16/04/2020 1 DC Machines Dr. Mohammed H. Haque School of Engineering
[email protected] EEET 3032 – Electrical Machines 1 Topics • Ideal electrical machines • Induced voltage and torque equations • Commutation and armature reaction • Construction and operating principle • Classification • Equivalent circuit, power flow diagram, losses and efficiency • Characteristics of various dc machines • Speed control of dc motors • Motor and load torque matching • Dynamics of dc machines EEET 3032/M. H. Haque/2019 2 1 2 16/04/2020 2 Electric Machines An electric machine converts electrical energy into mechanical energy or mechanical energy into electrical energy Generator: Converts mechanical energy into electrical energy Motor: Converts electrical energy into mechanical energy Electromechanical energy conversion EEET 3032/M. H. Haque/2019 3 In Electrical System: Primary quantities are voltage (V or E) and current (I) In Mechanical System: Primary quantities are torque (T) and speed (m or n) AC Machines: Electrical system is AC DC Machines: Electrical system is DC EEET 3032/M. H. Haque/2019 4 3 4 16/04/2020 3 Coupling Magnetic Field The coupling medium between the electrical and mechanical systems is magnetic field and is essential in all electromechanical energy conversion processes Coupling Magnetic Field Mechanical System Electrical System E and I T and m Coupling magnetic field between electrical & mechanical systems EEET 3032/M. H. Haque/2019 5 For an ideal (or lossless) machine Electrical energy = Mechanical energy Electrical power = Mechanical power Electrical power: Pe = EI (W) in DC circuit Mechanical power: Pm = Tm (W) For an ideal machine Pe = Pm EI = Tm Machine speed is usually measured (or given) in revolution per minute (rpm). However, in power calculation, machine angular velocity (in radian/sec) is used. The relationship between speed n (in rpm) and angular velocity m (in rad/sec) is nm = 60 2 That is m and n are linearly related and the proportionality constant is 2/60 EEET 3032/M. H. Haque/2019 6 5 6 16/04/2020 4 Electromagnetic Energy Conversion Two basic electromagnetic phenomena are: • A moving conductor in a magnetic field induces voltage. This is called generator action. • A current carrying conductor in a magnetic field produces force. This is called motor action. In all electric machines, both actions/effects are taken place simultaneously EEET 3032/M. H. Haque/2019 7 Induced Voltage The induced voltage or emf (e) in a moving conductor in the presence of a magnetic field is given by e = BLv (V) when B, L and v are mutually perpendicular B = magnetic flux density, T or Wb/m2 L =conductor length in magnetic field, m v = relative velocity between field and conductor, m/s The polarity or direction of induced voltage (e) can be determined by the Right Hand Rule (RHR) as shown in the figure Electric machines are designed in such a way that B, L and v are mutually perpendicular EEET 3032/M. H. Haque/2019 8 7 8 16/04/2020 5 A Simple Generator Consider a coil rotates in a magnetic field produced by two poles (N and S) as shown in the figure. Coil ends are connected to two rotating slip rings. Stationary brushes are placed on the rotating slip rings to extract the internal induced voltage. External load is connected between the stationary brushes. The resultant voltage appears between the slip rings A and B is alternating (but not DC). EEET 3032/M. H. Haque/2019 9 Generation of Unidirectional Voltage • Commutators are used to convert the internal AC induced voltage into unidirectional output voltage. • Commutators can be considered as mechanical rectifier. EEET 3032/M. H. Haque/2019 10 Commutator 9 10 16/04/2020 6 • Coil sides and commutators change position simultaneously because they are on the same structure • Connection between brushes and commutators changes whenever the polarity of the induced voltage is revered • Thus, the polarity of output (or load) voltage remains unchanged. The output voltage is unidirectional but pulsating as shown in the figure. It has high ripples. EEET 3032/M. H. Haque/2019 11 How to Reduce Ripples? Ripples can be reduced by using a large number of armature coils connected in series • First consider that only two coils A-B and C-D are placed at right angle and connected in series • The phase shift between EAB and ECD is 90 0 and the resultant voltage ER is EAB + ECD as shown in the following figure EEET 3032/M. H. Haque/2019 12 11 12 16/04/2020 7 Comparison of Output Voltage (a) 1-coil; (b) 2-coil; (c) 8-coil EEET 3032/M. H. Haque/2019 13 Construction of DC Machines Stator: Provides the physical support and magnetic poles Rotor: The main winding (where the voltage is induced) is placed in the rotor. The rotor of a DC machine is also called armature Three essential elements in DC machines are • Production of magnetic field or flux. Usually electromagnets are used • Rotating coils/conductors where the voltage is induced • Brush-commutator arrangement (to convert AC voltage to DC voltage) Field coil on a pole piece A complete armature EEET 3032/M. H. Haque/2019 14 13 14 16/04/2020 8 Field Windings and Representation There are two sets of field windings • Shunt field winding: Consists of a large number of turns with fine wire and carries less current (usually a few % of rated current). It has high resistance • Series field winding: Consists of less number of turns with heavy wire and carries large current (usually the load current). It has low resistance To produce magnetic field, it is not necessary to use both field windings simultaneously. Representation of Field Windings A field winding is represented by a coil as shown in schematic diagram. It consists of resistance (R) and inductance (L). For steady state analysis, L is ignored and thus only R is considered. Schematic diagram Electrical equivalent circuit R EEET 3032/M. H. Haque/2019 15 Representation of Armature Armature consists of a large number of conductors where voltage is induced. For steady state analysis, the armature inductance (L) is ignored and thus its equivalent circuit consists of induced voltage E and the resistance Ra. Note that E is not constant. E R a Brush Schematic diagram Electrical equivalent circuit EEET 3032/M. H. Haque/2019 16 15 16 16/04/2020 9 Classification of DC Machines The characteristic of a DC machine depends on the field winding(s) used (series or shunt) to produce flux. Classification of DC machines is based on mutual connection between the armature and field windings. (a) Separately excited The shunt field winding is connected to a separate DC source Vf. The series field winding is not used. V t E Shunt field V f (b) Self-excited The armature induced voltage/current is used to excite the field circuit(s). A separate DC source is not required. Self-excited machines can further be classified into three categories (i) Shunt, (ii) Series and (iii) Compound EEET 3032/M. H. Haque/2019 17 (i) Shunt: The shunt field winding is connected in parallel with the armature. The armature terminal voltage is the same as the shunt field voltage. Series field winding is not used in a shunt machine. VtE S h u n t fi e ld (ii) Series: The series field winding is connected in series with the armature. Thus, the armature current is the same as the series field current. Shunt field winding is not used in a series machine V t E Series field (iii) Compound: In compound machines, both series and shunt field windings are used. There are two possible connections of compound machines (1) short-shunt and (2) long-shunt. Vt E Series field S h u n t fi e ld Vt E Series field S h u n t fi e ld (1) Short-shunt (2) long-shunt EEET 3032/M. H. Haque/2019 18 17 18 16/04/2020 10 DC Generators EMF or Induced Voltage Equation The emf or induced voltage in the armature of a DC machine is directly proportional to the flux produced by the poles and the angular velocity m of the rotor or armature. The induced voltage (E) can be expressed as E = kφωm Volt Here k is a constant and it depends on the rotor or armature of the machine. Induced voltage (E) depends on the following three factors: • Flux produced by the poles • Angular velocity m of the rotor or armature • Constant k representing the size of the machine EEET 3032/M. H. Haque/2019 19 This is called voltage equation or EMF equation Magnetization Curve The flux is usually produced by passing current through the field winding(s). The flux induces voltage in the armature when it rotates. The variation of induced voltage against the field winding current (at constant speed) is called magnetization curve. The procedure of generating the magnetization curve is as follows • Run the machine at a constant speed (rated speed) as a separately excited generator without any load (or open circuit condition) • Measure the terminal voltage Vt for different values of fields current If. Note that at no load, the armature current is zero and thus the internal voltage E is the same as the terminal voltage Vt VtE If Rf Rrh Vf EEET 3032/M. H. Haque/2019 20 • Plot the induced voltage E against the field current If and is called the magnetization curve of the machine • The magnetization curve is also known as Open Circuit Characteristic (OCC) or no- load (NL) characteristic 19 20 16/04/2020 11 Magnetization Curve Magnetization curve has three distinct regions: • Linear region (E is directly proportional to If, i.e. E = k1If) • Transition region (relationship between E & If is nonlinear) • Saturation region (E constant, independent of If) Magnetization curve or OCC or no-load characteristic Residual Voltage The induced voltage E for If = 0 is called residual voltage. The residual voltage is the induced due to residual flux in the poles and is usually very small (about 5% of rated value) EEET 3032/M. H. Haque/2019 21 Effects of Speed and Flux on Voltage Voltage equation: E = km The ratio of the voltage (for two different conditions, say ‘1’ and ‘2’) is given by = = 2 1 2 1 2 1 2 1 2 1 2 1 n n E E E E m m This is a general equation and valid for all DC machines (a) For a Shunt Machine Flux is produced by the shunt field current If Thus, = f(If) In the linear region: = kf If (kf is a constant). Thus, the ratio of flux is the same as ratio of If VtE S h u n t fi e ld = = 2 1 2 1 2 1 2 1 2 1 2 1 n n I I E E or I I E E f f m m f f EEET 3032/M. H. Haque/2019 22 21 22 16/04/2020 12 (b) For a series machine Flux is produced by the series field current Is which is the same as armature current Ia Thus, = f(Is) or = f(Ia) In the linear region: = ksIs or = ksIa (ks is a constant). Thus, the ratio of flux is the same as ratio of Ia V t E Series field = = 2 1 2 1 2 1 2 1 2 1 2 1 n n I I E E or I I E E a a m m a a Example: The induced voltage E in a separately excited DC generator is found to be 150 V at 1450 rpm when the field current is adjusted to 2.8 A. Find the induced voltage E (a) for a field current of 3.0 A at 1600 rpm. (b) for a field current of 2.2 A at 1300 rpm. Assume that the field flux is directly proportional to the field current If. EEET 3032/M. H. Haque/2019 23 Solution: Induced voltage equation E = km ; Here = kfIf (kf = constant) Therefore, E= k(kfIf)m Volts Thus, the voltage ratio is given by = = 2 1 2 1 2 1 2 1 2 1 n n I I I I E E f f m m f f Here E1 = 150 V, If1 = 2.8 A and n1 = 1450 rpm VE E 34.177 1600 1450 0.3 8.2150 2 2 = = VE E 66.105 1300 1450 2.2 8.2150 2 2 = = (a) (b) EEET 3032/M. H. Haque/2019 24 23 24 16/04/2020 13 Voltage Build-up in a Shunt Generator In a shunt generator, the armature induced voltage is used to excite the field circuit. Only under certain conditions, the generator will build up voltage. The process of building up voltage in a shunt generator is described below • Assume that the field switch SW is initially opened and the armature is driven at a constant speed • A small voltage Ear will induce due to residual pole flux • When the switch SW is closed, the field circuit is connected to the armature circuit. A small field current If1 will flow due to Ear and that will produce an additional field flux • If the additional field flux is in the same direction of the residual flux, the resultant flux and hence the induced voltage will increase (to Ea1) • Higher voltage causes higher If and . This in turns induces more voltage. • The above voltage build-up process continues until the voltage reaches the steady state operating point ‘P’ VtE SWIf Rf EEET 3032/M. H. Haque/2019 25 At no-load, E = (Rf + Ra)If RfIf (because Rf >> Ra). Thus, E ≈ Vt = RfIf and it is represented by a straight line as shown in the figure. Since the armature and field are in parallel, they must have the same voltage and which can be obtained from the point of intersection of E vs. If curve and E ≈ Vt = RfIf line. Effect of speed on induced voltage For a given field current If, the induced voltage E is directly proportional to speed n or m (as shown) n1 > n2 > n3 > n4 > n5 EEET 3032/M. H. Haque/2019 26 25 26 16/04/2020 14 Adjustment of No-Load Voltage The no-load voltage can be adjusted by changing the point of intersection ‘P’. That is, by changing the slope of the line or field resistance. This can be achieved by adding a variable resistance in series with the shunt field circuit and is called field rheostat. The total field circuit resistance Rft is the sum of field winding resistance Rf and field rheostat resistance Rrh → (Rft = Rf + Rrh). The no-load voltage decreases when the field circuit resistance is increased Critical Field Circuit Resistance (Rfcr) When the field circuit resistance line coincides with the linear part of the magnetization curve, the voltage become unstable and the corresponding field circuit resistance is called the critical field circuit resistance Rfcr VtE If Rf Rrh EEET 3032/M. H. Haque/2019 27 When Rft < Rfcr : Voltage will build-up (normal operation) Rft > Rfcr : Voltage will not build-up (only up to Vt4) Rft = Rfcr : Unstable situation Conditions to be Satisfied to Build-up Voltage in Shunt Generators • Existence of residual flux • Flux produced by the field current must be in the same direction of the residual flux • Total field circuit resistance must be less than the critical field circuit resistance Rfcr EEET 3032/M. H. Haque/2019 28 27 28 16/04/2020 15 In a DC Generator The internal voltage E is always greater than the terminal voltage Vt. The difference between E and Vt is called armature voltage drop (AVD). Thus (E – Vt) = AVD; or E = (Vt + AVD) At no-load, the load current is zero and thus the AVD is zero or negligible. Note that in a DC generator, E is the source of electrical power. For analysis, write appropriate equations for various connections of the machine (using KVL and KCL) V t E Shunt field V f In a Separately Excited Generator The field circuit is isolated from the armature circuit. The field current If can be adjusted independently. The induced voltage E depends on If and which can directly be obtained from the magnetization curve EEET 3032/M. H. Haque/2019 29 Load Characteristic When the load current is increased, the load voltage or terminal voltage decreases because of AVD. Variation of load voltage (VL) or terminal voltage (Vt) against load current (IL) is called load characteristic Causes of voltage drop • Armature circuit resistance drop or AVD • Armature reaction voltage drop • Cumulative effects (for self excited generators) Armature Reaction • When the armature delivers current, it produces another flux called armature flux • The interaction of field flux and armature flux is called armature reaction (AR) • The AR reduces the resultant flux and hence E • The AR can be minimized/cancelled by using compensating windings connected in series with the armature EEET 3032/M. H. Haque/2019 30 29 30 16/04/2020 16 Voltage Regulation (VR) The voltage regulation (VR) is defined as %100%100 − = − = t t FL FLNL V VE V VV VR VNL : No-load voltage VFL = Full load voltage At no-load, Vt ≈ E (because of no or negligible AVD) Example A DC shunt generator has a field circuit resistance of 50 . The internal induced voltage E is 274.6 V when it delivers an output power of 50 kW at a terminal voltage of 250 V. Neglect the armature reaction. (a) Determine the armature resistance (b) Determine the internal voltage when it delivers an output power of 30 kW at a terminal voltage of 250 V. EEET 3032/M. H. Haque/2019 31 Solution (a) Here Rf = 50 , Vt = 250 V, E = 274.6 V, and Pout = 50 kW Load current IL = Pout/Vt = 50,000/250 =200 A Field current If = Vt/Rf = 250/50 = 5 A Armature current Ia = (IL + If) = (200 + 5) = 205 A But E = (Vt + IaRa) 274.6 = (250 + 205Ra) Ra = 0.12 (b) Load current IL = Pout/Vt = 30,000/250 =120 A Field current If = Vt/Rf = 250/50 = 5 A Armature current Ia = (IL + If) = (120 + 5) = 125 A But E = (Vt + IaRa)= (250 + 1250.12) = 265 V VtE ILIa If Rf Load Vt IL IaIf Rf Ra E Load EEET 3032/M. H. Haque/2019 32 31 32 16/04/2020 17 Losses and Efficiency The efficiency () of a generator is the ratio of output power to input power %100%100 powerInput powerOutput == in out P P But input power = output power + losses Pin = (Pout + Ploss) Pin = Input power to the generator is mechanical power and is given by = Tinm W. Pout = Output power of the generator is electrical power and is given by = VLIL W Note: For a separately excited machine, field circuit electrical input power is VfIf W Losses There are two major components of losses: Mechanical or rotational losses (Prot) It consists of bearing and brush frictions, and windage losses. The iron losses are also considered in the rotational losses. The rotational losses Prot are usually considered as constant losses. Electrical or copper losses (Pcu) Copper losses are I2R loss in various windings (armature, series field and shunt field). The electrical losses depend on current and hence load. Thus, the electrical losses are considered as variable losses. EEET 3032/M. H. Haque/2019 33 Power Flow Diagram The input power to a generator is mechanical power and the output power is electrical power. The portion of mechanical power which is converted into electrical power is called developed power (Pd). It is the internal power of the machine. Efficiency: %100= in out P P Pin = (Pd + Prot) Pd = (Pout + Pcu) EEET 3032/M. H. Haque/2019 34 33 34 16/04/2020 18 Example: A shunt generator operates at an efficiency of 85% when delivers 50 A of load current at 110 V. The rotational losses are 480 W and shunt field resistance is 65 . Calculate the armature resistance. Solution: Output power Pout = Vt IL = 110 50 = 5500 watts Input power Pin = Pout/ = 5500/0.85 = 6470.6 watts Total losses = (Pin – Pout) = (6470.6 – 5500) = 970.6 watts Electrical losses = (total losses – rotational losses) = (970.6 – 480) = 490.6 watts Shunt field current If = Vt/Rf = 110/65 =1.69 A Armature current Ia = (IL + If) = (50 + 1.69) = 51.69 A VtE ILIa If Rf Load Vt IL IaIf Rf Ra E Load Therefore Electrical losses = ffaa RIRI 22 + 490.6 = 51.692Ra + 1.69 265 Ra = 0.114 EEET 3032/M. H. Haque/2019 35 Construction wise, a DC motor is similar to a DC generator. The same machine can be operated either as a generator or as a motor. However the direction of power flow is opposite. In generators: Mechanical power is supplied to the armature. E > Vt; power flows from E to Vt; E act as a source; E = (Vt + AVD); terminal voltage Vt is the same as load voltage VL In motors: Armature delivers mechanical power. E < Vt; power flows from Vt to E; E act as a load; E = (Vt – AVD); or Vt = (E + AVD); terminal voltage Vt is the same as input voltage Vin Motor induced voltage (E) is usually called back emf DC Motors Pm Generator E Ra Ia Motor Pm E Ra Ia EEET 3032/M. H. Haque/2019 36 35 36 16/04/2020 19 Developed Force The force developed in a current carrying conductor in the presence of a magnetic field is given by Force F = BLI Netwon The direction of the force can be determined by the Left Hand Rule (LHR) Applications of DC Motors DC motors are widely used for accurate control of speed, position, torque, etc. Typical applications are: • Vehicles (starter, wiper, fan and other accessories) • Machine tools, robots, etc. • Applications that require high starting torque Classification of DC Motors Classification of DC motors is exactly the same as described for DC generatorsEEET 3032/M. H. Haque/2019 37 Torque Equation For an ideal or lossless machine, the power at the electrical port is the same as at the mechanical port. Thus, EIa = Tdm a m am m a d Ik IkEI T === )( ad IkT = This is called torque equation Expression of induced voltage or back emf (E) of a motor is exactly the same as that for a generator ( E = km) EEET 3032/M. H. Haque/2019 38 37 38 16/04/2020 20 Speed Equation Consider a separately excited motor is running at a speed of n (or m) while taking an armature current of Ia KVL in the armature circuit Vin = (E + AVD) = (km + AVD) Pm Vin Iin Rf If Vf k AVDVin m − = This is called speed equationThus, Speed Regulation (SR) Speed regulation (SR) of a motor is very similar to the voltage regulation of a generator %100%100 − = − = mFL mFLmNL FL FLNL n nn SR NL No-load and FL Full-load Speed adjustment should not be considered in the regulation EEET 3032/M. H. Haque/2019 39 Power Flow Diagram The input power to a motor is electrical while the output power is mechanical. The portion of electrical power which is converted into mechanical power is called developed power. This is the internal power of the machine. Efficiency: %100= in out P P Pin = (Pd + electrical losses) Pd = (Pout + rotational losses) EEET 3032/M. H. Haque/2019 40 For a separately excited machine, field circuit electrical input power is VfIf W 39 40 16/04/2020 21 How to Find Rotational Losses? The technique of determining the rotational losses is given below • Run the motor at no-load (NL) condition and measure the input voltage Vin and the input current Iin • At no-load, output power Pout is zero and thus the input power Pin (= VinIin) is the total losses in the machine Pin = (rotational losses + copper losses) Pin = Prot + Pcu Calculate Pcu and subtract it from Pin to get the rotational losses Prot = (Pin – Pcu) • From power flow diagram (Pin – Pcu) = Pd = EIa = Prot (at no load or when Pout = 0) The above calculations is to be done at no-load condition to get Prot. EEET 3032/M. H. Haque/2019 41 EEET 3032/M. H. Haque/2019 42 Example: A 250-V shunt motor has Ra = 0.25 and Rf = 125 . At no-load, Iin = 5 A and n = 1200 rpm. At full-load, Iin = 52 A. (a) Find the rotational losses (b) Also find speed, speed regulation, developed torque and efficiency of the motor at full load. Solution (a) At no-load Here If = Vin/Rf = 250/125 = 2 A Armature current Ia1 = (Iin – If) = (5 - 2) = 3 A Induced voltage E1 = (Vin – Ia1Ra) = (250 – 3 0.25) = 249.25 V Rotational losses Prot = Pd = EIa = 249.253 = 747.75 W Speed n1 = 1200 rpm (given) Pm Vin Iin Rf If Ia Iin Rf Vin Ia If Ra E (b) At full-load Ia2 = (Iin – If) = (52 - 2) = 50 A E2 = (Vin – Ia2Ra) = (250 – 50 0.25) = 237.5 V Speed n2 is unknown 41 42 16/04/2020 22 EEET 3032/M. H. Haque/2019 43 But rpmn nn n E E 1143 1200 1 5.237 25.249 2 22 1 2 1 2 1 === (Note that for the same field current, the ratio of flux is 1) Speed Regulation %5%100 1143 11431200 = − Developed power Pd = EIa = 237.550 = 11,875 watts Nm P T m d d 2.99 )60/1143(2 875,11 === Pout = (Pd – Prot) = (11,875 – 747.75) = 11,127.25 Watts Pin = VinIin = 25052 = 13,000 Watts %6.85%100 000,13 25.127,11 %100 === in out P P Torque-Speed Characteristic EEET 3032/M. H. Haque/2019 44 The torque-speed characteristic of a DC motor can be derived from the speed and torque equations Td = kIa This is called torque eqn. (or Td–Ia characteristic) k AVDVin m − = This is called speed eqn. (or m–Ia characteristic) For a Shunt Motor AVD = IaRa; flux depends on shunt field current If; = f (If); = kfIf (in the linear region). If the input voltage Vin and field circuit resistance Rf are constant, the field current If and hence flux is also constant Pm Vin Iin Rf If Ia 43 44 16/04/2020 23 EEET 3032/M. H. Haque/2019 45 For a Series Motor AVD = Ia(Ra+Rs); flux depends on series field current which is the same as the armature current Ia; = f (Ia); = ksIa (in the linear region). The flux depends on the armature current which in turn depends on load. Thus, of a series motor is not constant. Vin Iin Pm Rs To get the torque speed characteristic, plug in the value of Ia from torque equation into the AVD term of the speed equation Example: A 400 V series motor has Ra = 0.35 and Rs = 0.15 . When Iin = 44 A, n = 650 rpm. (a) Find the motor speed n for Iin = 36 A. (b) Obtain the speed-current characteristic and find the speed n for Iin = 36 A from the characteristic. Assume that the motor operates in the linear part of magnetization curve. EEET 3032/M. H. Haque/2019 46 Solution When Iin = Ia = 44 A E1 = Vin – Ia (Ra + Rs) = 400 – 44 (0.35 +0.15) = 378 V n1 = 650 rpm m1 = 2n1/60 = 68.07 rad/sec (a) When Iin = Ia = 36 A E2 = Vin – Ia (Ra + Rs) = 400 – 36 (0.35 +0.15) = 382 V But E = km For a series motor = f(Ia) = ksIa Therefore, E = k(ksIa)m Vin Iin Pm Rs Iin Vin Ra E Rs rpmn nn n I I E E a a 85.802 650 36 44 382 378 2 22 1 2 1 2 1 = = = (b) Speed-current characteristic E = k(ksIa)m 378 = kks4468.07 kks = 0.1262 Speed equation is as saainin m Ikk RRIV k AVDV )( +− = − = sec/96.3 6.3169 1262.0 )15.035.0(400 rad I I I a m a a m −= +− = When Ia = 36 A, m = (3169.6/36) – 3.96 = 84.08 rad/sec or, n = 802.8 rpm (same as obtained in part (a)). 45 46 16/04/2020 24 EEET 3032/M. H. Haque/2019 47 Example: A 240-V DC shunt motor has Ra = 0.15 and Rf = 120 . At full load, Iin = 100 A and n = 1150 rpm. Derive the torque-current, speed-current and torque-speed characteristics of the motor. Also find Td and n when Iin = 50 A. Solution At full load current of Iin = 100 A n = 1150 rpm m = 120.43 rad/sec If = Vin/Rf = 2 A Ia = (Iin – If ) = (100 – 2) = 98 A E = (Vin–IaRa)=(240 - 980.15) = 225.3 V But E = km k = E/m = 225.3/120.43 = 1.87 For a shunt machine, field current If and hence is constant Pm Vin Iin Rf If Ia Iin Rf Vin Ia If Ra E Torque-current characteristic Td = kIa Td = 1.87Ia sec/)0802.034.128( 87.1 15.0240 radI I k IRV am aaain m −= − = − = Speed-current characteristic EEET 3032/M. H. Haque/2019 48 Torque-speed characteristic 87.1 15.0)87.1/(240)/( − = − = − = dadinaainm T k RkTV k RIV m = (128.34 – 0.0429Td) rad/sec At 50 A of input current Iin = 50 A; Ia = (Iin – If) = 48 A Td = 1.87Ia = 1.8748 = 89.76 Nm m = (128.34–0.0802Ia) = (128.34–0.080248) =124.5 rad/sec n = 1189 rpm 47 48 16/04/2020 25 Speed Control EEET 3032/M. H. Haque/2019 49 The speed equation of a DC motor is k AVDVin m − = Three possible methods of speed control are: (a) Armature resistance or AVD control; (b) Field or flux control; (c) Input voltage control (a) Armature Resistance or AVD Control Add a variable external resistance Rext in series with the armature circuit to change the armature voltage drop Ia(Ra+Rext) and hence to change the speed. Here AVD = Ia(Ra+Rext) Main Features • Speed can only be decreased • Ineffective at no load (when Ia 0) • Higher losses and lower efficiency Vin Iin Rf If Ia Rext EEET 3032/M. H. Haque/2019 50 (b) Field Control Add a field rheostat Rrh in series with the shunt field circuit to reduce the field current If (or ) and hence to increase the speed. Vin Iin Rf IfIa Rrh frh in f RR V I + = ; = f(If) Main Features • Speed can only be increased • Speed can be controlled at all load levels (c) Input Voltage Control Connect a variable voltage source to the armature and a fixed voltage source to the field circuit (separately excited) Main Features • Speed can only be decreased (because Vin < Vrated) • Speed can be controlled at all load levels Iin Rf If Vf Variable voltage 49 50 16/04/2020 26 EEET 3032/M. H. Haque/2019 51 In many cases, it is required to control the speed over a wide range (below and above the normal speed). For such a case, a combination of field and input voltage controls can be used. Iin Rf If Vf Rrh Variable voltage Combination of field and voltage controls Control of dc motor by Power Electronics EEET 3032/M. H. Haque/2019 52 Example: A 240-V shunt motor has Ra = 0.05 and Rf = 60 . At no-load, Iin = 7 A and n = 1120 rpm. At full load, Iin = 46 A. The motor operates in the linear part of magnetization curve. Find the motor speed at full load input current (a) without any extra resistance; (b) with an extra field resistance of 20 , and (c) with an extra armature resistance of 1 and no extra field resistance. Solution At no load If = 240/60 = 4 A; Ia = (Iin – If) = (7 – 4) = 3 A E = (Vin–IaRa) = (240 – 30.05) = 239.85 V Pm Vin Iin Rf If Ia 2 1 2 1 2 1 2 1 2 1 n n I I I I E E f f m m f f == But 51 52 16/04/2020 27 EEET 3032/M. H. Haque/2019 53 Pm Vin Iin Rf If Ia (a) At full load (without any extra resistance) If = 240/60 = 4 A; Ia = (Iin – If) = (46 – 4) = 42 A E = (Vin – IaRa) = (240 – 420.05) = 237.9 V rpmn n 1111 1120 4 4 9.237 85.239 2 2 == (b) At full load (with 20 extra resistance in field circuit) If = 240/(60+20) = 3 A; Ia = (Iin – If) = (46 – 3) = 43 A E = (Vin–IaRa) = (240 – 430.05) = 237.85 V Vin Iin Rf IfIa Rrh rpmn n 1481 1120 3 4 85.237 85.239 2 2 == (c) At full load (with 1.0 extra resistance in armature circuit) Here If = 4 A and Ia = 42 A E = 240 – 42(0.05+1.0) = 195.9 V rpmn n 915 1120 4 4 9.195 85.239 2 2 == Vin Iin Rf If Ia Rext Starting of DC Motors EEET 3032/M. H. Haque/2019 54 KVL in the armature circuit is Vin =(E + IaRa) a in a R EV I − = At starting; n = 0 E = 0 (because E = km) Ia (start) = Vin / Ra Starting current is extremely high Pm Vin Iin Rf If Ia Consequences • Armature is subject to a severe mechanical shock • Blow fuses 53 54 16/04/2020 28 Starting of DC Motors EEET 3032/M. H. Haque/2019 55 How to overcome high current? Add an external variable resistance in series with the armature and is called starter startera in a RR V I + =start)( Power Electronics Power electronics devices can also be used to reduce the starting current. This can be achieved by applying reduced voltage to armature at starting. Most of the modern dc motors use this technique. Motor and Load Torque Matching EEET 3032/M. H. Haque/2019 56 Torque-speed characteristic of a motor and a load can be expressed as Motor: Td = f1(ωm) (1) Load: TL = f2(ωm) (2) Decelerating Torque Accelerating Torque Speed T o rq u e Motor torque Load torque Equilibrium point For steady state operation (when Prot = 0) Motor developed torque = Load torque Td = TL Steady state operating condition (m & T) can be obtained • By solving eqns. (1) and (2) simultaneously • From the point of intersection of eqns. (1) & (2) When Td > TL : motor accelerates When Td < TL : motor decelerates Torque-speed characteristic of motor and load 55 56 16/04/2020 29 EEET 3032/M. H. Haque/2019 57 Example: A 120 V DC shunt motor has Ra = 0.1 , Rf = 120 and Prot = 0. When Iin = 119.4 A, the motor runs at 1100 rpm. (a) Find the developed torque (b) Establish the torque-speed characteristic (c) Estimate the no-load speed and starting torque (d) Find the speed when the load torque is (e) Calculate Td, Ia and for the load condition of part (d) mLT 2778.6= Solution: (a) Here Ra = 0.1 , Rf = 120 , and n = 1100 rpm when Iin = 119.4 A m = 2n/60 = 115.2 rad/sec If = Vin/Rf = 120/120 = 1A Ia = (Iin – If) = (119.4 - 1) = 118.4 A E = (Vin – IaRa) = (120 – 118.40.1) = 108.16 V Pd = EIa = 108.16118.4 = 12.8 kW But Pd = Tdm Td = Pd/m = 111.2 Nm Also E = km k = E/m = 0.939 Pm Vin Iin Rf If Ia Iin Rf Vin Ia If Ra E EEET 3032/M. H. Haque/2019 58 (b) Speed equation: d ainadinaain m T k R k V k RkTV k RIV 2)( )/( −= − = − = ddm TT 1134.08.127 939.0 1.0 939.0 120 2 −=−= Therefore, the torque-speed characteristic of the motor is: NmTor radT md dm )82.81127(, sec/)1134.08.127( −= −= (1) (c) At NL: Td = 0; m = 127.8 rad/sec (or n =1220.4 rpm) At starting: n = 0 (or m = 0); Td = 1127 Nm (d) The torque-speed characteristic of the load is mLT 2778.6= (2) 57 58 16/04/2020 30 EEET 3032/M. H. Haque/2019 59 At steady state: Td = TL (1127 – 8.82m) m2778.6= 016.16327062.2562 =+− mm Therefore, m = 136.08 rad/sec or 119.98 rad/sec. Consider m = 119.98 rad/sec n = 1145.7 rpm (m = 136.08 rad/sec provides negative value of Td and thus not feasible) (e) From eqn. (1): Td = (1127 – 8.82119.98) = 68.78 Nm Also, Td = kIa Ia = Td/ k = 68.78/0.939 = 73.25 A Pd = Tdm = 68.78119.98 = 8252.2 W = Pout (note Prot = 0) If = Vin/Rf = 120/120 = 1A Iin = (Ia + If) = (73.25 + 1) = 74.25 A Pin =Vin Iin = 12074.25 = 8910 watts %62.92%100 8910 5.8252 Input Output === in out P P Efficiency = Dynamics of Armature Voltage Controlled DC Motors EEET 3032/M. H. Haque/2019 60 Consider the armature of a separately excited dc motor is connected to a voltage source through a switch ‘S’ as shown in the following figure. Sudden application of armature voltage would cause a transient whose duration is governed by the parameters of the machine and the load. Assume that the field circuit has already reached the steady state condition and its current remains constant. Under this case, the variation of armature current can be represented by the following first order differential equation )( )( )( 1 tk dt tdi LtiRV m a aaas ++= Note that the back emf E = (kφ)ωm = k1ωm (for a constant value of field current If or φ) 59 60 16/04/2020 31 EEET 3032/M. H. Haque/2019 61 The torque equation of the motor under transient condition can be expressed by the following first order differential equation. dt td JtDTtikT m mLad )( )()( 1 ++== Here TL = load torque D = viscous friction coefficient of rotating members J = inertia of the rotating members k1 = kφ The above two equations can be written in the following form saam a a Lam m VtiRtk Ldt tdi TtiktD Jdt td +−−= −+−= )()( 1)( )()( 1)( 1 1 EEET 3032/M. H. Haque/2019 62 − + −− − = s L a a m a a a a m V T L J ti t L R L k J k J D dt tdi dt td 1 0 0 1 )( )( )( )( 1 1 The above equation can be rearrange in matrix form as In general form: BUAXX += Here X is the state variable X = [ωm(t) ia(t)] T and U is the input variables U = [TL Vs] T Solutions of the above equation would provide the variation of motor speed ωm(t) and armature current ia(t) as function of time. 61 62 16/04/2020 32 Block Diagram of a Separately Excited DC Motor EEET 3032/M. H. Haque/2019 63 Dynamic equations: saam a a Lam m VtiRtk Ldt tdi TtiktD Jdt td +−−= −+−= )()( 1)( )()( 1)( 1 1 Laplace Transform of the above equations with zero initial condition )()()(1)( )()()( 1 )( 1 1 sVsIRsk L ssI sTsIksD J ss saam a a Lamm +−−= −+−= EEET 3032/M. H. Haque/2019 64 The above equations can be represented by the following block diagram in Matlab/Simulink 63 64 16/04/2020 33 EEET 3032/M. H. Haque/2019 65 + - + - k1 k1 Vs(s) TL(s) Ωr(s)Td(s)Ia(s) s 1 s 1 aL 1 J 1 aR D - - The combined block diagram is EEET 3032/M. H. Haque/2019 66 Typical variation of armature current under no-load condition 65 66 16/04/2020 34 EEET 3032/M. H. Haque/2019 67 Typical variation of speed under no-load condition Acknowledgement The figures and diagrams of this lecture note are taken from the following sources: • B. S. Guru and H. R. Hiziroglu, “Electric Machines and Transformers”, 3rd Edition, Oxford University Press, 2001. • G. B. Shrestha and M. H. Haque, “AC Circuits and Machines”, Pearson Prentice Hall, 2006. • P. C. Sen, “Principles of Electric Machines and Power Electronics”, 2nd Edition, John Wiley & Sons, 1997. • P. F. Ryff, D. Platnick and J. A. Karnas, “Electrical Machines and Transformers”, Prentice Hall, 1987. EEET 3032/M. H. Haque/2019 68 67 68
