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DC Generator
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DC Generator

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  1. DC Generator Types of Generators: Generators are usually classified according to the way in which their fields are excited. The classification of DC generator is shown as follows:

  2. Fig. 26.41. Equivalent Circuit of separately-excited generator. Generators may be divided into: Separately-excited generator: Separately-excited generators are those whose field magnets are energized from an independent external source of DC current.It is shown diagrammatically in Fig. 26.41 ( or equivalent circuit). Self-excited generator: Self-excited generators are those whose field magnets are energized by the current produced by the generators themselves. Due to the residual magnetism, there is always present some flux in the poles. When the armature is rotated, some e.m.f. and hence some induced current is produced which is partly or fully passed through the field coils thereby strengthening the residual pole flux.

  3. Fig. 26.42. Fig. 26.43 There are three types of self-exited generators named according to the manner in which their field coils (or winding) are connected to the armature. (a) Shunt Wound Generator: The field windings are connected across or in parallel with the armature conductors and have the full voltage of the generator applied across them. It is shown diagrammatically in Fig. 26.42. (b) Series Wound Generator: In this case, the field windings are joined in series with the armature conductors. It is shown diagrammatically in Fig. 26.43.

  4. Short-Shunt Compound Wound Generator: It is a combination of a few series and as a few shunt windings. In a compound generator shunt field is stronger than the series field. Compound wound generators are two types: (a) Short-Shuntwhere shunt field and armature are connected in parallel and that connection is in series with the series field. (b) Long-Shuntwhere series filed and armature are connected in series and that connection is in parallel with the shunt field.

  5. When series field aids the shunt the shunt field, the generator is said to be commutative-compound. On the other hand, if series field opposes the shunt field, the generator is said to be, differentially compound. Depending on load characteristics and the relative additional aiding flux produced by the series field, the cumulative compound generator (whether long-shunt or short-shunt) are three types. These types are called:(i) over-compound, (2) flat-compound, and (iii) under-compound. Most commercial compound dc dynamos are normally supplied by the manufacturer as over-compound machine. The degree of compounding (over, flat, and under) may be adjusted by means of diverter that shunts the series field.

  6. Fig. (a) Long-shunt cumulative compound generator with a diverter. Fig. (b) Short-shunt cumulative compound generator with a diverter. Diverter: A diverter is a variable resistance shunting the series field of compound generator to adjust the degree of compounding to produce a desired voltage regulation. A diverter is used to control and produce a sufficient voltage rise at the generator to compensate for the voltage drop in the lines at full load. The following figures show the connection of diverter in the long-shunt and short-shunt cumulative compound generator.

  7. Brush Contact Drop In the voltage drop over the brush contact resistance when current passes from commutator segments to brushes and finally to the external load. Its value depends on the amount of current and the value of contact resistance. This drop is usually small and includes brushes of both polarities. However, in practice, the brush contact drop is assumed to have following constant values for all loads. 0.5 V for metal-graphite brushes. 2.0 V for carbon brushes.

  8. Generated E.M.F. or E. M. F. Equation of a Generator Let, =flux/pole in weber, Z= total number of armature conductors or Z= No. of slots  No. of conductors/slot, P= No. of generator poles, A= No. of parallel paths in armature, For lap winding, A=mP, For wave winding= 2m, m= The multiplicity (such m=3 for triplex winding) N= armature rotation in revolutions per minute (rpm) E= e.m.f. induced in any parallel path in armature Eg= e.m.f generated in any one of the parallel paths i.e. E.

  9. Average e.m.f. generated/conductor= volt (since n=1) Now, flux cut/conductor in one revolution Wb E.M.F. generated/conductor: volt In general, generated e.m.f.: volt No. of revolution/second= N/60 So, time for one revolution, dt=60/N second Hence, according to Faraday’s Law of Electromagnetic Induction, No. of conductors (in series) per parallel path= Z/A For a simplex lap-wound generator:m=1 and A=mP=P then

  10. For a duplex lap-wound generator:m=2 and A=mP=2P then For a triplex lap-wound generator:m=3andA=mP=3Pthen For a simplex wave-wound generator:m=1 and A=2m=2 then For a duplex wave-wound generator:m=2andA=2m=4then For a triplex wave-wound generator:m=3 and A=2m=6 then

  11. The angular velocity or speed can be written as: Thus we obtain from the above equation as: where, N is in rpm where, N is in rps The generated e.m.f. in terms of angular velocity can be written as: For a given DC machine, Z, P, and A are constant. Hence, puttingKa=ZP/A, we get, It is seen from the above equation that the generated emf is directly proportional to the flux () and the speed (N).

  12. Fig. 26.46. Example 26.3 A shunt generators delivers 450 A at 230 V and the resistance of the shunt field and armature are 50 and 0.03 respectively. Calculate the generated e.m.f.. Solution: Generator circuit is shown in Fig. 26.46. Given, Vt=230V, IL= 450A, Rsh=50, Ra=0.03. Eg=? Current through shunt field winding is Ish=230/50=4.6 A. Load current IL= 450 A So, armature current, Ia=IL+Ish= 450+4.6=454.6 A Armature voltage drop, IaRa= 454.60.03=13.6 V Now, e.m.f. generated in the armature: Eg= terminal voltage (Vt) + armature drop (IaRa) = 230+13.6=243.6 V

  13. Example 26.4 A long-shunt compound generator delivers a load current of 50A at 500V and has armature, series field and shunt field resistances of 0.05, 0.03 and 250 respectively. Calculate the generated voltage and the armature current. Allow 1 V per brush for contact drop. Fig. 26.47 Solution: Generator circuit is shown in Fig. 26.47. Given, Vt=500V, IL= 50A, Ra=0.05, Rse=0.03, Rsh=250 and total brush drop= 21=2 V. Eg=? and Ia=?. Current through shunt field winding is Ish=500/250=2 A. Current through the armature and series field is Ia=50+2=52 A Voltage drop on armature and series field winding =52(0.05+0.03)=4.16 V Voltage drop at brushes= 21= 2 V Now, Eg= Vt+(IaRa+ Series drop) + brush drop=500+4.16+2=506.16 V

  14. Fig. 26.48 short-Shunt Example 26.5 A short-shunt compound generator delivers a load current of 30A at 220 V, and has armature, series field and shunt field resistances of 0.05 ohm, 0.3 ohm and 200 ohm respectively. Calculate the induced e.m.f. and the armature current. Allow 1.0 V per brush for contact drop. Solution: Generator circuit is shown in Fig. 26.48. Given, Vt=220V, IL= 30A, Ra=0.05 ohm, Rse=0.3 ohm, Rsh=200 ohm and total brush drop= 21=2 V. Eg=? and Ia=?. Voltage drop in series winding = 300.3=9V Voltage drop across shunt winding =220+9=229 V Ish=229/200=1.145 A; Ia= 30+1.145=31.145 A IaRa= 31.145  0.05 = 1.56 V; Brush drop =21=2 V Now, Eg= Vt+IaRa+ series drop + brush drop Eg =220 + 1.56+9+2= 232.56 V

  15. Example 26.6 In a long-shunt compound generator, the terminal voltage is 230 V when generator delivers 150 A. Determine (i) induced e.m.f., (ii) total power generated, and (iii) distribution of this power. Given that shunt field, series field, divertor and armature resistances are 92 ohm, 0.015 ohm, 0.03 ohm and 0.032 ohm respectively. Fig. 26.49 Solution: Generator circuit is shown in Fig. 26.49. Given, Vt=230V, IL= 150A, Rsh=92 ohm, Rse=0.015 ohm, Rd=0.03 ohm, Ra=0.032 ohm. Eg=? Total power generated=? and Distribution of the geneated power=?. Ish= 230/92= 2.5 A; Ia=150+2.5=152.5 A Since series field resistance and divertor resistance are in parallel their combined resistance is =0.030.015/(0.03+0.015) =0.01 ohm. Total resistance is= 0.032+0.01=0.042 ohm; Voltage drop= 152.50.042=6.4 V (i) voltage generated by armature, Eg= 230+6.4=236.4 V (ii) total power generated in armature =EgIa=236.4152.5=36,051 W

  16. Fig. 26.49 (iii) Total loss= power lost in armature (Ia2Ra) + power lost in series field and divertor (152.520.01) + Power dissipated in shunt winding ( VtIsh) + Power delivered to load (230150) =152.520.032+152.520.01+2300.01+230150 =36,051 W

  17. Example 26.7 The following information is given for a 300 kW, 600 V, long-shunt compound generator, the shunt field resistance=75, armature resistance including brush resistance =0.03, commutating field winding resistance =0.011, series field resistance =0.012, divertor resistance =0.036 . When the machine is delivering full load, calculate the voltage and power generated by the armature. Fig. 26.50 Solution: Generator circuit is shown in Fig. 26.50. Given, Vt=600V, Output power: VtIL= 300kW, Rsh=75, Ra=0.03 ohm, Rcom=0.011 ohm, Rse=0.012, Rd=0.036 ,. Eg=? Power generated= EgIa=? Output current,IL= 300000W/600V=500A; Ish= 600/75=8A; Ia=500+8=508 A Since the series field resistance and divertor resistance are in parallel their combined resistance is [(0.0120.036)/0.048]=0.009 . Total armature circuit resistance= 0.03+0.011+0.009=0.05  Voltage drop = 5080.05= 25.4 V Voltage generated by armature= 600+ 25.4=625.4 V Power generated= 625.4508=317,700 W = 317.7 kW.

  18. We know that Thus Example 26.8 A four-pole generator having wave-wound armature winding has 51 slots, each slot containing 20 conductors. What will be the voltage generated in the machine when driven at 1500 rpm assuming the flux per pole to be 7.0 mWb and what will be the generated voltage if the generator is the triplex lap wonding? Solution:Given, =07.0 mWb=710-3 Wb; Z=5120=1020; A=2m=2 1=2; P=4; N=1500 rpm We know for the lap winding that,A=mP=34=12 Thus for triplex lap-winding generator:

  19. Fig. 26.53. We know that thus Example 26.11 An 8 pole DC shunt generator with 778 armature conductors and running at 500 rpm. Supplies a load 12.5 ohm resistance at terminal of 250 V. The armature resistance is 0.24 and the field resistance is 250. Find the armature current, the induced e.m.f. and the flux per pole for (a) wave-connected winding and (b) Triplex lap-connected winding. Solution: Generator circuit is shown in Fig. 26.53. Given, P=8, Z=778, N= 500 rpm, Vt=250V, Ra=0.24 ohm, Rsh=250,Ia=?, Eg=? =? Load current, IL= Vt/Ra= 250/12.5= 20A; Shunt current, Ish= Vt/Rsh=250/250= 1 A. Armature current, Ia=IL+Ish=20+1=21 A; Induced e.m.f.= 250+(210.24)=255.04 V (a) For wave-connected winding, A=2m= 2 (m=1), Thus (b) For triplex lap-connected winding, A=mP= 38=24, Thus

  20. Fig. 26.54. Example 26.12: A separately excited generator, when running 1000 rpm supplied 200 A at 125 V. What will the load current when the speed drops to 800 rpm if If(field current) is unchanged? Given that the armature resistance = 0.04 and brushes drop= 2V. Solution: Generator circuit is shown in Fig. 26.54. Given, N1= 1000 rpm, Vt=125V, IL=200A, Ra=0.04, Brushes drop =2V,Eg2(at 800 rpm)=? The load resistance, RL= 125/200=0.625 Eg1(at 1000 rpm)=125+2000.04+2= 135 V; N1= 1000 rpm

  21. Fig. 26.54. According to the voltage generated equation, we obtain that Thus, Eg2(at 800 rpm) = 135800/1000=108 V If IL2 is the new load current, then terminal voltage is given by Vt2= Eg-(IL2Ra + Brushes drop)108-0.04 IL2-2= 106-0.04 IL2. So, IL2= Vt2/RL=(106-0.04 IL2)/0.625; 0.625IL2=106-0.04 IL2; (0.625+0.04)IL2=106; 0.665IL2=106; IL2=106/0.665; IL2= 159.398 A

  22. Backward Force or Magnetic Drag In the case of DC generator as shown in Fig. 29.2, it is seen that the flux due the armature current carrying conductor a force is produced. This force is in a direction opposite to that of armature rotation. Hence, it is known as backward force or magnetic drag on the conductors. It is against this drag action on all armature conductors that the prime mover has to work. The work done in overcoming this opposition is converted into electrical energy.

  23. Armature reaction The current in the armature produces a flux. So, the interaction between this flux and the main field flux is called armature reaction. The armature magnetic field has two effects: (i) it demagnetizes or weakens the main flux which leads to reduced generated voltage, and (ii) it cross-magnetizes, which leads to the sparking at the brushes, or distorts it. When there is no load connected to the generator, the current in the armature conductors is zero. Under these conditions there is only one magnetic field in the generator, and that field is produced by the main-field poles of the generator.

  24. The main field is represented by an arrow, which indicates the direction of the magnetic flux from the north pole to south pole as shown in Fig. 6.1. A load is now connected to the generator, and of course current flow exists. The current to the load is the current in the armature conductors and is equal to the sum of the currents from the parallel paths in the armature. Consider Fig. 6.2, which allows the armature rotating in the magnetic field and the resulting armature current when a load is connected to the generator. When the current flows through a conductor a magnetic field is set up around the conductor as indicated in Fig. 6.2.

  25. The flux from the conductors on the left side of the armature and the flux from the conductors on the right side of the armature cause a resultant flux in the center of the armature that is downward in direction. This resultant flux can be represented by an arrow as indicated, noting that the arrow passes through both top and bottom brushes. There are now two fluxes inside the generator, one produced by the main field poles of the generator and the other by the current in the armature conductors. These two fluxes now combine to form a new resultant flux as shown in Fig. 6.3. This new resultant flux is not in the same direction as the original main field flux but runs from the tip of one of the poles, across the armature, to the tip of the other pole.

  26. The armature conductors are now cutting this new resultant flux that is not the same direction as the main field flux which conductors were originally cutting. The brushes are supposed to be located at the point of minimum flux, which of courses at right angles to the direction of the flux. Since the brushes were at right angles to the main-field flux, they certainly cannot be at right angles to the new resultant flux. With the brushes in their present location they will be short-circuiting coils in which there is a voltage induced, thereby producing sparking at the brushes, undue brushes wear, and other unfavorable conditions.

  27. Effect of Brush Shifting If the brushes are no longer at the points of minimum flux, or magnetic neutral, as the points of minimum flux are known, it might appear to be a simple solution to shift the brushes until they do fall on the magnetic neutral, and then the brushes will once again be at the points of minimum flux. It has just been seen that the direction of resultant flux depends upon both the flux from the main-field poles and the flux produces by the current in the armature conductors. The flux from the main-field poles is fairly constant and will remain constant even though the generator is supplying current to a load. If the load current is small, the armature current will be small and the flux produced by the armature conductors will be small; hence the shift in the resultant flux will be small as compared with the main-field flux. The greater the current delivered by the generator, the greater the current in the armature conductors, and therefore the greater the flux produced by the armature conductors, ending with greater shift in the direction of the resultant flux. Hence if the brushes are to be moved to a new neutral position, the new position will depend upon the load. With the load on a generator constantly varying, it would be impossible to preset the position of the brushes and expect satisfactory result.

  28. Fig. 6.4 shows a new position of brushes at the minimum flux point. Referring Fig. 6.4, it is seen by Fleming’s right-hand rule that the conductors under the north pole carry current away from the observer and the conductors under the south pole carry current toward the observer. The flux from these conductors is indicated on the diagram, and of course the combined flux from all the conductors is still in a direction from the top brush to the bottom brush. But the flux from the armature is not at right angles to the flux from thee main-field poles. The effects of the armature flux in the new position of brushes is illustrated in Fig. 6.5.

  29. It is seen from Fig. 6.5 that there have two components of armature flux that are at right angles to each other. One component is at right angles to the main field, and because this component crosses the main-field flux, it is known as cross-magnetizing component of the armature flux. The second component is in the same plane as the main-field flux. The direction of this component is opposite to the direction of the main-field flux, with the result that it tends to reduce the effect of the main-field flux. This component of the armature flux is known as the demagnetizing component of the armature flux. It now appears that the shifting of the brushes has not improved the situation. In fact, it seems to have become worse.

  30. Before the brushes were shifted, the armature flux was at right angles to the main-field flux and therefore produced only a cross-magnetizing field. With the brushes shifted to the new position, there is still a cross-magnetizing field, some what reduced in magnitude, but in addition there is now a demagnetizing field which tends to reduce the main-field flux, resulting in a lower generated voltage. This demagnetizing component was obtained only after the brushes were shifted, and the brushes were shifted because of the change in direction of the resultant flux, which was due to the armature conductors carrying current. The application of some means to prevent the shift of the resultant flux would eliminate the necessity of shifting the brushes, and hence no demagnetizing field would be produced.

  31. Demagnetizing and Cross-magnetizing Conductors The exact conductors which produce these distorting and demagnetizing effects are shown in Fig. 27.6 where the brush axis has been given a forward lead of  so as to lie along the new position of magnetic neutral axis (M. N. A). All conductors lying with in the angles AOC=BOD=2 at the top and bottom of the armature, are carrying current in such a direction as to send the flux through the armature from right to left. It is these conductors which act in direct opposition to the main-field and are hence called the demagnetizing armature conductors. Now consider the remaining armature conductors lying between angles AOD and COB as shown in Fig. 27.7. These conductors carry current in such a direction as to produce a combined flux pointing at right angles to the main-field flux. This results in distortion of the main field. Hence, these conductors are known as cross-magnetizing conductors and constitute distorting ampere conductors.

  32. Without shifting the brush the armature effect can be minimized by using the following methods: (a) High-Reluctance Pole Tips (b) Horizontal Slots in Main-Field Pole (c) Compensating Windings High-Reluctance Pole Tips It can be seen that the flux from the current in the armature conductors causes the main-field flux to shift from the center of the main-field poles to the tips of the poles. Applying the knowledge that most of the flux flows the path of least reluctance, poles are designed where the reluctance at the ends of the poles is greater than the reluctance at the center.

  33. The variation of reluctance is obtained by constructing the poles that the greater the distance from the center of the pole, the greater the air gap between the pole and the armature. The greater the air gap, the greater the reluctance. With no current in the armature conductors, the flux will be concentrated at the center of the pole, and when current flows in the armature conductors, the flux will tend to shift the end of the pole. The air gap will offer an increase in reluctance to the flux as it moves from the center of the pole, thereby tending to keep the flux in the same original position. The shape of this pole is shown in Fig. 6.6. The reluctance of the center of pole can be reduced by using lamination.

  34. Horizontal Slots in Main-Field Pole The high reluctance pole-tip construction reduced the effects of armature reaction by not allowing the flux to shift. The shift in the resultant flux was caused by the armature flux. If the armature flux could be reduced to a negligible value, then its cross-magnetizing effect upon the main-field flux would be small and the brushes would not have to be shifted. Fig. 6.8 shows that part of the path of the magnetic flux is through the field poles. By cutting horizontal slots in the poles, several air gaps are introduced in the path of the flux. These slots increase the reluctance to the armature flux while having very little effect on the main-field flux. The armature is materially reduced, and the brushes need not be shifted.

  35. Compensating Windings The compensating windings are placed in the pole faces of the filed pole can run parallel to the armature conductors. A connection is made from one of the brushes to one end of this winding, so that current from armature must first pass through the winding before going to the load. The direction of current through the winding is opposite to that of the current in the armature conductors under that pole. The location of the compensating winding and of the connection to the winding is shown in Fig. 27.8. Figure 27.8 indicates that the current in the armature conductors located under the north pole carry current away from the observer; therefore the direction of current flowing in that part of the compensating winding situated in the north pole is toward the observer. Since the current in the compensating winding is opposite in direction to the current in the armature conductors, the flux produced by the current in the compensating winding will be opposite in direction to the armature flux. The compensating flux, being opposite in direction to the armature flux, tends to cancel the armature flux. If the armature current increase, the compensating current increases, so that the armature flux is canceled for al load conditions.