STEADY STATE OPERATION OF DC MACHINES

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Electric DC machines, as indeed any other type of electric machine, can be used to either produce electric energy from the input mechanical energy, or to convert electric energy into output mechanical energy. These two possible operating regimes are called generation and motoring.

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ENGNG2024 Electrical Engineering STEADY STATE OPERATION OF DC MACHINES 1. INTRODUCTION Electric DC machines, as indeed any other type of electric machine, can be used to either produce electric energy from the input mechanical energy, or to convert electric energy into output mechanical energy. These two possible operating regimes are called generation and motoring. As already mentioned, DC machines used to be in the past the major source of DC power. In order to produce electric power DC machines were operated as generators. Nowadays, however, use of DC generators is becoming more and more rare. DC power is obtained instead by means of power electronic converters. The remaining applications of DC machines are today restricted to motoring. In this operating regime a DC machine is operated as a DC motor: it consumes DC power, while delivering at its shaft mechanical power. The shaft drives a certain load, that is characterised with the load torque. A DC machine consists of stationary part, called stator, and rotating part, called rotor. Both stator and rotor are equipped with one winding. Stator winding is supplied from a DC voltage source and the role of this winding is to produce magnetic flux in the air gap of the machine. This flux is stationary in space. Rotor winding is again supplied from a DC voltage source (for motoring): DC supply is connected to the rotor winding through a special assembly, that is composed of brushes and commutator. Brushes are stationary, while commutator is fixed to the rotor and hence rotates together with the rotor. This assembly enables supply of electric power from stationary power supply to rotating winding on rotor. Principle of operation of this assembly is illustrated by means of Fig. 1. Rotor winding is shown in a very simplified manner, as consisting of just one coil, connected to two segments of the commutator. Motoring action is assumed and the current is therefore delivered to the rotor winding through the stationary brushes and rotating commutator. Two positions of the rotor winding are shown in Fig. 1. Terminal current (current brought to the brushes) and the winding current are illustrated in Fig. 2. As can be seen from these two figures, current inside the rotor winding is reversed (commutated) after each half-revolution of the rotor. Current inside the rotor is therefore AC, while the terminal current is DC. Frequency of the current inside the rotor winding equals frequency of rotation. θ θ A B ia = Ia ia ia = Ia ia i=Ia i=−Ia B A Fig. 1 - Current reversal (commutation) in rotor coil by means of the commutator.  E Levi, 2001 1
ENGNG2024 Electrical Engineering ia Ia i Ia 0 θ = ωt 0 π 2π θ=ωt − Ia Fig. 2 - Terminal current and current through rotor coil. Note that such a situation regarding frequencies in the two windings is the only possible one that satisfies the condition of average torque existence. Since the stator winding is supplied with pure DC current of zero frequency, the machine can develop an average torque if and only if the rotor winding frequency equals the frequency of rotation. This means that it is not possible to realise an electric machine with DC currents flowing in both stator and rotor windings. Such a situation would result in the possibility of developing an average torque at zero speed only. At zero speed however converted power equals zero and therefore such a machine could not do the process of electromechanical energy conversion. Let the stator winding, which is called excitation or field winding as well, be supplied with constant DC voltage equal to Vf . Current that flows through this winding is in steady- state operation determined with If = Vf /Rf (1) Flux produced in the air gap of the machine is, neglecting saturation of the magnetic circuit, proportional to this current. Hence Φf = c1 If (2) It has to be emphasised that the excitation winding can be replaced with permanent magnets. Many of the modern DC motors rely on permanent magnet excitation and in such a case there is not any possibility of changing the excitation flux, since it is fixed by the magnet properties. Rotor rotates at certain angular speed ω [rad/s], in the stationary flux produced by the excitation winding. Thus, according to the basic law of electro-magnetic induction, e = B l v (where v is linear speed, B is flux density and l is length of the conductor), there is an induced electromotive force (emf) in the rotor winding E = c2 Φf ω = k If ω (3) Induced emf is proportional to the flux (i.e., excitation current) and to the speed of rotation, through a constant determined with constructional features of the machine. Speed of rotation ω is the so-called angular speed of rotation in [rad/s] and it is correlated with the speed n in revolutions per minute [rpm] through ω = ( 2π 60) n . As rotor winding (often called armature winding) is supplied from a DC voltage source and as the winding is of certain resistance, then the voltage equilibrium equation for the armature winding (index a) is Va = Ra Ia + E (4) According to the basic law of electromagnetic force creation, if a conductor that carries current I moves in the flux of flux density B, then an electromagnetic force F = B I l acts on the conductor. As rotor winding rotates in the flux density produced by the excitation winding, an electromagnetic torque is produced, equal to Te = c2 Φf Ia = k If Ia (5) This electromagnetic torque is the reason why the rotor rotates. If there is load connected to the shaft of the machine, then this load opposes rotation with its torque, called load torque TL.  E Levi, 2001 2
ENGNG2024 Electrical Engineering In any steady-state condition electromagnetic torque and load torque are equal and act in the opposite direction. While electromagnetic torque acts in the direction of rotation, load torque acts in the opposite direction, as illustrated in Fig. 3 for two possible directions of rotation, and expressed with the following relation: Te = TL (6) Angular speed of rotation in [rad/s] is correlated with speed in [rpm] through the already given scaling factor n [rpm] = (60/2 π) ω [rad/s] (7) Te Te ω ω rotor rotor TL TL forward motoring reverse motoring Fig. 3 - Directions of electromagnetic torque, load torque and speed in motoring. Note that constant of proportionality in expressions for induced emf and electromagnetic torque, k, is the same as long as speed in expression for emf is given in rad/s. Input and output powers of the motor are electrical and mechanical powers, respectively, and are given with the following expressions: Pin = VaIa + Vf If (8) Pout = Te ω (9) The difference of the two powers represents power loss in the machine, which includes mechanical loss, iron loss and loss in the windings. In what follows mechanical loss and iron loss will be frequently neglected, but copper loss will always be accounted for. Power loss in the machine and the efficiency are given with Ploss = Pin - Pout (10) η = Pout / Pin (11) Equations (1)-(11) completely describe steady-state operation of a DC motor. Standard data that are given for a DC motor on its nameplate are so-called rated values of output power, armature voltage and current, excitation voltage and current, and speed of rotation in rpm. Rated values will always be identified in examples with index n. Note that, as rated power is always output power, rated power for a motor is mechanical output power. Load torque that an electric motor drives can be either speed independent (say, crane lifting a load) or speed dependent. Frequently met speed dependent load torques are either linearly proportional to speed or proportional to speed squared (pumps, compressors, fans, etc.). These three types of load torques are illustrated in Fig. 4. If the load torque is constant, then it follows from (5) that product of excitation current and armature current is constant as well. This is the simplest case and all the examples will assume that a DC motor drives a constant speed independent load torque. Depending on how excitation winding and armature winding are supplied from DC voltage sources, various types of DC motors may be identified. Two of the types that are nowadays in wide application are separately excited DC motor and series excited DC motor. The third one, rarely used nowadays, is the shunt excited DC machine. In a separately excited  E Levi, 2001 3
ENGNG2024 Electrical Engineering DC machine, two voltage DC sources are used. Shunt and series excited DC machines require only one voltage source, with the stator and rotor winding connected in parallel and in series, respectively. These three DC machine types are considered in more detail in what follows. TL = kω TL TL = const. TL = kω2 ω Fig. 4 - Illustration of various types of load torques. 2. SEPARATELY EXCITED DC MOTOR Excitation winding and armature winding of a separately excited DC motor are supplied from two independent DC power sources. A separately excited DC motor is described in steady-state operation with the following set of equations, that essentially only summarise again (1)-(6) Va = E + Ra I a Vf = Rf I f Φ f = c1I f (12) E = c2 Φ f ω = kI f ω Te = c2 Φ f I a = kI f I a TL = Te Equivalent circuit of a separately excited DC motor is shown in Fig. 5. Mechanical, speed- torque characteristic of the motor (speed against electromagnetic torque) can be derived from (12) as follows: Va = E + Ra Ia = kI f ω + Ra Ia Þ ( ) I a = Va − kI f ω Ra ( ) Te = kI f I a = kI f Va − kI f ω Ra Va Ra ω= − Te kI f (kI f )2 (13) If both voltages have rated values then I fn = V fn R f Van Ra ω= − Te kI fn (kI fn )2 Speed-torque characteristic, with rated voltages applied to both windings, is the so-called natural operating characteristic and is shown in Fig. 5 with bold trace. Equation (13) enables examination of available speed control methods for a separately excited DC motor, that are beyond the scope here. As can be seen from the speed torque characteristic, speed slightly decreases, in a linear manner, as load is increased. The highest value of the operating speed is under no-load conditions (i.e., load torque equal to zero). Rated operating point is indicated in  E Levi, 2001 4
ENGNG2024 Electrical Engineering Fig. 5 as well. Speed drop in a separately excited DC machine from no-load condition to full (rated) load conditions is typically a few tens of rpms. Example 1: A separately excited DC motor has the following rated data: 500 V, 100 A, 1000 rpm. Armature resistance is 0.5 Ω. Excitation flux is constant and equal to rated. Calculate rated torque and rated power of the motor and evaluate efficiency of the motor in the rated operation if power loss in the excitation winding is 5 kW. Solution: As rated voltage and rated armature current are known, then from (12) one can calculate induced electro-motive force in rated operating conditions: an = E n + R a I an E n = V an − Ra I an = 500 − 0.5x100 = 450 V Þ 2π En E n = kI fn ω n = kI fn nn Þ kI fn = = 30 x 450 / (1000π ) = 4.3 60 2π nn 60 Rated torque is then Ten = kI fn I an = 4.3 x100 = 430 Nm Rated power is mechanical (output)power, 2π Pn = Ten ω n = 430 x x1000 = 45030 W 60 Input power is the power delivered to the armature winding (500 V times 100 A) plus the power delivered to the excitation winding (5 kW). The efficiency in rated operation is ratio of output to input power. Hence η n = Pn / Pin = 45030 / (50000 + 5000) = 45.03 / 55 = 0.819 Example 2: Separately excited DC motor, 230 V, armature resistance = 0.2 Ω, operates under no-load conditions at 1200 rpm. Under rated conditions armature current is 40 A. Find the rated speed and rated electromagnetic torque of the motor. Excitation flux is constant and rated. Solution: Speed is in this example known for no-load operating conditions. If there is not load connected to the shaft of the motor, then load torque is zero. From (12) it follows that electromagnetic torque is zero as well. As electromagnetic torque is proportional to the armature current, then zero torque indicates that armature current is zero. No-load point is denoted with index ‘0’. Hence E 0 = Van − Ra I a 0 = Van 2π E0 V E 0 = kI fn n0 Þ kI fn == an = 30x 230 / (1200π ) = 183 . 60 2π 2π n0 n0 60 60 Then for rated operating conditions E n = Van − Ra I an = 230 − 0.2 x 40 = 222 V 2π En E n = kI fn nn Þ nn = = 30x 222 / (183π ) = 1150 rpm . 60 2π kI fn 60 Ten = kI fn I an = 183x 40 = 73.2 Nm .  E Levi, 2001 5
ENGNG2024 Electrical Engineering Example 3: A separately excited DC motor, whose rated data are 440 V, 120 A, 970 rpm, armature resistance = 0.16 Ω, is loaded with such a load torque that the armature current is 40 A. Find the speed and torque of the motor for this operating condition. Excitation flux is constant and equal to rated. Solution: Rated data for voltage and current apply to the armature. When excitation flux is constant, as the case is here, excitation winding data are not needed and are hence not given. The motor in this question operates in an operating point other than rated. However, initial calculations always have to deal with rated operating point, in order to find necessary values for subsequent calculations. For rated operating point Van = E n + Ra I an Þ E n = Van − Ra I an = 440 − 0.16 x120 = 420.8 V 2π En E n = kI fnω n = kI fn nn Þ kI fn = = 30 x 420.8 / (970π ) = 4.14 60 2π nn 60 Ten = kI fn I an = 4.14 x120 = 497 Nm 2π Pn = Tenω n = 497 x x 970 = 50484 W 60 In new operating point E1 = Van − Ra I a1 = 440 − 0.16 x 40 = 433.6 V 2π E1 E1 = kI fnω 1 = kI fn n1 Þ n1 = = 30x 433.6 / (4.14π ) = 1000.13 rpm 60 2π kI fn 60 Te1 = kI fn I a1 = 4.14 x 40 = 165.6 Nm 2π P1 = Te1ω 1 = 165.6 x x1000.13 = 17343.8 W 60 The two operating points are illustrated in Figure 5. Ra Ia Va = E + Ra Ia Rf Vf = Rf If Va E If E = k If ω Vf Te = k If Ia ω ω1 ωn Te1 Ten Te = TL Fig. 5 - Equivalent circuit and natural speed torque curve of a separately excited DC motor.  E Levi, 2001 6
ENGNG2024 Electrical Engineering Example 4: A separately excited DC motor has the following rated data: 500 V, 100 A, 1000 rpm. Armature resistance is 1 Ω. Excitation flux is constant and equal to rated. a) Calculate rated torque and rated power of the motor. b) Evaluate efficiency of the motor in rated operation if power loss in the excitation winding is 5 kW. c) The motor drives a load such that the armature current is 40 A. Calculate torque and speed for this operating regime. Determine efficiency of the motor in this operating point. d) Plot speed - torque curve and denote the two calculated points. Solution: Note that parts a) and b) are the ‘exam’ version of the Example 1, with minor changes and additions! a) As rated voltage and rated armature current are known, then from (12) one can calculate induced electro-motive force in rated operating conditions: V an = E n + R a I anÞE n = V an − R a I an = 500 − 1x100 = 400 V 2π En E n = kI fn ω n = kI fn nn Þ kI fn = = 30 x 400 /(1000π ) = 3.82 60 2π nn 60 Rated torque is then Ten = kI fn I an = 3.82 x100 = 382 Nm Rated power is mechanical (output) power, 2π Pn = Ten ω n = 382 x x1000 = 40 kW 60 b) Input power is the power delivered to the armature winding (500 V times 100 A) plus the power delivered to the excitation winding (5 kW). The efficiency in rated operation is ratio of output to input power. Hence η n = Pn / Pin = 40000 /(50000 + 5000) = 40 / 55 = 0.73 = 73% c) The load torque is now smaller, since current is only 40 A. As torque is directly proportional to the armature current through the constant kIfn (which is 3.82) then the new torque is 40 A/100 A times the rated torque, i.e. 152.8 Nm. Further E = V an − R a I a = 500 − 1x 40 = 460 V 2π E = kI fn n n = 60 E /(2πkI fn ) = 30 x 460 /(πx3.82) = 1150 rpm Þ 60 Output and input power are, respectively 2π Pout = Te ω = 152.8 1150 = 18.4 kW 60 Pin = V an I a + 5000 = 500 x 40 + 5000 = 25 kW η = 18.4 / 25 = 0.736 = 73.6% d) The two operating points are as indicated in Fig. 5. 3. SHUNT EXCITED DC MOTOR A shunt excited DC motor is the motor where the excitation winding and the armature winding are connected in parallel and supplied from the same voltage source. The equivalent circuit is illustrated in Fig. 6. As long as the supply voltage is constant and rated (this is the assumption valid here; voltage variation is a mean of speed control, that is beyond the scope of interest at present), all the equations given for a separately excited DC motor remain valid. It is  E Levi, 2001 7
ENGNG2024 Electrical Engineering only necessary to replace individual voltages for the excitation and the armature winding in (12)-(13) with one and the same value V of the supply voltage. The torque-speed curve of the shunt motor is therefore the same as the one for a separately excited DC motor and remains to be as given in Fig. 5. Ra Ia V Rf E If Fig. 6 – Equivalent circuit of a shunt excited DC machine. Example 5: A shunt excited DC motor, whose rated data are 440 V, 122 A, 970 rpm, armature resistance = 0.16 Ω, excitation winding resistance = 220 Ω, is loaded with such a load torque that the terminal current is 42 A. Find the speed and torque of the motor for this operating condition. Solution: Note that in the case of the shunt motor the rated current is the terminal current. That is , the given rated current is the sum of the rated armature and rated excitation winding current. Since the rated voltage and excitation winding resistance are known, the rated excitation current is 440/220 = 2 A and is the same regardless of the motor loading. Thus the rated armature current is 122 – 2 = 120 A, and the armature current for the new operating point is 42 – 2 = 40 A. Note that the rated motor data and the armature currents are now the same as in the Example 3 for a separately excited motor. Hence the solution is the same as well, with rated armature voltage symbol being replaced with the rated voltage symbol. V n = E n + R a I an Þ E n = V n − R a I an = 440 − 0.16 x120 = 420.8 V 2π En E n = kI fn ω n = kI fn nn Þ kI fn = = 30 x 420.8 /(970π ) = 4.14 60 2π nn 60 Ten = kI fn I an = 4.14 x120 = 497 Nm 2π Pn = Ten ω n = 497 x x970 = 50484 W 60 In new operating point E1 = V n − R a I a1 = 440 − 0.16 x 40 = 433.6 V 2π E1 E1 = kI fn ω 1 = kI fn n1 Þ n1 = = 30 x 433.6 /(4.14π ) = 1000.13 rpm 60 2π kI fn 60 Te1 = kI fn I a1 = 4.14 x 40 = 165.6 Nm 2π P1 = Te1ω 1 = 165.6 x x1000.13 = 17343.8 W 60 The two operating points remain to be as illustrated in Figure 5.  E Levi, 2001 8
ENGNG2024 Electrical Engineering 4. SERIES EXCITED DC MOTOR In a series excited DC motor, as the name suggests, excitation winding and armature winding are connected in series and fed from a single DC power source. Equivalent circuit of a series excited DC motor is illustrated in Fig. 7. Due to series connection of the two windings, equations (1)-(6) now become ( V = E + Ra + R f I ) Ia = I f = I E = kI f ω = kIω (14) Te = kI f I a = kI 2 Te = TL As excitation winding current is one and the same as armature current, then operation of a series motor with variable load torque results in variable flux in the machine. Electromagnetic torque is therefore proportional to the current squared, while emf is proportional to the product of current and speed. Mechanical torque-speed characteristic is derived using the same procedure as for a separately excited DC motor: ( ) V = Ra + R f I + E = Ra + R f I + kIω ( ) Þ I= V ( Ra + R f ) + kω kV 2 Te = kI 2 = [( R ) ] 2 a + R f + kω (15) For rated voltage supply kVn 2 Te = [( R ) ] 2 a + R f + kω Torque-speed curve is shown in Fig. 7 as well, with bold trace (rated operating point is indicated in the Figure 7). It substantially differs from the curve valid for a separately excited motor. Small variation of the torque produces large variation in operating speed. Note that series DC motor must not be ever allowed to run unloaded. As follows from (15), operation with zero electromagnetic torque results when speed approaches infinity. This means that, if a series motor is allowed to run without load, its rotor speed will reach such a high speed that the motor will eventually disintegrate. Rf Ra I Te V Vf Va E Ten Vn ωn ω Fig. 7 - Equivalent circuit and torque-speed characteristics of a series excited DC motor.  E Levi, 2001 9
ENGNG2024 Electrical Engineering It should be noted that all the calculations regarding operation of a series DC motor remain the same as in the case of a separately excited DC motor as long as operation with the constant load torque is considered. However, if the load torque is speed dependent, then change of load torque implies quadratic change of armature current, while in the case of the separately excited motor the change is linear. Example 6: A series DC motor has armature and excitation winding resistances of 0.2 Ω and 0.1 Ω, respectively. Rated motor date are 1000 rpm, 40 A and 450 V. Calculate rated torque, rated power and efficiency in rated operation. Solution: From rated data E n = Vn − ( R a + R f ) I n = 450 − (0.1 + 0.2) x 40 = 438 V E n = kI n ω n Þ k = En ( I n πnn / 30) = 0.10456 2 2 Ten = kI n = 0.10456 x 40 = 167.3 Nm Pn = Ten ω n = 167.3xπx1000 / 30 = 17520 W Pin = V n I n = 450 x 40 = 18000 W η n = Pn / Pin = 17520 / 18000 = 0.9733 Difference between input and output power is 480 W and this must 2 equal loss on resistances of the two windings: (0.1 + 0.2)x40 = 480 W. Example 7: A series DC motor operates under rated conditions with 161.2 Nm torque, 1000 rpm speed, 41 A current and 420 V voltage. Resistance of excitation and armature winding is 0.2 Ω. Find the speed and current of the motor if the torque is now 87 Nm. Solution: Note that this example does not involve speed control and serves the purpose of illustrating calculations for the case when load torque changes with speed. From the given data En = Vn − RI n = 420 − 0.2x 41 = 4118 V . Te1 I2 Ten = kIn = 161.2 Nm 2 Te1 = kI1 = 87 Nm 2 Þ = 87 / 161.2 = 12 Ten In I1 = I n Te1 Ten = 41 87 / 161.2 = 30.12 A E1 = Vn − RI1 = 420 − 0.2x 30.12 = 413.98 V E1 nI En = kI nω n E1 = kI1ω 1 Þ = 11 En nn I n E1I n 413.98x 41 n1 = nn = 1000 = 1368.4 rpm E n I1 4118x 30.12 . Example 8: A series DC motor, 220 V, 1500 rpm, 270 A has combined resistance of armature and field winding of 0.11 Ohms. The motor drives a load that is characterised with constant load torque, equal to the motor rated torque. Find the rated torque and rated power of the motor.  E Levi, 2001 10
ENGNG2024 Electrical Engineering Solution: E n = Vn − ( Ra + R f ) I n = 220 − 0.11x 270 = 190 V E n = kI nωn Þ kI n = E n ( πnn / 30) = 1.21 2 Ten = kI n = 1.21x 270 = 326.6 Nm Pn = Tenωn = 326.6 xπx1500 / 30 = 51.3 kW Pin = Vn I n = 220 x 270 = 59.4 kW η n = Pn / Pin = 51.3 / 59.4 = 0.86 5. TUTORIAL QUESTIONS Q1. A series DC motor operates under rated conditions with 161.2 Nm torque, 1000 rpm speed, 41 A current and 420 V voltage. Combined resistance of excitation and armature winding is 0.2 Ω. a) Sketch the equivalent circuit of the series DC motor, derive its torque-speed characteristic and show graphically toque-speed curve. b) Calculate rated power of the motor and its efficiency in rated operation. c) Find the speed and current of the motor when the load torque is 87 Nm. Determine output power and efficiency in this operating point. Q2. A series DC motor, 420 V, 41 A, 1000 rpm, has armature and excitation winding resistances of 0.12 Ω and 0.08 Ω, respectively. a) Calculate rated torque and rated power of the motor. b) The motor drives a load whose torque is proportional to the speed of rotation squared, TL = 0.015 ω2. Determine operating speed, current and torque of the motor for this load torque. Q3. a) Sketch equivalent circuits of (i) separately and (ii) series excited DC motors and write the basic steady-state equations for each type. b) A separately excited DC motor has the following rated data: 500 V, 100 A, 1000 rpm. Armature resistance is 0.5 Ω. Excitation flux is constant and equal to rated. i) Calculate rated torque and rated power of the motor. ii) The motor drives a load such that the armature current is 40 A. Calculate torque and speed for this operating regime. iii) Plot speed - torque curve and denote the two calculated points.  E Levi, 2001 11
ENGNG2024 Electrical Engineering Q4. a) Sketch equivalent circuits of (i) separately and (ii) series excited DC motors and write the basic steady-state equations for each type. b) Explain the role of commutator in DC machines using simple representation of rotor winding with just one turn and sketch the waveform of current in the turn assuming that current at rotor terminals is constant DC. c) A separately excited DC motor has the following rated data: 400 V, 100 A, 950 rpm. Armature resistance is 1 Ω. Excitation flux is constant and equal to rated. Mechanical losses and iron core losses can be neglected and armature voltage is constant and equal to rated. Calculate rated torque and rated power of the machine whose data are given. If the motor now drives a load such that the armature current is 50 A, calculate speed of rotation, torque and output power for this operating regime. Plot to scale torque- speed curve and denote the two calculated points. Q5. a) A series DC motor has armature resistance of 0.12 ohms and excitation winding resistance of 0.08 ohms. When the motor operates at 1000 rpm the current is 41 A and the torque is 176 Nm. The motor is supplied from a constant 420 V DC source. Calculate the current and the speed when the motor operates with torque equal to 87 Nm. b) Sketch equivalent circuits of (i) separately and (ii) series excited DC motors and write the basic steady-state equations for each type. c) Explain the role of commutator in DC machines using simple representation of rotor winding with just one turn and sketch the waveform of current in the turn assuming that current at rotor terminals is constant DC. Q6. A 220-V dc shunt motor draws 10-A at 1800-rpm. The armature circuit resistance is 0.2-Ω and the field winding resistance is 440-Ω. a) Calculate the torque developed by the motor under the above conditions. b) If the field current is unchanged, determine the speed and line current when the motor produces a torque of 20-Nm. c) Derive an expression representing the motor speed [rpm]-torque [Nm] characteristic and then determine the machine no-load speed.  E Levi, 2001 12