2 Pc Copper loss of generator = Iag ²Ra Copper loss og generator field = Ishg ² Rshg ή = VIag/ ( VIag + Iag ²Ra + Ishg ² Rshg + 1⁄2 Pc ) Efficiency as motor ή =Out put / In put = (Input – Losses )/ input Input V Im = ( V ( Iam + Ishm) Copper loss+ constant losses = Iam² Ra + Ishm² Rshm + 1⁄2 Pc ή = { V (Iam +Ishm) – (Iam² Ra + Ishm² Rshm + 1⁄2 Pc)} V (Iam +Ishm)
  28. 28. APPLICATION OF DC MACHINES:- Presently the DC generators are overtaken by AC to DC rectifiers which are static equipments. Thus DC generator are superseded by rectified ac supply. The main application of DC machines are follows SERIES MOTOR:- These motors are required where high starting torque is required and speed can vary like traction, cranes. SHUNT MOTOR :- These motors are used where constant speed is required and starting condition are not severe for example, Lifts, Fans, Blowers, Lathe etc. COMPOUND MOTORS These motors are used where high starting torque and fairly constant speed is required, presses, shears, conveyors, elevators, rolling machines. Small DC machines ( In fractional KW ) are used primarily as control device like tacho generator for speed sensing and servomotor for positioning and tracking.