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Chapter 4 : AC Machine Lecturer : Puan Nordiana Binti Mohamad Saaid Slide prepared/edited by: Dr Rosemizi Abd Rahim/Puan Nordiana Click here to watch the ac machine animation video http://rmz4567.blogspot.my/2013/02/electrical-engineering.html AC Machine Alternating current (ac) is the primary source of electrical energy. It is less expensive to produce and transmit alternating current (ac) than direct current (dc). For this reason, and because ac voltage is induced into the armature of all generators, ac machines are generally more practical. It is very easy to step up or step down the voltage for AC electricity through the use of transformers. It is not very straight forward to step up or step down the voltage for DC electricity- for power loads, it is typically done through inversion to AC- stepping up/down using a transformer- and rectification back to DC. May function as a generator (mechanical to electrical) or a motor (electrical to mechanical) DC Machine & AC machine • DC motor - ends of the coil connect to a split ring to 'rectify' the emf produced • AC motor - no need rectification, so don't need split rings. AC Motor As in the DC motor case, • a current is passed through the coil, generating a torque on the coil. • Since the current is alternating, the motor will run smoothly only at the frequency of the sine wave. AC Generator • This process can be described in terms of Faraday's law . The rotation of the coil continually changes the magnetic flux through the coil and therefore generates a voltage Generator and Motor How Does an Electric Generator Work? Classification of AC Machines Two major classes of machines; i) Synchronous Machines: • Synchronous Generators: A primary source of electrical energy. • Synchronous Motors: Used as motors as well as power factor compensators (synchronous condensers). ii) Asynchronous (Induction) Machines: • Induction Motors: Most widely used electrical motors in both domestic and industrial applications. • Induction Generators: Due to lack of a separate field excitation, these machines are rarely used as generators. Synchronous Machine Synchronous Machine Origin of name: syn = equal, chronos = time Synchronous machines are called ‘synchronous’ because their mechanical shaft speed is directly related to the power system’s line frequency. the rotating air gap field and the rotor rotate at the same speed, called the synchronous speed. Synchronous machines are ac machine that have a field circuit supplied by an external dc source. – DC field winding on the rotor, – AC armature winding on the stator Synchronous Machine Synchronous machines are used primarily as generators of electrical power, called synchronous generators or alternators. They are usually large machines generating electrical power at hydro, nuclear, or thermal power stations. Synchronous motors are built in large units compare to induction motors (Induction motors are cheaper for smaller ratings) and synchronous motors are used for constant speed industrial drives Application of synchronous machine as a motor: pumps in generating stations, electric clocks, timers, and so forth where constant speed is desired. Synchronous Machine The frequency of the induced voltage is related to the rotor speed by: where ; P = the number of magnetic poles fe = the power line frequency. Typical machines have two-poles, four-poles, six-poles etc. Synchronous Machine Construction • Energy is stored in the inductance • As the rotor moves, there is a change in the energy stored • Either energy is extracted from the magnetic field (and becomes mechanical energy – motor) • Or energy is stored in the magnetic field and eventually flows into the electrical circuit that powers the stator – generator Synchronous Machine Construction • DC field windings are mounted on the (rotating) rotor - which is thus a rotating electromagnet • AC windings are mounted on the (stationary) stator resulting in three-phase AC stator voltages and currents The main part in the synchronous machines are i) Rotor ii) Stator Synchronous Machine Rotor There are two types of rotors used in synchronous machines: i) cylindrical (or round) rotors ii) salient pole rotors Machines with cylindrical rotors are typically found in higher speed higher power applications such as turbogenerators. Using 2 or 4 poles, these machines rotate at 3600 or 1800 rpm (with 60hz systems). Salient pole machines are typically found in large (many MW), low mechanical speed applications, including hydrogenerators, or smaller higher speed machines (up to 1-2 MW). Salient pole rotors are less expensive than round rotors. Synchronous Machine – Cylindrical rotor D1m Turbine L 10 m Steam High speed 3600 r/min 2-pole Stator winding N Uniform air-gap Stator 1800 r/min 4-pole Direct-conductor cooling (using hydrogen or water as coolant) d-axis q-axis Rotor winding Rotor Rating up to 2000 MVA S Turbogenerator Synchronous Machine – Cylindrical rotor Stator Cylindrical rotor Synchronous Machine – Salient Pole 1. Most hydraulic turbines have to turn at low speeds (between 50 and 300 r/min) 2. A large number of poles are required on the rotor d-axis Non-uniform air-gap N D 10 m q-axis S S Turbine Hydro (water) Hydrogenerator N Synchronous Machine – Salient Pole Stator Synchronous machine rotors are simply rotating electromagnets built to have as many poles as are produced by the stator windings. Dc currents flowing in the field coils surrounding each pole magnetize the rotor poles. The magnetic field produced by the rotor poles locks in with a rotating stator field, so that the shaft and the stator field rotate in synchronism. Salient poles are too weak mechanically and develop too much wind resistance and noise to be used in large, high-speed generators driven by steam or gas turbines. For these big machines, the rotor must be a solid, cylindrical steel forging to provide the necessary strength. Axial slots are cut in the surface of the cylinder to accommodate the field windings. Since the rotor poles have constant polarity they must be supplied with direct current. This current may be provided by an external dc generator or by a rectifier. In this case the leads from the field winding are connected to insulated rings mounted concentrically on the shaft. Stationary contacts called brushes ride on these slip rings to carry current to the rotating field windings from the dc supply. The brushes are made of a carbon compound to provide a good contact with low mechanical friction. An external dc generator used to provide current is called a “ brushless exciter “. Synchronous Machine Stator The stator of a synchronous machine carries the armature or load winding which is a three-phase winding. The armature winding is formed by interconnecting various conductors in slots spread over the periphery of the machine’s stator. Often, more than one independent three phase winding is on the stator. An arrangement of a three-phase stator winding is shown in Figure below. Notice that the windings of the three-phases are displaced from each other in space. Synchronous Machine Construction Stator Synchronous Machine Magnetomotive Forces (MMF’s) and Fluxes Due to Armature and Field Windings Flux produced by a stator winding Synchronous Machine Magnetomotive Forces (MMF’s) and Fluxes Due to Armature and Field Windings Synchronous Machine Magnetomotive Forces (MMF’s) and Fluxes Due to Armature and Field Windings Two Cycles of mmf around the Stator Synchronous Generator Equivalent circuit model – synchronous generator If the generator operates at a terminal voltage VT while supplying a load corresponding to an armature current Ia, then; In an actual synchronous machine, the reactance is much greater than the armature resistance, in which case; Among the steady-state characteristics of a synchronous generator, its voltage regulation and power-angle characteristics are the most important ones. As for transformers, the voltage regulation of a synchronous generator is defined at a given load as; Synchronous Generator Phasor diagram of a synchronous generator The phasor diagram is to shows the relationship among the voltages within a phase (Eφ,Vφ, jXSIA and RAIA) and the current IA in the phase. Unity P.F (1.0) Synchronous Generator Lagging P.F Leading P.F. Synchronous Generator Power and Torque In generators, not all the mechanical power going into a synchronous generator becomes electric power out of the machine The power losses in generator are represented by difference between output power and input power shown in power flow diagram below Synchronous Generator Losses Rotor - resistance; iron parts moving in a magnetic field causing currents to be generated in the rotor body - resistance of connections to the rotor (slip rings) Stator - resistance; magnetic losses (e.g., hysteresis) Mechanical - friction at bearings, friction at slip rings Stray load losses - due to non-uniform current distribution Synchronous Generator The input mechanical power is the shaft power in the generator given by equation: The power converted from mechanical to electrical form internally is given by The real electric output power of the synchronous generator can be expressed in line and phase quantities as and reactive output power Synchronous Generator In real synchronous machines of any size, the armature resistance RA is more than 10 times smaller than the synchronous reactance XS (Xs >> RA). Therefore, RA can be ignored Synchronous Motor Synchronous Motor Power Flow Example : Synchronous Generator. A three-phase, wye-connected 2500 kVA and 6.6 kV generator operates at full-load. The per-phase armature resistance Ra and the synchronous reactance, Xd, are (0.07+j10.4). Calculate the percent voltage regulation at (a) 0.8 power-factor lagging, and (b) 0.8 power-factor leading. Solution. Induction Machine The machines are called induction machines because of the rotor voltage which produces the rotor current and the rotor magnetic field is induced in the rotor windings. Induction generator has many disadvantages and low efficiency. Therefore induction machines are usually referred to as induction motors. Induction Machine • Three-phase induction motors are the most common and frequently encountered machines in industry – – – – simple design, rugged, low-price, easy maintenance wide range of power ratings: fractional horsepower to 10 MW run essentially as constant speed from no-load to full load Its speed depends on the frequency of the power source • not easy to have variable speed control • requires a variable-frequency power-electronic drive for optimal speed control Construction Slip rings Cutaway in a typical woundrotor IM. Notice the brushes and the slip rings Brushes Construction • An induction motor has two main parts i) a stationary stator • consisting of a steel frame that supports a hollow, cylindrical core • core, constructed from stacked laminations (why?), – having a number of evenly spaced slots, providing the space for the stator winding Stator of IM Construction ii) a revolving rotor • composed of punched laminations, stacked to create a series of rotor slots, providing space for the rotor winding • conventional 3-phase windings made of insulated wire (wound-rotor) » similar to the winding on the stator • aluminum bus bars shorted together at the ends by two aluminum rings, forming a squirrel-cage shaped circuit (squirrel-cage) Construction • Two basic design types depending on the rotor design – squirrel-cage: conducting bars laid into slots and shorted at both ends by shorting rings. – wound-rotor: complete set of three-phase windings exactly as the stator. Usually Y-connected, the ends of the three rotor wires are connected to 3 slip rings on the rotor shaft. In this way, the rotor circuit is accessible. Construction 1. Squirrel cage – the conductors would look like one of the exercise wheels that squirrel or hamsters run on. Construction 2. Wound rotor – have brushes and slip rings at the end of rotor Notice the slip rings Rotating Magnetic Field • Balanced three phase windings, i.e. mechanically displaced 120 degrees form each other, fed by balanced three phase source • A rotating magnetic field with constant magnitude is produced, rotating with a speed 120 f e nsync P (rpm) Where fe is the supply frequency and P is the no. of poles and nsync is called the synchronous speed in rpm (revolutions per minute) Synchronous speed P 50 Hz 60 Hz 2 3000 3600 4 1500 1800 6 1000 1200 8 750 900 10 600 720 12 500 600 Principle of operation • This rotating magnetic field cuts the rotor windings and produces an induced voltage in the rotor windings • Due to the fact that the rotor windings are short circuited, for both squirrel cage and wound-rotor, and induced current flows in the rotor windings • The rotor current produces another magnetic field • A torque is produced as a result of the interaction of those two magnetic fields ind kBR Bs Where ind is the induced torque and BR and BS are the magnetic flux densities of the rotor and the stator respectively Induction motor speed • At what speed will the IM run? – Can the IM run at the synchronous speed, why? – If rotor runs at the synchronous speed, which is the same speed of the rotating magnetic field, then the rotor will appear stationary to the rotating magnetic field and the rotating magnetic field will not cut the rotor. So, no induced current will flow in the rotor and no rotor magnetic flux will be produced so no torque is generated and the rotor speed will fall below the synchronous speed – When the speed falls, the rotating magnetic field will cut the rotor windings and a torque is produced Induction motor speed • So, the IM will always run at a speed lower than the synchronous speed • The difference between the motor speed and the synchronous speed is called the Slip (s) nslip nsync nm Where nslip= slip speed nsync= speed of the magnetic field nm = mechanical shaft speed of the motor The Slip s nsync nm nsync Where s is the slip Notice that : if the rotor runs at synchronous speed s=0 if the rotor is stationary s=1 Slip may be expressed as a percentage by multiplying the above eq. by 100, notice that the slip is a ratio and doesn’t have units Induction Motors and Transformers • Both IM and transformer works on the principle of induced voltage – Transformer: voltage applied to the primary windings produce an induced voltage in the secondary windings – Induction motor: voltage applied to the stator windings produce an induced voltage in the rotor windings – The difference is that, in the case of the induction motor, the secondary windings can move – Due to the rotation of the rotor (the secondary winding of the IM), the induced voltage in it does not have the same frequency of the stator (the primary) voltage Frequency • The frequency of the voltage induced in the rotor is given by Pn fr 120 Where fr = the rotor frequency (Hz) P = number of stator poles n = slip speed (rpm) P (ns nm ) fr 120 P sns sf e 120 Frequency • What would be the frequency of the rotor’s induced voltage at any speed nm? f r sf e • When the rotor is blocked (s=1) , the frequency of the induced voltage is equal to the supply frequency • On the other hand, if the rotor runs at synchronous speed (s = 0), the frequency will be zero Torque • While the input to the induction motor is electrical power, its output is mechanical power and for that we should know some terms and quantities related to mechanical power • Any mechanical load applied to the motor shaft will introduce a Torque on the motor shaft. This torque is related to the motor output power and the rotor speed load Pout m ( N .m) , where m 2nm 60 (rad / s) Horse power • Another unit used to measure mechanical power is the horse power • It is used to refer to the mechanical output power of the motor • Since we, as an electrical engineers, deal with watts as a unit to measure electrical power, there is a relation between horse power and watts 1 hp 746 watts Example A 208-V, 10hp, four pole, 60 Hz, Y-connected induction motor has a full-load slip of 5 percent 1. What is the synchronous speed of this motor? 2. What is the rotor speed of this motor at rated load? 3. What is the rotor frequency of this motor at rated load? 4. What is the shaft torque of this motor at rated load? Solution 120 f e 120(60) 1800 rpm P 4 1. nsync 2. nm (1 s)ns 3. f r sfe 0.05 60 3Hz 4. (1 0.05) 1800 1710 rpm Pout m 2 nm 60 10 hp 746 watt / hp 41.7 N .m 1710 2 (1/ 60) load Pout Equivalent Circuit • The induction motor is similar to the transformer with the exception that its secondary windings are free to rotate As we noticed in the transformer, it is easier if we can combine these two circuits in one circuit but there are some difficulties Equivalent Circuit • When the rotor is locked (or blocked), i.e. s =1, the largest voltage and rotor frequency are induced in the rotor. • On the other side, if the rotor rotates at synchronous speed, i.e. s = 0, the induced voltage and frequency in the rotor will be equal to zero. ER sE R 0 Where ER0 is the largest value of the rotor’s induced voltage obtained at s = 1(locked rotor) Equivalent Circuit • The same is true for the frequency, i.e. f r sf e • It is known that X L 2fL • So, as the frequency of the induced voltage in the rotor changes, the reactance of the rotor circuit also changes X r r Lr 2f r Lr Where Xr0 is the rotor reactance at the supply frequency (at blocked rotor) 2sf e Lr sX r 0 Equivalent Circuit • Then, we can draw the rotor equivalent circuit as follows Where ER is the induced voltage in the rotor and RR is the rotor resistance Equivalent Circuit • Now we can calculate the rotor current as ER IR ( RR jX R ) sER 0 ( RR jsX R 0 ) • Dividing both the numerator and denominator by s so nothing changes we get ER 0 IR ( RR jX R 0 ) s Where ER0 is the induced voltage and XR0 is the rotor reactance at blocked rotor condition (s = 1) Equivalent Circuit • Now we can have the rotor equivalent circuit Equivalent Circuit • Now as we managed to solve the induced voltage and different frequency problems, we can combine the stator and rotor circuits in one equivalent circuit Where 2 X 2 aeff X R0 2 R2 aeff RR I2 IR aeff E1 aeff ER 0 aeff NS NR Power losses in Induction machines • Copper losses – Copper loss in the stator (PSCL) = I12R1 – Copper loss in the rotor (PRCL) = I22R2 • Core loss (Pcore) • Mechanical power loss due to friction and windage • How this power flow in the motor? Power flow in induction motor Power relations Pin 3 VL I L cos 3 Vph I ph cos PSCL 3 I12 R1 PAG Pin ( PSCL Pcore ) PRCL 3I 22 R2 Pconv PAG PRCL Pout Pconv ( Pf w Pstray ) ind Pconv m Equivalent Circuit • We can rearrange the equivalent circuit as follows Actual rotor resistance Resistance equivalent to mechanical load Power relations Pin 3 VL I L cos 3 Vph I ph cos PSCL 3 I12 R1 PAG Pin ( PSCL Pcore ) Pconv PRCL R2 3I s 2 2 PRCL 3I 22 R2 Pconv PAG PRCL 3I 22 Pconv (1 s) PAG R2 (1 s ) s Pout Pconv ( Pf w Pstray ) ind PRCL s PRCL (1 s ) s Pconv m (1 s ) PAG (1 s )s Power relations PAG Pconv 1 1-s PRCL s PAG : PRCL : Pconv 1 : s : 1-s Example A 480-V, 60 Hz, 50-hp, three phase induction motor is drawing 60A at 0.85 PF lagging. The stator copper losses are 2 kW, and the rotor copper losses are 700 W. The friction and windage losses are 600 W, the core losses are 1800 W, and the stray losses are negligible. Find the following quantities: 1. 2. 3. 4. The air-gap power PAG. The power converted Pconv. The output power Pout. The efficiency of the motor. Solution 1. Pin 3VL I L cos 3 480 60 0.85 42.4 kW PAG Pin PSCL Pcore 42.4 2 1.8 38.6 kW 2. Pconv PAG PRCL 700 38.6 37.9 kW 1000 3. Pout Pconv PF &W 600 37.9 37.3 kW 1000 Solution 37.3 Pout 50 hp 0.746 4. Pout 100% Pin 37.3 100 88% 42.4 Example A 460-V, 25-hp, 60 Hz, four-pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: R1= 0.641 R2= 0.332 X1= 1.106 X2= 0.464 XM= 26.3 The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2.2 percent at the rated voltage and rated frequency, find the motor’s 1. 2. 3. Speed Stator current Power factor 4. Pconv and Pout 5. ind and load 6. Efficiency Solution 1. n 120 f e 120 60 1800 rpm sync P 4 nm (1 s)nsync (1 0.022) 1800 1760 rpm R2 0.332 j 0.464 2. Z 2 jX 2 s 0.022 15.09 j 0.464 15.11.76 1 1 Zf 1/ jX M 1/ Z 2 j 0.038 0.0662 1.76 1 12.9431.1 0.0773 31.1 Solution Ztot Z stat Z f 0.641 j1.106 12.9431.1 11.72 j 7.79 14.0733.6 4600 V 3 I1 18.88 33.6 A Ztot 14.0733.6 3. PF cos 33.6 0.833 lagging 4. Pin 3VL I L cos 3 460 18.88 0.833 12530 W PSCL 3I12 R1 3(18.88)2 0.641 685 W PAG Pin PSCL 12530 685 11845 W Solution Pconv (1 s) PAG (1 0.022)(11845) 11585 W Pout Pconv PF &W 11585 1100 10485 W 5. ind load 10485 = 14.1 hp 746 PAG 11845 sync 2 1800 Pout 10485 m 2 1760 62.8 N.m 60 56.9 N.m 60 6. Pout 100% 10485 100 83.7% Pin 12530 Induction Motor – Power and Torque The output power can be found as Pout = Pconv – PF&W – Pmisc The induced torque or developed torque: Example A two-pole, 50-Hz induction motor supplies 15kW to a load at a speed of 2950 rpm. 1. What is the motor’s slip? 2. What is the induced torque in the motor in N.m under these conditions? 3. What will be the operating speed of the motor if its torque is doubled? 4. How much power will be supplied by the motor when the torque is doubled? Solution 1. n 120 f e 120 50 3000 rpm sync s 2. P nsync nm nsync 2 3000 2950 0.0167 or 1.67% 3000 no Pf W given assume Pconv Pload and ind load ind Pconv m 15 103 48.6 N.m 2 2950 60 Solution 3. In the low-slip region, the torque-speed curve is linear and the induced torque is direct proportional to slip. So, if the torque is doubled the new slip will be 3.33% and the motor speed will be nm (1 s )nsync (1 0.0333) 3000 2900 rpm 4. Pconv ind m 2 (2 48.6) (2900 ) 29.5 kW 60 Torque-slip characteristics • Slip, s nsync nm nsync • Slip at maximum torque, SmaxT • SmaxT = 𝑅2 ′ 𝑋2 ′ = 𝑅2 𝑋2 = 𝑟𝑜𝑡𝑜𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑠𝑡𝑎𝑛𝑑𝑠𝑡𝑖𝑙𝑙 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 Torque • Maximum torque, 𝑇𝑚𝑎𝑥 • 𝑇𝑚𝑎𝑥 = 3 𝑉1 2 𝜔𝑠 2𝑋2 Torque • Full load torque, Tfl 𝑉1 • Current, 𝐼2 = • 𝑇𝑓𝑙 = 𝑃𝐺 𝜔𝑠 = 3 𝜔𝑠 𝑅2 /𝑠 2 +𝑋2 2 = 𝑉1 2 𝑅2 /𝑠 𝑅2 /𝑠 2 +𝑋2 2 Example • A 5kV, 50Hz, 20-pole, 3-phase star-connected induction motor has a rotor resistance of 0.15Ω and standstill reactance of 1.15Ω. The motor has a speed of 292.5rpm at full load. Calculate the following:i) slip (slip at full load) ii) slip at maximum torque iii) the ratio of maximum torque to full-load torque Solution • Slip, s nsync nm nsync 292.5 1 0.025 300 • Slip at maximum torque, SmaxT= 𝑅2 𝑋2 = 0.15 1.15 = 0.13 • Ratio of Tmax to Tfl • 𝑇𝑚𝑎𝑥 = • 𝑇𝑓𝑙 = 3 𝑉1 2 𝜔𝑠 2𝑋2 3 𝜔𝑠 • Ratio of = = 3 𝑉1 2 𝜔𝑠 2 1.15 𝑉1 2 𝑅2 /𝑠 𝑅2 /𝑠 2 +𝑋2 2 𝑇𝑚𝑎𝑥 𝑇𝑓𝑙 = = 2.7045 3 𝑉1 2 0.15/0.025 𝜔𝑠 0.15/0.025 2 +1.152