EEE313 Basic Electrical Machines 1 Lecture Notes PDF

EEE 313 BASIC ELECTRICAL MACHINES 1 [2 UNITS] SYNCHRONOUS MACHINES Synchronous machines are one of three of the most popular types of electric machines (that is devices that carries out energy conversion from one form to another majorly between electrical and mechanical forms). This is so because t...

EEE 313 BASIC ELECTRICAL MACHINES 1 [2 UNITS] SYNCHRONOUS MACHINES Synchronous machines are one of three of the most popular types of electric machines (that is devices that carries out energy conversion from one form to another majorly between electrical and mechanical forms). This is so because they operate at constant speeds and frequencies under steady state (does not change with time). Synchronous machines are capable of operating either as a generator or a motor. It therefore means that a machine in which the following relation is maintained for its satisfactory operation is called a synchronous machine. 𝑷 𝑵𝒔 f= / (No. of cycle made per revolution / No. of revolutions made per second) 𝟐 𝟔𝟎 𝑷𝑵𝒔 = (this is the relationship between f, p and Ns) in cycles/s or Hz 𝟏𝟐𝟎 where NS is the synchronous speed in rpm, f is the supply frequency and P is the number of poles of the machine. When the machine is to work as a generator, it has to run at synchronous speed (NS) to generate power at certain frequency (f), called power frequency. In Nigeria & some countries such as india, its value is 50Hz, whereas in USA it is kept at 60 Hz. Mechanical energy (Input) Electrical Energy (Output) = Generator Meanwhile if the machine works as a motor, it can rotate only at synchronous speed (NS) since the magnetic poles are locked with the revolving field. If the machine fails to rotate at synchronous speed, it is palled out of step and stops. Electrical Energy (Input) Mechanical energy (Output) = Motor Thus, a synchronous machine (generator or motor) is a machine which only runs at synchronous speed and maintains the above relation. E.g1 Calculate the synchronous speed of a 12-pole synchronous machine supplied with power in Nigeria. Solution 𝑃𝑁 f= 𝑠 120 Ns = (50*120)/12 = 500rpm E.g2 What number of poles will a synchronous machine running at 750rpm in India have? Solution 𝑃𝑁𝑠 f= 120 p = (50*120)/750 = 8 E.g3 Calculate the frequency of a 2-pole synchronous machine running at 3000rpm. Solution 𝑃𝑁𝑠 f= 120 2∗3000 = = 50Hz 120 Basic Principles of a Synchronous Machine A synchronous machine converts mechanical energy into electrical energy or electrical energy into mechanical energy. The fundamental phenomenon which makes these conversions possible are: (a) Law of electromagnetic induction: This relates to the production of emf, i.e., emf is induced in a conductor whenever it cuts across the magnetic field (Fig. A). This is called Faraday’s first law of electromagnetic induction. (b) Law of interaction: This law relates to the phenomenon of production of force or torque i.e., whenever a current carrying conductor is placed in the magnetic field, by the interaction of the magnetic fields produced by the current carrying conductor and the main field, force is exerted on the conductor and torque is developed (Fig. B). Figure A: Generator Principle Figure B: Motor Principle Generator and Motor Action Comparison Supply  NB: Two terms commonly used to describe the windings on a machine are field windings and armature windings. In general, the term “field winding” applies to the windings that produce the main magnetic field in a machine, and the term “armature windings” applies to the windings where the main voltage is induced. For synchronous machines, the field windings are on the rotor, so the terms “rotor windings” and “field windings” are used interchangeably. Similarly, the armature windings are on the stator, so the terms “stator windings” and “armature windings” are used interchangeably. Figure C: Power angle curve of synchronous machine Further explanation: In generator action, an emf is induced in the armature conductors when they cut across the magnetic field. On closing the circuit, current flows through the armature conductors which produces another field. By the interaction of this field and main field, a force is exerted on the conductor which acts in opposite direction to that of rotation. It is this force against which the relative motion of conductors has to be maintained by the mechanical power supplied by the prime-mover (turbine/engine…), thus the mechanical power is converted into electrical power. In motor action, a current is supplied to the machine which flows through the armature conductors. The armature conductors produce a field which interacts with the main field. Thus, a force is exerted on the conductors and rotation takes place (i.e., torque is developed). Once rotation occurs, an emf is induced in the conductors due to relative motion. This emf acts in opposite direction to the flow of current. The flow of current has to be maintained against this emf by applying external voltage source, thus electrical power is converted into mechanical power. Armature Winding The stationary part of large synchronous machine is called the armature. On the inner periphery of the stator core, number of slots (mostly open parallel sided slots) are provided. In these slots armature winding is placed. Types of Armature Winding a) Single-phase and poly-phase windings: When only one winding is placed on the armature and only one emf is obtained at the output, winding is called single-phase winding. When more than one winding are placed on the armature and emfs induced are more than one, displaced from each other by some angle, the winding is called poly-phase winding. b) Concentrated and distributed windings: When one slot per pole or slots equal to the number of poles are employed, the windings thus obtained are called concentrated windings. When number of slots per poles are more than one, the windings thus obtained are called distributed windings. c) Single layer and double layer windings: When only one coil side is placed in a slot, the winding is called single layer winding. However, when two coil sides are placed in one slot, one over the other, the winding is called double layer winding. d) Full pitched and short pitched windings: When the two coil sides of the same coil are 180 electrical degrees apart, the winding is called full pitch winding. When the two sides of the same coil are less than 180 electrical degrees apart, the winding is called short pitch winding. e) Concentric (or spiral), Lap and Wave windings: When each group of coils under a pole is arranged into a sort of concentric shape, such winding is called concentric or chain or spiral winding. In case of lap winding, coils or coil sides overlap the other consecutively and connections are made. Whereas in wave winding, the coils are always forward connected.  Concentric Winding Lap Winding Wave Winding Common Terms Used in Armature Winding i. Electrical angle: When a conductor passes through a pair of poles, one cycle of emf is induced in it. Thus, a pair of poles represents an angle of 360 electrical degrees. ii. Pole pitch: Distance between two neutral axis (or similar points) of adjacent poles is called poles pitch. If S is the number of slots on the whole periphery of armature and P is the number of poles. Then, Pole pitch = No. of slots per pole = S/P. iii. Coil pitch or coil span: The distance between two active sides of a coil is called coil span. iv. Slot pitch: The distance between centre points (or similar points) of two consecutive slots or teeth is called slot pitch. 180° 𝑆𝑙𝑜𝑡 𝑝𝑖𝑡𝑐ℎ = 𝑁𝑜. 𝑜𝑓 𝑠𝑙𝑜𝑡𝑠/𝑝𝑜𝑙𝑒 v. Phase spread: The angle or space of pole face over which coil sides of the same phase are spread is called phase spread. In three phase winding, Phase spread = No. of slots/pole/phase Constructional Features of Synchronous Machines The Stator and the Rotor are the two most important parts of a synchronous machine although other parts exist. Thus, I. Stator The outer stationary part of the machine is called stator and it has the following important parts: (i) Stator frame: It is the outer body of the machine made of cast iron and it protects the inner parts of the machine. It can be also made of any other strong material since it is not to carry the magnetic field. Cast iron is used only because of its high mechanical strength. (ii) Stator Core: The stator core is made of silicon steel material. It is made from number of stampings which are insulated from each other. Its function is to provide an easy path for the magnetic lines of force and accommodate the stator winding. (iii) Stator Winding: Slots are cut on the inner periphery of the stator core in which three-phase or one-phase winding is placed. Enamelled copper is used as winding material. II. Rotor The rotating part of the machine is called rotor. From construction point of view, there are two types of rotors named as (i) Salient pole type rotor and (ii) Non-salient pole type (cylindrical) rotor. i. Salient pole type rotor In this case, projected poles are provided on the rotor. The cost of construction of salient pole type rotors is low, moreover sufficient space is available to accommodate field winding but these cannot bear high mechanical stresses at high speeds. Therefore, salient pole type construction is suited for medium and low speeds and are usually employed at hydro-electric and diesel power plants as synchronous generators. Since the speed of these machines (generators) is quite low, to obtain the required frequency, the machines have large number of poles. To accommodate such a large number of poles, these machines have larger diameter and small length. Spider, pole core and pole shoe, field winding or exciting winding, damper winding are parts of a salient pole rotor. Fig IIa: Salient-pole type rotor in different views ii. Cylindrical rotor In this case, there are no projected poles but the poles are formed by the current flowing through the rotor (exciting) winding. Non-salient pole type construction is suited for the high speeds. The steam turbines rotate at a high speed (3000 rpm). When these turbines are used as prime-mover for this machine working as a generator, a small number of poles are required for given frequency. Hence, these machines have smaller diameter and larger length. Cylindrical rotors have the following parts: rotor core, and rotor winding or exciting winding. Fig IIb: Cylindrical rotor in different views The following are other parts of a synchronous machine III. Brushes: are made of carbon and these are just slip over the slip rings. DC supply is given to the brushes. From the brushes, current flows to the slip rings and then to the exciting winding. IV. Bearings: are provided between the shaft (rod) and outer stationary body to reduce the friction. V. Shaft: is made of mild steel. Mechanical power is taken or given to the machine through the shaft. What are the applications of salient-pole type and cylindrical rotor respectively?....Check it out Characteristic Features of Salient & Cylindrical Structures For non-salient structures, we have: The non-salient field structure has the following special features. They are of smaller diameter and of very long axial length. Robust construction and noiseless operation. Less windage (air-resistance) loss. Better in dynamic balancing.  High operating speed (3000 rpm). Nearly sinusoidal flux distribution around the periphery, and therefore, gives a better emf waveform than that obtainable with salient poles field structure. There is no need of providing damper windings (except in special cases to assist in synchronising) because the solid field poles themselves act as efficient dampers. While for salient structures, we have: These are of larger diameter and shorter length. Usually, 2/3rd of the pole pitch is covered by the pole shoes. To reduce eddy current losses, the poles are laminated. The machine having such structure are employed with hydraulic turbines or with diesel engines which are usually operated at low speeds (100 to 375 rpm). Rotating Magnetic Field (RMF) For synchronous machines, the rotating magnetic field is generated in the stator winding. The speed of its rotation is called NS, and it depends on f and P. Their relationship is as below: 𝑷𝑵𝒔 f= 𝟏𝟐𝟎 Synchronous Reactance & Impedance Synchronous reactance: It is fictitious reactance which has the effect equivalent to the combined effects of both the leakage reactance and armature reaction reactance. It is represented by Xs. Xs = XL + Xa Synchronous impedance: This is the vector sum of armature resistance and synchronous reactance. It is represented by Zs. ZS = Ra+ jXS = √𝑅 2 + 𝑋𝑆2 Open circuit test and short circuit test are utilized to determine the value of synchronous impedance of an alternator experimentally. Short Circuit Ratio of a Synchronous Machine The ratio of field current (𝐼𝑓1 ) to produce rated voltage on open-circuit (O.C) to the field current (𝐼𝑓2 ) required to circulate rated current on short-circuit (S.C) while the machine is driven at synchronous speed is called short-circuit ratio (SCR) of a synchronous machine. The short circuit ratio can be calculated from the open-circuit characteristic (O.C.C) at rated speed and the short circuit characteristic (S.C.C) of a three-phase synchronous machine (alternator) as shown in the figure below. [NB: PU is per unit – dimensionless quantity] From the above figure, the short circuit ratio is given by the equation shown below: 𝐼𝑓1 𝑓𝑜𝑟 𝑟𝑎𝑡𝑒𝑑 𝑂.𝐶 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑂𝐴 𝐴𝐸 𝐴𝐸 1 Thus, SCR = = = = = 𝐴𝐵1⁄ = 𝐼𝑓2 𝑓𝑜𝑟 𝑟𝑎𝑡𝑒𝑑 𝑆.𝐶 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑂𝐷 𝐷𝐶 𝐴𝐵 𝐴𝐸 𝑋𝑆 = reciprocal of per unit synchronous reactance XS of the machine The value of synchronous reactance depends upon saturated conditions of the machine whereas, SCR is specific and defined at rated voltage. Significance of SCR: For Smaller value of SCR, larger is the value of synchronous reactance which limits the short circuit current to a smaller value. Larger value of SCR increases the stability of the machine and improves its voltage regulation. Usually, the SCR of a high-speed non-salient pole alternators lies between 0.5 and 0.75 whereas it lies between 1.0 and 1.5 for low-speed salient pole type alternators. Therefore, the salient pole type alternators are more stable than non-salient pole type alternators. Excitation System of a Synchronous Machine The word Excitation means the production of flux by passing current in the field winding. The arrangement or the system used for the excitation of the synchronous machine is known as Excitation System. The amount of excitation required depends on the load current, load power factor and speed of the machine. The excitation system is mainly classified into three types. They are: 1). DC Excitation System; 2). AC Excitation System (this is subdivided into two: Rotor Excitation System and Brushless Excitation System); 3). Static Excitation System SYNCRONOUS GENERATORS EMF Equation of Synchronous Alternator Let: P = No. of poles; Ø = Flux per pole in Wb; N = Speed in rpm; f = frequency in Hz; Zph = No. of conductors connected in series per phase Tph = No. of turns connected in series per phase Kc = Coil span factor; Kd = Distribution factor Flux cut by each conductor during one revolution = P Ø (Wb) Time taken to complete one revolution = 60/N (second) 𝑃Ø 𝑃ØN Average emf induced per conductor = = 60/𝑁 60 Average emf induced per phase equals 𝑃ØN 𝑃ØN = * Zph = * 2Tph 60 60 𝑃N = 4*Ø* Tph * 120 = 4*Ø*f* Tph R.M.S. (Root Mean Square) value of emf induced per phase: Eph = Average value*form factor Eph = 4 Ø f Tph*1·11 = 4·44 Ø f Tph (volt) Taking into consideration the coil span factor (KC) and distribution factor (Kd) of the winding. Actual emf induced per phase is given as: Eph = 4·44 Kc Kd Ø f Tph (volt) Eph = 4·44 Kw Ø f Tph (volt) Note that: Coil Span Factor: is defined as the ratio of the induced emf in a coil when the winding is short-pitched to the induced emf in the same coil when the winding is full pitched. Distribution Factor: is defined as the ratio of induced EMF in the coil group when the winding is distributed in a number of slots to the induced EMF in the coil group when the winding is concentrated in one slot. Then, Winding factor, Kw = KC *Kd E.g4 A 3-phase, 50 Hz, star connected alternator has 200 conductors per phase and the flux per pole is 0.0654 Wb. Calculate the following i. EMF induced per phase ii. EMF between the line terminals (Assume the winding to be full pitched and the distribution factor to be 0.86) Solution KC is 1 for full pitched winding i. Eph = 4·44 Kw Ø f Tph = 4.44*1*0.86*0.0654*50*(200/2) = 1248.6V ii. 𝐸𝐿 = √3 *1248.6 = 2162.6V E.g5 Calculate the no-load (line) terminal voltage of a 3-phase, 8-pole, star connected alternator running at 750 rpm having the following data: Sinusoidally distributed flux per pole = 55mWb Total No. of armature slots = 72 Number of conductors/slot = 10 Distribution factor = 0·96 Assume full pitch windings. Solution No. of poles, P = 8; Speed, Ns = 750 rpm, Flux, Ø = 55*10−3 Wb; No. of slots = 72 No. of conductors/slot = 10; Distribution factor, Kd = 0·96. For full pitch winding (Coil span factor, Kc = 1) No. of turns/phase, Tph = Z/6 (Z = total no. of stator conductors) Tph = (10*72)/6 = 120 𝑃Ns Supply frequency, f = = (8*750)/120 = 50Hz 120 Emf induced per phase, Eph = 4·44KcKd f Ø Tph = 4.44*1*0.96*50*55m*120 = 1406·6 V Since the alternator is star connected; No-load terminal voltage, EL = √3Eph = = √3 *1406.6 = 2436·3 V Equivalent Circuit & Phasor Diagrams of an Alternator A- Equivalent Circuit Fig.1: Equivalent circuit Fig.2: Equivalent circuit in a simpler form From the above, we have that: R = Armature resistance, XL = Leakage reactance, Xa = Armature reaction reactance, I = Load current E0 = Induced emf in the armature (voltage at no-load) V = Terminal voltage (PS: for three-phase machines, all quantities are phase values) B- Phasor Diagram The phasor diagram of an alternator depends upon the type of load (hence, we will look at its operation at different pf). The load may be non-inductive/resistive (I & V are in phase), inductive (I lags V) or capacitive (V lags I). In drawing phasor diagram, the below steps are followed: (a) Terminal voltage V is taken as the reference vector. (b) Load current I is drawn in phase with voltage vector V for non-inductive load (resistive). For inductive load, it is drawn so that it lags behind the voltage vector by an angle Ø (power factor angle). Whereas for capacitive load it leads the voltage vector by an angle Ø. (c) Draw voltage drop vector IR parallel to current vector from point A. (d) Draw voltage drop vector IXS from point B perpendicular (vertical) to vector AB since it is in quadrature to current vector. (e) Join point O and C, where OC is the emf E0 or the terminal voltage at no-load and is the vector sum of V, IR and IXS. (A): Phasor diagram for resistive load @ cosØ = 1(unity pf) (B): Phasor diagram for inductive load @ lagging pf (C): Phasor diagram for capacitive load @ leading pf Expression for No-Load Terminal Voltage (EMF) of an Alternator Note that the no-load terminal voltage expression here is the per phase value 1) For resistive loads From its phasor diagram above, OC2 = OB2 + BC2 E0 = √(𝑉𝑐𝑜𝑠Ø + IR)2 + (𝑉𝑠𝑖𝑛Ø + IXs)2 For 𝑐𝑜𝑠Ø = 1 E0 = √(𝑉 + IR)2 + (IXs)2 2) For Inductive loads E0 = √(𝑉𝑐𝑜𝑠Ø + IR)2 + (𝑉𝑠𝑖𝑛Ø + IXs)2 3) For capacitive loads E0 = √(𝑉𝑐𝑜𝑠Ø + IR)2 + (𝑉𝑠𝑖𝑛Ø − IXs)2 Voltage Regulation This is defined as the rise in the terminal voltage when the load is decreased from full-load rated value to zero with the excitation (field current) and speed remaining constant. Or as the change in terminal voltage from no-load to full-load (the speed and field excitation being constant) divided by full-load voltage. Thus, if E0 = no-load terminal voltage and V = terminal voltage at a given load. Therefore 𝐸𝑂 −𝑉 Voltage regulation = (the answer gotten is in per unit and thus, unitless) 𝑉 In percentage value, the above is multiplied by 100% The voltage regulation is positive both at unity and lagging p.f. because this causes rise in terminal voltage when the load is thrown off (removed). However, in case of leading p.f., the terminal voltage may fall when the load is thrown off. Therefore, at leading p.f. voltage regulation may be negative. Voltage regulation can be determined through direct load test and indirect method (Synchronous impedance method or EMF method, Ampere-turn method or MMF method and Zero power factor method or Potier method). E.g6 A single-phase 100 kVA, 600V, 50 Hz alternator has effective armature resistance and leakage reactance of 0·072 and 0·18 ohm respectively. At rated terminal voltage and kVA load, determine internal induced emf at (i) unit p.f. (ii) 0·75 p.f. lagging; (iii) 0·75 p.f. leading. Solution Rated power = 100 kVA, Terminal voltage, V = 600 V, Armature resistance, Ra = 0·072 ohm; Leakage reactance, XS = XL = 0·18 ohm, Rated current, I =100k/600 = 166.67A (from I = S/V) (i) When the p.f., cosØ = 1, sinØ = sin cos–1 1 = 0 E0 = √(𝑉 + IRa)2 + (IXs)2 = √(600 + 166.67 ∗ 0.072)2 + (166.67 ∗ 0.18)2 = 612·73 V (ii) When the p.f., cos Ø = 0·75 lagging, sin Ø = sin cos–1 0·75 = 0·6614 E0 = √(𝑉𝑐𝑜𝑠Ø + IR)2 + (𝑉𝑠𝑖𝑛Ø + IXs)2 E0 = √(600 ∗ 0.75 + 166.67 ∗ 0.072)2 + (600 ∗ 0.6614 + 166.67 ∗ 0.18)2 = 629 V (iii) When the p.f., cos Ø = 0·75 leading; sin Ø = sin cos-1 0.75 = 0·6614 E0 = √(𝑉𝑐𝑜𝑠Ø + IR)2 + (𝑉𝑠𝑖𝑛Ø − IXs)2 E0 = √(600 ∗ 0.75 + 166.67 ∗ 0.072)2 + (600 ∗ 0.6614 − 166.67 ∗ 0.18)2 = 590V E.g7 A three-phase star connected 1200 kVA, 3300 V, 50 Hz, alternator has armature resistance of 0·25 ohm per phase. A field current of 40 A produces a short circuit current of 200 A and an open circuit emf of 1100 V between lines. Calculate regulation on full load 0·8 power factor lagging. Solution Rated power = 1200 kVA, Terminal line voltage, VL = 3300 V (star connected), Armature resistance, R = 0·25ohm At field current of 40 A; Short circuit current, Isc = 200A, Open circuit emf (phase value), Eph = 1100/√3= 635.1V Synchronous impedance, Zs = Eph/ Isc = 3.175ohm Synchronous reactance, Xs = √𝑧𝑠2 − 𝑅 2 =√3.1752 − 0.252 = 3.175ohm Full load, current, I = 1200k/(√3 ∗ 3300) = 210A (S = √3 VLIL) Terminal phase voltage, V = VL/√3 = 3300/√3 = 1905.2V Power factor, cosØ = 0.8; sinØ= sin cos–1 0·8 = 0·6 Open circuit terminal voltage (phase value), E0 = √(𝑉𝑐𝑜𝑠Ø + IR)2 + (𝑉𝑠𝑖𝑛Ø + IXs)2 E0 = √(1905.2 ∗ 0.8 + 210 ∗ 0.25)2 + (1905.2 ∗ 0.6 + 210 ∗ 3.175)2 = 2400V  𝐸𝑂 −𝑉 2400−1905.2 %Voltage regulation = = ∗ 100 = 25·98% 𝑉 1905.2 Power and Torque of Synchronous Generator A- POWER For 1-Ø system: P=VI For 3-Ø system: Output Power, with respect to line values is given as: P = √3 VLILcosØ (P is real/active power) Q = √3 VLILsinØ (Q is reactive power) Output Power, with respect to phase values is given as: P = 3VphIphcosØ Q = 3VphIphsinØ Working with the phase value of real power output, we have that, P = 3VphIacosØ (1) Where: P = power output, Vph = phase voltage, Ia = armature current, cosØ = power factor Figure: Phasor diagram of a synchronous machine (@ lagging pf) From the above diagram, b is added and thus will be used as a guide. Therefore, taking the sine of the b vector triangle section, we have that, / sin(90-Ø) = b IaXs b= Ia*Xs*sin(90-Ø) = Ia*Xs*cosØ Also, taking the sine of the E triangle section, we have that, Sinα = b/E Ia∗Xs∗cosØ = 𝐸 E∗sinα Ia*cosØ = (2) Xs Combining eqn. 1 and 2, we have that 3∗𝑉𝑝ℎ ∗E∗sinα P = 3VphIacosØ = (this is the power equation) Xs At maximin power, sinα =1 (that is α = 90º). Hence, 3∗𝑉𝑝ℎ ∗E P= Xs Diagram of power Vs Load angle (@ α = 90º) B- TORQUE P = ωT 3∗𝑉𝑝ℎ ∗E 3∗𝑉𝑝ℎ ∗E T = P/ω = /ω= / 2πNs (this is torque equation @ maximum power) Xs Xs Losses in Synchronous Machine and Efficiency A synchronous machine is used to convert mechanical energy into electrical energy or vice-versa. While doing so, the whole of input energy does not appear at the output but a part of it is lost in the form of heat in the surroundings. This wasted energy is called losses in the machine. These losses affect the efficiency of the machine. A reduction in these losses leads to higher efficiency. Thus, the major objective in the design of a synchronous machine is to reduce these losses. The various losses occurring in a synchronous machine can be sub-divided as: 1. Copper losses. 2. Iron losses. others are 3. Mechanical losses 4. Stray losses 1. Copper losses: The various windings of the synchronous machine such as armature and field windings are made of copper and have some resistance. When current flows through them, there will be power loss proportional to the square of their respective currents. These power losses are called copper losses. In general, the various copper losses in a synchronous machine are: (i) Armature copper loss = I2R (ii) Field winding copper loss = 𝐼𝑓2 Rf (iii) Brush contact loss = I2Rb The brush contact loss is generally included in field winding copper losses. 2. Iron losses: The losses which occur in the iron parts of the machine are called iron losses or core losses or magnetic losses. These losses consist of the following: (i) Hysteresis loss: Whenever a magnetic material is subjected to reversal of magnetic flux, this loss occurs. It is due to retentivity (a property) of the magnetic material. It is expressed with reasonable accuracy by the following expression: 1.6 Ph = Kh V f 𝐵𝑚 Where; Kh = hysteresis constant in J/m3 i.e., energy loss per unit volume of magnetic material during one magnetic reversal, its value depends upon nature of material; V = volume of magnetic material in m3. f = frequency of magnetic reversal in cycle/second and Bm = maximum flux density in the magnetic material in tesla. It occurs in the armature (stator core). To minimise this loss, the armature core is made of silicon steel which has low hysteresis constant. (ii) Eddy current loss: When flux linking with the magnetic material changes (or flux is cut by the magnetic material) an emf is induced in it which circulates eddy currents through it. These eddy currents produce eddy current loss in the form of heat. It is expressed with reasonable accuracy by the expression: Pe = KeVf 2t2𝐵𝑚 2 Where; Ke = constant called co-efficient of eddy current; its value depends upon the nature of magnetic material; t = thickness of lamination in m; V, f and Bm are the same as above. The major part of this loss occurs in the armature core. To minimise this loss, the armature core is laminated into thin sheets (0·3 to 0·5 mm) since this loss is directly proportional to the square of thickness of the laminations. 3. Mechanical losses: As the field system of a synchronous machine is a rotating part, some power is required to overcome: (i) Air friction of rotating field system (windage loss). (ii) Friction at the bearing and friction between brushes and slip rings (friction loss). These losses are known as mechanical losses. To reduce these losses proper lubrication is done at the bearings. 4. Stray losses. In addition to the iron losses, the core losses are also caused by distortion of the magnetic field under load conditions and losses in insulation of armature and field winding, these losses are called stray losses. These losses are also included while determining the efficiency of synchronous machines. Stray losses = Iron losses + Mechanical losses. Efficiency of a Synchronous Generator The ratio of output power to the input power of a synchronous generator is called its efficiency. Output Input−Losses Losses Output Efficiency, ŋ = = = 1- = 𝑂𝑢𝑡𝑝𝑢𝑡+𝐿𝑜𝑠𝑠𝑒𝑠 𝐼𝑛𝑝𝑢𝑡 𝐼𝑛𝑝𝑢𝑡 𝐼𝑛𝑝𝑢𝑡 Power Flow Diagram of Synchronous Generator The power flow diagram for a synchronous machine working as a generation is shown below Power flow diagram of a Synchronous Generator Methods of Cooling: Cooling methods of synchronous machine are broadly divided into; (i) Open-circuit cooling (ii) Closed-circuit cooling (check them out in your leisure time) What are the applications of synchronous generators……Check it out Check this out as well: A 1-phase 60 kVA, 220 V, 50 Hz, alternator has an effective armature leakage reactance of 0·07 ohm and negligible armature resistance. Calculate the voltage induced in the armature when the alternator is delivering rated current at a load power factor of 0·7 lagging. A single-phase, 500 V, 50 Hz alternator produces a short-circuit current of 170 A and an open circuit emf of 425 V when a field current of 15A passes through its field winding. If its armature has an effective resistance of 0.2 ohm, determine its full-load regulation at unity pf and at 0.8 pf lagging. A three-phase star-connected alternator has an armature resistance of 0·1 ohm per phase. When excited to 173·3 V line voltage and on short circuit the alternator gave 200 A. What should be the emf (in line terms) the alternator must be excited to, in order to maintain a terminal potential difference of 400 V with 100 A armature current at 0·8 power factor lagging? PARALLEL OPERATION OF ALTERNATORS To meet with the ever-increasing demand of electrical power, it is economical and advisable to run number of generating units in parallel. Although a single larger unit used to meet with the demand is more economical but it reduces the reliability. In fact, there are a number of good reasons to use number of smaller units operating in parallel to meet with the existing demand. In the present scenario, not only the number of units placed at one generating station are operated in parallel rather all the other units placed at the other generating stations which are interconnected are also operating in parallel with each other. Necessity of Parallel Operation of Alternators To meet with the demand of electrical power, a variety of generating stations having large capacity can be erected. In the modern generating stations, where huge power is generated, several alternators are operated in parallel. When number of alternators are connected to same bus-bars, they are said to be connected in parallel. Such practice is considered necessary for the following reasons: 1. Physical size: The output power of modern power stations is so high that it is difficult to build a single unit (large generator) of that capacity. 2. Reliability or continuity of service: Several small units are more reliable than a single large unit because if one unit fails, the continuity of supply can be maintained by operating the other units. 3. Repair and maintenance: Repair and maintenance of a unit is more convenient and economical if a large number of smaller units are installed at the power station. 4. Size and cost of stand-by unit: Since each unit is of smaller size, the cost of stand by unit is small. 5. Extension of power plant: The additional unit can be installed as and when the load demand increases. 6. Operating efficiency: Moreover, the load on the power station varies greatly both during day and night as well as during the different seasons. Thus, the number of units operating at a particular time can be varied depending upon the load at that time. This keeps the machines loaded up to their rated capacity and hence results in increase in efficiency of operation as the efficiency of an electrical machine is maximum at or near rated capacity. Requirements for Parallel Operation of Alternators The procedure of connecting an alternator in parallel with another or with common bus-bars to which a number of alternators are already connected, is called synchronising of alternators. 1. Output voltage rating: The output voltage rating of all the alternators must be the same 2. Output frequency: The rated speed of all the machines should be such that they produce the same frequency (f = PN/120). 3. Output wave shape: The output wave form of all the alternators must be the same, although their kVA rating may be different. 4. Speed-load characteristics: The drooping speed-load characteristics of the prime-movers of the alternators should be the same so that alternators share the load in their proportion as per their output (kVA) rating. 5. Impedance triangles: The impedance triangles of the alternators should be identical for successful parallel operation. Synchronising Single-Phase Alternators Synchronising of single-phase alternators is generally done by using lamps called lamp methods. There are two methods, namely, (i) Dark lamp method (ii) Bright lamp method (check them out in your leisure time) Synchronising Three-Phase Alternators For synchronising of three-phase alternators, two methods are used i.e., (i) By using lamps, (ii) By using synchroscope. SYNCHRONOUS MOTORS The same synchronous machine can be used as a motor. When it converts electric power or energy into mechanical power or energy, it is called a synchronous motor. Working Principle of a Synchronous Motor A synchronous motor operates on the principle of magnetic locking. It is a situation in a synchronous motor where the rotor is locked in synchronism with the rotating magnetic field produced by the stator. In this condition, the rotor’s magnetic field is aligned with the stator’s magnetic field, allowing the motor to operate at synchronous speed. Here’s a detailed breakdown of its working principle [First recall: Stator - the stationary part of the motor that contains coils of wire (windings) through which alternating current (AC) is passed, creating a rotating magnetic field. And Rotor - the rotating part of the motor]: When three-phase AC power is supplied to the stator windings, a rotating magnetic field is created. The speed of this magnetic field is known as the synchronous speed. The rotor is designed to follow this rotating magnetic field. The rotor rotates at the same speed as the stator’s magnetic field, hence the term ‘synchronous’ (This synchronization occurs due to the magnetic forces acting on the rotor). And as long as the rotor maintains this synchronization, the motor operates efficiently. The interaction between the magnetic field of the rotor and the magnetic field of the stator produces torque, which causes the rotor to turn and perform mechanical work Equivalent Circuit of a Synchronous Motor Simplified Equivalent Circuit of a 3-Ø Synchronous Motor Let: V = Terminal voltage per phase applied to the motor; Ef = Excitation voltage or back emf (𝐸𝑏 ); Ia = Armature current per phase drawn by the motor from the supply; Ra = Effective armature resistance per phase; XS = Synchronous reactance per phase of the motor armature winding; ZS = Synchronous impedance per phase of the armature By applying KVL (Kirchhoff’s Voltage Law) to the equivalent circuit of the motor above, we have, V - IaZs - Ef = 0 V=Ef +IaZs (1) The synchronous impedance of the motor is ZS=Ra+jXS (2) Combining 1 and 2, we have that: V=Ef + Ia (Ra+jXS) Therefore, Ef = 𝐸𝑏 =V−IaRa−jIaXS = V-IaZS Thus, armature current per phase is given by; 𝑉−𝐸𝑓 𝐸 Ia = = 𝑟 𝑍𝑆 𝑍𝑆 Where 𝐸𝑟 = resultant voltage in the armature circuit And if: Er = V, then the synchronous motor is said to be normally excited. Er < V, then the synchronous motor is said to be under-excited. Er > V, then the synchronous motor is said to be over-excited. Note that: IaRa is the voltage drop per phase in the armature resistance. IaXS is the voltage drop per phase due to armature reactance. What is this back emf? A motor has coils turning inside magnetic fields, and a coil turning inside a magnetic field induces an emf. This emf, known as the back emf, acts against the applied voltage that's causing the motor to spin in the first place, and reduces the current flowing through the coils of the motor. Phasor Diagrams of a Synchronous Motor (Specifically, Cylindrical Rotor Type) Given that: Ef = Excitation voltage; V = Terminal voltage; Ia = Armature current per phase drawn by the motor from the supply; Ra = armature resistance; XS = Synchronous reactance; Cosφ = pf; δ = Torque angle i. For unity power factor The phasor diagram is as below: Suppose that the synchronous motor is drawing the current (Ia) from the supply at unity power factor. Here, the supply voltage (V) is taken as the reference phasor along OA such that OA = V. For unity power factor, the armature current (Ia) drawn by the motor is in phase with the supply voltage (V) and is represented by OB, where OB = Ia. The voltage drop in the armature resistance is IaRa which is in phase with the armature current. The phasor IaRa is represented by CD. The voltage drop per phase in the synchronous reactance is IaXS. The phasor is IaXS in a direction perpendicular to the phasor IaRa and is represented by DA. Therefore, the phasor V is equal to the phasor sum of Ef, IaRa and jIaXS. The angle δ between V and Ef is called the torque angle. ii. For lagging power factor The phasor diagram is as below: Consider the synchronous motor is taking a lagging current from the supply. Here, the supply voltage (V) is taken as the reference phasor along OA such that OA = V. For lagging power factor cosφ, the armature current (Ia) lags behind the supply voltage (V) by an angle φ along OB where OB = Ia. The voltage drop in the armature resistance is IaRa which is in phase with the armature current. The phasor IaRa is represented by CD. The voltage drop per phase in the synchronous reactance is IaXS. The phasor is in a direction perpendicular to the phasor IaRa and is represented by DA. Therefore, the phasor V is equal to the phasor sum of Ef, and IaRa. The angle δ between V and Ef is called the torque angle. iii. For leading power factor The phasor diagram is as below: Suppose that the synchronous motor is drawing the current (Ia) from the supply at leading power factor cosφ. Here, the supply voltage (V) is taken as the reference phasor along OA such that OA = V. For leading power factor, the armature current (Ia) drawn by the motor leads the supply voltage (V) by the phase angle φ and is represented by OB where OB = Ia. The voltage drop in the armature resistance is IaRa which is in phase with the armature current. The phasor IaRa is represented by CD. The voltage drop per phase in the synchronous reactance is IaXS. The phasor is IaXS in a direction perpendicular to the phasor IaRa and is represented by DA. Therefore, the phasor V is equal to the phasor sum of Ef, IaRa and jIaXS. Expression for No-Load Terminal Voltage (EMF) of a Synchronous Motor Note that the EMF equation here is the per phase value a. For leading PF E = √(𝑉𝑐𝑜𝑠Ø − IR)2 + (𝑉𝑠𝑖𝑛Ø + IXs)2 b. For lagging PF E = √(𝑉𝑐𝑜𝑠Ø − IR)2 + (𝑉𝑠𝑖𝑛Ø − IXs)2 c. For unity PF E = √(𝑉𝑐𝑜𝑠Ø − IR)2 + (𝑉𝑠𝑖𝑛Ø + IXs)2 For 𝑐𝑜𝑠Ø = 1 E = √(𝑉 − IR)2 + (IXs)2 Mechanical Power and Gross Torque Developed in a Synchronous Motor The mechanical power developed (Pm) by a three-phase synchronous motor will be derived herein. Here, we will neglect the armature resistance Ra of the synchronous motor. Then, the armature copper loss will be zero and hence the mechanical power developed by the motor is equal to the input power (Pin) to the motor, i.e., Pm (output) = Pin (input) [Recall that Input = output + losses (in this case copper loss)] We will consider an under-excited (i.e. Er