EE 261 Asynchronous and DC Machines PDF

EE 261 ASYNCHRONOUS AND DC MACHINES Ing. Dr. Francis Boafo Effah Senior Lecturer Department of Electrical & Electronic Eng. Faculty of Computer & Electrical Eng. College of Engineering Room BKB 13, Bamfo Kwakye Building Email: [email protected] 1...

EE 261 ASYNCHRONOUS AND DC MACHINES Ing. Dr. Francis Boafo Effah Senior Lecturer Department of Electrical & Electronic Eng. Faculty of Computer & Electrical Eng. College of Engineering Room BKB 13, Bamfo Kwakye Building Email: [email protected] 1  UNIT 3 ASYNCHRONOUS (INDUCTION) MACHINES 2  AC Machines - Rotating Field Principle MMF of Single and Three-phase windings Rotating Field Distributed windings 3 AC Machines – Principle of Operation (1) In AC machines, the production of electromagnetic torque is still a result of current and flux density distribution in the air gap of the motor in the same way as in a DC machine. The difference is mainly due to how the rotating field is produced and the terminal currents and voltages. In its most basic form, an AC machine: Is supplied by sinusoidally varying (in time) currents and voltages Has sinusoidally distributed (in space) winding configuration Has sinusoidally varying airgap field (in time and space) 4 AC Machines – Principle of Operation (2) There is no need for a commutator as the space- varying fields occur naturally due to the nature of the winding distribution and supplied voltage and current waveforms. We will look at AC-Induction Motors within this course. The stator windings of these machines are wound and supplied to generate a rotating sinusoidal field. We will look at winding configuration first. 5  Winding MMF due to one coil Assume the coil carries a time-varying current 𝑖𝑎 𝑡 , then: ∞ 2𝑁 ∙ 𝑖𝑎 𝑡 cos(𝑛𝜃) 𝑀𝑀𝐹 𝜃 = 𝑁 𝜃 ∙ 𝑖𝑎 𝑡 = ෍ 𝜋 𝑛 𝑛=2𝑘+1 If 𝑖𝑎 𝑡 is sinusoidal, neglecting higher harmonics of 𝑁 𝜃 , we obtain: 𝑀𝑀𝐹𝑎 𝑡, 𝜃 = 𝑁𝑎 𝜃 ∙ 𝑖𝑎 𝑡 = 𝐼 sin(𝜔𝑡) ∙ 𝑁 cos(𝜃) 𝑁𝐼 𝑀𝑀𝐹𝑎 𝑡, 𝜃 = sin 𝜃 + 𝜔𝑡 − sin(𝜃 − 𝜔𝑡) 2 6  Three-phase winding Consider a set of three-phase currents: 𝑖𝑎 𝑡 = 𝐼 sin(𝜔𝑡) 2𝜋 𝑖𝑏 𝑡 = 𝐼 sin 𝜔𝑡 − 3 2𝜋 𝑖𝑐 𝑡 = 𝐼 sin 𝜔𝑡 + 3 Assuming a sinusoidal winding distribution: 𝑁𝑎 𝜃 = 𝑁 cos(𝜃) 2𝜋 𝑁𝑏 𝜃 = 𝑁 cos 𝜃 − 3 4𝜋 𝑁𝑐 𝜃 = 𝑁 cos 𝜃 − 7 3  Three-phase winding MMF Now the MMF generated by each winding (MMF=NI) 𝑀𝑀𝐹𝑎 𝑡, 𝜃 = 𝑁𝑎 𝜃 ∙ 𝑖𝑎 𝑡 = 𝐼 sin(𝜔𝑡) ∙ 𝑁 cos(𝜃) 𝑁𝐼 𝑀𝑀𝐹𝑎 𝑡, 𝜃 = sin 𝜃 + 𝜔𝑡 − sin(𝜃 − 𝜔𝑡) 2 2𝜋 2𝜋 𝑀𝑀𝐹𝑏 𝑡, 𝜃 = 𝑁𝑏 𝜃 ∙ 𝑖𝑏 𝑡 = 𝐼 sin 𝜔𝑡 − ∙ 𝑁 cos 𝜃 − 3 3 𝑁𝐼 4𝜋 𝑀𝑀𝐹𝑏 𝑡, 𝜃 = sin 𝜃 + 𝜔𝑡 − − sin(𝜃 − 𝜔𝑡) 2 3 4𝜋 4𝜋 𝑀𝑀𝐹𝑐 𝑡, 𝜃 = 𝑁𝑐 𝜃 ∙ 𝑖𝑐 𝑡 = 𝐼 sin 𝜔𝑡 − ∙ 𝑁 cos 𝜃 − 3 3 𝑀𝑀𝐹𝑐 𝑡, 𝜃 = 𝑁𝐼 sin 𝜃 + 𝜔𝑡 − 2𝜋 − sin(𝜃 − 𝜔𝑡) 8 2 3  Three-phase winding MMF Total MMF generated is: 𝑀𝑀𝐹𝑇 𝑡, 𝜃 = 𝑀𝑀𝐹𝑎 𝑡, 𝜃 + 𝑀𝑀𝐹𝑏 𝑡, 𝜃 + 𝑀𝑀𝐹𝑐 𝑡, 𝜃 3𝑁𝐼 𝑀𝑀𝐹𝑇 𝑡, 𝜃 = sin(𝜔𝑡 − 𝜃) 2 9  Air gap Flux and MMF Note that the field shifts in space by the same amount as the phase angle in the current waveforms. Thus, for a 2 –pole field, the mechanical spatial speed is the same as the supply frequency. 10  Synchronous Speed – Number of poles In general, for a p-pole-pair motor, the mechanical speed of the rotating field is related to the electrical frequency by: 𝜔𝑒 𝜔𝑠 = 𝑝 Poles p 𝝎𝒔 [𝒓𝒂𝒅/𝒔] 𝒏𝒔 [𝒓𝒆𝒗/𝒔] 𝑵𝒔 [𝒓𝒆𝒗/𝒎𝒊𝒏] For: 2 1 314 50 3000 𝜔𝑒 = 2 ∙ 𝜋 ∙ 𝑓𝑒 = 2 ∙ 𝜋 ∙ 50 4 2 157 25 1500 𝑁𝑠 = 𝜔𝑒 ∙ 60 6 3 105 16.7 1000 𝑝 2∙𝜋 8 4 78.5 12.5 750 10 5 62.8 10 600 11 Winding – MMF space harmonics The resultant spatial waveform has some harmonics, mainly 5th, 7th, 11th, 13th, 17th, and 19th. These rotate at different speeds. Triplen harmonics cancel out. 12  Distributed windings AC windings are usually distributed to improve the MMF quality Fractional pitch windings are often adopted to reduce/cancel some MMF harmonics ∞ 𝑁𝑎 𝜃 = ෍ 𝑁𝑛 cos 𝑛𝜃 𝑛=2𝑘+1 2𝑁𝑝ℎ 𝑁𝑛 = ∙ 𝐾𝑤𝑑𝑛 ∙ 𝐾𝑤𝑓𝑝𝑛 𝑛∙𝜋 13  Distributed winding - example Example of a distributed stator winding of an Induction Machine. 14  Induction Machine Characteristics Principle of Operation Advantages of Induction Machines (IMs) Electrical Equivalent Circuit 15  IM - characteristics Characterised by: No saliency Having a stator and rotor winding Stator winding is connected to an AC supply The rotor winding is short-circuited. Two forms of machine dependent on rotor: Squirrel cage The rotor is short-circuited on itself. Slip ring The rotor has electrical sliding connections The short circuit, or adding extra resistance, to the rotor is made external to the machine. 16  IM - characteristics Squirrel Cage Induction Motor 17  IM - characteristics Three-phase stator winding 18  IM - characteristics The rotor shows single bars short-circuited by “End Rings.” 19  IM - characteristics Similar squirrel cage rotor to the previous slide showing bars just below the surface of the rotor laminations 20  IM - characteristics Stator windings of previous motor 21  IM - characteristics Slip ring rotor, note skewed rotor as for the cage machine 22  IM - characteristics Slip rings from a 250 kW, 8-pole machine 23  Principle of Operation (1) A balanced 3-phase supply applied to the stator windings generates a sinusoidal distributed rotating field. As we have already seen, the fundamental field will rotate in the air gap at a synchronous speed of: 𝜔𝑒 𝜔𝑠 = 𝑝 where 𝑝 is the number of pole pairs The rotor windings will see a field rotating at a frequency: 𝜔′ = 𝜔𝑠 − 𝜔𝑟 The rotor is short-circuited, so the current will flow at 𝜔′. 24  Principle of Operation (2) In turn, these currents will produce a field rotating at 𝜔𝑠 relative to the rotor. Thus, the rotor currents will produce a field rotating at: 𝜔 ′ + 𝜔𝑟 = 𝜔𝑠 No torque at synchronous speed as there will not be any current induced in the rotor winding. This is because the rotor conductors are not moving RELATIVE to the stator field. Consequently, they have no current induced in them at this speed. The torque always acts in a direction to move the speed towards synchronous. 25  Advantages of IM (1) Self-starting capability No special starting equipment Speed reversal ‘plugging’ by reversing phase sequence to stator (swapping over any two of the phase connections) The squirrel cage machine has no rotor electrical connections Slip ring machine can use external resistance to increase starting torque and decrease the starting current (shorted out when full speed is reached) 26  Advantages of IM (2) Advantages of cage IM It's cheaper and more robust. Slightly higher efficiency and power factor. Explosion-proof, since the absence of slip-rings and brushes eliminates the risk of sparking. Advantages of slip-ring IM The starting torque is much higher, and the starting current is much lower. The speed can be varied by means of an external rotor resistor. 27  Steady State Operation (1) In induction machines, only the stator winding is excited. This implies that both the load (torque) producing current (𝐼𝑞 or 𝐼𝑎 in DC motors) and the excitation current (𝐼𝑑 or 𝐼𝑓 in DC motors) flow in the stator. Consequently, independent control of “B” and “I” is not easy! The machine has some similarities to the compensated DC machine in that The stator has an excitation current and a mirror image of the load current. The rotor has the load current. Armature reaction flux is neutralized. 28  Steady State Operation (2) At steady-state, a one-phase, induction motor equivalent circuit will be utilised to analyse the machine's steady-state behaviour. Since the motor operates under balanced three-phase conditions, we can use one phase for analysis (multiply powers by 3). The rotor and stator look like two sets of coupled coils. All parameters can be referred to the stator winding as we have no external rotor power supply (rotor values marked with ‘ to indicate they are referred). 29  Steady State Operation (3) ❖To aid our analysis, we define a normalised difference between the synchronous and the actual speed: slip, s 𝑁𝑠 − 𝑁𝑟 𝜔𝑠 − 𝜔𝑟 𝑠= = 𝑁𝑠 𝜔𝑠 𝜔′ = 𝜔𝑠 − 𝜔𝑟 = 𝑠𝜔𝑠 𝜔𝑟 = 𝜔𝑠 − 𝑠𝜔𝑠 = (1 − 𝑠)𝜔𝑠 30  Steady State Operation (4) 𝑑𝑖𝑆 (𝑡) 𝑣𝑠 𝑡 = 𝑅𝑠 𝑖𝑠 + 𝐿𝑆 + 𝑒𝑠 𝑡 𝑑𝑡 𝑑𝑖𝑟 (𝑡) 0 = 𝑅𝑟 𝑖𝑟 + 𝐿𝑟 + 𝑒𝑟 𝑡 𝑑𝑡 In steady state: 𝑉ത𝑠 = 𝑅𝑠 𝐼𝑠ҧ + 𝑗𝐿𝑆 𝜔𝑒 𝐼𝑠ҧ + 𝐸ത𝑠 0 = 𝑅𝑟 𝐼𝑟ҧ + 𝑗𝑠𝐿𝑟 𝜔𝑒 𝐼𝑟ҧ + 𝐸ത𝑟 ➔ 𝑉ത𝑠 = 𝑅𝑠 𝐼𝑠ҧ + 𝑗𝐿𝑆 𝜔𝑒 𝐼𝑠ҧ + 𝐸ത𝑠 𝑅𝑟 𝐸ത𝑟 0= 𝐼𝑟ҧ + 𝑗𝐿𝑟 𝜔𝑒 𝐼𝑟ҧ + 𝑠 𝑠 31  Rotor frequency The frequency of the voltage induced in the rotor depends on the slip. It is given by: 𝑓𝑟 = 𝑠𝑓 Where 𝑓 = frequency of the source connected to the stator. 32 The Equivalent Circuit (1) 33  The Equivalent Circuit (2) Where, 𝐼𝑠 is the stator phase current, 𝐼𝑟 ′ is the referred rotor phase current, 𝐼𝑚 is the magnetising current 𝑉𝑠 is the stator phase voltage j is the square root of -1 and represents a 90o phase shift 34  The Equivalent Circuit (3) The circuit will consist of: Stator and rotor winding resistances (Rs and Rr’) Stator and rotor leakage reactances (Xs and Xr’) Mutual, airgap or magnetising reactance (Xm) Speed EMF induced in the stator by the rotor currents (since the rotor current produces this EMF, it looks like a rotor resistance when referred to the stator) The speed EMF It is induced in the stator by the referred rotor current. It appears as a resistance Rr’(1-s)/s Is zero at zero speed (s=1) and infinite at synchronous speed (s=0) Is responsible for power output and torque 35  Power The rotating magnetic field produces a torque 𝑇𝑑 on the rotor and does work at the rate: 𝑇𝑑 𝜔𝑠 – this represents input power to the rotor. Rotor output power is 𝑇𝑑 𝜔𝑟 - this represents the mechanical output power of our motor (friction and windage to be deducted) The difference between the two represents the energy lost in the resistance of the rotor. Rotor electromagnetic input: 𝑃𝑒𝑚 = 𝜔𝑠 𝑇𝑑 Rotor output: 𝑃𝑚𝑒𝑐ℎ = 𝜔𝑟 𝑇𝑑 = (1 − 𝑠)𝜔𝑠 𝑇𝑑 Rotor loss: 𝑃𝑙𝑜𝑠𝑠 = 𝜔𝑠 − 𝜔𝑟 𝑇𝑑 = 𝑠𝜔𝑠 𝑇𝑑 = 𝑠𝑃𝑒𝑚 Net (Shaft) Power output: 𝑃𝑜 = 𝑃𝑚𝑒𝑐ℎ − 𝑃𝑓𝑟𝑖𝑐+𝑤𝑖𝑛 36 Slipping Clutch Analogy 37  Torque The power input to the slipping clutch is the power crossing the air gap from the stator to the rotor at synchronous speed: ′ 𝑅𝑟 𝑃𝑒𝑚 = 3 ∙ 𝐼𝑟′2 ∙ 𝑠 Thus, the actual torque 𝑇𝑑 can be found by dividing this power by synchronous speed following the slipping clutch analogy: ′ 𝑃𝑒𝑚 ′2 𝑅𝑟 𝑝 𝑇𝑑 = = 3 ∙ 𝐼𝑟 ∙ ∙ 𝜔𝑠 𝑠 𝜔𝑒 𝐼𝑟 ′ and 𝐼𝑠 are obtained by solving the equations for the per-phase equivalent circuit 38 Torque – simple 4 kW motor model 39 Line Current – simple 4 kW motor model 40 Efficiency– simple 4 kW motor model 41  Summary For high full load efficiency, 𝑅𝑟′ should be low. For a high starting torque relative to a full load, 𝑅𝑟′ should be high. High rotor resistance reduces the starting current. A wound rotor machine with an external starting resistance, which can be reduced to zero as the machine speeds up, is ideal. However, slip rings, brushes, and rotor starters all increase the cost and introduce maintenance problems. For a squirrel cage motor, the design is a compromise. 42 PERFORMANCE EVALUATION 43  Performance Evaluation Approximate equivalent circuit Parameters from experimental tests Loss and efficiency Starting cage induction motor Double cage rotor 44  Evaluating IM performance Calculating the developed torque. As we have already seen, the torque developed can be expressed as: 𝑃𝑒𝑚 𝑅𝑟′ 𝑝 𝑃𝑚𝑒𝑐ℎ 𝑅𝑟′ (1−𝑠) 𝑇𝑑 = =3∙ 𝐼𝑟′2 ∙ ∙ or 𝑇𝑑 = =3∙ 𝐼𝑟′2 ∙ 𝜔𝑠 𝑠 𝜔𝑒 𝜔𝑟 𝑠 This is a particularly handy expression for calculating torque at zero speed! Calculating the peak, breakdown or pull-out torque: 𝑑𝑇𝑑 𝑅𝑟′ Have a go at = 0, 𝑠𝑝𝑜 ≈ deriving this from the 𝑑𝑠 𝑅𝑠′2 +(𝑋𝑟′ +𝑋𝑠′ )2 equivalent circuit 45  Approximate Equivalent Circuit (1) Approximate equivalent circuit: The equivalent circuit of the IM may be approximated by moving the magnetizing branch to the input terminals, as shown below. This simplified model is much easier to use and is useful to determine the parameters from experimental tests. 46 Approximate Equivalent Circuit (2) 47  Approximate Equivalent Circuit Parameters from Experimental Tests Stator resistance measurement: This is obtained by DC measurement. The DC current is adjusted to an approximately rated value so that the temperature of the winding approximates that of the running condition. R1 = RLL/2 where RLL = the stator resistance measured between any two terminals A 200 V dc V STATOR ROTOR Fig 10. Measurement of stator resistance 48  Approximate Equivalent Circuit Parameters from Experimental Tests No-Load Test: Rated voltage is applied to the machine’s terminals and allowed to run at no load. Input power, voltage and current are measured. 𝑅𝑚 and 𝑋𝑚 are obtained from 𝑉𝑠2 𝑉𝑠2 𝑅𝑚 = 𝑋𝑚 = 𝑃𝑜 𝑉𝑠2 𝐼𝑠2 −𝑃𝑜2 where 𝑃𝑜 is the power (real) per phase, corrected for friction and windage 49  Approximate Equivalent Circuit Parameters from Experimental Tests Locked-Rotor Test: In this test, the rotor is locked (using a metal bar clamped onto the shaft), thus s=1. Reduced voltage from a VARIAC or variable voltage power electronic supply is applied until rated current flows through the winding. Input power, voltage and current are measured. In this test, the iron losses and the magnetizing current are assumed to be negligible. 𝑃𝑜 𝑉𝑠2 𝐼𝑠2 −𝑃𝑜2 𝑅𝑠 + 𝑅𝑟′ = 𝑋𝑠 + 𝑋𝑟′ = 𝐼𝑠2 𝐼𝑠2 𝑅𝑠 can be directly measured, thus 𝑅𝑟′ can be found. There is no simple method to distinguish between the stator and rotor leakage reactances. These can be assumed equal in most circumstances. ❖ Note that in these tests no loading test rig is required. 50  Loss and Efficiency (1) 𝑃𝑜𝑢𝑡 Efficiency: 𝜂= ∙ 100% 𝑃𝑖𝑛 𝑃𝑙𝑜𝑠𝑠 = 𝑃𝑖𝑛 − 𝑃𝑜𝑢𝑡 = 𝑃𝐶𝑢𝑠 + 𝑃𝐶𝑢𝑟 + 𝑃𝐹𝑒 + 𝑃𝑣 + 𝑃𝑠𝑙𝑙 where: 𝑃𝐶𝑢𝑠 is the stator copper loss 𝑃𝐶𝑢𝑟 is the rotor copper loss 𝑃𝐹𝑒 is the iron loss (mainly in stator) 𝑃𝑣 is the friction and windage loss 𝑃𝑠𝑙𝑙 is the stray loss 51  Loss and Efficiency (2) Stray load loss is a small additional power loss dependent on load. It is mainly due to torque pulsations and higher order fields. It is very difficult to measure and is sometimes assumed to be just a percentage of the input power. For example, in IEC 4999, it’s taken to be 0.5% of input power at the rated load. With an experimental setup, power input can be calculated to say 0.5% accuracy whilst output power to 0.2% - thus, for a 95% efficient machine – the error on loss is approx. 14%!! 52  Exercise A 3-phase, 415 V, delta-connected, 8-pole, 50 Hz induction motor runs at a speed of 738 rev/min when operated at its rated load. The phase equivalent circuit parameters are 𝑅𝑠 = 0.4 Ω, 𝑅𝑟 ′ =0.7 Ω, 𝑥𝑠 = 𝑥𝑟′ =2.5 Ω, 𝑋𝑚 = 40 Ω, 𝑅𝑚 =298 Ω. Using the simplified equivalent circuit: Determine, for rated load, the values of the line current and power factor, torque, output power and efficiency. The mechanical loss amounts to 100 watts at full load. Calculate the starting torque and current and thus determine if the machine is able to start up with the rated load. Will the machine start up with rated load if it is connected through a star-delta starter? 53 STARTING OF CAGE MOTORS 54  Starting cage IM In cage motors, 𝑅𝑟 cannot be varied. A star-delta starter may be used to reduce the starting current if the machine is to be directly operated from the mains. Star – Delta starter Star connection reduces phase voltage on starting This reduces starting current and starting torque but provides less stress on the power supply: Phase current reduces by 3 Line current is reduced by 3 Torque reduces by 3 (torque proportional to 𝐼2 ) 55  Variable rotor resistance in Cage rotors Varying the rotor resistance gives rise to high starting torque (high 𝑅𝑟 ′) and high efficiency at rated conditions (low 𝑅𝑟 ′). Resistor across the slip rings at starting and shorting it out at running. We can obtain a similar effect in cage rotors by utilizing a “double cage” or a “deep bar cage” rotor. It is advantageous because it provides a way of improving IM starting from a fixed grid voltage. The double cage or deep bar concept is based on the skin effect, where the current distribution is not uniform and changes with the frequency of the current. 56  Skin Effect Non-uniform current distribution. The rotor bar currents have a frequency equal to the supply frequency at motor start-up. At high rotor current frequencies, the current will take the path of least impedance. Consequence: variable rotor resistance with rotor speed: At low speed, most current will go through the outer cage. This gives the current flow a high rotor resistance and a low leakage inductance. Resulting current-distribution at start-up (i.e. rotor At high speed, current is distributed evenly in the slot, and currents at 50 Hz). Red = highest current density. thus, the rotor resistance is lower and leakage inductance higher. 57  Double Cage Rotor - Lamination The outer cage is high-resistance Often, brass in built-up rotors The inner cage is low-resistance Copper bars in built-up rotors A deeper slot gives it more leakage inductance 58  Skin Effect – Deep bar cage Simulation of an 11 kW, 3-phase, 50 Hz, 400 V, delta-connected, 4-pole Induction Machine 59  Skin Effect – parameter variation The effective rotor leakage inductance and resistance values will change as a function of rotor speed. The graphs on the right-hand side illustrate the change in rotor resistance and leakage inductance of the 11 kW IM, as shown on the previous slide, as a function of rotor frequency. At 50 Hz, representing start-up, the resistance is twice that at rated value. Rotor resistance will decrease as speed increases while the effective leakage inductance will increase. The increased leakage reactance of the inner cage to force current into the high-resistance outer cage at starting will reduce maximum torque capability. This is because the machine has a higher leakage reactance Variation of rotor resistance Rr’ (top) and than a single cage design at full load speed. leakage inductance Lr’ (bottom) with rotor frequency for the 11 kW, 3-phase IM machine presented in the previous slide 60 Double Cage Rotors – Equivalent circuit 61  Double Cage Rotors The slot shape and/or bar material can be designed to shape the torque-speed behaviour of the induction motor. Torque dips (saddle) may occur. 62  Double Cage Rotors 4-kW design 63  Double Cage Rotors 4-kW design 64  Double Cage Rotors The double cage allows good starting torque with a low starting current. Torque dips (saddle) may appear in the mid-speed range if the design concentrates only on starting and full load conditions. Some designs have a third cage inside the other two to give three degrees of freedom, giving good starting, full load and acceleration characteristics. Third cages (Tri slot) are not popular and are rare. Different rotor slot designs: 65  Double Cage Rotors – Summary (1) Having an increased leakage reactance of the inner cage to force current into the high resistance outer cage at starting will reduce maximum torque capability. This is due to the machine having a higher leakage reactance than a single cage design at full load speed Having a double cage will imply either: the rotor diameter has to be bigger than the equivalent single- cage design or some of the available rotor conductor area has to be sacrificed to fit the double cage slots into the lamination 66  Double Cage Rotors – Summary (2) This results in the design generally being either: larger than the equivalent single-cage or less efficient than it However, the equivalent single-cage design may not start effectively direct-on-line. Such single-cage machines work well with power electronic variable frequency supplies where the slip is controlled, and operation is always above pull-out speed (which decreased with decreasing frequency) 67 Induction Machine Operation with Power Electronics  IM operation with PE IM supplied by Power Electronic Converters Control Methods 69  Power Electronic Drives Power electronics allow us to tailor the supply to our machine fully. Advantages: Improves machine dynamic performance Better utilisation of energy (higher efficiency at different loads) Extended speed range Variable speed Solves starting problems, etc. Disadvantages: Cost Less reliable system 70  Drive System Drive systems can be used to control position, speed or torque. Drive components: converter, motor, gearing/mechanical load, sensors and control. 71  IM operation (1) Till now, we have looked at direct-on-line (DOL) starting and operation. This is advantageous because it does not require any power electronics or sensors. Most IMs operate this way in industrial environments where accurate speed control is unnecessary. The main issues when operating in this mode are: Limited speed range – speed determined by the applied load. Large starting currents and stresses. Slow dynamic response. 72  IM operation (2) There are other ways to operate induction machines, namely: Constant frequency, variable voltage – operating the machine from a variable source will help limit starting currents and provide limited speed control. This is, however, not ideal due to increased copper loss. Variable frequency, variable voltage – good speed regulation. Good steady-state performance, however, poor dynamic behaviour. Vector (field) control performs best during steady speed and transient operation. We have decoupled control over the field- producing and torque currents using vector control. 73  IM operation (3) Although motors designed for DOL can generally be used with other control methods, one needs to realise that these might not be the optimal solution for controlled, converter-supplied operation: Cooling fan directly connected to shaft designed to cool the motor at rated speed Insulation may be damaged (shorter lifetime) if not designed for “inverter-duty.” Double cage rotors are not required if the motor is not connected directly to the supply. 74 A) Constant frequency, variable voltage Speed control Common load characteristics for pumps and fans Constant frequency, variable voltage (i.e., 𝜔𝑒 , constant) Peak torque proportional to 𝑉𝑠2 : 𝑝 3𝑉𝑠2 𝑇𝑚𝑎𝑥 = ∙ 2 2𝜔𝑒 𝑋𝑟′ 𝑉𝑠 adjusted by employing TRIACS or anti-parallel THYRISTOR pairs for voltage phase control It is not generally used for speed control since operation at increased slip means increased copper loss and reduced efficiency, whilst a distorted stator current waveform gives rise to parasitic torques and increased loss. Used for “soft-starting” when low 𝑉𝑠 is required at high slip and thus reduces starting current. 75 B) Variable frequency, variable voltage Until now, it has been assumed that the excitation frequency is fixed (i.e. the machine is driven from a fixed frequency source). The figure shows characteristics for a selected number of different excitation frequencies, 𝜔𝑒 < 𝜔𝑟𝑎𝑡𝑒𝑑. As 𝜔𝑒 is reduced, the characteristic retains its shape but is shifted. To retain its shape, the voltage must be varied in proportion to the frequency so that the magnetizing flux and, thus, 𝐼𝑚 is kept constant. 𝑉𝑠 Recall: 𝐼𝑚 = 𝜔𝑒 𝐿𝑚 If 𝑉𝑠 = 𝑘𝜔𝑒 , i.e., the supply voltage and frequency are held at a constant ratio: 𝑘 𝐼𝑚 = 𝐿𝑚 Therefore, motor flux is approximately constant: 3𝑝𝑅𝑟′ 𝑘 2 1 𝑇= ∙ 2 2𝑠𝜔𝑒 𝑅𝑠 𝑅𝑟′ + + 𝐿𝑠 + 𝐿′𝑟 2 𝜔𝑒 𝑠𝜔𝑒 𝑅𝑟′ 𝑅𝑠 For practical values of s, ≫ unless 𝜔𝑒 is very small. 𝑠𝜔𝑒 𝜔𝑒 DC-link H-bridge inverter allows Therefore, Torque expression is dependent only on slip frequency, 𝑠𝜔𝑒 us to supply variable voltage and variable frequency waveform to the IM 76 B) Variable frequency, variable voltage Concept of field weakening. Consider once again the torque equation: 3𝑝𝑅𝑟′ 𝑘 2 1 𝑇= ∙ 2 2𝑠𝜔𝑒 𝑅𝑠 𝑅𝑟′ + + 𝐿𝑠 + 𝐿′𝑟 2 𝜔𝑒 𝑠𝜔𝑒 Since 𝑉𝑠 = 𝑘𝜔𝑒 , 𝑘 is such that 𝑉𝑟𝑎𝑡𝑒𝑑 (e.g. 415 V) occurs at 𝜔𝑒_𝑟𝑎𝑡𝑒𝑑 (e.g. 50 Hz). If extended speed operation is required 𝜔𝑒 > 𝜔𝑒_𝑟𝑎𝑡𝑒𝑑 , then since the maximum supply voltage to the converter is fixed at 𝑉𝑟𝑎𝑡𝑒𝑑 , (𝐼𝑚 = 𝑉𝑟𝑎𝑡𝑒𝑑 Τ𝜔𝑒 𝐿𝑚 ), then the field must reduce. 𝑅𝑟′ 𝑅𝑠 Neglecting 𝐿𝑠 and 𝐿′𝑟 then, with ≫ and noting that 𝑉𝑠 = 𝑉𝑟𝑎𝑡𝑒𝑑 The rated point, P, is the point at 𝑠𝜔𝑒 𝜔𝑒 which rated 𝐼𝑠 is drawn. If 𝑉𝑠 = constant, 𝐼𝑠_𝑟𝑎𝑡𝑒𝑑 = constant, it can be 3𝑝𝑉𝑠2 𝑝 1 𝑉𝑠2 shown that 𝑇 × 𝜔𝑒 = constant. 𝑇= 𝑠 = 3 ∙ ∙ ′ 2 𝑠𝜔𝑒 2𝜔𝑒 𝑅𝑟′ 2 𝑅𝑟 𝜔𝑒 Note: Although using the variable Peak (pull-out torque) decreases with 𝜔𝑒2 frequency, variable voltage control method, we can operate the 𝑝 3𝑉𝑠2 machine at various speeds in steady Recall 𝑇𝑚𝑎𝑥 = ∙ state, we do not have control over 2 2𝜔𝑒2 𝐿′𝑟 the dynamic transients from one Slope of Torque/speed curve decreases with 𝜔𝑒. steady state point to the other. 77 Summary of V/f control The transient response is poor due to a lack of direct current control. Low-speed performance is poor (below about 5 Hz) V/f is not good at rapid changes and can lose the plot if the frequency changes too rapidly. Field (flux) weakening (recall DC machine operation) occurs automatically since the motor terminal voltage effectively defines the reducing flux in the machine as the frequency rises. Better dynamic performance is achieved with vector control. 78 Induction Machine Application Exercises Exercise 1 We consider the circuit of the figure below (equivalent circuit of induction machine). A sinusoidal voltage supplies the circuit. 1. Determine the total impedance of the circuit. 2. Determine the complex representations of currents and voltages for each impedance. 3. Determine the active and reactive powers in each circuit branch. 𝑉𝑠 = 200 V , 𝑍1 = 𝑅1 = 1 Ω, 𝑍2 = 𝑅2 + 𝑗𝑋2 = 20 + 𝑗1 Ω, 𝑍𝑚 = 𝑗𝑋𝑚 = 200 Ω 80 Exercise 2 At rated load, a 3-phase induction machine supplied by a balanced three-phase sinusoidal power supply at 50 Hz rotates at 1430 rpm. 1. What is the number of pole pairs of the machine? 2. What is the value of the normalised slip? 3. What is the frequency of the induced currents in the rotor? 81 Exercise 3 A 3-phase induction machine has the following characteristics: ▪ Supply voltage: 115/200 V. Squirrel cage ▪ Frequency: 400 Hz. ▪ Rated speed: 11500 rpm. ▪ Absorbed power at rated operation: 4200 W, cos 𝜙 = 0.6. ▪ Phase resistance: 0.16 Ω. The machine is supplied by a balanced 3-phase power supply of 200 V, 400 Hz. The machine operates at full load. 1. What is the connection that must be adopted? 2. What is the value of the normalised slip? 3. What is the RMS value of the absorbed current? 4. What is the value of stator copper loss? 5. Knowing that the stator iron loss is equal to 350 W and that we can neglect the rotor iron and mechanical losses, what is the machine's efficiency? 6. What is the rated torque? 82 Exercise 4 The maximum torque of a 3-phase induction machine, obtained for a normalised slip of 20%, equals 3 times its nominal (rated) torque. Neglecting the stator resistance: 1. What is the normalised slip corresponding to the nominal torque? 2. What is the starting torque as a percentage of the nominal torque? 83 THANK YOU 84