l.txt

D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                                    Magnetic fields

 A permanent magnet is a piece of ferromagnetic material (such as iron, nikel or cobalt)
 which has properties of attracting other pieces of these materials. A permanent magnet will
 position itself in a north and south direction when freely suspended. The north – seeking
 end of the magnet is called the North Pole, N, and the south seeking end the South Pole, S.
 The area around a magnet is called the magnetic field and it is in this area that the effects
 of the magnetic force produced by the magnet can be detected. A magnetic field cannot be
 seen, felt, smelt or head and therefore is difficult to represent. Michael Faraday suggested
 that the magnetic field could be represented pictorially by imagining the field to consist of
 lines of magnetic flux, which enables investigation of the distribution and density of the
 field to be carried out.
 The distribution of a magnetic field can be investigated by using some iron filings. A bar
 magnet is placed on a flat surface covered by, say, cardboard, upon which is sprinkled
 some iron filings. If the cardboard is gently tapped the filings will assume a pattern similar
 to that shown in Figure (1). If a number of magnets of different strength are used, it is
 found that the stronger the field the closer are the line of magnetic flux and vice versa.
 Thus a magnetic field has the property of exerting a force, demonstrated in this case by
 causing the iron filings to move into pattern shown. The strength of the magnetic field
 decreases as we move away from the magnet. It should be realized, of course, that the
 magnetic field is three dimensional in its effect, and not acting in one plane as appears to
 be the case in this experiment.




                                          Figure (1)

 If a compass is placed in the magnetic field in various positions, the direction of the lines
 of flux may be determined by noting the direction of the compass pointer. The direction of
 a magnetic field at any point is taken as that in which the north-seeking pole of a compass
                                              1
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 needle points when suspended in the field. The direction of a line of flux is form the north
 pole to the south pole on the outside of the magnet and is then assumed to continue through
 the magnet back to the point at which it emerged at the north pole. Thus such lines of flux
 (1) always form complete closed loops or paths; (2) they never intersect and (3) always
 have a definite direction.
 The laws of magnetic attraction and repulsion can be demonstrated by using two bar
 magnets. In Figure (2-a), with unlike adjacent, attraction takes place.
 Lines of flux are imagined to contract and the magnets try to pull together. The magnetic
 field is strongest in between the two magnets shown by the lines of flux being close
 together. In Figure (2-b), with similar poles adjacent (i.e. two north poles), repulsion
 occurs, i.e. the two north pole try to push each other a part, since magnetic flux line
 running side by side in the same direction repel.




                                            Figure (2)

 Magnetic Flux and Flux Density
 Magnetic flux is the amount of magnetic field (or the number of line of force) produced by
 a magnetic source.
 The symbol for magnetic flux is Φ (Greek letter, 'phi').
 The unit of magnetic flux is the Weber, wb which equal to one hundred million line force.
 Magnetic flux density is the amount of flux passing through a defined area that is
 perpendicular to the direction of the flux:
 Magnetic flux density  magnetic flux
                                         area
 The symbol of magnetic flux density is B. the unit of magnetic flux density is the tesla, T,
 where
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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 1 T = 1 wb/m2. Hence
      
 B       tesla ………………….. (1)
      A
 Where A (m2) is the area.
 Magnetic Field due to an Electric Current
 Magnetic field can be setup not only by permanent magnets, but also by electric currents.
 Let a piece of wire be arranged to pass vertically through a horizontal sheet of cardboard
 on which is placed some iron filings, as shown in figure (3-a). If a current is now passed
 through the wire, then the iron filings will form a definite circular field pattern with the
 wire at the center, when the cardboard is gently tapped. By placing a compass indifferent
 positions the lines of flux are seen to have a definite direction as shown in figure (3-b).




                                           Figure (3)

 If the current direction is reversed, the direction of the lines of flux is also reversed. The
 effect on both the iron filings and the compass needle disappears when the current is
 switched off. The magnetic field is thus produced by the electric current.
 The magnetic flux produced has the same properties as the flux produced by a permanent
 magnet. If the current is increased the strength of the filed increases and, as for the
 permanent magnet, the field strength decreases as we move away from the current-
 carrying conductor.
 In Figure (3) the effect of a small part of the magnetic field is shown. If the whole length of
 the conductor is similarly investigated it is found that the magnetic field round a straight
 conductor is in the form of concentric cylinders as shown in Figure (4), the field direction
 depending on the direction of the current flow.


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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                                         Figure (4)

 When dealing with magnetic fields formed by electric current it is usual to portray the
 effect as shown in figure (5). The convention adopted is:
 i- Current flowing away from the viewer, i.e. into the paper, is indicated by  . This may
 be thought of as the feathered end of the shaft of the arrow. See Figure (5-a)
 ii- Current flowing towards the viewer, i.e. out of paper, is indicated by  . This may be
 thought of as the point of an arrow. See Figure (5-b).




                   (a) Current flowing away           (b) Current flowing
                       from viewer                       towards viewer
                                        Figure (5)

 The direction of the magnetic lines of flux is best remembered by the screw rule which
 states that:
 If a normal right-hand thread screw is screwed along the conductor in the direction of the
 current, the direction of rotation of the screw is in the direction of magnetic field.
 For example with current flowing away from the viewer Figure (5-a) a right –hand thread
 screw driven into the paper has to be rotated clock wise. Hence the direction of the
 magnetic field is clock wise.
 A magnetic field setup by a long coil, or solenoid, is shown in Figure (6-a) and is seen to
 be similar to that of a bar magnet. If the solenoid is wound on an iron bar, as shown in

                                             4
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Figure (6-b), an even stronger magnetic field is produced, the iron becoming magnetized
 and behaving like a permanent magnet. The direction of the magnetic field produced by the
 current I in the solenoid may be found by either of two methods, i.e. the screw rule or the
 grip rule.
 a- The screw rule states that if a normal right-hand thread screw is placed along the axis of
 the solenoid and is screwed in the direction of the current it moves in the direction of the
 magnetic field inside the solenoid. The direction of the magnetic field inside the solenoid is
 from south to north.
 Thus in Figure (6-a&b) the north pole is to the right.
 b- The grip rule states that if the coil is gripped with the right hand, with the fingers
 pointing in the direction of the current, then the thumb, out stretched parallel to the axis of
 the solenoid, point in the direction of the magnetic field inside the solenoid.




            (a) Magnetic field of a solenoid          (b) magnetic field of an iron cored solenoid
                                               Figure (6)

 Magneto motive Force and Magnetic Field Strength
 Magneto motive force (m.m.f) is the cause of the existence of a magnetic flux in a
 magnetic circuit,
 m.m. f .    Fm  NI       amperes
 Where N is the number of conductor (or turns) and I is the current in amperes. The unit of
 m.m.f. is sometimes expressed as 'ampere-turns'. However since 'turns' have no
 dimensions, the S.I. unit of m.m.f is the ampere.
 Magnetic field strength (or magnetic field intensity or magnetizing force),
     NI
 H         ampere per         meter
       L
 Where L is the mean length of the flux path in meters. Thus
 m.m. f  NI  HL                amperes
 Permeability and B-H curves


                                                  5
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 For air, or any non-magnetic medium, the ratio of magnetic flux density to magnetizing
 force is a constant, i.e. B/H=constant. This constant is  0 , the permeability of free space (or
 the magnetic space constant) and is equal to 4  10 7 H/m, i.e. for air, or any non-magnetic
 medium, the ratio
  B
      0 ………………………….. (2)
  H
 (Although all non-magnetic materials, including air, exhibit slight magnetic properties,
 these can effectively be neglected)
 For all media other than free space
  B
     0 r
  H
 Where  r is the relative permeability, and is defined as
        flux density in material
 r 
         flux density in vacuum
  r varies with the type of magnetic material and since it is a ratio of flux densities, it has
 no unit. From its definition,  r for a vacuum is 1.
 0 r      Called the absolute permeability.
 By plotting measured value of flux density B against magnetic field strength H , a
 magnetization curve (or B  H curve) is produced. For non-magnetic materials this is
 straight line. Typical curves for four magnetic materials are shown in Figure (7).
 The relative permeability of a ferromagnetic material is proportional to the slope of the
 B  H curve and thus varies with the magnetic field strength.




                                                6
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                                             Figure (7)
 The four principal section of a typical magnetization curve are illustrated in Figure (7-a).
 The curve is concave up for ''low'' values of magnetic field intensity, exhibits a somewhat
 (but not always) linear characteristic for ''medium'' field intensities, and then is concave
 down for ''high'' field intensities, eventually flattening to an almost horizontal line for ''very
 high'' intensities. The part of the curve that is concave down is known as the knee of the
 curve, and the ''flattened'' section is the saturation region. Magnetic saturation is complete
 when all of the magnetic domains of the material are oriented in the direction of the
 applied magneto motive force. Saturation begins at the start of the knee region and is
 essentially complete when the curve starts to flatten.
 Depending on the specific application, the magnetic core of an apparatus may be operated
 in the linear region, and/or the saturation region. For example, transformers and AC
 machines are operated in the linear region and lower end of the knee; self-excited DC
 generator and DC motors are operated in the upper end of the knee region, extending into
 the saturation region; separately excited DC generators are operated in the linear and lower
 end of the knee region.
 Magnetization curves supplied by manufacturers for specific electrical steel sheets or
 casting are usually plotted on semi log paper, and often include a curve of relative
 permeability vs. field intensity as shown in Figure (7-b).

                                                7
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                                      Figure (7-a)




                                      Figure (7-b)




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Magnetic Circuits
 It is possible to analyze the operation of electromagnetic devices such as the one depicted
 in Figure (8), the means of magnetic equivalent circuits, similar in many respects to the
 equivalent electric circuits. Before we can present this technique, however, we need to
 make a few simplifying approximations. The first of these approximations assumes that
 there exists a mean path for the magnetic flux, and that the corresponding mean flux
 density is approximately constant over the cross-sectional area of the magnetic structure.
 Using equation (1) we see that a coil wound around a core with cross-sectional area A will
 have flux density.
      
 B
      A
 Where A is assumed to be perpendicular to the direction of flux lines. Figure (8), illustrates
 such a mean path and the cross-sectional area A . knowing the flux density, we obtain the
 field intensity:
      B       
 H             …………………………… (3)
             A
 But then, knowing the field intensity, we can relate the m.m.f of the coil F to the product
 of the magnetic field intensity H and the length of magnetic (mean) path L ; we can use
 equation (2) and (3) to

 F  Ni  HL …………………………… (4)




                                          Figure (8)

 In summary, the m.m.f. is equal to the magnetic flux times the length of the magnetic path,
 divided by the permeability of the material times the cross-sectional area:
           L
 F         …………………………… (5)
          A


                                              9
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 A review of this formula reveal that the magneto motive force F may be viewed as being
 analogous to the voltage source in a series electric circuit, and that the flux  is then
                                                                           L
 equivalent to the electric current in a series circuit and the term           to the magnetic
                                                                          A
                                                                              L
 resistance of one leg of the magnetic circuit. You will note that the term      is very similar
                                                                             A
 to the term describing the resistance of a cylindrical conductor of length L and cross-
                                                                                             L
 section area A , where the permeability  is analogous to the conductivity  . The term
                                                                                            A
 occurs frequently enough to be assigned the name of reluctance and the symbol  it is the
 (magnetic resistance) of a magnetic circuit to the presence of magnetic flux.
      F       NI   HL      L       L
                          
                 BA    
                         B
                           H
                              A 0 r A

 The unit     of reluctance is 1 H or H 1   or A wb .
 Ferromagnetic materials have a low reluctance and can be used as magnetic screens to
 prevent magnetic field affecting materials within the screen.
 In summary, when an N -turn coil carrying a current i wound around a magnetic core such
 as the one indicated in Figure (8), the m.m.f., F generated by the coil produces a flux 
 that is mostly concentrated within the core and is assumed to be uniform across the cross
 section.
 Within this simplified picture, then, the analysis of a magnetic circuit is analogous to that
 of resistive electric circuits. This analogy illustrated in Table (1) and in the examples in
 this section.

 Table (1) Analogy between electric and magnetic circuits

      Electric quantity                            Magnetic quantity
      Electric field intensity E. V/m              Magnetic field intensity H . A-turns/m
      Voltage v. V                                 Magneto motive force F . A-turns
      Current i. A                                 Magnetic flux  . Wb
      Current density J. A/m2                      Magnetic flux density B . Wb/m2
      Resistance R,                               Reluctance  A-turns/Wb
      Conductivity  . 1/  .m                     Permeability  . Wb/A.m

 The usefulness of the magnetic circuit analogy can be emphasized by analyzing a magnetic
 care similar to that of Figure (8), but with a slightly modified geometry. Figure (9), depicts
 the magnetic structure and its equivalent-circuit analogy. In the figure, we see that the
 m.m.f., F  NI excites the magnetic circuit, which is composed of four leg; two of mean
 path length L1 and cross-sectional area A1  d1 w and the other two of mean length L2 and


                                                     11
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 cross-sectional are A2  d 2 w . Thus, the reluctance en countered by the flux in its path
 around the magnetic core is given by the quantity  series , with
             series  21  2 2
                      L1              L2
 and          1        ,    2 
                     A1             A2




                                           Figure (9)

 It is importance at this stage to review the assumptions and simplifications made in
 analyzing the magnetic structure of figure (9).
 1- All the magnetic flux is linked by all the turns of the coil.
 2- The flux is confined exclusively within the magnetic core.
 3- The density of the flux is uniform across the cross-sectional area of the core
 You can probably see intuitively that the first of these assumptions might not hold true near
 the ends the coil, but that it might be more reasonable if the coil is tightly wound. The
 second assumption is equivalent to stating that the relative permeability of the core is
 infinitely higher than that of air (presuming that this is the medium surrounding the core);
 if this were the case, the flux wound indeed be confined within the core. It is worthwhile to
 note that we make a similar assumption when we treat wires in electric circuits as perfect
 conductors: The conductivity of copper is substantially greater than that of free space by a
 factor of approximately 1015. In the case of magnetic materials, however, even for the best
 alloys, we have a relative permeability only on the order of 10 3 to 104. Thus, an
 approximation that is quite appropriate for electric circuits is not nearly as good in the case
 of magnetic circuits. Some of the flux in a structure such as those of Figure (8) and (9)
 would thus not be confined within the core (this is usually referred to as leakage flux).
 Finally, the assumption that the flux is uniform across the core cannot hold for a finite-
 permeability medium, but it is very helpful in giving an approximate mean behavior of the
 magnetic circuit.
                                              11
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Magnetic Circuit with Air gaps.
 Consider the analysis of the some simple magnetic structure when an air gap is present. Air
 gaps are very common in magnetic structure; in rotating machines; for example, air gaps
 are necessary to allow for free rotation of the inner core of the machine. The magnetic
 circuit of Figure (10-a) differs from the circuit analyzed in Figure (8) simple because of the
 presence of an air gap; the effect of the gap is to break the continuity of the high-
 permeability path for the flux, adding a high reluctance component to the equivalent
 circuit. It should be evident from Figure (10-a) that the basic concept of reluctance still
 applies, although now two different permeability's must be taken into account.
 The equivalent circuit for the structure of Figure (10-a) may by drown as shown in
 Figure(10-b) where  is the reluctance of path Ln , for n = 1,2,….,5 and  g is the
 reluctance of the air gap. The reluctances can be expressed as follows, it we assume that
 the magnetic structure has a uniform cross-section area A :
          L1                   L2                  L3
 1              ,   2              ,    3 
        r 0 A              r 0 A            r 0 A
          L4                   L5                   
 4              ,   5                    g 
        r 0 A              r 0 A               0 Ag




                                                      Figure
                                                   (10)
 Not that in computing  g the length of the gap is given by  and the permeability is given
 by  0 , as expected. But Ag is different from the cross-sectional area A of the structure.
 This is so because the flux lines exhibit a phenomenon known as fringing (see figure
 below) as they cross an air gap. The flux lines actually bow out of the gap defined by the
 cross-sectional A , not being contained by the high-permeability material any longer. Thus
 it is customary to define an area Ag that is greater than A , to account for this phenomenon.




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                                          13
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                               Hysteresis and hysteresis loss
 Hysteresis loop:
 Let a ferromagnetic material which is completely demagnetized, i.e. one in which
  B  H  0 be subjected to increasing values of magnetic field strength H and the
 corresponding flux density B measured. The resulting relationship between B and H is
 shown by the curve oab in Figure (11). At a particular value of H , shown as oy, it becomes
 difficult to increase the flux density any further. The material is said to be saturated. Thus
 by is the saturation flux density.
 If the value of H is now reduced it is found that the flux density follows curve bc. When H
 is reduced to zero, flux remains in the iron. This remnant flux density is shown as oc in
 Figure (11). When H is increased in the opposite
 direction, the flux density decrease until at a value
 shown as od, the flux density has been reduced to
 zero. The magnetic field strength od required to
 remove the residual magnetism, i.e. reduce B to
 zero is called the coercive force.
 Further increase of H in the reverse direction
 causes the flux density to increase in the reverse
 direction until saturation is reached as shown by
 curve de. If H is varied backwards from ox to oy the
 flux density follows the curve efgb, similar to curve
 bcde.
 It is seen from figure (11) that the flux density
 changes lag behind the changes in the magnetic field
 strength. This effect is called hysteresis. The closed
  Figure bcdefgb is called the hysteresis loop (or the B H loop)         Figure (11)

 Hysteresis loss
 A disturbance in the alignment of the domains (i.e.
 groups of atoms) of a ferromagnetic material
 causes energy to be expended in taking it through a
 cycle of magnetization. This energy appears as
 heat in the specimen and is called the hysteresis
 loss.
 The power loss due to hysteresis for a given type                         remnant flux density

 and volume of core material varies directly with
 the frequency and the nth power of the maximum
 value of the flux density wave.
 Expressed mathematically,
 Ph  k h . f .Bmax
                n
                    .v
 Ph =   hysteresis loss (w)
                                              14
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


  f = frequency of flux wave (Hz)
 Bmax =  maximum value of flux density wave (T)
  k h = constant
 n =Steinmetz exponent
 v =volume of core in m3
 The Steinmetz exponent varies with the core material and has an average value of 1.6 for
 silicon steel sheets.
 The constant k h is dependent on the magnetic characteristics of the material.
 The energy loss associated with hysteresis is proportional to the area of the hysteresis loop.
 The area of a hysteresis loop varies with the type of material. The area, and thus the energy
 loss, is much greater for hard materials than for soft materials.
 1- Hard magnetic material
 Magnetic materials, which have large hysteresis loop area and hence large energy loss per
 cycle of magnetization, are classified as hard magnetic materials. The magnetization curve
 for hard magnetic materials is shown in Figure (12). Carbon steel, tungsten steel, cobalt
 steel and hard ferrites are categorized as the hard magnetic materials. These materials are
 suitable for making the instruments and devices, which require permanent magnets.
 2- Soft magnetic material
 Some magnetic materials have steep magnetization curve as given in Figure (13). These
 materials have relatively small and narrow hysteresis loop and hence small energy loss per
 cycle of magnetization. These materials are called soft magnetic materials. Silicon steel,
 nickel iron alloys and soft ferrites are the soft magnetic materials. These materials can be
 used for the construction of cores of electrical machines, transformers, electromagnets,
 reactors, relates etc.




                   Figure (12)                              Figure (13)



                                              15
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Hysteresis losses caused by rotation
 Hysteresis losses are also produced when a piece of iron rotates in a constant magnetic
 field. Consider for example, an armature AB, made of iron that revolves in a field
 produced by permanent magnets N S figure (14). The magnetic domains in the armature
 tend to line up with the magnetic field, irrespective of the position of the armature.
 Consequently, as the armature rotates, the N poles of the domains point first toward A and
 then toward B. A complete reversal occurs therefore every half revolution, as can be seen
 in figure (14-a, 14-b) consequently, the magnetic domains in the armature reverse
 periodically, even though the magnetic field is constant. Hysteresis losses are produced just
 as they are in an ac magnetic field.




                   (a)                                        (b)
                                     Figure (14)
 Eddy currents
 Consider an ac flux  that links a rectangular-shaped conductor Figure (16). According to
 Faraday's law, an ac voltage E, is induced across it terminals.
 If the conductor is short-circuited, a substantial alternating current I1, will flow, causing the
 conductor to heat up. If a second conductor is placed inside the first, a smaller voltage is
 induced because it links a smaller flux. Consequently, the short-circuit current I2 is less
 than I1 and so, too, is the power dissipated in this loop. Figure (15) shows four such
 concentric loops carrying currents I1, I2, I3 and I4.
 The currents are progressively smaller as the area of loops surrounding the flux decreases.




          Figure (15)                                            Figure (16)
                                               16
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class



 In figure (17) the ac flux passes through a solid metal plate. It is basically equivalent to a
 densely packed set of rectangular conductors touching each other. Current swirl back and
 forth inside the plate, following the paths shows in the figure. These so-called eddy current
 (or Foucault currents) can be very large, due to the low resistance of the plate.
 Consequently, a metal plate that is penetrated by an ac flux can become very hot. In this
 regard, special core has to be taken in transformers so that stray leakage flux does not
 cause sections of the enclosing tanks to overheat.
 The flux  in figure (16&17) is assumed to be increasing. As a result, due to the Lenz's
 law, the eddy currents flow in such a way as to oppose in the increasing flux.




                                 Figure (17)

 Eddy current in a stationary iron core
 The eddy current problem becomes particularly important when iron has to carry an ac
 flux. This is the cause in all ac motors and transformers. Figure (18) shows a coil carrying
 an ac current that produces an ac flux in a solid iron core. Eddy currents are set up as
 shown and they flow throughout the entire length of the core. A large core could eventually
 because red hot due to these eddy-current losses.
 We can reduce the losses by splitting the core in two along its length, taking care to
 insulate the two sections from each other Figure (19). The voltage induced in each section
 is one half of what it was before, with the result that the eddy currents, and the
 corresponding losses, are considerably reduced. If we continue to subdivide the core, we
 find that the losses decrease progressively. In practice, the core is composed of stacked
 laminations, usually a fraction of a millimeter think. Furthermore, small amount of silicon
 is alloyed with the steel to increase its resistively; there by reducing the losses still more
 Figure (20).




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                Figure (18)             Figure (19)                     Figure (20)
 The cores of ac motors and generators are therefore always laminated. A thin coating of
 insulation covers each lamination to prevent electrical contact between them. The stacked
 laminations are tightly held in place by bolts and appropriate end-pieces. For a given core,
 the eddy-current losses decrease in proportion to the square of the number of laminations.

 Eddy-current losses in a revolving core
 The stationary field in direct-current motors and generators produces a constant dc flux.
 This constant flux induces eddy currents in the revolving armature. To understand how
 they are induced, consider a solid cylindrical iron care that revolves between the poles of a
 magnet Figure (21-a). As it turns, the core cuts flux lines and, according to Faraday's law, a
 voltage is induced along its length having the polarities shown. Owing to this voltage, large
 eddy currents flow in the core because its resistance is very low Figure (21-b).
 These eddy currents produce large I2R losses which are immediately converted into heat.
 The power loss is proportional to the square of the speed and the square of the density.
 Eddy current loss is given by equation.
 We  kBm2 f 2t 2V         watts
 Where k is constant for a given material, B m is the peak value of the flux density, f is
 supply frequency, t is the thickness of each stamping, and V is the volume of the core.
 To reduce the eddy-current losses, we laminate the armature using thin circular laminations
 that are insulated from each other. The laminations are tightly stacked with the flat side
 running parallel to flux line Figure (22).




                                              18
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                     Figure (21)                           Figure (22)
 1 Part of a magnetic circuit is made from steel of length 120 mm, cross sectional area 15
 cm2 and relative permeability 800. Calculate (a) the reluctance and (b) the absolute
 permeability of the steel. [(a) 79580/H (b) 1 mH/m]

 2 A mild steel closed magnetic circuit has a mean length of 75 mm and a cross-sectional
 area of 320.2mm2. A current of 0.40 A flows in a coil wound uniformly around the circuit
 and the flux produced is 200mWb. If the relative permeability of the steel at this value of
 current is 400 find (a) the reluctance of the material and (b) the number of turns of the coil.
 [(a) 466000/H (b) 233]

 3 A magnetic circuit of cross-sectional area 0.4 cm2 consists of one part 3 cm long, of
 material having relative permeability 1200, and a second part 2 cm long of material having
 relative permeability 750. With a 100 turn coil carrying 2 A, find the value of flux existing
 in the circuit. [0.195 m Wb]

 4 (a) A cast steel ring has a cross-sectional area of 600 mm2 and a radius of 25 mm.
 Determine the m.m.f. necessary to establish a flux of 0.8 mWb in the ring. Use the B-H
 curve for cast steel shown on Figure 7. (b) If a radial air gap 1.5 mm wide is cut in the ring
 of part (a) find the m.m.f. now necessary to maintain the same flux in the ring. [(a) 270A
 (b) 1860 A]



                                              19
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 5 A closed magnetic circuit made of silicon iron consists of a 40 mm long path of cross -
 sectional area 90 mm2 and a 15 mm long path of cross-sectional area 70 mm2. A coil of 50
 turns is wound around the 40 mm length of the circuit and a current of 0.39 A flows. Find
 the flux density in the 15 mm length path if the relative permeability of the silicon iron at
 this value of magnetizing force is 3000.
 [1.59T]

 6 For the magnetic circuit shown in Figure below find the current I in the coil needed to
 produce a flux of 0.45 m Wb in the air-gap. The silicon iron magnetic circuit has a uniform
 cross - sectional area of 3 cm2 and its magnetization curve is as shown on
 Figure (7) [0.83 A]




 7 A ring forming a magnetic circuit is made from two materials; one part is mild steel of
 mean length 25 cm and cross-sectional area 4 cm2, and the remainder is cast iron of mean
 length 20 cm and cross-sectional area 7.5 cm2. Use a tabular approach to determine the
 total m.m.f. required to cause a flux of 0.30 mWb in the magnetic circuit. Find also the
 total reluctance of the circuit. Use the magnetization curves shown on figure (7) . [550A,
 18.3 x 105/H]

 8 Figure below shows the magnetic circuit of a relay. When each of the air gaps are 1.5
 mm wide find the m.m.f. required to produce a flux density of 0.75 T in the air gaps. Use
 the B-H curves                                   shown on figure 7 [2970 A]




                                             21
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 9. A section through a magnetic circuit of uniform cross-sectional area 2 cm2 is shown in
 figure below. The cast steel core has a mean length of 25 cm. The air gap is 1mm wide and
 the coil has 5000 turns. The B–H curve for cast steel is shown on Figure (7). Determine the
 current in the coil to produce a flux density of 0.80 T in the air gap, assuming that all the
 flux passes through both parts of the magnetic circuit.[0165 A]




                       Further problems on magnetic circuits

 1 What is the flux density in a magnetic field of cross-sectional area 20cm having a flux of
 3mWb? [1.5T]

 2 Determine the total flux emerging from a magnetic pole face having dimensions 5cm by
 6 cm, if the flux density is 0.9T [2.7mWb]

 3 The maximum working flux density of a lifting electromagnet is 1.9T and the effective
 area of a pole face is circular in cross-section. If the total magnetic flux produced is 611
 mWb determine the radius of the pole face. [32cm]

 4 An electromagnet of square cross-section produces a flux density of 0.45T .If the
 magnetic fluxis720 µWb finds the dimensions of the Electromagnet cross-section. [4cm by
 4cm]

 5 Find the magnetic field strength applied to a magnetic circuit of mean length 50 cm when
 a coil of 400 turns is applied to it carrying a current of 1.2A [960A/m]

 6 A solenoid 20 cm long is wound with 500 turns of wire. Find the current required to
 establish a magnetizing force of 2500A/m inside the solenoid.[1A]

 7 A magnetic field strength of 5000A/m is applied to a circular magnetic circuit of mean
 diameter 250 mm. If the coil has 500 turns find the current in the coil. [7.85A]


                                             21
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


    Electromechanical Energy Conversion
    The conversion of electrical energy into mechanical energy or vice versa is known
    as electromechanical energy conversion.
    Electromechanical energy conversion involves the interchange of energy between
    an electrical system and a mechanical system through the medium of a coupling
    electric field or magnetic field.
    Therefore, an electromechanical conversion system has three essential parts viz., an
    electrical system, a mechanical system and a coupling field (electric or magnetic)
    figure (1) shows the block diagram of an electromechanical energy.




                   Figure (1) Electromechanical Energy Conversion System

   (i) Electric field as coupling medium. Electromechanical energy conversion can
        take place when electric field is used as the medium.
   (ii) Magnetic field as coupling medium. Electromechanical energy conversion can
        also take place more effectively when magnetic field is used as medium

    Energy Balance Equation
    An electromechanical energy conversion system has three essential parts viz., an
    electrical system, a mechanical system and a coupling magnetic field as shown in
    figure (2). Since conversion of energy from one form into another form satisfies the
    principle of conservation of energy, the energy transfer equation is as under:

              Electrical energy     Mechanical        increase in energy        Energy
         (                      )=(             )+(                          )+(        )
             input from source     energy output    stored in coupling field     losses

    Eq. (i) is applicable to all conversion devices. For motor action, the electrical and
    mechanical energy terms have positive values. For generator action, the electrical
    and mechanical energy terms have negative values.




                                         Figure (2)

                                             22
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


    During this energy conversion, energy loss occurs due to three causes viz., (i) i 2R
    loss in the winding of the energy converter (ii) core or field loss due to changing
    magnetic field and (iii) mechanical loss is the friction and windage loss due to the
    motion of moving parts. All these losses are converted to heat. If the energy losses
    in the electrical system, the coupling magnetic field and the mechanical system are
    grouped with the corresponding terms in eq. (i) above, the energy balance equation
    can be written as under:


          Electrical energy  Mechanical energy        increase in stored
         ( input minus ) = (output plus friction ) + ( field energy plus )                   (𝑖𝑖)
          resistance losses and windage losses             core losses

    Now consider a differential time 𝑑𝑡 during which an increment of electrical energy
    𝑑𝑊𝑒𝑙𝑒𝑐𝑡 (excluding 𝑖 2 𝑅 loss) flows to the system. During this time 𝑑𝑡, let 𝑑𝑊𝑓𝑙𝑑 be
    the energy supplied to the field (either stored or lost, or part stored and part lost)
    and 𝑑𝑊𝑚𝑒𝑐ℎ the energy converted to mechanical form (in useful form or as loss, or
    part useful and part as loss). In differential form, eq. (ii) can be expressed as
                              𝑑𝑊𝑒𝑙𝑒𝑐𝑡 = 𝑑𝑊𝑚𝑒𝑐ℎ + 𝑑𝑊𝑓𝑙𝑑                      (𝑖𝑖𝑖)

    If no mechanical work is down [i.e. 𝑑𝑊𝑚𝑒𝑐ℎ = 0], then eq. (iii) becomes:
                                      𝑑𝑊𝑒𝑙𝑒𝑐𝑡 = 𝑑𝑊𝑓𝑙𝑑

    In this case, electrical energy input is stored in the magnetic field (neglecting core
    losses).

    Energy In Magnetic System
    Consider singly-excited magnetic system
    shown in figure (3). It is the magnetic
    system of an attracted armature relay.
    Here a coil of N turns wound on the
    magnetic core is connected to an electric
    source. Let us assume that the armature is
    held stationary at some air gap and the
    current is increased from zero to some
    value i. As a result, flux 𝜙 will be
    established in the magnetic system.
    Total flux linkages,                 𝜆 = 𝑁𝜙                       Figure (3)


                                             23
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                                           𝑁𝑑𝜙 𝑑        𝑑𝜆
         induced e. m. f.             𝑒=       = (𝑁𝜙) =
                                            𝑑𝑡  𝑑𝑡      𝑑𝑡
    For the coupling device to absorb energy from the electric circuit, the coupling field
    must produce a reaction in the circuit. This reaction is the e.m.f. e produced by the
    magnetic field.
    The incremental electrical energy (𝑑𝑊𝑒𝑙𝑒𝑐𝑡 ) due to the flow of current i in time dt is

                                        𝑑𝑊𝑒𝑙𝑒𝑐𝑡 = 𝑒 𝑖 𝑑𝑡
    The energy balance equation indifferent form is
                                      𝑑𝑊𝑒𝑙𝑒𝑐𝑡 = 𝑑𝑊𝑚𝑒𝑐ℎ + 𝑑𝑊𝑓𝑙𝑑

    Since we have assumed that the armature is held stationary, there is no mechanical
    output i.e., 𝑑𝑊𝑚𝑒𝑐ℎ = 0. Therefore, all the increment electrical input energy is
    stored as incremental field energy i.e.
                                           𝑑𝑊𝑒𝑙𝑒𝑐𝑡 = 𝑑𝑊𝑓𝑙𝑑
    or
                                         𝑒 𝑖 𝑑𝑡 = 𝑑𝑊𝑓𝑙𝑑
    or
                                        𝑑𝜆
                                           𝑖 𝑑𝑡 = 𝑑𝑊𝑓𝑙𝑑
                                        𝑑𝑡
    or
                                   𝑑𝑊𝑓𝑙𝑑 = 𝑖 𝑑𝜆 = 𝑁 𝑖 𝑑𝜙

    The relationship between coil flux linkage 𝜆 and current i for
    a particular air gap length is shown in figure (4). The
    incremental field energy 𝑑𝑊𝑓𝑙𝑑 is shown as crosshatched
    area in this figure. When the flux linkage is increased from
    zero to 𝜆 (or flux from zero to 𝜙), the energy stored in the
    field is
                                   𝜆            𝜙

                            𝑊𝑓𝑙𝑑 = ∫ 𝑖𝑑𝜆 = 𝑁 ∫ 𝑖𝑑𝜙
                                  0            0
                                                                                    Figure (4)

    The integral represents the area between the 𝜆 axis and 𝜆_𝑖 characteristic and is
    equal to the entire shaded area shown in figure (4).
    We can also derive another useful expression for the energy stored in the magnetic
    field. If l and A are the length and cross-section of the magnetic circuit respectively
    and B is the magnetic flux density, then,

                                               24
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                          𝑁𝑖 = 𝐻𝑙 𝑎𝑛𝑑 𝑑𝜙 = 𝐴 𝑑𝐵
    Here H is the magnetic field intensity.


                                        𝜙               𝐵          𝐵

         ∴                  𝑊𝑓𝑙𝑑 = 𝑁 ∫ 𝑖𝑑𝜙 = ∫ 𝐻𝑙𝐴𝑑𝐵 = 𝐴𝑙 ∫ 𝐻𝑑𝐵
                                       0            0             0

                                           𝐵
                                                                                    Figure (5)
         𝑜𝑟                 𝑊𝑓𝑙𝑑 = 𝐴𝑙 ∫ 𝐻𝑑𝐵
                                       0

    Figure (5) shows B-H curve for magnetic circuit.
    Energy and Coenergy
    The 𝜆 − 𝑖 characteristic of an electromagnetic system depends on the air gap length
    and B-H characteristic of the magnetic material. These 𝜆 − 𝑖 characteristics are
    shown in figure (6) for three values of air gap length. If the air length is large, the
    characteristic is essentially linear. The characteristic become nonlinear as the air gap
    length decreases.




                        Figure (6)                                     Figure (7)
    For a particular value of air gap length, the energy stored in the field is represented
    by the area A between the 𝜆-axis and the 𝜆 − 𝑖 curve as shown in figure (7). The
    area B between the i-axis and the 𝜆 − 𝑖 curve is known as the coenergy and is
    defined as:
                                                        𝑖
                                          ′
                                        𝑊𝑓𝑙𝑑 = ∫ 𝜆 𝑑𝑖
                                                    0



                                               25
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


    This quantity has no physical significance. However, it can be used to find force
    developed in an electromagnetic system. We use coenergy to determine force in
    terms of the current i and energy to determine force in terms of flux linkage 𝜆 from
    figure (7), we have,
                                                  ′
                                         𝑊𝑓𝑙𝑑 + 𝑊𝑓𝑙𝑑 =𝜆𝑖
                                                   ′
    If the 𝜆-i characteristic is linear, 𝑊𝑓𝑙𝑑 = 𝑊𝑓𝑙𝑑 . However, if 𝜆-i characteristic is non-
               ′
    linear, 𝑊𝑓𝑙𝑑 > 𝑊𝑓𝑙𝑑 .
    Field Energy and Mechanical Force
    Consider the electromagnetic system shown in figure (8). Let the current through
    the coil be i when a voltage source v is applied across its terminals. The current i
    sets up magnetic flux 𝜙 in the magnetic circuit. The flux linkages induce an e.m.f. e
    in the coil.




                                          Figure (8)
    Suppose the movable part moves from one position (say x=x1) to another position
    (x=x2) so that at the end of the movement, the air gap decreases. The 𝜆 − 𝑖
    characteristics of the system for these two positions are shown in figure (9-i). note
    that operating points of the system are a when x=x1 and b when x=x2. The current
    𝑖 = (𝑣⁄𝑅) will remain the same at both the positions in the steady state.




                     Figure (9-i)                             Figure (9-ii)
                                              26
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


    If the movable part has moved slowly, the current has remained essentially constant
    during the motion. Therefore, the operating point moves upward from point a to
    point b as shown in figure (9-i).
    During the motion,
                                                𝜆2

                       𝑑𝑊𝑒𝑙𝑒𝑐𝑡 = ∫ 𝑒 𝑖 𝑑𝑡 = ∫ 𝑖 𝑑𝜆 = 𝐴𝑟𝑒𝑎 𝑎𝑏𝑐𝑑
                                               𝜆1


                              𝑑𝑊𝑓𝑙𝑑 = 𝐴𝑟𝑒𝑎 𝑜𝑏𝑐 − 𝐴𝑟𝑒𝑎 𝑜𝑎𝑑
    Now,
                                𝑑𝑊𝑚𝑒𝑐ℎ = 𝑑𝑊𝑒𝑙𝑒𝑐𝑡 − 𝑑𝑊𝑓𝑙𝑑

                                           = (𝐴𝑟𝑒𝑎 𝑎𝑏𝑐𝑑) − (𝐴𝑟𝑒𝑎 𝑜𝑏𝑐 − 𝐴𝑟𝑒𝑎 𝑜𝑎𝑑)

                                         = 𝐴𝑟𝑒𝑎 𝑎𝑏𝑐𝑑 + 𝐴𝑟𝑒𝑎 𝑜𝑎𝑑 − 𝑎𝑟𝑒𝑎 𝑜𝑏𝑐

                                          = 𝐴𝑟𝑒𝑎 𝑜𝑎𝑏

    Thus if the motion occurs under constant-current conditions, the mechanical work
    done (𝑑𝑊𝑚𝑒𝑐ℎ ) is represented by the shaded area in figure (9-i). in fact, this area is
    the increase in the coenergy.
                                                               ′
         ∴                                    𝑑𝑊𝑚𝑒𝑐ℎ = 𝑑𝑊𝑓𝑙𝑑
    If 𝑓𝑚 is the mechanical force causing the different displacement 𝑑𝑥, then,
                                                     ′
                                 𝑓𝑚 𝑑𝑥 = 𝑑𝑊𝑚𝑒𝑐ℎ = 𝑑𝑊𝑓𝑙𝑑
                                                 ′
                                              𝜕𝑊𝑓𝑙𝑑 (𝑖, 𝑥)
         or                              𝑓𝑚 =              |                           (𝑖)
                                                  𝜕𝑥         𝑖=constant

    Lets us now consider that the movement has occurred very quickly. It may be
    assumed that during the motion, the flux linkages have remained essentially as
    shown in figure (9-ii). It can be shown that during the motion, the mechanical work
    done is represented by the shaded area oap. In fact, this area is decrease in field
    energy.
         ∴                                  𝑓𝑚 𝑑𝑥 = 𝑑𝑊𝑚𝑒𝑐ℎ = −𝑑𝑊𝑓𝑙𝑑
                                                      𝜕𝑊𝑓𝑙𝑑 (𝜆, 𝑥)
         or                                    𝑓𝑚 =                |                   (𝑖𝑖)
                                                         𝜕𝑥         𝜆=consyant

    The following point may be noted from this analysis:
                                              27
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


    (i)     Eqs. (i) and (ii) express the mechanical force of electrical origin in terms of
                                                                               ′
            partial derivatives of coenergy and energy functions vis. 𝑊𝑓𝑙𝑑        (𝑖, 𝑥) and
            𝑊𝑓𝑙𝑑 (𝜆, 𝑥).
    (ii)    In the limit when the differential displacement 𝑑𝑥 is small, the area oab in
            figure (9-i) and area oap in figure (9-ii) will be essentially the same.
            Therefore, the force computed from eq. (i) and eq. (ii) will be the same. The
            selection of the energy or coenergy function as a basis for analysis is a matter
            of convenience. The choice depends upon the initial description of the system
            and the desired variables in the result.
    (iii)   The algebraic signs in eq. (i) and (ii) show that the mechanical force acts in a
            direction to increase the coenergy at constant current or to decreases the
            magnetic field stored energy at constant flux.

    Liner System
    Refer back to the electromagnetic system shown un figure (8). If the reluctance of
    the magnetic core path is negligible as compared to that of air gap path, then 𝜆-i
    relation becomes linear.
    For this idealized system, we have,

                                           𝜆 = 𝐿(𝑥)𝑖

    Where 𝐿(𝑥) is the inductance of the coil whose value depends on the air gap length
    (x).
    The field energy of the system is given by;
                                                     𝜆

                                         𝑊𝑓𝑙𝑑 = ∫ 𝑖 𝑑𝜆
                                                    0
                                                    𝜆
                                                          𝜆                       𝜆
                                              =∫              𝑑𝜆        (∵ 𝑖 =        )
                                                         𝐿(𝑥)                    𝐿(𝑥)
                                                 0

                                  𝜆2     [𝐿(𝑥)𝑖]2 1
                               =       =         = 𝐿(𝑥)𝑖 2
                                 2𝐿(𝑥)    2𝐿(𝑥)   2

    The mechanical force in terms of energy function is
                                       𝜕   𝜆2
                                 𝑓𝑚 = − (      )|
                                       𝜕𝑥 2𝐿(𝑥) 𝜆=constant


                                               28
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                                          𝜆2 𝑑𝐿 (𝑥)
                                       = 2
                                        2𝐿 (𝑥) 𝑑𝑥

                                      1 𝑑𝐿(𝑥)                                       𝜆2   1
         ∴                        𝑓𝑚 = 𝑖 2                   (𝑖)                [∵      = 𝐿(𝑥)𝑖 2 ]
                                      2    𝑑𝑥                                      2𝐿(𝑥) 2

    For a linear system, energy and coenergy are equal i.e.,
                                                  ′
                                         𝑊𝑓𝑙𝑑 = 𝑊𝑓𝑙𝑑
    ∴ Mechanical force in terms of coenergy function is
                                       𝜕 1
                                𝑓𝑚 =     ( 𝐿(𝑥)𝑖 2 )|
                                       𝜕𝑥 2           𝑖=constant

                                              1 𝑑𝐿(𝑥)
         or                               𝑓𝑚 = 𝑖 2                                                 (𝑖)
                                              2    𝑑𝑥
    Eqs. (i) and (ii) show that the same expressions are obtained for force whether
    analysis is based on energy or coenergy functions.
    Another useful expression for 𝑓𝑚 . If the reluctance of the magnetic core path in
    figure (8) is neglected, then,
                                                       𝐵g
                                       𝑁𝑖 = 𝐻g × 2g= × 2g
                                                       𝜇𝑜
    Energy stored in the field is given by;
                                              𝐵g                   𝐵g
                                                                        𝐵g
                               𝑊𝑓𝑙𝑑 = 𝐴g 𝑙g ∫ 𝐻g 𝑑𝐵g = 𝐴g 𝑙g ∫             𝑑𝐵
                                                                        𝜇𝑜 g
                                              0                0

         [Here lg and Ag are the length and area of air gap]
                                 𝐵g2
                               =     × volume of the air gap
                                 2𝜇𝑜
                                           𝐵g2
                                         =     × 𝐴g × 2g                             (∵ 𝑙g = 2g)
                                           2𝜇𝑜
    ∴ Mechanical force in terms of energy function is
                                      −𝜕 Bg2
                                 𝑓𝑚 =   (    × 𝐴g × 2g)
                                      𝜕g 2𝜇𝑜


                                                  29
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                                                    𝐵g2
         or                                  𝑓𝑚 = −     (2𝐴g )
                                                    2𝜇𝑜
    note that total area of cross-section of the air gap is 2𝐴g .




 Construction of D.C. Machines
 A d.c. machine consists of the following essential parts:
 1- Yoke or magnetic frame.
 2- Poles:
  i) Pole core
  ii) Pole shoes
 3- Pole coils or field coils
 4- Armature core
 5- Armature windings or conductors
 6- Commutator
 7- Brushes and Bearings
 A d.c. machine has a magnetic circuit and an electric circuit the magnetic circuit provides
 the needed magnetic flux and consists of:
 1- Yoke
 2- Pole cores
 3- Armature core
 4- The air gaps between the pole and the armature
 The electric circuit consists of:
 1- The armature winding
 2- The field coils
 3- The commutator
 4- The brushes.
 Figure (23) shows the general
 arrangement of the various parts of
 a 4-pole d.c. machine
 Stator = Yoke + Pole + Field
 windings
 Rotor = Armature + Commutatore

 The Yoke:-
 Yoke are normally made of cast
 iron (in small machines) but (in

                                               31
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 large machine) made of solid cast steel or roll steel is employed because of it has higher
 permeability than cast iron. Yoke serves in:-
 - forming part of the flux path
 - Providing mechanical support for the poles
 - Protecting the whole machine




                        Figure (23)
 Pole cores and pole shoes:-
 The field magnets consist of pole cores and pole shoes there are two main types of pole
 construction:
 1-Solid pole and laminated pole shoe.
 2-Laminated steel core and pole shoe.
 Laminated construction is necessary because of the pulsations of field strength that result
 when the notched armature rotor magnetic structure passes the pole shoe. Variations in
 field strength result in internal eddy currents being generated in a magnetic structure.
 The thickness of laminations varies from (0.25 mm to 1.5 mm) the laminated poles may be
 secure to the yoke either by rivets or by steel bar. Pole shoes are fastened to the pole core
 counter sunk screws
 The pole shoes are used for:
 -Provide mechanical support for the field coils.
 -Reduce the reluctance of the magnetic circuit (magnetic path)
           Reluctance  L
                            A
 Hence spread out the flux in the air gap and also being of large cross-section.
 i.e the area is increased and the reluctance of the air gap is reduced
 -Obtain better flux distribution.




                                             31
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class



 Pole coil (field coils)
 Which consists of copper wire, the coils consist of certain number of turns wound around
 the pole core to produce the magneto motive force (m.m.f) which yields the required flux.
 The coils are wound in such a way that when current is passed in them they produce
 opposite polarities in adjacent poles. Figure (24) shows the magnetic circuit and the
 magnetic path.




                                          Figure (24)
 Shunt field
 Where the armature and field windings are connected in
 parallel. The field winding in these machines carries a
 separated current, solely for the purpose of producing
 excitation, so it is better to produce the required flux by
 a small current and large number of turns in the field
 winding. Thus in D.C. shunt machine, one normally
 finds a large number of turns of thin wire on the poles.
 The field winding resistance is very high.


 Series field
 This winding is connected series with the armature. It
 use a few turns of heavy wire, because the field coil
 carries the same current what the armature carries.
 Resistance of this field winding is very small, normally

                                               32
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 less than that of armature winding. Flux in this poles depends upon the load current in its
 armature.


 Compound field
 When both the series and
 shunt field connected with
 the armature. The net field
 in the pole is the algebraic
 sum of these two fields. so
 a compound machine can
 be connected to make any
 one of the following types.
 i) Long shunt compound.
 ii) Short shunt compound
 short shunt compound           long shunt compound
 The Air Gap
 It is the space between the pole face and the armature core. It should be big enough to
 allow safe mechanical rotation and small enough to minimize the reluctance of the
 magnetic circuit. Usually 0.5 mm> lg >5mm

 Armature Core
 The armature has a cylindrical or drum shape and it is made of a stack of circular steel
 laminations to reduce the eddy current loss.
 It houses the armature conductors or coils and causes them to rotate and hence cut the
 magnetic flux of the field magnets. In addition to this, its most important function is to
 provide a path of very low reluctance to the flux through the armature from N- pole to P-
 pole.
 The laminations are punched to form slots and teeth parallel to the shaft. Cores of large
 size machines may contain ventilating ducts (Ventilating channels) to improve the cooling
 process.
 The laminations are perforated for air ducts which permit axial flow of air through the
 armature for cooling purposes.
 Armature Windings
 It is that part of the machine in which voltage is generated and torque is produced.
 The windings are usually former – wound from insulated wire and placed in the slots of the
 armature.
 Each junction point between coils in connected to a commutator segment.




                                            33
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




 Commutator
 Its rectifies i.e converts the alternating current induced in the armature conductors into
 unidirectional current in the external load circuit.
 It has cylindrical structure and is built up of wedge- shape segments of high conductivity.
 These segments are insulated from each other by thin layers mica.

 No. of segments = No. of armature coils

 Each commutator segment is connected
 to the armature conductor by means of a
 copper lag or strip.
 The segments have V-grooves so that
 they cannot fly out under the centrifugal
 force. These grooves being insulated by
 conical mica rings.



 Brushes and Bearings
 The brushes whose function is to collect
 current from commutator. The brushes
 are rectangular in shape and are made of
 carbon or graphite.
 Brushes at equal potential are connected
 by internal wires. The current is led to or
 from the brushes by flexible wires. The
 brushes are held in contact with the
 commutator by means of brush holders
 and springs, whose tension may be

                                               34
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 adjusted. Usually the brush holders are of the box type and it is always mounted on an
 insulating material and attached to the stator frame.
 The brush holder is located such that the brush always makes contact with a commutator
 segment connected to a coil in which no voltage is induced.

 Introduction to electromagnetic induction
 When a conductor is moved across a magnetic field so as to cut through the lines of force
 (or flux), an electromotive force (e.m.f.) is produced in the conductor. If the conductor
 forms part of a closed circuit then the e.m.f. produced causes an electric current to flow
 round the circuit. Hence an e.m.f. (and thus current) is 'induced' in the conductor as a result
 of its movement across the magnetic field. This effect is known as 'electromagnetic
 induction'

 Figure (25.a) shows a coil of wire
 connected to a centre-zero galvanometer,
 which is a sensitive ammeter with the
 zero-current position in the centre of the
 scale.
 (a) When the magnet is moved at
 constant speed towards the coil (Figure
 (25.a)), a deflection is noted on the
 galvanometer showing that a current has
 been produced in the coil.
 (b) When the magnet is moved at the
 same speed as in (a) but away from the
 coil the same deflection is noted but is in
 the opposite direction (see Figure (25.b))
 (c) When the magnet is held stationary, even within the coil, no deflection is recorded.
 (d) When the coil is moved at the same speed as in (a) and the magnet held stationary the
 same galvanometer deflection is noted.
 (e) When the relative speed is, say, doubled, the galvanometer deflection is doubled.
 (f) When a stronger magnet is used, a greater galvanometer deflection is noted.
 (g) When the number of turns of wire of the coil is increased, a greater galvanometer
 deflection is noted.
 Figure (25.c) shows the magnetic field associated with the magnet. As the magnet is
 moved towards the coil, the magnetic flux of the magnet moves across, or cuts, the coil. It
 is the relative movement of the magnetic flux and the coil that causes an e.m.f. and thus
 current, to be induced in the coil. This effect is known as electromagnetic induction. The
 laws of electromagnetic induction stated in section below evolved from experiments such
 as those described above.


                                              35
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Laws of electromagnetic induction
 Faraday's laws of electromagnetic induction state:
 (i) An induced e.m.f is set up whenever the magnetic field linking that circuit changes.
 (ii) The magnitude of the induced e. m.f in any circuit is proportional to the rate of change
 of the magnetic flux linking the circuit.
 Lenz's law states:
 The direction of an induced e.m.f is
 always such that it tends to set up a current
 opposing the motion or the change of flux
 responsible for inducing that e.m.f
 An alternative method to Lenz's law of
 determining relative directions is given by
 Fleming's Right-hand rule (often called
 the generator rule) which states:
 Let the thumb first finger and second finger of the right hand
 be extended such that they are all at right angles to each other
 (as shown in Figure (26)). If the first finger points in the
 direction of the magnetic field and the thumb points in the
 direction of motion of the conductor relative to the magnetic
 field, then the second finger will point in the direction of the
 induced e.m.f
 Summarizing:
 First finger- Field
 ThuMb – Motion                                  Figure (26)
 SEcond finger - E.m.f
 In a generator, conductors forming an electric circuit are
 made to move through a magnetic field. By Faraday's law an
 e.m.f. is induced in the conductors and thus a source of e.m.f is created. A generator
 converts mechanical energy into electrical energy. The induced e.m.f. E set up between the
 ends of the conductor shown in Figure (27) is given by:
  E = B.l.v     volts
 Where B. the flux density, is measured in teslas, l, the length of conductor in the
 magnetic field is measured in meters, and v, the conductor velocity, is measured in meters
 per second.

 If the conductor moves at an angle  to the magnetic field (instead of at 90° as assumed
 above) then

     E = B.l.v sin (   volts



                                             36
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                                                                            Figure (27)

 Generating an ac voltage
 Irrelevant as it may seem. The study of a direct-current (dc) generator has to begin with
 knowledge of the alternating-current (ac) generator. The reason is that the voltage
 generated in any dc generator is inherently alternating and only becomes dc after it has
 been rectified by the commutator.
 Figure (28) shows an elementary ac-generator composed of a coil that revolves at 60 r/min
 between the N. S poles of a permanent magnet. The rotation is due to an external driving
 force, such as a motor (not shown). The coil is connected to two slip rings mounted on the
 shaft. The slip rings are connected to an external load by means of two stationary brushes x
 and y.




                                          Figure (28)

 As the coil rotates, a voltage is induced (e.m.f) between its terminals A and D. This voltage
 appears between the brushes and, therefore, across the load. The voltage is generated
 because the conductors of the coil cut across the flux produced by the N, S poles. The
 induced voltage is therefore maximum (20 V. say) when the coil is momentarily in the
 horizontal position, as shown. No flux is cut when the coil is momentarily in the vertical
 position; consequently the voltage at these instants is zero. Another feature of the voltage is
 that its polarity changes every time the coil makes half a turn. The voltage can therefore be
 represented as a function of the angle of rotation Figure (29). The wave shape depends
 upon the shape of the N, S poles. We assume the poles were designed to generate the
 sinusoidal wave shown.
 The coil in our example revolves at uniform speed; therefore each angle of rotation
 corresponds to a specific interval of time. Because the coil makes one turn per second, the

                                              37
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 angle of 3600 in Figure (29) corresponds to an interval of one second. Consequently, we
 can also represent the induced voltage as a function of time Figure (30)




                  Figure (29)                                   Figure (30)

 Direct-current generator
 If the brushes in Fig.(28) could be switched from one slip ring to the other every time the
 polarity was about to change, we would obtain a voltage of constant polarity across the
 load. Brush x would always be positive and brush y negative. We can obtain this result by
 using a commutator Figure (31). A commutator in its simplest form is composed of a slip
 ring that is cut in half, with each segment insulated from the other as well as from the shaft.
 One segment is connected to coil-end
 A and the other to coil-end D. The
 commutator revolves with the coil and
 the voltage between the segments is
 picked up by two stationary brushes x
 and y .
 The voltage between brushes x and y
 pulsates but never changes polarity
 Figure (32).
 The alternating voltage in the coil is
 rectified by the commutator which acts
 as a mechanical reversing switch.
                                                               Figure (31)

 Due to the constant polarity between the brushes, the current in the external load always
 flows in the same direction. The machine represented in Figure (31) is called a direct-
 current generator, or dynamo.




                                              38
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                                          Figure (32)

 Difference between ac and dc generators
 The elementary ac and dc generators in Figure (28) and (31) are essentially built the same
 way. In each case, a coil rotates between the poles of a magnet and an ac voltage is induced
 in the coil. The machines only differ in the way the coils are connected to the external
 circuit Figure (33) ac generators carry slip rings Figure (33-b) while dc generators require a
 commutator Figure (33-a). We sometimes build small machines which carry both slip rings
 and a commutator Figure (33-c). Such machines can function simultaneously as ac and dc
 generators




                                          Figure (33)

 Improving the wave shape
 Returning to the dc generator, we can improve the pulsating dc voltage by using four coils
 and four segments, as shown in Figure (34). The resulting wave shape is given in Figure
 (35). The voltage still pulsates but it never falls to zero: it is much closer to a steady dc
 voltage.




                                              39
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 By increasing the number of coils and segments, we can obtain a dc voltage that is very
 smooth. Modern dc generators produce voltages having a ripple of less than 5 percent. The
 coils are lodged in the slots of a laminated iron cylinder. The coils and the cylinder
 constitute the armature of the machine. The percent ripple is the ratio of the RMS value of
 the ac component of voltage to the dc component, expressed in percent.




                Figure (34)                                      Figure (35)

 It is important to understand the physical meaning of Figure (34) because we will be using
 similar drawings to explain the behavior of dc machines. The four coils in the figure are
 identical to the coil shown in Figure (34).At the instant shown, coil A is not culling any
 flux and neither is coil C. The reason is that the coil sides, of these two coils are midway
 between the poles. On the other hand, coils B and D are cutting flux coming from the
 center of the N and S poles. Consequently, the voltage induced in these coils is at its
 maximum possible value (20 V, say). That is also the voltage across the brushes at this
 particular instant.
 A schematic diagram such as Figure (34) Tells us where the coil sides of the individual
 coils are located: between the poles, under the poles, near the pole tips, and so on. But we
 must remember that the coil sides (a1, a2, b1, b2, etc.) of each coil are actually located at
 1800 to each other and not side by side as Figure (34) seems to indicate.
 The actual construction of this armature is shown in Figure (36). The four coils are placed
 in four slots. Each coil has two coil sides, and so there are two coil sides per slot. Thus,
 each slot contains the conductors of two coils.
 For reasons of symmetry, the coils are wound so that one coil side is at the bottom of a slot
 and the other is at the top. For example, in Figure (34) coil side a1 is in the top of slot 1,
 while coil side a2 is in the bottom of slot 3. The coil connections to the commutator
 segments are easy to follow in this simple armature. The reader should compare these

                                              41
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 connections with those in Figure (36) to verify that they are the same, Note also the actual
 position and schematic position of the brushes with respect to the poles.
 Figure (37) shows the position of the coils when the armature has moved through 45°. The
 sides’ a1, a2, of coil A are now sweeping past pole tip 1 and pole tip 4. The sides of coil C
 are experiencing the same flux because they are in the same slots as coil A. Consequently,
 the voltage ea induced in coil A is exactly the same as the voltage e c induced in coil C.
 Note, however that coil A is moving downward, while coil C is moving upward, The
 polarities of ea and ec are. Therefore, opposite as shown,
 The same reasoning leads us to conclude that eb and ed are equal and opposite in polarity,
 This means that ea + eb + ec + ed = 0 at all times, Consequently, no current will flow in the
 closed loop formed by the four coils, This is most fortunate, because any such circulating
 current would produce I2 R losses,
 The voltage between the brushes is equal to eb + ec (or ea + ed) at the instant shown, It
 corresponds to the minimum voltage shown in Figure (35)
 The armature winding we have just discussed is called a lop winding. It is the most
 common type of winding used in direct-current generators and motors.




                Figure (36)                                      Figure (37)

 Neutral zones
 Neutral zones are those places on the surface of the armature where the flux density is zero.
 When the generator operates at no-load, the neutral zones are located exactly between the
 poles. No voltage is induced in a coil that cuts through the neutral zone. We always try to
 set the brushes so they are in contact with coils that are momentarily in a neutral zone.




                                             41
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class



 Armature Winding Terms
 The different terms related to armature windings are conductor, turn, coil, coil side,
 overhang coil span etc. These terms are briefly described below.
 (a) Conductor
 The length of winding wire across the axial length of machine is called conductor as shown
 in Figure (38).
 Number of conductor = 2 × Number of turn
 (b) Turn
 A turn consists of two conductors’ accommodated in the armature slots approximately a
 pole pitch apart as shown in Figure (38). If two conductors of a turn are placed in the slots
 at a pole pitch apart, the resultant emf induced in the two conductors will be the sum of
 emf’s induced in each conductor.
 (c) Pole pitch or pole span
 It is the distance between the centers of two adjacent poles and is equivalent to 180ed and is
 measured in terms of number of slots i.e.
                                                 𝑁𝑜. 𝑜𝑓 𝑠𝑙𝑜𝑡𝑠
                                  𝑃𝑜𝑙𝑒 𝑃𝑖𝑡𝑐ℎ =
                                                 𝑁𝑜. 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠
                                 One pole pitch = 180ed = 360md /p
 (d) Coil
 A coil consists of one or more turns. A coil with single turn is called single turn coil or
 turn as in Figure.(38) and coil with more turns is called as multiturn coil or coil and several
 connecting coils in series called winding as in Figure (38).




                                  Figure (38)

 (e) Coil Side
 The coil has two sides. The two sides of a coil are placed in the different armature slots.
 One side of a coil is placed in the upper portion of one slot and the other side is placed in
 the lower portion of the other slot at a pole pitch apart in a double layer winding normally
 used in d.c. machines. The coil side placed on upper portion of slot is known as upper coil
 side and the coil side placed on lower portion of slot is known lower coil side.

                                                42
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


  (f) Over Hang
 The winding length required to connect the two conductors at the end of machine is known
 as overhang of the winding.
 (g) Coil Span or coil pitch (Ys)
 The peripheral distance between two slots in which the two sides of a coil are placed is
 called coil span. It is generally expressed in terms of number of slots or conductors. Thus if
 the coil span is 9 slots, it means one side of the coil is in slot 1 and the other side in slot 10.
 (h) Full Pitch Coil
 When the two coil sides of a coil are placed in the slots over the periphery of armature
 exactly a pole pitch apart then the coil is called full pitch coil. In full pitch coil, the coil
 span is equal to the pole pitch of the machine. It means that coil span is 180 electrical
 degrees. In this case, the coil sides lie under opposite poles, hence the induced e.m.f in
 them are additive.
 (i) Short Pitched or Chorded Coil
 When the coil span is less than the pole pitch of the machine, the coil is called a chorded
 coil or short pitched coil.
 (j) Commutator Pitch
 The number of commutator segments spanned from one end of a coil to the other end of the
 same coil.
 In figure (38-1), one side of the coil is connected to commutator segment 1 and the other
 side connected to commutator segment 2. Therefore, the number of commutator segments
 spanned by the coil is 1 i.e. YC =1 in figure (38-2), one side of the coil is connected to
 commutator segment 1 and the other side to commutator segment 8. Therefore, the number
 of commutator segments spanned by the coil = 8-1 = 7 segment i.e. YC = 7
 For simplex lap winding, YC = 1
 For simplex wave winding, YC ≃ 2 pole pitch (segments).




               Figure (38-1)                             Figure (38-2)

                                                43
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Single Layer and Double Layer Winding
 (a) Single Layer Winding
 When the whole slot of the machine is accommodated by one side coil of only one coil, the
 winding is known as single layer winding. Single layer winding is normally not used due to
 commutation problems.
 (b) Double Layer Winding
 When upper portion of a slot (top layer and nearer to air gap) is occupied by the upper coil
 side of a coil and the lower portion (bottom layer) of the same slot is occupied by the lower
 coil side of another coil, winding is known as double layer winding. The two layers of the
 windings in the same slot are insulated by an insulation strip' called separator. Double layer
 windings are most commonly used in d.c. machines. It gives satisfactory arrangement of
 end connections. For double layer winding, the number of armature slots is equal to the
 number of coils in the armature winding. But in larger machines, the number of coils may
 become larger and number of slots may be limited due to design considerations. In such
 machines, the slot may accommodate two or more than two coil sides. The numbering of
 conductors should be done in such a way that odd numbered conductors should occupy top
 layers of the slot portion and even numbers should occupy the bottom layer of slot portion
 Figure. (39) Shows the slots with 2, 4 and 6 conductors.




                               Armature slots and conductors
                              Figure (39) Winding arrangement


 Types of Armature Winding
 All the coils in one parallel path are connected in series and so the emfs induced in these
 coils are added to get the resultant emf, which is equal to voltage rating of machine.
 Generally, two types of windings are used in d.c. machines and they are lap winding and
 wave winding.

 (a) Lap winding
 When end of a coil one is connected to the starting of the coil two like coil one is lapping
 over the coil two. The other coils of the winding are connected in similar way to form a
 closed winding. Since one coil is lapping over the next coil, the winding is known as lap
 winding. Figure (40) shows a portion of lap winding with single turn coil. Coil a and b
 consists of coil sides a1, a2 and b1 , b2 respectively. The end of coil side a2 is connected to
 b1 which is under the same pole as the starting of coil a. Other coils of the winding are
 connected in similar way and the end of the last coil is connected to the starting of the first
 coil and closing the winding. The lap winding arrangement can be done by two way i.e.
 - Simple lap winding and
                                              44
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 - Duplex lap winding.




          (a) Lap winding                         (b) Wave winding
                            Figure (40) Lap and wave winding
 Simple Lap winding: The armature lap winding, which forms only one closed circuit is
 called simple lap winding. The commutator pitch of simple lap winding is unity. The
 number of parallel circuits in simple lap winding (a) is equal to the number of poles (P) in
 the machine. Generally, simple lap windings are being used in d.c. machines.
 Number of parallel paths (a) = P = Number of brushes
 Multiplex Lap Winding: Armature lap winding with several closed circuits is called
 multiplex lap winding. It is used in the machines in which commutator pitch is ±2 and is
 known as duplex lap winding. Thus a duplex lap winding have two closed circuits with the
 parallel circuits equal to twice of parallel circuits in simplex lap winding. In case when
 single lap winding is not suitable double lap winding is used. Since the numbers of parallel
 paths are increasing in duplex lap winding, it may be used for heavy current machines.
 Duplex lap windings are classified as singly re-entrant and double re-entrant. In singly re-
 entrant the starting point is reached after the entire winding and in doubly re-entrant the
 starting point is reached after one half of the winding. Duplex lap winding with odd
 number of commutator bars are singly re-entrant and those with even number of
 commutator bars are double re-entrant.

 Number of parallel paths (a) = mP.

 Where (m) is the multiplicity of the winding and (P) the number of poles.
 (b) Wave Winding
 Wave windings can be classified into single wave winding and multiplex wave winding.
 Figure (40) shows a portion of wave winding with single turn coil with coils a and b. The
 end a2 of coil a which lies in south pole is connected to b 1 the starting of coil b, which lies
 in north pole. Other coils are connected in similar fashion.
 Simple wave winding: Armature winding with wave winding and having only one closed
 circuit is known as simple wave winding. The commutator pitch for simple wave winding
 depends upon the number of poles and commutator segments. The number of parallel
                                               45
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 circuits (parallel paths) in simple wave winding is always 2 irrespective of number of
 poles.

 Number of parallel paths (a) = 2.

 Multiplex wave winding: Armature wave winding with several closed circuits is termed
 as multiplex wave winding. In a multiplex wave winding after making one round of the
 commutator, one arrives at the commutator segment which is away from the initial
 segment by 2, 3, ... m segments, thus obtaining 2, 3 ... , m windings, each simple wave
 winding.
 Duplex winding equivalent to two simple wave winding is used in d.c. machines where
 simplex-winding is not suitable. The number of parallel circuits (parallel path) in duplex
 wave winding is always 4 irrespective of number of poles (double of parallel path in simple
 wave winding).

 Number of parallel paths = 2m

 Where (m) is the multiplicity of the winding
 Duplex wave winding is single re-entrant when commutator bars are odd and it is doubly
 re-entrant if the commutator bars are even.
 Winding Pitches
 Different winding pitches are explained below.
 Back Pitch (YB)
 It is the distance measured in terms of armature conductors between the two side of a coil
 at the back of the armature in figure (40) back pitch for lap and wave wound is shown. It is
 denoted by YB for example, if a coil is formed by connecting conductor 1 (upper conductor
 in a slot) to conductor 12 (bottom conductor in another slot) at the back of the armature,
 then back pitch is YB = 12-1 = 11 conductors.
 Front Pitch (YF)
 Front pitch is the distance measured in terms of armature conductors between the end of a
 coil (conductor) and the starting of the next coil (conductor) connected to the same
 commutator segment as shown in figure (40). For example, if coil side 12 and coil side 3
 are connected to the same commutator segment, then front pitch is Y F = 12 – 3 = 9
 conductors.

 Resultant Pitch (Y)
 The distance ( measured in terms of armature conductors ) between the beginning of one
 coil and the beginning of the next coil to which it is connected figure(40). It is denoted by
 (Y or YR).therefore, the resultant pitch is the algebraic sum of the back and front pitches.



                                             46
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 General rules for D.C. armature windings
 i-   The back pitch (YB) as well as front pitch (YF) should be nearly equal to pole pitch.
      This will result in increased e.m.f. in the coils.
 ii-  Both pitches (YB and YF) should be odd. This will permit all end connections (back
      as well as front connection) between a conductor at the top of a slot and one at the
      bottom of a slot.
 iii- The number of commutator segments is equal to the number of slots or coils (or half
      the number of conductors).
      No. of commutator segments = No. of slots = No. of coils
 iv- The winding must close upon itself i.e., it should be closed circuit winding.

 Relations between pitches for simplex lap winding
 i-      The back and front pitches are odd and are of opposite signs. They differ
         numerically by 2.
                            ∴                   𝑌𝐵 = 𝑌𝐹 ± 2
                                              𝑌𝐵 = 𝑌𝐹 + 2          for progressive winding
                                               𝑌𝐵 = 𝑌𝐹 − 2         for retrogressive winding
 ii-     Both YB and YF should be nearly equal to pole pitch.
                              𝑌 +𝑌
 iii- Average pitch 𝑌𝐴 = 𝐵 𝐹 . It equals pole pitch (= 𝑍⁄𝑃).
                                2
 iv- Commutator pitch, 𝑌𝐶 = ±1
         𝑌𝐶 = +1 for progressive winding
         𝑌𝐶 = −1 for retrogressive winding
 v-      The resultant pitch (YR) in even, being the arithmetical difference of two odd
         numbers viz., YB and YF.
 vi- If Z = number of armature conductors and P = number of poles, then
                       𝑍
         Pole pitch =
                       𝑃
         Since YB and YF both must be about one pole pitch and differ numerically by 2,
                                    𝑍
                               𝑌𝐵 = + 1
                                    𝑃    } 𝑓𝑜𝑟 𝑝𝑟𝑜𝑔𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑤𝑖𝑛𝑑𝑖𝑛𝑔
                                    𝑍
                               𝑌𝐹 = − 1
                                    𝑃
                                   𝑍
                              𝑌𝐵 = − 1
                                   𝑃    } 𝑓𝑜𝑟 𝑟𝑒𝑡𝑟𝑜𝑔𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑤𝑖𝑛𝑑𝑖𝑛𝑔
                                   𝑍
                              𝑌𝐹 = + 1
                                   𝑃
 It is clear that Z/P must be even number to make the winding possible.
 Example
 A 4-pole, simplex lap – wound armature contains 16 slots and has two coil sides per slot.
 Find back pitch, front pitch and commutator pitch for (i) progressive winding
 (ii) retrogressive winding.

                                            47
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




 Solution
              𝑍   16×2
 Pole pitch = =        = 8 conductors
              𝑃     4
 (i)   For progressive winding
                        𝑍
       Back pitch, 𝑌𝐵 = + 1 = 8 + 1 = 9 conductors.
                          𝑃
                          𝑍
        Front pitch,𝑌𝐹 = − 1 = 8 − 1 = 7 conductors
                         𝑃
        Commutator pitch, 𝑌𝐶 = +1 segment
 (ii)   For retrogressive winding
                          𝑍
        Back pitch, 𝑌𝐵 = − 1 = 8 − 1 = 7 conductors.
                          𝑃
                          𝑍
        Front pitch,𝑌𝐹 = + 1 = 8 + 1 = 9 conductors
                        𝑃
        Commutator pitch, 𝑌𝐶 = −1 segment

 Example
 Design a 4-pole, simplex lap winding suitable for an armature containing 20 slot. Assume
 single turn coils with 2 conductors per slot.
 Solution
 Number of commutator segments = number of slots =20
 Number of armature coils = number of slots = 20
 Number of armature conductors = 2×20 = 40
               𝑍    40
 Pole pitch = = = 10 conductors
               𝑃     4
 Let us design the progressive lap winding. For progressive winding:
                          𝑍
        Back pitch, 𝑌𝐵 = + 1 = 10 + 1 = 11 conductors.
                          𝑃
                          𝑍
       Front pitch,𝑌𝐹 = − 1 = 10 − 1 = 9 conductors
                       𝑃
 Figure below shows the winding table

 Back connections                                       front connections
 1 to ( 1 + 11 ) = 12            ⟶                      12 to ( 12 – 9 ) = 3
 3 to ( 3+ 11 ) = 14             ⟶                      14 to ( 14 – 9 ) = 5
 5 to ( 5 + 11 ) = 16            ⟶                      16 to ( 16 – 9 ) = 7
 7 to ( 7 + 11 ) = 18            ⟶
                        18 to ( 18 - 9 ) = 9
 ………………………………………………………………………….
 ………………………………………………………………………….
 ………………………………………………………………………….
 29 to ( 29 + 11 ) = 40                      ⟶          40 to ( 40 – 9 ) = 31

                                           48
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 31 to ( 31 + 11 ) = 42 = ( 42 – 40 ) = 2       ⟶            2 to (42-9)=33
 33 to ( 33 + 11 ) = 44 = ( 44 – 40 ) = 4       ⟶            4 to (44-9)=35
 35 to ( 35 + 11 ) = 46 = ( 46 – 40 ) = 6       ⟶                 6 to ( 46 – 9 ) = 37
 37 to ( 37 + 11 ) = 48 = ( 48 – 40 ) = 8       ⟶                 8 to ( 48 – 9 ) = 39
 39 to ( 39 + 11 ) = 50 = ( 50 – 40 ) = 10      ⟶                 10 to ( 10 – 9 ) = 1




 Note that last conductor 10 joins to the free end of conductor 1 at which the winding
 started. Therefore, the winding closes upon itself.
 Developed diagram. Developed diagram is obtained by imagining the cylindrical surface
 of the armature to be cut by an axial plane and then flattened out. Figure (41) shows the
 developed diagram of the winding. Note that full lines represent the top coil sides (or
 conductors and dotted lines represent the bottom coil sides (or conductors).
 The winding goes from commutator segment 1 by conductor 1 across the back to
 conductor 12 and at the front to commutator segment 2, thus forming a coil. Then from
 commutator segment 2, through conductors 3 and 14 back to commutator 3 and so on till
 the winding returns to commutator segment 1 after using all the 40 conductors.
 Position and number of brushes. We now turn to fined the position and the number of
 brushes required. The brushes, like field poles, remain fixed in space as the commutator
 and winding revolve. It is very important that brushes are in correct position relative to the
 field poles. The arrowhead marked “rotation” in figure (41) shows the direction of motion
 of the conductors. By right-hand rule, the direction of e.m.f. in each conductor will be as
 shown.
 In order to find the position of brushes, the ring diagram shown in figure (42) is quite
 helpful. A positive brush will be placed on that commutator segment where the currents in
 the coil are meeting to flow out of the segment. A negative brush will be placed on that
 commutator segment where the currents the coils are meeting to flow in referring to figure
 (41) there four brushes-two positive and two negative. Therefore, we arrive at a very
 important conclusion that in a simplex lap winding, the number of brushes is equal to the
 number of poles. If the brushes of the same polarity are connected together, then all the
 armature conductors are connected in four parallel paths; each path containing an equal
 number of conductors in series. This is illustrated in figure (43)

                                              49
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Since segments 6 and 16 are connected together through positive brushes and segments 11
 and 1 connected together through negative brushes, there are four parallel paths, each
 containing 10 conductors in series. Therefore, in simples lap winding, the number of
 parallel paths is equal to the number of poles.




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                                          51
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                   Figure (41)                                      Figure (42)

                                         Figure (43)

 Relations between pitches for wave winding
 i-    The distance measured in terms of armature conductors between the two sides of a
       coil at the back of the armature is called back pitch YB figure (40). The YB must be
       an odd integer so that a top conductor and a bottom conductor will be joined.
 ii-   The distance measured in terms of armature conductor between the coil sides
       attached to any one commutator segment is called front pitch YF figure (40). The YF
       must be an odd integer so that a top conductor and a bottom conductor will be
       joined.
 iii- Resultant pitch, YR = YB + YF figure (40)
       The resultant pitch must be an even integer since Y B and YF are odd. Further YR is
       approximately two pole pitch because YB as well as YF is approximately one pole
       pitch
                              𝑌 +𝑌
 iv- Average pitch, 𝑌𝐴 = 𝐵 𝐹
                                2
       When one tour of armature has been completed, the winding should connect to the
       next top conductor (progressive) or to the preceding top conductor (retrogressive). In
       either case, the difference will be of 2 conductors or one slot. If P is the number of
       poles and Z is the total number of armature conductors, then,
                                  𝑃 × 𝑌𝐴 = 𝑍 ± 2                  (𝑖)
                                     𝑍±2
       or                       𝑌𝐴 =
                                       𝑃
       since P is always even and 𝑍 = 𝑃𝑌𝐴 ± 2, Z must be even. It means 𝑍 ± 2⁄𝑃 must be
       an integer. In eq. (i), plus sign will give progressive winding and the negative sign
       retrogressive winding.
 v-    The number of commutator segment spanned by a coil is called commutator pitch
       (YC) figure (40). Suppose in a simplex wave winding,
       P = number of poles; NC = number of commutator segments: YC = commutator pitch
       ∴ number of pair of poles = P/2
 If YC× P/2 = NC, then the winding will close on itself in passing once around the armature.
 In order to connect to the adjacent conductor and permit the winding to proceed,


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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class

                                               𝑃
                                        𝑌𝐶 ×     = 𝑁𝐶 ± 1
                                               2
                         2𝑁𝑐 ± 2 𝑁𝐶 ± 1 𝑁𝑜. 𝑜𝑓 𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑜𝑟 𝑠𝑒𝑔. ±1
                  𝑌𝐶 =          =      =
                           𝑃      𝑃 ⁄2   𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑖𝑟 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠
                                        2𝑁𝐶 ± 2 𝑍 ± 2
                                 𝑌𝐶 =          =      = 𝑌𝐴
                                          𝑃       𝑃
                                                                           𝑌𝐵 + 𝑌𝐹
                ∴                   𝐶𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑜𝑟 𝑝𝑖𝑡𝑐ℎ, 𝑌𝐶 = 𝑌𝐴 =
                                                                              2
 In simples wave winding YB, YF and YC may be equal. Note that YB, YF and YA are in
 terms of armature conductors where as YC is in terms of commutator segments.

 Design of simplex wave winding
 i-   Both pitch YB and YF are odd and are of the same sign.
                             𝑍±2
 ii-  𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑖𝑡𝑐ℎ, 𝑌𝐴 =
                               𝑃
 iii- Both YB and YF are nearly equal to pole pitch and may be equal or differ by 2. If
      they differ by 2, they are one more and one less than YA.
 iv- Commutator pitch is given by;
                                  𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑜𝑟 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 ± 1
                       𝑌𝐶 = 𝑌𝐴 =
                                         𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑖𝑟 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠
      The plus sign of progressive winding and negative for retrogressive winding.
             𝑍±2
 v-   𝑌𝐴 =
              𝑃
      Since YA must be a whole number, there is a restriction on the value of Z. With
      Z=180, this winding is impossible for a 4-pole machine because YA is not a whole
      number.
 vi- 𝑍 = 𝑃𝑌𝐴 ± 2
                                                                𝑍 𝑃𝑌𝐴 ± 2
                   ∴                       𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑖𝑙𝑠 = =
                                                                2       2
 vii- The even number of coils are not used for wave winding

 Example
 A 4-pole armature has 30 armature conductors. The armature is to be simplex wave-wind
 with single-turn coils. Determine for a retrogressive winding (i) back pitch (ii) front pitch
 (iii) commutator pitch.
 Solution
                             𝑍±2
 (i)     Average pitch, 𝑌𝐴 =
                             𝑃
                                           𝑍−2          30−2
       For a retrogressive winding, 𝑌𝐴 =            =          = 7 conductors
                                               𝑃         4


                                                   53
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


         ∴                            𝑌𝐵 = 7 conductors
                𝑌𝐵 +𝑌𝐹
 (ii)    𝑌𝐴 =
                  2
                           7+𝑌𝐹
         or           7=              ∴       𝑌𝐹 = 7 conductors
                            2

 (iii)   YC = YA = 7 conductor segments
         Alternatively, for a retrogressive winding,
                         𝑁𝑜. 𝑜𝑓 𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑜𝑟 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑠 − 1 15 − 1
                  𝑌𝐶 =                                 =       = 7 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑠
                                𝑁𝑜. 𝑜𝑓 𝑝𝑎𝑖𝑟 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠      2

 Example
 A 4-pole, simple wave-wound armature has 21 slots and 21 segments. Determine the
 commutator pitch.
 Solution
                        2𝑁 ±2  2×21±2
 𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑜𝑟 𝑝𝑖𝑡𝑐ℎ, 𝑌𝐶 = 𝐶 =           = 10 𝑜𝑟 11 𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑜𝑟 𝑠𝑒𝑔𝑚𝑒𝑛𝑡
                                  𝑃       4

 Example
 Design a 4-pole wave winding for an armature with 21 slots. Assume single turn coils and
 two conductors per slot.
 Solution
                     𝑧±2    42±2
 Average pitch, 𝑌𝐴 =      =      = 11 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠
                      𝑃       4
 We have taken +2 on the assumption that progressive winding is desired. Since Y B is
 whole number, the wave winding is possible. Taking YB = 11 conductors, we have YF = 11
 conductors. commutator pitch, YC = YA = 11 segments
 Winding table
 The winding table for the winding is given below:
 Back connections                                      front connections
 1 to ( 1 + 11 ) = 12                         ⟶                       12 to ( 12 + 11 ) = 23
 23 to ( 23 + 11 ) = 34                       ⟶        34 to ( 34 + 11 ) = 45 = 45 – 42 = 3
 3 to ( 3 + 11 ) = 14                         ⟶                       14 to ( 14 + 11 ) = 25
 25 to ( 25 + 11 ) = 36                       ⟶
                        36 to ( 36 + 11 ) = 47 = 47 – 42 = 5
 …………………………………………….………………………………………
 ……………………………………………………….……………………………
 ………………………………………………………………….…………………
 19 to ( 19 + 11 ) = 30                       ⟶                       30 to ( 30 + 11 ) = 41
 41 to ( 41 + 11 ) = 52 = 52 – 42 = 10        ⟶                       10 to ( 10 + 11 ) = 21

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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 21 to ( 21 + 11 ) = 32                    ⟶         32 to ( 32 + 11 ) = 43 = 43 – 42 = 1

 Since we have come back to conductor No. 1 from where we started, the winding gets
 closed at this stage.

 Develop diagram. Figure (45) shows the developed diagram for the winding. Note that
 full lines represent the top coil sides (or conductors and dotted lines represent the bottom
 coil sides (or conductors). The two conductors which lie in the same slot are drawn near to
 each other than to those in the other slots.
 Referring to figure (45), conductor 1 connects at the back to conductor 12 (1+11) which in
 Turn connects at the front to conductor 23 (12+11) and so on round the armature until the
 winding is complete. Note that the commutator pitch Y C = 11 segments. This means that
 the number of commutator segment spanned between the start end and finish end of any
 coil in 11 segments.
 Position and number of brushes. We now turn to find the position and the number of
 brushes. The arrowhead marked “rotation” in figure (45) shows the direction of motion of
 the conductors. By right hand rule, the direction of e.m.f. in each conductor will be as
 shown.
 On order to find the position of brushes, the ring diagram shown in figure (46) is quite
 helpful. It is clear that only two brushes (one positive and one negative) required (though
 two positive and two negative brushes can also be used). We find that there are two parallel
 paths between the positive brush and the negative brush. This is illustrated in figure (44)
 therefore, we arrive at a very important conclusion that in simplex wave winding, the
 number of parallel paths is two irrespective of the number of poles. Note that the first
 parallel path has 11 coils (or 22 conductors while the second parallel path has 10 coils (or
 20 conductors). This fact is not important as it may appear at first glance. The coils in the
 smaller group should supply less current to the external circuit. But the identity of the coils
 in either parallel path is rapidly changing from moment to moment. Therefore, the average
 value of current through any particular coil is the same.




                                          Figure (44)
                                              55
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                                          56
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                          Figure (45)                                  Figure (46)
 Dummy coil in wave. Winding
 Sometimes an armature of standard stampings is to be wound in wave windings but the
 number of slots does not fulfill the conditions of wave winding, so instead of leaving any
 slot vacant, a coil is inserted onto it. This additional coil is known as dummy coil.
 This coil is used on the proper place to maintain the mechanical balance of the armature.
 This coil is completely insulated from other coils. If it is possible, the dummy coil should
 not be used because it creates unbalance effect on the armature.
 e.g.           suppose No. of poles               =4
                No. of armature slots              = 65
                No. of conductor per slot           =4
 Now            Total coil sides (conductor)         = 65×4 = 260
                                                     260±2         1      1
 Average pitch                                     =       = 65 𝑜𝑟 64
                                                        4         2       2
 But the average pitch should be an integer. This winding is not possible with 260
 conductors because in 65 slots, we cannot insert more than 260 conductors.
 Now if we take coil sides or conductors = 258
                                               258±2
 Then average pitch                         =        = 65 𝑜𝑟 64
                                                 4
 Now there is possibility of two windings i.e. Progressive and retrogressive but in both type
 of windings, one coil will remain inactive that is one coil will not be considered the part of
 the windings. It will work as filler in the slot and the coil will be known as dummy coil. In
 this condition, creeping is observed in the winding and the winding is known as forced
 winding.
 The ends of these dummy coils are tapped and these coils only help the armature for
 maintaining the mechanical balance. At the time of numbering the coils, these dummy coils
 are not numbered.

 Equalizer connections (equalizer rings.)
 In case of lap winding, the number of parallel paths is equal to the number of poles. Each
 of these parallel paths consists of coil sides which are placed under two adjacent poles. If
 flux under all poles is equal then the e.m.f. induced in all parallel paths will be equal. But
 practically, it may happen that e.m.f. induced in different parallel paths may not be same.
 This difference in e.m.f. is due to difference in reluctance values of corresponding iron
 parts of magnetic circuits which causes due to defects in castings and structural
 irregularities. It may be possible that length of air gap may not be same under all the poles
 due to machining or assembling defects which results in difference in e.m.f. induced. The
 third reason is that the poles may have different strengths due to error in putting field
 winding.
 The effect of this difference in e.m.f. is that as the e.m.f.s in different parallel paths are
 different there will be resultant e.m.f. acting across armature winding and will force heavy
 current through armature winding as its resistance is very small. This will cause inequality

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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 in brush arm currents. It will give rise to I2R loss in the winding on no load and at full load
 will add to the losses both in windings and brushes. It will also cause commutation
 difficulties resulting in overheating and sparking.
 To overcome the objectionable effects of the circulating currents, it is customary to employ
 equalizer connections in all lap-wound armatures. These are extremely low resistance
 copper wires that connect together points on the armature winding that are exactly 360
 electrical degrees (two pole pitches) apart; these points should, under ideal conditions, be
 at the same potential at all times but because of electrical, magnetic, and mechanical
 differences, are not. If there is unequal potential difference, equalization of potential will
 be there due to flow of current through these low resistance conductors. This bypassing of
 circulating currents relieves the brushes from extra loading. These equalizer connections
 may be in the form of rings which are known as equalizer rings.
 Figure (i) illustrate rather simply how the circulating current is made to avoid the brush
 contacts when an equalizer connection is used. In the schematic 4-pole lap winding,
 assume that point a is at a higher potential than point b and that points c and d are at the
 same potential. In Fig. (i-a), where no equalizer connection is employed, the circulating
 current is seen to pass across brush contacts x and y and through the winding as indicated.
 With an equalizer connection, however, as in Fig. (i-b), the circulating-current paths by-
 pass brush contacts x and y; this improves the operating performance of the machine.




    Figure (i) Sketches illustrating how circulating currents flow in an armature winding
                           without and with an equalizer connection
 Such equalizer connection performs two important functions:


                                              58
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 i- They relieve the brushes of existing circulating currents by providing paths of low
     resistance that by-pass the brush contacts;
 ii- They create electromagnetic effects that tend to equalize the flux distributions under all
     the poles, strengthening the weaker ones and weakening the stronger ones.


 Generation E.M.F.
 Generated e.m.f Eg = e.m.f generated in any one of the parallel paths.
                                        𝑑𝜙
 Average e.m.f generated / conductor =        volt
                                         𝑑𝑡
 Flux cut / conductor in one revolution 𝑑𝜙 = 𝜙𝑝 (Wb)
 n = No. of revolutions / second = N/60
 Time for one revolution dt = 60/N (second).
 Hence according to Faraday's Lows.
                                𝑑𝜙    𝜙𝑝    𝜙𝑝𝑁
 E.M.F generated /conductor =      = 60 =         volt.
                                𝑑𝑡            60
                                         𝑁
 𝜙 = flux/pole in Weber.
 Z = Total number of armature conductors.
   = No. of slot × No. of conductors /slot.
 p = No. of generator poles.
 a = No of parallel paths in armature.
 N = armature rotation in revolutions per minute (r.p.m).
 E = e.m.f induced in any parallel path in armature.

 For a simplex wave wound generator
 No. of parallel paths = 2
 No. of conductors (in series) in one path = Z/2
    E.M.F. generated / path =  pN/60 × Z/2 =  ZpN/120 volt
 For a simplex lap – wound generator.
 No. of parallel paths = p
 No. of conductors (in series) in one path = Z/p
    E.M.F. generated / path = p  N/60 × Z/p =Z  N/60 volt
 In general generated e.m.f Eg =  ZN/60 × (p/a) volt
 Where        a = 2 - for simplex wave – winding
                 = p - for simplex lap – winding
                1      2𝜋𝑁            𝑝     𝜔𝜙𝑍   𝑝
 Also, Eg = × ( ) × 𝜙𝑍 × ( ) =                  ×( )   volt- 𝜔 in rad/s
               2𝜋     60             𝑎       2𝜋    𝑎

 The equivalent armature resistance (Ra)
 Rt The resistance of the used wire at operating temperature.
 Rt =  l/A

                                              59
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class

  = resistivity at working temperature ohm/m.
 L = total length of the wire used.
 A = cross sectional area of the wire.

 Rp = Rt/a      so      rp = Rt / 2   for wave &     rp = Rt / p   for lap

 Rp = resistance of any parallel path of that machine.

 Ra = rp / a = Rt /a2        so Ra = Rt / 4 for wave Ra = Rt / p2 for lap

 Example:
 Rt = 12 ohm
 p=6

 Ra = 12/4 = 3 ohm                     wave wound.
 Ra = 12/36 = 0.333 0hm                lap wound.

 TORQUE EQUATION
 Let E and Ia be the armature induced voltage and armature current respectively.
 :. Armature power = E Ia.
 Since, P = 2nT where n = N/60. N is the speed in rpm and T is the torque of the
 armature. Therefore, T= P/ 2n = EIa/ 2n = 60 EIa/ 2  N.
 Substituting here for E from Equation above, we have:
 T = 60× 𝜙 × ZNPIa/60 × 𝑎 × 2𝜋𝑁
 T = 𝜙ZPIa/2𝜋𝑎

 For a given machine Z, P and a are constants.
 Therefore, T ∝ ϕIa
 It is also sometimes used as T= kt If Ia where kt is called torque constant. If and Ia are the
 field and armature winding currents respective




                                              61
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                                  Type of d.c. generator

 (a) Separately – excited generator
 A typical separately – excited generator circuit is shown in figure (47). We can use a pair
 of electromagnets, called field poles, as shown in figure (47). When the d.c. current in such
 a generator is supplied by an independent source (such as a storage battery or another
 generator, called an exciter, the generator is said to be separately excited. Thus in figure
 (47) the dc source connected to terminal a and b causes an exciting current If to flow. If the
 armature is driven by a motor or a diesel engine, a voltage E appears between brush
 terminals x and y
 When the load is connected across the armature terminals, a load current I a will flow. The
 terminal voltage V will fall from its open – circuit emf due to a voltage drop caused by
 current flowing through the armature resistance, shown as Ra
 Terminal voltage V = E – Ia Ra
 Generated emf      E = V + Ia Ra




                                          Figure (47)

 (b) Shunt wound generator
 In a shunt wound generator the field winding is connected in parallel with the armature so
 that the generator can be self-excited as shown in figure (48). The principal advantage of
 this connection is that it eliminates the need for an external source of excitation
 When a shunt generator is started up, a small voltage is induced in the armature, due to the
 remnant flux in the poles. This voltage produces a small exciting current If in the shunt
 field. The resulting small mmf acts in the same direction as the remnant flux, causing the
 flux per pole to increase. The increased flux increases E which increases I f which increases
 the flux still more, which increases E even more, and so forth. This progressive buildup
 continues until E reaches a maximum value determined by the field resistance and the
 degree of saturation.
 The field winding has a relatively high resistance and therefore the current carried only a
 fraction of the armature current.

                                              61
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




                                         Figure (48)

 For the shown in figure (48),
 Terminal voltage, V = E – Ia Ra
 Generated emf,          E = V + Ia Ra
 Ia = If + I from Kirchhoff's current law, where
 Ia = armature current,
 If = field current (If = V/Rf)
 I = load current.

 (c) Series – wound generator
 In the series – wound generator the field winding is connected in series with the armature
 as shown in figure (49)




                                        Figure (49)

 Terminal voltage     V = E – Ia (Ra + Rf)
 Generated emf        E = V + Ia (Ra + Rf)

                                             62
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




 (d) Compound – wound generator
 In the compound – wound generator two methods of connection are used, both having a
 mixture of shunt and series windings, designed to combine the advantages of each. Figure
 (50-a) shows what is termed a long – shunt compound generator, and figure (50-d) shows a
 short – shunt compound generator. The latter is the most generally used from of dc
 generator.




                                          Figure (50)

 For long – shunt compound
 Terminal voltage V = E – Ia(Ra +Rf.se)
 or                 V = IfRf.sh

 and for short – shunt compound
 terminal voltage     V = E – (IaRa + IRf.se)
 or                   V = IRf.se + IfRf.sh



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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                                    D.C machine losses
 A generator is a machine for converting mechanical energy into electrical energy and a
 motor is a machine for converting electrical energy into mechanical energy. When such
 conversions take place, certain losses occur which are dissipated in the form of heat.
 The principal losses of machines are:
 (i) Copper loss, due to I2R heat losses in the armature (Ia2Ra) and field windings (If2Rf).
 (ii) Iron (or core) loss: due to hysteretic and eddy-current losses in the armature. This loss
 can be reduced by constructing the armature of silicon steel laminations having a high
 resistively and low hysterics loss. At constant speed, the iron loss is assumed constant.
 (iii) Friction and windage losses (mechanical losses): due to bearing and brush contact
 friction and losses due to air resistance against moving parts (called windage). At constant
 speed, these losses are assumed to be constant.
 (iv) Brush contact loss: between the brushes and commutate. The voltage drop across the
 brushes, called brush contact drop or brush drop is relatively small. It includes brushes of
 both polarities and is assumed to have the following constant values for all loads.
 0.5 V for metal – graphite brushes.
 2.0 V for electro graphitic and graphite brushes.
 This loss is approximately proportional to the load current.
 (v) Rotational losses (stray-losses)
 Rotational losses occur in the rotational part of the generator and are made up of the
 following two components:
 a) Mechanical losses are made up of the following components:
 1- Bearing friction losses and brushes friction on the commutator losses
 2- Wind friction, usually called windage
 b) Iron losses or magnetic losses are made up of the following components:
 1- Hysteresis losses
 2- Eddy current losses
 The total losses of a machine can be quite significant and operating efficiencies of between
 80 per cent and 90 per cent are common.
 In the case of a shunt or compound generator, shunt field copper losses and stray losses are
 constant and armature and series field losses are variable losses.
 Efficiency of a d.c. generator
 The efficiency of an electrical machine is the ratio of the output power to the input power
 and is usually expressed as a percentage. The Greek letter,  (eta) is used to signify
 efficiency and since the units are, power/power, then efficiency has no units. Thus

 Efficiency  = (output power/input power)×100%

 If the total resistance of the armature circuit (including brush contact resistance) is Ra then
 the total loss in the armature circuit is Ia2 Ra

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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 If the terminal voltage is V and the current in the shunt circuit is If, then the loss in the
 shunt circuit is If V.
 If the sum of the iron, friction and windage losses is C (mechanical losses) then the total
 losses is given by:
 I2a Ra + If V+ C       (I2a Ra + If V is, in fact, the 'copper loss').
 If the output current is I then the output power is VI.
 Total input power = VI + I2a Ra + If V + C.          Hence

          = (VI/ VI + I2a Ra + If V + C) ×100%

 The efficiency of a generator is a maximum when the load is such that:
 I2a Ra = VIf + C
 i.e. when the variable loss = the constant loss
 ARMATURE REACTION AND COMMUTATION
 Armature Reaction
 By armature reaction is meant the effect of magnetic field set up by armature current on the
 distribution of flux under main poles of a generator. The armature magnetic field has two
 effects:
 (i) It demagnetises or weakens the main flux and
 (ii) It cross-magnetises or distorts it
 The first effect leads to reduced generated voltage and the second to the sparking at the
 brushes.
 These effects are well illustrated in Figure (45) which shows the flux distribution of a
 bipolar generator when there is no current in the armature conductors. For convenience,
 only two poles have been considered, though the following remarks apply to multicolor
 fields as well. Moreover, the brushes are shown touching the armature conductors directly,
 although in practice, they touch commutator segments, It is seen that
 (a) The flux is distributed symmetrically with respect to the polar axis, which is the line
 joining the centers of NS poles.
 (b) The magnetic neutral axis or plane (M.N.A.) coincides with the geometrical neutral
 axis or plane (G.N.A)
 Magnetic neutral axis may be defined as the axis along which no e.m.f. is produced in the
 armature conductors because they then move parallel to the lines of flux.
 Or M.N.A. is the axis which is perpendicular to the flux passing through the armature.
 The brushes are always placed along M.N.A Hence M.N.A. is also called 'axis of
 commutation' because reversal of current in armature conductors takes place across this
 axis. In Figure (45) is shown vector OFm which represent, both in magnitude and direction,
 the m.m.f. producing the main flux and also M.N.A which is perpendicular to OFm
 In Figure (46) is shown the field (or flux) set up by the armature conductors alone when
 carrying current, the field coils being unexcited. The direction of the armature current is
 the same as it would actually be when the generator is loaded. It may even be found by

                                             65
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 applying Fleming's Right-hand Rule. The current direction is downwards in conductors
 under N-pole and upwards in those under S-pole. The downward flow is represented by
 crosses and upward flow by dots.
 As shown in Figure (46) the m.m.fs of the armature conductors combine to send flux
 downwards through the armature. The direction of the lines of force can be found by
 applying cork-screw rule. The armature m.m.f. (depending on the strength of the armature
 current) is shown separately both in magnitude and direction by the vector OFA which is
 parallel to the brush axis.
 So far, we considered the main m.m.f. and armature m.m.f. separately as if they existed
 independently, which is not the case in practice. Under actual load conditions, the two exist
 simultaneously in the generator as shown in Figure (47)
 It is seen that the flux through the armature is no longer uniform and symmetrical about the
 pole axis, rather it has been distorted. The flux is seen to be crowded at the trailing pole
 tips but weakened or thinned out at the leading pole tips (the pole tip which is first met
 during rotation by armature conductors is known as the leading pole tip and the other as
 trailing pole tip). The strengthening and weakening of flux is separately shown for a four-
 pole machine in Figure (48). As seen, air-gap flux density under one pole half is greater
 than that under the other pole half.
 Figure (47) is shown the resultant m.m.f. OF which is found by vectorially combining OFm
 and OFA.
 The new position of M.N.A., which is always perpendicular to the resultant m.m.f. vector
 OF, is also shown in the figure. With the shift of M.N.A, say through an angle  brushes
 are also shifted so as to lie along the new position of M.N.A. Due to this brush shift (or
 forward lead), the armature conductors and hence armature current is redistributed. Some
 armature conductors which were earlier under the influence of N-pole come under the
 influence of S-pole and vice-versa. This regrouping is shown in Figure (49), which also
 shows the flux due to armature conductors. Incidentally, brush position shifts in the same
 direction as the direction of armature rotation.
 All conductors to the left of new position of M.N.A. but between the two brushes carry
 current downwards and those to the right carry current upwards. The armature m.m.f. is
 found to lie in the direction of the new position of M.N.A. (or brush axis). The armature
 m.m.f. is now represented by the vector OFA which is not vertical (as in Figure (46)) but is




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                Figure (45)                                     Figure (46)
 inclined by an angle  to the left. It can now be resolved into two rectangular components,
 OFd parallel to polar axis and OFc perpendicular to this axis. We find that
 (i) Component OFc is at right angles to the vector OFm (of Figure (45)) representing the
 main m.m.f. It produces distortion in the main field and is hence called the cross-
 magnetising or distorting component of the armature reaction.
 (ii) The component OFd is in direct opposition of OFm which represents the main m.m.f. It
 exerts a demagnetising influence on the main pole flux. Hence, it is called the
 demagnetising or weakening component of the armature reaction.
 It should be noted that both distorting and demagnetising effects will increase with increase
 in the armature current.




              Figure (47)                                   Figure (48)

 Demagnetising and Cross-magnetising Conductors
 The exact conductors which produce these distorting and demagnetising effects are shown
 in Figure (50) where the brush axis has been given a forward lead of  so as to lie along
 the new position of M.N.A. All conductors lying within 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. This fact may be checked by applying crock screw
 rule. It is these conductors which act in direct opposition to the main field and are hence
 called the demagnetising armature conductors.
 Now consider the remaining armature conductors lying between angles AOD and COB. As
 shown in Figure (51), these conductors carry current in such a direction as to produce a
 combined flux pointing vertically downwards i.e. at right angles to the main flux. This
 results in distortion of the main field. Hence, these conductors are known as cross-
 magnetising conductors and constitute distorting ampere-conductors.



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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Demagnetising AT per Pole
 Since armature demagnetising ampere-turns are neutralized by adding extra ampere-turns
 to the main field winding, it is essential to calculate their number. But before proceeding
 further, it should be remembered that the number of turns is equal to half the number of
 conductors because two conductors-constitute one turn.
 Let
 Z = total number of armature conductors
 I = current in each armature conductor
   = Ia/2          for simplex wave winding
   = Ia/P           for simplex lap winding
  m = for word lead in mechanical or geometrical or angular degrees.
 Total number of armature conductors in angles AOC and BOD = (4  m / 360) × Z
 As two conductors constitute one turn,
  Total number of turns in these angles = (2  m / 360) × Z
  demagnetizing amp-turns per pair of poles = (2  m / 360) ×ZI
 .demagnetizing amp-turns / pole = (  m / 360) × ZI
 .ATd per pole = ZI × (  m / 360)




                   Figure (49)                               Figure (50)




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




 Figure (51)                                                     Figure (52-i)
 Cross-magnetising AT per pole
 The conductors lying between angles AOD and HOC constitute what are known as
 distorting or cross-magnetising conductors. Their number is found as under:
 Total armature-conductors/pole both cross and demagnetising = Z/P
 Demagnetising conductors / pole = Z ×(2  m/360) (found above)
  cross-magnetising conductors/pole = (Z/P) – Z × (2  m / 360) = Z (1/P - 2  m/360)
  cross-magnetising amp-conductor/pole = ZI × (1/P - 2  m / 360)
  cross-magnetising amp-turns/pole = ZI × (1/2P -  m / 360)
  ATc/pole = ZI × (1/2P -  m / 360)
 Note. (i) For neutralizing the demagnetising effect of armature-reaction, an extra number
 of turns may be put on each pole.
 No. of extra turns/pole = ATd/Ish                         for shunt generator.
                                  = ATd/Ia                 for series generator.
 If the leakage coefficient  is given, then multiply each of the above expressions by it.
 (ii) If lead angle is given in electrical degrees, it should be converted into mechanical
 degrees by the following relation
  (mechanical) =  (electrical) / pair of poles or  m =  e/ (P/2) = 2  e / P

 Compensating Windings
 These are used for large direct current machines which are subjected to large fluctuations
 in load i.e. rolling mill motors and turbo-generators etc. Their function is to neutralize the
 cross magnetizing effect of armature reaction. In the absence of compensating windings,
 the flux will be suddenly shifting backward and forward with every change in load. This
 shifting of flux will induce statically induced e.m.f. in the armature coils. The magnitude of
 this e.m.f. will depend upon the rapidity of changes in load and the amount of change. It
 may be so high as to strike an arc between the consecutive commutator segments across the


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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 top of the mica sheets separating them. This may further develop into a flash-over around
 the whole commutator thereby short-circuiting the whole armature.
 These windings are embedded in slots in the pole shoes and are connected in series with
 armature in such a way that the current in them flows in opposite direction to that flowing
 in armature conductors directly below the pole shoes. An elementary scheme of
 compensating winding is shown in Figure (52-i, ii).




                        Figure (52-ii)
 It should be carefully noted that compensating winding must provide sufficient m.m.f so as
 to counterbalance the armature m.m.f. Let
 Zc = No. of compensating conductors/pole face
 Za = No. of active armature conductors/pole.
 Ia = total armature current
 Ia / A= current/armature conductor
 ... Zc Ia = Za (Ia/a)   or      Zc = Za/a
 Owing to their cost and the room taken up by them, the compensating windings are used in
 the case of large machines which are subject to violent fluctuations in load and also for
 generators which have to deliver their full-load output at considerable low induced voltage
 as in the Ward-Leonard set.

 No. of Compensating Windings
  No. of armature conductors/pole = Z/P
  No. of armature turns/pole = Z/2P
  No. of armature-turns immediately under one pole
 = Z/2P × pole arc/pole pitch = 0.7 Z/2P (approx.)
  No. of armature amp-turns/pole for compensating winding
 = 0.7 × ZI/2P = 0.7 ×armature amp-turns/pole.




                                            71
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Commutation process
 When a generator is under load, the individual coils
 on the armature carry one-half the load current
 carried by one brush. The currents flowing in the
 armature windings next to a positive brush are
 shown in Figure (53-a). Note that the currents in the
 coils flow toward the brush, coming both from the
 right and the left. If the load current is 80 A, the
 coils all carry 40 A.
 If the commutator segments are moving from right
 to left, the coils on the right-hand side of the brush
 will soon be on the left-hand side. This means that
 the current in these coils must reverse. The reversal
 takes place during the millisecond interval that a
 coil takes to move from one end of the brush to the
 other. The process whereby the current changes
 direction in this brief interval is called
 commutation.
 To understand how commutation takes place, we
 refer to Figures (53-a to 53-e).
 In Figure (53-a) the brush is in the middle of
                                                            Figure (53)
 segment 1, and the 40 A from the coils on the right
 and the left of the brush unite to give the 80 A output. The contact resistance between the
 segment and brush produces a voltage drop of about 1 V.
 In Figure (53-b) the commutator has moved a short distance, and 25 percent of the brush
 surface is now in contact with segment 2, while 75 percent is in contact with segment 1.
 Owing to the contact resistance, the conductivity between the brush and commutator is
 proportional to the contact area. The area in contact with segment 2 is only one-fourth of
 the total contact area, and so the current from segment 2 is only one-fourth of the total
 current, namely 0.25 × 80 = 20. By the same token, the current from segment 1 to the
 brush is 0.75 × 80 = 60 A.
 If we now apply Kirchhoff's current law, we discover that the current flowing in coil 1
 must be 20 A. Thus, by coming in contact with the brush, the current in this coil has
 dropped from 40 A to 20 A.
 In Figure (53-c) the commutator has moved a little further, and the brush area in contact
 with segments 1 and 2 is now the same. Consequently, the conductivities are the same and
 so the currents are equal. This means that the current in coil 1 is zero at this instant.
 In Figure (53-d) the commutator has moved still farther to the left. Segment 2 is now in
 contact with 75 percent of the brush, and so the currents divide accordingly: 60 A from
 segment 2 and 20 A from segment 1. Applying Kirchhoff's current law, we find that the
 current in coil 1 is again 20 A, but it flows in the opposite direction to what it did before!

                                              71
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 We can now understand how the brush contact resistance forces a progressive reversal of
 the current as the segments slide over the brush.
 In Figure (53-e) the current reversal in coil 1 is complete and the current in coil 2 is about
 to be reversed.
 In this ideal commutation process, it is important to note that the current density (amperes
 per square centimeter) remains the same at every point across the brush face. Thus, the heat
 produced by the contact resistance is spread uniformly across the brush surface.
 Unfortunately, such ideal commutation is not possible in practical machines, and we now
 investigate the reason why?
 The practical commutation process
 The problem with commutation is that it takes place in a very short time; consequently, the
 current cannot reverse as quickly as it should. The reason is that the armature coils have
 inductance and it strongly opposes a rapid change in current.
 Suppose, for example, that the commutator in Figure (53) has 72 bars and that the armature
 turns at 600 r/min. One revolution is, therefore, completed in 1/10 of a second and during
 this short period 72 commutator bars sweep past the brush. Thus, the time available to
 reverse the current in coil 1 is only 1/10  1/72 = 1/720 s or 1.39 ms
 The voltage induced by self-induction is given by
 e  LI t
 In which
 e = induced voltage [V]
 L = inductance of the coil [H]
 I t = rate of change of current [A/s]
 If coil 1 has an inductance of, say, 100  H, the induced voltage is
 e  LI t or e  2 L.I c tc where I c is the conductor current and tc time of commutation
    100  10 6   40   40 
 
            1.39  10 3
 = 5.75 V
 It is the presence of this induced voltage (attributable to L), that opposes the change in
 current.
 Figures (54-a to 54-e) illustrate the new currents that flow in coil 1 when the self-
 inductance of the coil is considered. We have assumed plausible values for these currents
 in order to determine the resulting current flows in the brush. The currents should be
 compared with those in Figure (53).




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 In Figure (54-a) the brush is in the middle of
 segment 1, and the currents in the coils are
 neither increasing nor decreasing. As a result, the
 coil inductance does not come into play.
 In Figure (54-b) the current in coil 1 is changing
 due to the contact resistance effect. However, the
 induced voltage e prevents the current from
 dropping to its ideal value of 20 A. Suppose the
 coil current is 35 A. From Kirchhoff's current
 law, the currents flowing from segments 1 and 2
 into the brush are then respectively 75 A and 5
 A, instead of 60 A and 20 A. Note that the
 current density is no longer uniform over the
 brush face. The density is low where the brush
 touches segment 2, and high where it touches
 segment 1.
 In Figure (54-c) the brush is momentarily
 symmetrically placed as regards segments 1 and
 2. But the current in coil 1 has not fallen to zero,
 and is still, say 30 A. As a result, the current in
 segment 1 is 70 A while that in segment 2 is only
 10 A. The current density on the left-hand side of
 the brush is, therefore, 7 times greater than on the
 right-hand side. The left-hand side of the brush
 will tend to overheat.
 In Figure (54-d) segment 1 has moved beyond
 the midpoint of the brush and the current in coil 1
 has still not reversed. Assuming it has a value of
 20 A. the current flowing from segment 1 to the
 brush is now 60 A. despite the fact that the             Figure (54)
 contact area is getting very small. The resulting
 high current density causes the brush to overheat at the tip. Because 720 coils are being
 commutated every second, this overheating raises the brush tip to the incandescent point
 and serious sparking will result.
 How to Improve Commutation
 This can be done by following methods:
 1- Reduction of inductance and increase of resistance of the commutating coils:
 Reduction of inductance is possible by using single or minimum turns coils and the
 resistance can be increased by increasing contact resistance in riser or using high resistivity
 brushes. Thus the relative effect of inductance on the commutation is decreased.
 2- Shifting brushes from g.n.a.:

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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 The brushes can be usefully shifted from the g.n.a. as discussed in the armature reaction
 section of this chapter. The main advantage of this shifting is that now the commutating
 coil comes in the main pole field which gives a dynamic emf to help commutation. That is
 why the shifting of brushes in the other direction is not allowed, in that case the
 commutating coil comes under the influence of the opposite pole which will provide a
 dynamic emf in it to severe the commutation.
 3- Using inter-pole this is the most commonly used and the best method of improving
 commutation. When a machine is provided with inter pole at the geometrical axis the
 brushes are also kept on the same axis. No shifting of brushes is done in case of inter pole
 machines.
 The winding of the inter pole is in series with the armature winding so the same armature
 current flows through the inter poles. This is also the commutating current. When this
 current is large i.e. the commutation is likely to be severe this current produces a large flux
 at the inter pole i.e. in the commutating zone. The winding in the inter pole is done in such
 a direction so that the dynamic emf produced due to the cutting of this inter pole flux by
 the commutating coil sides is in a direction helping the commutation. The number of turns
 on the inter pole is as per the degree of remedial action required. The polarity of these inter
 poles should be similar in effect to the polarity which we get when we improve the




                                              74
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 commutation by shifting of brushes. That is in case of generating mode the polarity of the
 inter pole should be the same as that of the main pole being approached and in motor mode
 it should be that of the main pole being left behind.
 4- Compensation winding
 Compensation winding is placed in the main pole shoes and connected in series with the
 armature winding such that the mmf. of the two windings oppose and cancel each other.
 This makes the operation of the machine much more reliable and practically independent
 of load variations but the design of the machine becomes complicated and its cost goes
 high.

 Other Causes of Sparking and Flash-over on Commutator
 The sparking on commutator may still be there, although the commutation in the machine
 may be good, because of some of the following other reasons.
 1- Mechanical defects: Non circular form of commutator and its eccentricity due to
 improper manufacturing process or caused by over-heating and centrifugal forces, mica
 protruding between the bars, poor surface finish. shaking and vibration of brushes in the
 brush holder, improper fixing of brush holder, loose mounting of the brush rockers, poor
 quality brush material are all the mechanical reasons of sparking on the commutator.
 2- Voltage surges between commutator segments: The commutator design is based on
 some maximum voltage coming across the two adjacent commutator segments. The
 voltage coming across these two adjacent commutator segment is the emf induced in the
 armature coil connected to these segments. Because of heavy distortion or abrupt
 fluctuation of flux in the gap due to, sudden load changes or faults etc may induce such a
 heavy emf in one or more armature coil which causes the spark between the segments.
 3- Dirt and oil on commutator: Dirt or carbon dust particles when impregnated with oil,
 due to oily surface, bridges the insulation gaps between the commutator segments. These
 particles when come under the brushes and get heated up, create conductive bridge
 between the segments. This causes large leakage current between the segments, which
 enhance the heating, creating a highly ionized space around the commutator which may
 cause a flash over, short circuiting the brushes around the commutator. Regular cleaning of
 commutator prevents this fault




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


  Characteristic of D.C generators
  1- No-load saturation characteristic (E&If). It is also known as Saturation curve,
  Magnetic characteristics, & Open – circuit characteristic (O.C.C)
  It shows the relation between the no load generator e.m.f in armature (E) and the field or
  exciting current If at a given fixed speed.
  No-load curve characteristics
  The o.c.c or no-load saturation curves for self-excited generators whether shunt or series
  connected, are obtained in similar way.
  1- The field winding of the generator (whether shunt or series wound) is disconnected from
  the machine and connected to an external source of direct current.
  2- The field or exciting current If is varied rheostatically and its value read on the ammeter.
  3- The machine is driven at constant speed by the prime mover and the generated e.m.f on
  no load is measured by the voltmeter connected across the armature
  4- If is increased by suitable steps (starting from zero) and the corresponding value of E are
  measured.
  5- On plotting the relation between If and E a curve is obtained.




   Building of the voltage of self-excited shunt generator
   One of the simples of self-excited generator is the shunt-
   wound machine, the connection diagram (without load) of
   which is shown in figure (55). The manner in which a
   self-excited generator manages to excite its own field and
   build a d.c. voltage across its armature is described with
   reference to figure (56) in the following steps:
i- Assume that the generator starts from rest, i.e., prime-
    mover speed is zero. Despite a residual magnetism, the
    generated e.m.f. E is zero.
ii- As the prime-mover rotates the generator armature and
    the speed approaches rated speed, the voltage due to
    residual magnetism and speed increases.
iii- At rated speed, the voltage across the armature due to

                                               76
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


  residual magnetism is small, E1, as shown in the figure. But this voltage is also across the
  field circuit whose resistance is Rf. Thus, the current which flows in the field circuit I1, is
  also




iv- When I1 flow in the field circuit of the generator of figure (55), an increases in m.m.f.
   results (due to IfTf, Tf being field turns, which aids the residual magnetism in increasing
   the induced voltage to E2 as shown in figure (56)
v- Voltage E2 is now impressed across the field, causing a large current I 2 to flow in the field
   circuit. I2Tf is an increased m.m.f., which produces generated voltage E3.
vi- E3 yields I3 in the field circuit, producing E4. But E4 causes I4 to flow in the field
   producing E5; and so on, up to E8, the maximum value.
vii- The process continues until that point where the field resistance line crosses the
   magnetization curve in figure (56) here the process stops. The induced voltage produce
   when impressed across the field circuit produces a current flow that in turn produces an
   induced voltage of the same magnitude, E8, as shown in the figure.

  Critical resistance:
 In the above description a particular value of field resistance R f was used for building up
  of self-exited shunt generator. If the field resistance were reduced by means of adjusting
  the field rheostat of figure (55) to a lower value say R f1 shown in figure (56), the build-up
  process would take place along field resistance line Rf1, and build-up a somewhat higher

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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


  value than E8, i.e. the point where Rf1 intersects the magnetization curve. E9. Since the
  curve is extremely saturated in the vicinity of E9, reducing the field resistance (to its
  limiting field winding resistance) will not increase the voltage appreciably. Conversely,
  increasing the field rheostat resistance and the field circuit resistance (to a value having a
  higher slop than Rf in the figure) will cause a reduction of the maximum value to which
  build-up can possibly occur.
 The field resistance may be increased until the field circuit reaches a critical field
  resistance. Field circuit resistance above the critical field will fail to produce build-up

  Condition for build up voltage of a shunt generator
  1-There must be some residual magnetism in the generator poles.
  2-The shunt field coil should be correctly connected to the armature for a given direction of
  rotation i.e. they should be so connected that the induced current reinforced the e.m.f
  produced initially due to residual magnetism.
  3-Shunt field resistance should be less than the critical resistance (Rc < Rf)
  4-The speed must be greater than critical speed to generated the e.m.f (n <nc).
  How to find critical resistance (Rc)
  1-First o.c.c is plot from the given data.
  2-Then, tangent is drowning to its initial portion.
  3-The slope of this curve gives the critical resistance for the speed at which the data was
  obtained.
  Rc = OX/OZ = Volt/Ampere = Slope = ( ) ohm.

  How to draw o.c.c at different speeds
  1-Given o.c.c at speed N1.
  2-Since  for any fixed excitation E = ZPN  /a 60 = k N 
  Where k = ZP/a 60 constant.
  3-If fixed excitation 1  2 so    E2/E1=N2/N1 then E2 = E1×(N2/N1)
  If = OH
  E1 = HC             E2 = E1×(N2/N1)
  E2 = HD             E2 = HC×(N2/N1) = HD         so point D is located.
  4-In such way, other such point can be found and new o.c.c at N2 drawn.




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Critical speed Nc
 1-Draw o.c.c at (N).
 2-Draw OT tangent to o.c.c (curve 1)
 3-Draw field line (Rf = Rsh line) corresponds to critical speed because Rsh line is tangential
 to it.
 4-From point (C) at If draw perpendicular line such that cut Rsh = Rf at B and cut OT at A.
 5-Then Nc=BC/AC × N




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Characteristic of d.c. generator
 2-Internal and external characteristics:
 The behavior of E or V of a generator as its load changes at constant speed is known as
 internal or external load characteristics respectively. That is, internal load characteristic
 is E vs IL and external load characteristic is V vs IL or (E- IaRa) vs IL.
 a- Separately – excited generator
 When the field is given the normal supply the maximum e.m.f is generated at no load. As
 the machine is loaded the voltage drop at any particular load current, indicated by the
 vertical distance between the external characteristic and the no-load voltage is brought
 about by two causes:
  i) Armature reaction which has a demagnetizing effect upon the field.
  ii) Resistance drop, this being the product of the armature current and the total armature-
      circuit resistance, consisting of the armature resistance, interpole resistance and brush
      contact resistance.
 There is some demagnetizing effect of armature reaction because of which the value of
 e.m.f generated decreases slightly. As the load is more and more this effect is more and
 more.
                                           𝑉 = 𝐸 − 𝐼𝑎 𝑅𝑎
                                             𝜙 ∝ 𝐼𝑓
                                           𝑉
                                    𝐼𝑓 =      ⇒ 𝐶𝑂𝑁𝑆𝑇𝐴𝑁𝑇
                                           𝑅𝑓
                                       𝑃 𝑍𝑁𝜙
                                  𝐸=         ⇒ 𝐶𝑂𝑁𝑆𝑇𝐴𝑁𝑇
                                       𝑎 60

 Thus curve (1) in figure (57) represents internal characteristic of a separately excited d.c.
 generator. If we deduct armature resistance drop at every point from curve (1) we get curve
 (2) which is the external characteristics of the separately excited machine.
 A separately – excited generator is used only in special cases, such as when a wide
 variation in terminal p.d. is required, or when exact control of the field current is necessary.
 Its disadvantage lies in requiring a separate source of direct current




                                           Figure (57)

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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 b- Shunt generator:
 In case of d.c. shunt generator as the load increases there is decreases in e.m.f generated
 due to:
 i) Armature-circuit resistance (𝑅𝑎 )
 ii) Armature reaction
 iii) Reduction in field current
 The armature reaction as is the case in a separately excited machine. But in shunt machine
 there is an additional drop in e.m.f due to the fall in terminal voltage. As the load increases
 the terminal voltage decreases and the field winding gets slightly reduced voltage so the
 field current and the flux in the poles slightly decreases.
                                          𝑉 = 𝐸 − 𝐼𝑎 𝑅𝑎
                                           𝑃 𝑍𝑁𝜙
                                      𝐸=         ⇒ 𝑘𝜙
                                           𝑎 60
                                            𝜙 ∝ 𝐼𝑓
                                                   𝑉
                                            𝐼𝑓 =
                                                   𝑅𝑓
 When load (𝐼𝑎 ) ↑⟹ 𝑉 ↓⟹ 𝐼𝑓 ↓⟹ 𝜙 ↓⟹ 𝐸 ↓
 Thus there is fall of e.m.f generated on that account. Thus curve (i) in figure (58)
 represents internal characteristic of a shunt generator. Curve (ii) got after deducting
 armature resistance drop at every point from curve (i) represents its external characteristics.
 As shown in figure (58) further application of load causes the generator to reach a
 breakdown point beyond which further load causes it to ‘unbuild’ as it operation on the
 unsaturated portion of its magnetization curve. This unbuilding process continues until the
 terminal voltage is zero, at which point the load current is of such magnitude that the
 internal armature circuit voltage drop equals the e.m.f. generated on the unsaturated or
 linear portion of its magnetization curve
 The shunt –wound generator is the type most used in practice, but the load current must be
 limited to a value that is well below the maximum value. This then avoids excessive
 variation of the terminal voltage. Typical applications are with battery charging and motor
 car generators.




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                                          Figure (58)
 c- Series generators:
 The saturation curve (1) may be obtained in a manner exactly similar to that already
 described for the shunt machine. The series-connected generator is illustrated in figure (59).
 The series connected generator illustrated in figure (60) must be capable of safely carrying
 the maximum current to be used, or about 125 per cent of rated current. A plot of
 simultaneous readings of generated voltage and field current, take at rated speed, yields the
 magnetization curve 1. External curve 2 of figure (60). If OA is the resistance line for the
 circuit the terminal voltage is the ordinate to the curve at A. When the resistance of the
 circuit is gradually increases, the load falls off along the curve, and A approaches B. When
 the resistance line finally becomes tangent to the curve, however, operation become
 unstable, and any slight further increase in the resistance causes the machine to unbuild its
 voltage and also its load. The resistance that brings about this condition is called the critical
 resistance. Therefore, to begin with, the resistance of the circuit must be reduced below the
 critical value before the generator delivers any load.
 Internal characteristic curve 3 of figure (60). This curve is obtained by adding the resistance
 drop 𝐼𝑎 (𝑅𝑎 + 𝑅𝑠𝑒 ) to external characteristic curve; 𝑅𝑎 and 𝑅𝑠𝑒 being armature resistance
 and series field resistance respectively.
 The difference between curve 1 and 3 is the reduction in voltage caused by armature
 reaction.
                                      𝑉 = 𝐸 − 𝐼𝑎 (𝑅𝑎 + 𝑅𝑠𝑒 )
                                            𝑃 𝑍𝑁𝜙
                                       𝐸=         ⟹ 𝑘𝜙
                                            𝑎 60
                                             𝜙 ∝ 𝐼𝑓
                                             𝐼𝑓 = 𝐼𝑎
 When load current (𝐼𝑎 ) ↑⟹ 𝐼𝑓 ↑⟹ 𝜙 ↑⟹ 𝐸 ↑




          Figure (59)                                      Figure (60)

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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class



 It is worth noting that between A and C a considerable change in resistance brings about
 only a slight change in load current. Over this range the voltage decreases rapidly, owing to
 increasing armature reaction (particularly when the brushes are shifted forward), while the
 current remains nearly constant. Thus, between A and C the machine may be used to supply
 power to a constant current variable-voltage circuit, such as series arc circuit.
 Owing to initially rising characteristic, the generator is often used as a voltage booster to
 given an increase of voltage practically proportional to the current.
 A series generator also finds application in electric traction where ‘dynamic braking’ is
 employed. The connections of the series traction motors are changed by means of a
 controller so that they act as generators; the power absorbed in braking the vehicle being
 dissipated in resistances, which are also used for starting purposes when the machines are
 reconnected as motors.

 d- Cumulatively compounded generators:
 In compound generators, for the sake of simplicity we draw only the external load
 characteristics. One can easily get the internal load characteristics by adding armature and
 series winding drops at every point to the external characteristic curve. Figure (61)




                                         Figure (61)

                                      𝜙𝑡 = 𝜙𝑠ℎ. + 𝜙𝑠𝑒.
 One gets the load characteristics of a cumulatively compounded generator by
 superimposing, additively, the series characteristics over the shunt characteristics. The
 resultant curve depends upon the strength of the series field winding the shunt field
 winding. Thus we may have the three types of cumulatively compounded generators as
 shown in figure (62). When the series field strength is just enough, that it compensates all
 the voltage drop in the machine and the terminal voltage on full load is the same as on no
 load, the machine is called level compounded generator. When the series field is stronger
 and full load terminal voltage is higher than the no load terminal voltage the machine is

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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 called over compounded. If the full load terminal voltage is less than the no load voltage it
 is under compounded machine.
 Cumulatively compounded generators are normally used for generating d.c supplies to feed
 d.c machinery and apparatus. d.c. level compounded generator is especially suited to where
 a constant output voltage in a wide range is required.
 e- Differentially compounded generators:
 In these generators as the series field is subtractive from the shunt field, so when the load
 increases the voltage falls very rapidly as shown in figure (63).
                                        𝜙𝑡 = 𝜙𝑠ℎ. − 𝜙𝑠𝑒.
 Differentially compound – wound generators are used in electric arc welding. Its special
 highly drooping load characteristic especially matches with the arc characteristics required
 for the d.c arc welding. They have also been used with arc lamps and arc furnaces.




                    Figure (62)                                        Figure (63)

Application of d.c. generators
Separately excited generators:
i)    The separately excited generators are usually more expensive than self-excited
      generators as they require a separate source of supply. Consequently they are generally
      used where self-excited are relatively unsatisfactory. These are used in Ward Leonard
      systems of speed control, because self-excitation would be unstable at lower voltages.
ii)   These generators are also used where quick and requisite response to control is
      important (since separate excitation gives a quicker and more precise response to the
      changes in the resistance of the field circuit).
Shunt generators:
i)    These generators are used to advantage, in conjunction with automatic regulators, as
      exciters for supplying the current required to excite the fields of A.C. generators. The
      regulator controls the voltage of the exciter by cutting in and out some of the resistance
      of the shunt-field rheostat, thereby holding the voltage at whatever value is demanded


                                              84
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


       by operating conditions. This is one of the most important applications of shunt
       generators.
ii)    Shunt generator are used to charge batteries in this application the voltage should drop
       off slightly as the load increases, because the voltage of a lead battery is lower when
       battery is discharged than when battery is charged. When it is discharged, however, the
       battery can stand a large charging current than when it is charged. Because of its
       drooping characteristic the shunt generator is admirably suited to battery charging
       service, for, in a general way, the voltage curve of the generator has the same shape as
       the voltage curve of the battery itself. In both cases, as the load falls the voltage rises.
Series generators
i)     Series arc lighting.
ii)    Series incandescent lighting.
iii) As a series booster for increasing the voltage across the feeder carrying current
       furnished by some other sources.
Compound generators
i)     It may be built and adjusted automatically to supply an approximately constant voltage
       at the point of use, throughout the entire range of load. This is very great advantage. It
       is possible to provide a constant supply voltage at the end of a long feeder by the
       simple expedient of over compounding the generator, because the resistance drop in the
       line is compensated for by the rising characteristic of the generator.
ii)    Differentially compounded generator finds an useful application as an arc welding
       generator where the generator is practically short circuited every time the electrode
       touches the metal plates to be welded.

Terminal voltage of a generator on load
The terminal voltage of a generator on load is corresponding to the point of intersection
between its external characteristics and equivalent load resistance line as shown in figure (64)
It is clear from figure (65) that a series generator will not build up any voltage if the load
resistance is above the critical value.




                 Figure (64)                                         Figure (65)

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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class



 Voltage Regulation
 The voltage regulation of a D.C. generator is the percent change in terminal voltage from no
 load to rated load when run at constant speed with respect to rated voltage:
        VnL  Vrated
 VR                  100
           Vrated
 Where
            VR = voltage regulation (percent)
            VnL = no-load (open-circuit) voltage (V)
            Vrated = voltage indicated on name plate of machine (V)
 Example
 A 100-kW, 1800 r/min generator operating at rated load has a terminal voltage of 240 V. if
 the voltage regulation is 2.3 percent, determine the no-load voltage.
 Solution
       VnL  Vrated
 VR                 100  VnL  Vrated  (1  VR )
          Vrated
 VnL  240  (1  0.023)  245.5 volts




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                         Parallel operation of D.C. generators
 Need for parallel operation of D.C. generator
 In a d.c. power plant, power is usually supplied from several generators of small ratings
 connected in parallel instead of from one large generator. This is due to the following
 reasons:
 (i) Continuity of service. If a single large generator is used in the power plant, then in case
 of its breakdown, the whole plant will be shut down. However, if power is supplied from a
 number of small units operating in parallel, then in case of failure of one unit, the
 continuity of supply can be maintained by other healthy units.
 (ii) Efficiency. Generators run most efficiently when loaded to their rated capacity.
 Electric power costs less per kWh when the generator producing it is efficiently loaded.
 Therefore, when load demand on power plant decreases, one or more generators can be
 shut down and the remaining units can be efficiently loaded.
 (iii) Maintenance and repair. Generators generally require routine maintenance and
 repair. Therefore, if generators are operated in parallel, the routine or emergency
 operations can be performed by isolating the affected generator while load is being
 supplied by other units. This leads to both safety and economy.
 (iv) Increasing plant capacity. In the modern word of increasing population, the use of
 electricity is continuously increasing. When added capacity is required, the new unit can be
 simply paralleled with the old units.
 (v) Non-availability of single large unit. In many situations, a single unit of desired large
 capacity may not be available. In that case a number of smaller units can be operated in
 parallel to meet the load requirement.
                               Shunt generator in parallel
 Whenever generators are in parallel, their +ve and –ve terminals are respectively connected
 to the +ve and –ve side of the bus-bars. These bus-bars are heavy thick capper bars and
 they acts as +ve and –ve terminals for the whole power station. If the polarity of the
 incoming generator is not the same as the line polarity, a serious short-circuit will occur
 when S1 is closed.




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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class




 The procedure of connected the second generator
 1- The armature of generator No.2 is speeded by the prime-mover up to its rated value.
 2- Switch S2 is closed.
 3- The circuit is completed by putting a voltmeter V across the open switch S1
 4- The excitation of the incoming generator No.2 is changed till V reads zero. This mean
    the terminal voltage is the same as that of generator No.1 or bus-bar voltage.
 5- After this, switch S1 is closed and so the incoming machine is parallel to the system.
 Under this conditions generator No.2 is not taking any load because induced e.m.f is the
 same as bus-bar voltage and there can be no flow of current between two points at the same
 potential. The generator is said to be 'floating' on the bus-bar
                          EV
 Current supplied by I 
                            Ra
 - The induced e.m.f of the incoming generator is increased by strengthening its field and
    it may be found necessary to weaken the field of generator No.1 to maintain the bus-bar
    voltage V constant.
                                       Load Sharing
 - Shunt generators are most suited for stable parallel operation. Because of their slightly
   drooping voltage characteristics.
 - Any tendency on the part of a generator to take more or less than its proper share of
   load results in certain change of voltage in the system.
 - For taking a generator out of service, its field is weakened and that of the other
   generator is increased till the ammeter of the generator to be cleared read zero. After
   that, its breaker and then the switch are opened thus removing the generator out of
   service.
 - It is obvious that if the field of one generator is weakened too much, then power will be
   delivered to it and it will run in its original direction as a motor, thus driving its prime-
   mover.
                                              88
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 - If it is desired that two generators of different kW ratings automatically share a load in
   proportion to their ratings for example a 100-kW generator is working in parallel with a
   200 kW generator to supply a total load of (240-kW) then first generator will supply 80
   kW and the other 160 kW.




 - It is seen for a common terminal voltage (V)
 The generator No.1 delivers I1 amperes.
 The generator No.2 delivers I2 amperes.
 It is seen that generator No.1 having more drooping characteristic delivers less current.
 - When two generators have the same external characteristics then will sharing the load
       equally.
 - Les us discuss the load sharing of two generators which have unequal no-load voltages.
 E1, E2 = no-load voltages of the two generators
 Ra1, Ra2 = their armature resistance
 V = common terminal voltage
          E V                   E V
 I a1  1            ,     Ia 2  2
           R a1                    R a2
  I a 2 E 2  V R a 1 K 2  2 N 2  V R a1
                                     
  I a1 E1  V R a 2 K11 N1  V R a 2
 From the above equation, it is clear that:
 - bus-bar voltage can be kept constant
 - any load can be transferred from 1 to 2 by increasing  2 or N2 or by reducing N1 and 1
 - N2 and N1 are changed by changing the speed of driving engines
 - 1 and  2 are changed with the help of regulating shunt field resistance.
 It should be kept in mind that
 i- Two parallel shunt generators have equal no load voltages share the load in such ratio
       that the load current of each machine produces the same drop in each generator.
 ii- In the case of two parallel generators have unequal no-load voltage, the load currents
       produce sufficient voltage drops in each so as to keep their terminal voltage the same.

                                             89
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 iii- Paralleled generator with different power ratings but the same voltage regulation will
      divide any oncoming bus load in direct proportion to their respective power ratings.

                           Compound generators in parallel

 Because of the rising characteristic of the used over compound generators, it is obvious
 that in the absence of any corrective devices, the parallel operation of such generators is
 unstable let us suppose that, to begin with, each generator is taking its proper share of load.
 I1  I f 1R f 1  E1  take more load 
 difference E1 & V  I1  R L cons tan t  PL1 , PL 2 
 I 2  I f 2 R f 2  E 2  I 2  unstable  I1 , I 2  I L  I1  G 2  as Motor
 - Generator No.1 will therefore tend to take the entire load and finally drive generator No.2
 as a motor.
 - The circuit breaker of at least one of the two generators will open thus stopping their
 parallel operation.
 - For making the parallel operation of over-
 compound and level-compound generators
 stable, they are always used with an equalizer
 bar connected to the armature ends of the series
 coils of the generators.
 - The equalizer bar is a conductor of low
 resistance and its operation is as follows
 - Suppose that generators No.1 starts taking
 more than its proper share of load. Its series
 field current in increased.
 - But now this increased current passes partly
 through the series field coil of generator No.1
 and partly it flows via the equalizer bar through the series field winding of generator No.2
 - Hence, the generators are affected in a similar manner with the result that generator No.1
 cannot take the entire load.
 For maintaining proper division of load from no-load to full load it is essential that
 i)   The regulation of each generator is the same.
 ii)  The series field resistances are inversely proportional to the generator rating.
                              Series generators in parallel

 Suppose E1 and E2 are initially equal




                                              91
D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


 Suppose E1 increases slightly so that E1>E2 in that case I1 become greater than I2
 consequently field of machine 1 is strengthened thus increasing E 1 further 1st the field of
 machine 2 is weakened thus decreasing E2 further.
 A final stage is reached when machine 1 supplies not only the whole load but also supplies
 power to machine 2 which starts running as a motor.
 E1  I f 1R f 1  I 2  I1  E 2  E1
 I L  I1  G 2  as motor
 Obviously, the two machines will form a short-circuited loop and the current rise
 indefinitely this condition can be prevented by using equalizing bar because of which two
 similar    machines
 pass approximately
 equal currents to
 the load, the slight
 difference between
 the two currents
 being confined to
 the loop made by
 the armatures and
 the equalizer bar.
 Characteristic triangle as a tool for solving load distribution problems
 between paralleled
 The solution of problems involving load distribution between DC generators in parallel
 may be accomplished in a straightforward manner by defining a characteristic triangle for
 each machine, and then using similar triangles to determine a solution.
 Referring to Figure (66), the characteristic triangle is fixed for a given voltage droop and
 does not change with changes in the field-rheostat setting. The bus voltage is the terminal
 voltage (also called output voltage) of the generator.
 An increase in bus load causes a decrease in bus voltage, as shown by the broken line in
 Figure (66). The intersection of the new voltage line with the load-voltage characteristic
 establishes a new triangle similar to the characteristic triangle. From the geometry of
 similar triangles,
  Vbus Vnl  Vrated
        
   I         I rated
                   V  Vrated
 Where VR  nl                    100 %
                       Vrated
                      VR
  Vnl  Vrated             Vrated
                      100



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D.C. machine                          By Assistant lecture Mohammed Qassim Abbas
  Electrical Engineering Technical College / Depart of Electrical Power / second Class


                       VR
            Vrated 
  Vbus                100
        
   I           I rated


                                                         Figure (66)

 Where VR  voltage regulation.
       I  change in generator load current (A).
       Vbus  change in bus due to change in generator load current (V).




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