Construction of alternator or synchronous machines. Cylindrical type machines

Construction of alternator or synchronous machines. Cylindrical type machines

1. ELECTRICAL MACHINES II Unit I

2. SYLLABUS Unit – I Synchronous Generators Three Phase Synchronous Generators: Synchronous Generator Construction (both Cylindrical Rotor and Salient Pole type), the Speed of Rotation of a Synchronous Generator, Induced voltage in A.C Machines, The Equivalent Circuit of a Synchronous Generator (Armature Reaction Reactance, Synchronous Reactance and Impedance). Cylindrical Rotor type Three Phase Synchronous Generators: The Phasor Diagram of a Synchronous Generator, Power and Torque in Synchronous Generators (Power Angle Equation and Power Angle Characteristic), Measuring Synchronous Generator Model Parameters (Open Circuit and Short Circuit Tests and Determination of Synchronous Impedance and Reactance, The Short Circuit Ratio), Voltage Regulation and Speed Regulation. Voltage Regulation by Synchronous Impedance Method .

3. SYLLABUS Salient Pole type Three Phase Synchronous Generators: Two Reaction Concept, Development of the Equivalent Circuit of a Salient Pole type Three Phase Synchronous Generator (Direct axis and Quadrature axis Reactance, Phasor Diagram for various load power factors,), Torque and Power Equations of Salient Pole Synchronous Generator (Power Angle Equation and Power Angle Characteristic with stator resistance neglected). Slip Test for determination of Direct axis and Quadrature axis Reactance. Parallel operation of Three Phase A.C. Synchronous Generators: The Conditions Required for synchronization, The General Procedure for Paralleling Generators. Unit – II: Salient Pole type Three Phase Synchronous Generators

4. SYLLABUS

5. SYLLABUS

6. SYLLABUS

7. Three Phase Synchronous Generators An alternator operates on the same fundamental principle of electromagnetic induction as a d.c. generator i.e., when the flux linking a conductor changes, an e.m.f . is induced in the conductor. Like a d.c. generator, an alternator also has an armature winding and a field winding. But there is one important difference between the two. In a d.c. generator, the armature winding is placed on the rotor in order to provide a way of converting alternating voltage generated in the winding to a direct voltage at the terminals through the use of a rotating commutator . The field poles are placed on the stationary part of the machine. Since no commutator is required in an alternator, it is usually more convenient and advantageous to place the field winding on the rotating part (i.e., rotor) and armature winding on the stationary part

8. Advantages of stationary armature: It is easier to insulate stationary winding for high voltages for which the alternators are usually designed. This is because they are not subjected to Centrifugal forces and also extra space is available due to the stationary arrangement of the armature. The stationary 3-phase armature can be directly connected to load without going through large, unreliable slip rings and brushes. Only two slip rings are required for d.c. supply to the field winding on the rotor. Since the exciting current is small, the slip rings and brush gear required are of light construction. Due to simple and robust construction of the rotor, higher speed of rotating d.c. field is possible. This increases the output obtainable from a machine of given dimensions.

9. Construction of Alternator 1. Stator It is the stationary part of the machine and is built up of sheet-steel laminations having slots on its inner periphery. A 3-phase winding is placed in these slots and serves as the armature winding of the alternator. The armature winding is always connected in star and the neutral is connected to ground.

10. 2. Rotor   The rotor carries a field winding which is supplied with direct current through two slip rings by a separate d.c. source. This d.c. source (called exciter) is generally a small d.c. shunt or compound generator mounted on the shaft of the alternator. Rotor construction is of two types, namely; ( i ) Salient (or projecting) pole type (ii) Non-salient (or cylindrical) pole type

11. Salient pole type In this type, salient or projecting poles are mounted on a large circular steel frame which is fixed to the shaft of the alternator. The individual field pole windings are connected in series in such a way that when the field winding is energized by the d.c. exciter, adjacent poles have opposite polarities. Low and medium-speed alternators (120-400 r.p.m .) such as those driven by diesel engines or water turbines have salient pole type rotors due to the following reasons: (a) The salient field poles would cause an excessive windage loss if driven at high speed and would tend to produce noise. (b) Salient-pole construction cannot be made strong enough to withstand the mechanical stresses to which they may be subjected at higher speeds.

12. Non-salient pole type In this type, the rotor is made of smooth solid forged-steel radial cylinder having a number of slots along the outer periphery. The field windings are embedded in these slots and are connected in series to the slip rings through which they are energized by the d.c. exciter. The regions forming the poles are usually left un slotted. It is clear that the poles formed are non-salient i.e., they do not project out from the rotor surface. High-speed alternators (1500 or 3000 r.p.m .) are driven by steam turbines and use non-salient type rotors due to the following reasons: (a) This type of construction has mechanical robustness and gives noiseless operation at high speeds. (b) The flux distribution around the periphery is nearly a sine wave and hence a better e.m.f . waveform is obtained than in the case of salient-pole type.

13. Turbine - Alternator

14. Frequency and Speed The frequency of induced e.m.f . in the armature conductors depends upon speed and the number of poles. Let N = rotor speed in r.p.m . P = number of rotor poles f = frequency of e.m.f . in Hz Consider a stator conductor that is successively swept by the N and S poles of the rotor. If a positive voltage is induced when a N-pole sweeps across the conductor, a similar negative voltage is induced when a S-pole sweeps by. This means that one complete cycle of e.m.f . is generated in the conductor as a pair of poles passes it i.e., one N-pole and the adjacent following S-pole. The same is true for every other armature conductor.

15. No. of cycles/revolution = No. of pairs of poles = P/2 No. of revolutions/second = N/60 No. of cycles/second = (P/2)(N/60) = N P/120 But number of cycles of e.m.f . per second is its frequency. f = NP \ 120 It may be noted that N is the synchronous speed and is generally represented by Ns. For a given alternator, the number of rotor poles is fixed and, therefore, the alternator must be run at synchronous speed to give an output of desired frequency. For this reason, an alternator is sometimes called synchronous generator.

16. Armature Winding of Alternator: Winding Factors: The armature winding of an alternator is distributed over the entire armature. The distributed winding produces nearly a sine waveform and the heating is more uniform. Likewise, the coils of armature winding are not full-pitched i.e., the two sides of a coil are not at corresponding points under adjacent poles. The fractional pitched armature winding requires less copper per coil and at the same time waveform of output voltage is unproved. The distribution and pitching of the coils affect the voltages induced in the coils. We shall discuss two winding factors: ( i ) Distribution factor ( Kd ), also called breadth factor (ii) Pitch factor ( Kp ), also known as chord factor

17. Distribution factor ( K d ) A winding with only one slot per pole per phase is called a concentrated winding. In this type of winding, the e.m.f . generated/phase is equal to the arithmetic sum of the individual coil e.m.f.s in that phase. However, if the coils/phase are distributed over several slots in space (distributed winding), the e.m.f.s in the coils are not in phase (i.e., phase difference is not zero) but are displaced from each by the slot angle a (The angular displacement in electrical agrees between the adjacent slots is called slot angle).

18. The e.m.f ./phase will be the phasor sum of coil e.m.f.s . The distribution factor Kd is defined as:

19. Pitch factor ( K p ) A coil whose sides are separated by one pole pitch (i.e., coil span is 180 ° electrical) is called a full-pitch coil. With a full-pitch coil, the e.m.f.s induced in the two coil sides a in phase with each other and the resultant e.m.f . is the arithmetic sum of individual e.m.fs . However the waveform of the resultant e.m.f . can be improved by making the coil pitch less than a pole pitch. Such a coil is called short-pitch coil. This practice is only possible with double-layer type of winding The e.m.f . induced in a short-pitch coil is less than that of a fullpitch coil.

20. The factor by which e.m.f . per coil is reduced is called pitch factor Kp . It is defined as:

21. Advantages of Short Pitched Coil They save copper of end connections. They improve the wave-form of the generated e.m.f . i.e. the generated e.m.f . can be made to approximate to a sine wave more easily and the distorting harmonics can be reduced or totally eliminated. Due to elimination of high frequency harmonics, eddy current and hysteresis losses are reduced thereby increasing the efficiency. But the disadvantage of using short-pitched coils is that the total voltage around the coils is somewhat reduced. Because the voltages induced in the two sides of the short-pitched coil are slightly out of phase, their resultant vectorial sum is less than their arithmetical sum.

22. Calculate the pitch factor for the under-given windings : (a) 36 stator slots, 4-poles, coil-span, 1 to 8 (b) 72 stator slots, 6 poles, coils span 1 to 10 and (c) 96 stator slots, 6 poles, coil span 1 to 12. Sketch the three coil spans. Calculate the distribution factor for a 36-slots, 4-pole, single-layer three-phase winding.- Ans : 0.96 A part of an alternator winding consists of six coils in series, each coil having an e.m.f . of 10 V r.m.s . induced in it. The coils are placed in successive slots and between each slot and the next, there is an electrical phase displacement of 30º. Find graphically or by calculation, the e.m.f . of the six coils in series. ANS : 38.74

23. E.M.F. Equation of an Alternator Let Z = No. of conductors or coil sides in series per phase Φ = Flux per pole in webers P = Number of rotor poles N = Rotor speed in r.p.m .   In one revolution (i.e., 60/N second), each stator conductor is cut by P Φ webers i.e.,   d Φ = P Φ ; dt = 60/N Average e.m.f . induced in one stator conductor

24. Effect of Harmonics on Pitch and Distribution Factors If the short-pitch angle or chording angle is β degrees (electrical) for the fundamental flux wave, then its values for different harmonics are for 3rd harmonic = 3 α ; for 5th harmonic = 5 α and so on. ∴ pitch-factor, k p = cos β/2 — for fundamental = cos 3 β/2 — for 3rd harmonic = cos 5 β /2 —for 5th harmonic etc.

25. Similarly, the distribution factor is also different for different harmonics. Its value becomes k d = (sin m  /2) / m sin  /2 for fundamental, n = 1 , k d = (sin m  /2) / m sin  /2 for 3rd harmonic, n = 3, k d = (sin 3m  /2) / m sin 3  /2 for 5th harmonic, n = 5, k d = (sin 3m  /2) / m sin 5  /2

26. An alternator has 18 slots/pole and the first coil lies in slots 1 and 16. Calculate the pitch factor for ( i ) fundamental (ii) 3rd harmonic (iii) 5th harmonic and (iv) 7th harmonic. A 3-phase, 16-pole alternator has a star-connected winding with 144 slots and 10 conductors per slot. The flux per pole is 0.03 Wb , Sinusoidally distributed and the speed is 375 r.p.m . Find the frequency rpm and the phase and line e.m.f . Assume full-pitched coil. Find the no-load phase and line voltage of a star-connected 3-phase, 6-pole alternator which runs at 1200 rpm, having flux per pole of 0.1 Wb sinusoidally distributed. Its stator has 54 slots having double layer winding. Each coil has 8 turns and the coil is chorded by 1 slot. The stator of a 3-phase, 16-pole alternator has 144 slots and there are 4 conductors per slot connected in two layers and the conductors of each phase are connected in series. If the speed of the alternator is 375 r.p.m ., calculate the e.m.f . inducted per phase. Resultant flux in the air-gap is 5 × 10 − 2 webers per pole sinusoidally distributed. Assume the coil span as 150° electrical.

27. Alternator on Load When the load on the alternator is increased (i.e., armature current I a is increased), the field excitation and speed being kept constant, the terminal voltage V (phase value) of the alternator decreases. This is due to Voltage drop I a R a where R a is the armature resistance per phase. Voltage drop I a X L where X L is the armature leakage reactance per phase. Voltage drop because of armature reaction.

28. Armature Resistance (Ra) Since the armature or stator winding has some resistance, there will be an I a R a drop when current ( I a ) flows through it. The armature resistance per phase is generally small so that I a R a drop is negligible for all practical purposes.

29. Armature Leakage Reactance (X L ) When current flows through the armature winding, flux is set up and a part of it does not cross the air-gap and links the coil sides. This leakage flux alternates with current and gives the winding self-inductance. This is called armature leakage reactance. Therefore, there will be I a X L drop which is also effective in reducing the terminal voltage.

30. Armature reaction The load is generally inductive and the effect of armature reaction is to reduce the generated voltage. Since armature reaction results in a voltage effect in a circuit caused by the change in flux produced by current in the same circuit, its effect is of the nature of an inductive reactance. Therefore, armature reaction effect is accounted by assuming the presence of a fictitious reactance X AR in the armature winding. The quantity X AR is called reactance of armature reaction. The value of X AR is such that I a X AR represents the voltage drop due to armature reaction.

31. Equivalent Circuit

32. Equivalent Circuit The Given fig shows the equivalent circuit of the loaded alternator for one phase. All the quantities are per phase. Here E = No-load e.m.f . E = Load induced e.m.f . It is the induced e.m.f . after allowing for armature reaction. It is equal to phasor difference of E and I a X AR . V = Terminal voltage. It is less than E by voltage drops in X L and R a .

33. Synchronous Reactance ( X s ) The sum of armature leakage reactance (X L ) and reactance of armature reaction (X AR ) is called synchronous reactance X s

34. Phasor Diagram of a Loaded Alternator

35. A 3-phase, star-connected alternator supplies a load of 10 MW at p.f . 0.85 lagging and at 11 kV (terminal voltage). Its resistance is 0.1 ohm per phase and synchronous reactance 0.66 ohm per phase. Calculate the line value of e.m.f . generated. The effective resistance of a 2200V, 50Hz, 440 KVA, 1-phase, alternator is 0.5 ohm. On short circuit, a field current of 40 A gives the full load current of 200 A. The electromotive force on open-circuits with same field excitation is 1160 V. Calculate the synchronous impedance and reactance.

36. Voltage Regulation The voltage regulation of an alternator is defined as the change in terminal voltage from no-load to full-load (the speed and field excitation being constant) divided by full-load voltage. The factors affecting the voltage regulation of an alternator are: ( i ) IaRa drop in armature winding (ii) IaXL drop in armature winding (iii) Voltage change due to armature reaction

37. Determination of Voltage Regulation 1. Synchronous Impedance or E.M.F. Method. It is due to Behn Eschenberg . 2. The Ampere-turn or M.M.F. Method. This method is due to Rothert . 3. Zero Power Factor or Potier Method. As the name indicates, it is due to Potier . All these Methods Require— 1. Armature (or stator) resistance Ra 2. Open-Circuit / No-Load Characteristic 3. Short-Circuit Characteristic

38. Armature (or stator) resistance Ra The armature resistance R a per phase is determined by using direct current and the voltmeter-ammeter method. This is the d.c. value. The effective armature resistance ( a.c . resistance) is greater than this value due to skin effect. It is a usual practice to take the effective resistance 1.5 times the d.c. value (R a = 1.5 R dc ).

39. Open-Circuit / No-Load Characteristic Like the magnetization curve for a d.c. machine, the (Open-circuit characteristic of an alternator is the curve between armature terminal voltage (phase value) on open circuit and the field current when the alternator is running at rated speed. Fig. shows the circuit for determining the O.C.C. of an alternator. The alternator is run on no-load at the rated speed. The field current I f is gradually increased from zero (by adjusting field rheostat) until open-circuit voltage E (phase value) is about 50% greater than the rated phase voltage.

40. Open-Circuit / No-Load Characteristic The graph is drawn between open-circuit voltage values and the corresponding values of I f as shown in Fig

41. Short-circuit characteristic (S.C.C.) In a short-circuit test, the alternator is run at rated speed and the armature terminals are short-circuited through identical ammeters. Only one ammeter need be read; but three are used for balance. The field current I f is gradually increased from zero until the short-circuit armature current I SC is about twice the rated current.

42. Short-circuit characteristic (S.C.C.) The graph between short-circuit armature current and field current gives the short-circuit characteristic (S.C.C.) as shown in Fig.

43. Synchronous Impedance Method In this method of finding the voltage regulation of an alternator, we find the synchronous impedance Z s (and hence synchronous reactance X s ) of the alternator from the O.C.C. and S.S.C. For this reason, it is called synchronous impedance method. The method involves the following steps: Step 01 : Plot the O.C.C. and S.S.C. on the same field current base as shown in Fig. Step 02 : Consider a field current If. The open-circuit voltage corresponding to this field current is E1. The short-circuit armature current corresponding to field current If is I1. On short-circuit p.d . = 0 and voltage E1 is being used to circulate the snort-circuit armature current I1 against the synchronous impedance Zs . This is illustrated in Fig.

44. Step 03: Step 04: Determination of E for Lagging, Leading and Unity P.F. Step 05:

45. The effective resistance of a 2200V, 50Hz, 440 KVA, 1-phase, alternator is 0.5 ohm. On short circuit, a field current of 40 A gives the full load current of 200 A. The electromotive force on open-circuits with same field excitation is 1160 V. Calculate the synchronous impedance and reactance. A 60-KVA, 220 V, 50-Hz, 1- φ alternator has effective armature resistance of 0.016 ohm and an armature leakage reactance of 0.07 ohm. Compute the voltage induced in the armature when the alternator is delivering rated current at a load power factor of (a) unity (b) 0.7 lagging and (c) 0.7 leading. In a 50-kVA, star-connected, 440-V, 3-phase, 50-Hz alternator, the effective armature resistance is 0.25 ohm per phase. The synchronous reactance is 3.2 ohm per phase and leakage reactance is 0.5 ohm per phase. Determine at rated load and unity power factor : (a) Internal e.m.f . Ea (b) no-load e.m.f . E0 (c) percentage regulation on full-load (d) value of synchronous reactance which replaces armature reaction.

46. Find the synchronous impedance and reactance of an alternator in which a given field current produces an armature current of 200 A on short-circuit and a generated e.m.f . of 50 V on open-circuit. The armature resistance is 0.1 ohm. To what induced voltage must the alternator be excited if it is to deliver a load of 100 A at a p.f . of 0.8 lagging, with a terminal voltage of 200V.

47. Find the synchronous impedance and reactance of an alternator in which a given field current produces an armature current of 200 A on short-circuit and a generated e.m.f . of 50 V on open-circuit. The armature resistance is 0.1 ohm. To what induced voltage must the alternator be excited if it is to deliver a load of 100 A at a p.f . of 0.8 lagging, with a terminal voltage of 200V.

48. From the following test results, determine the voltage regulation of a 2000-V, 1-phase alternator delivering a current of 100 A at ( i ) unity p.f . (ii) 0.8 leading p.f . and (iii) 0.71 lagging p.f . Test results : Full-load current of 100 A is produced on short-circuit by a field excitation of 2.5A. An e.m.f . of 500 V is produced on open-circuit by the same excitation. The armature resistance is 0.8 Ω .

49. A 100-kVA, 3000-V, 50-Hz 3-phase star-connected alternator has effective armature resistance of 0.2 ohm. The field current of 40 A produces short-circuit current of 200 A and an open-circuit emf of 1040 V (line value). Calculate the full-load voltage regulation at 0.8 p.f . lagging and 0.8 p.f . leading. Draw phasor diagrams.

50. A 3-phase, star-connected alternator is rated at 1600 kVA, 13,500 V. The armature resistance and synchronous reactance are 1.5 Ω and 30 Ω respectively per phase. Calculate the percentage regulation for a load of 1280 kW at 0.8 leading power factor.

51. A 3-phase, 10-kVA, 400-V, 50-Hz, Y-connected alternator supplies the rated load at 0.8 p.f . lag. If arm. resistance is 0.5 ohm and syn. reactance is 10 ohms, find the power angle and voltage regulation.

52. The following test results are obtained from a 3-phase, 6,000-kVA, 6,600 V, star-connected, 2-pole, 50-Hz turbo-alternator: With a field current of 125 A, the open-circuit voltage is 8,000 V at the rated speed; with the same field current and rated speed, the short-circuit current is 800 A. At the rated full-load, the resistance drop is 3 per cent. Find the regulation of the alternator on full-load and at a power factor of 0.8 lagging.

53. A given 3-MVA, 50-Hz, 11-kV, 3- φ , Y-connected alternator when supplying 100 A at zero p.f . leading has a line-to-line voltage of 12,370 V; when the load is removed, the terminal voltage falls down to 11,000 V. Predict the regulation of the alternator when supplying full-load at 0.8 p.f . lag. Assume an effective resistance of 0.4 Ω per phase

54. Power and Torque in Synchronous Generators A sy n c hron o u s ge n era t o r i s a sy n c hron o u s ma c hin e u sed as a ge n e rator. It co n verts m ec hani ca l power t o thre e-p ha se e l ec tri ca l power. Th e so urce o f me c hani ca l power, the prime mover, ma y be a diesel e n g in e, a s t ea m turbine , a water turbine , o r any s imilar device. Whatev e r th e so ur ce, it mu s t have the basic property that it s speed i s almost co n s tant rega rdle ss o f th e power demand. I f that were n o t so, then th e resulting power sys tem s fr e qu e n cy would wander. Not all th e m ec hani ca l power go in g int o a sy n c hron o u s ge n era t o r becomes e le c tri ca l power o ut of the ma c hin e . The dif f erence between input power and o utput power represe nt s th e l osses of the ma c hin e.

55. T he input me c hani ca l power i s the s haft power in th e ge n era t o r P in = T app w m , While th e power co n ve rt ed from m ec hani ca l t o e l ec tri ca l form internally i s g iv e n by P conv = T ind w m The r ea l e le c tri ca l o utput power o f th e sy n c hron o u s ge n e rator ca n be ex pressed in lin e quantiti es as   P out = and in phase quantities as P out =  

56. In the above figure Due to which Maximum value of power when angle is 90 As we have considered Ra is zero so P conv = P out

57. Power angle Characteristic:

58. The Short-Circuit Ratio Another parameter u sed t o describe sy n c hronou s ge nerat ors i s th e s h o rt- c ir c uit ra ti o . The short-circuit ratio of a ge n era tor i s defined as th e rati o o f th e field current required for the rated voltage at open circuit t o the field current required for the rated armature current at short circuit.

59. Find the power angle when a 1500-kVA, 6.6 kV, 3-phase, Y-connected alternator having a resistance of 0.4 ohm and a reactance of 6 ohm per phase delivers full-load current at normal rated voltage and 0.8 p.f . lag. Draw the phasor diagram.

60. END OF UNIT – 01