النتائج 1 إلى 3 من 3

الموضوع: Direct Current Generators

  1. #1
    Junior Engineer
    تاريخ التسجيل
    Jul 2006
    الدولة
    iraq
    المشاركات
    6

    Direct Current Generators

    1
    Chapter 5
    Direct Current Generators
    5.1 INTRODUCTION
    The linear machine was introduced because of its simple construction and the fact that it served to
    demonstrate clearly the principles of electromechanics. It also allowed us to establish a model to show
    symbolically the relationships which exist with devices of this kind. At this point we must take a quantum
    jump to see how these principles have been implemented in rotating machines, specifically generators and
    motors. Historically, d-c machines came into being before a-c machines because the scientists of that time
    (about the middle of the 19th century) were only familiar with battery sources and consequently strived to make
    motors which operated from batteries, as well as generators to charge the batteries and operate arc lamps.
    Although superior in many ways, a-c machines have not completely replaced d-c machines, and will not in the
    foreseeable future, since the d-c motor offers a controllability not yet approached by a-c motors. The d-c
    generator, on the other hand, is declining rapidly in use since its functions have been largely taken over by solid
    state rectification of alternator outputs (in automobiles for example). Nevertheless a thorough study of d-c
    generators is worthwhile because the construction of motors and generators is the same, and the bilateral nature
    of the energy conversion process means their “inner workings” share much in common.
    5.2 CONSTRUCTION AND TERMINOLOGY
    No mention was made of the origin of the magnetic field in the linear machine; in the loudspeaker it
    was furnished by a permanent magnet. In d-c machines the magnetic field is supplied by field coils” of wire
    wound around “pole cores” which are part of the magnetic circuit. These terms and others relating to the
    construction of a typical d-c generator or motor are explained in the following table.
    Figure 5.1. Cross Section of a Typical d-c Machine. Numbers refer to items in Table 5.1.
    2
    1. FIELD COILS. Coils of insulated copper wire which provide the mmf for the magnetic
    field.
    2. POLE CORES. Steel cores around which field coils are wound. Adjacent poles alternate in
    polarity (N-S-N-S etc.)
    3. POLE SHOE. Part of the pole structure (steel) which conforms to the curvature of the
    armature core in order to provide a uniform air gap length.
    4. YOKE. Steel frame providing mechanical rigidity and also providing a path of low
    magnetic reluctance between poles.
    5. ARMATURE A stack of steel laminations mounted on the shaft of the machine. CORE
    CORE Copper armature conductors are placed in the slots. A major part of the
    magnetic circuit.
    6. ARMATURE. The collection of copper wires in which voltages are induced (BLu) and on
    which forces are produced by current (BLi). In the linear machine the ‘bar’
    is the armature.
    7. SLOTS. Rectangular openings around the periphery of the armature core into which
    armature conductors are placed.
    8. TEETH. Rectangular area between slots, around periphery of the armature core, i.e.
    material left after slots are cut out.
    9. COMMUTATOR. A ring of copper segments surrounding the shaft, which are insulated from
    each other by strips of mica. The ends of armature conductors are
    connected to commutator segments.
    10. SHAFT. A steel rod on which the armature core is mounted. The means by which
    mechanical power is delivered from a prime mover.
    11. BRUSHES. Stationary rectangular carbon and graphite blocks which make electrical
    contact with the rotating commutator for the purpose of completing the
    current path from the external terminals through the armature conductors
    and return.
    12. BRUSH RIG- Mechanical assembly which holds the brushes in place and which GING
    provides for adjusting tension of the springs which push the brushes against the commutator.
    13. END BELLS. Steel structures on both ends of the machine which provide support for the
    bearings and brush rigging.
    14. EXCITATION. A general term referring to the production of the magnetic field within the
    machine. “Separate excitation” refers to the supplying of field-coil current
    from an outside source such as a battery; whereas “self excitation” refers to
    the generator supplying field-coil current from its own armature.
    Table 5.1 MACHINE TERMINOLOGY
    A typical armature assembly is shown below in Figure 5.2. Part (a) shows a lamination used in the core.
    Multiple laminations are stacked to build the core. Laminations are used for ease of construction and to prevent
    eddy currents. Part (b) shows the laminations stacked together on a shaft and part (c) includes the wires and
    commutator segments.
    3
    Figure 5.3: The Elementary Generator
    (a) (b) (c)
    Figure 5.2. Components of a Typical Armature Assembly
    5.3 THE ELEMENTARY GENERATOR AND MECHANICAL COMMUTATION:
    A very simple elementary AC generator is discussed first because of the similarity in concepts and
    construction between AC and DC generators. Then the process of mechanical commutation is introduced.
    Commutation is used to change an AC into a DC machine be it a generator or a motor.
    Elementary Generator Construction
    The elementary generator of Figure 5.3 consists of a loop of wire free to rotate in a stationary
    magnetic field. Relative motion between the wire and the magnetic field will induce a potential difference
    between the ends of the loop. Sliding contacts are used to connect the rotating loop to an external circuit in
    order to use the induced voltage.
    4
    Figure 5.4: How the Elementary Generator Works
    The pole pieces are the north and south poles of the magnet which supply the magnetic field. The
    loop of wire which rotates through the field is called the "armature." The ends of the armature loop are
    connected to rings called "slip rings," which rotate with the armature. Brushes ride up against the slip rings
    providing a sliding electrical contact to pick up the electricity generated in the armature and carry it to the
    external circuit.
    In the de******ion of the generator action which follows, visualize the loop rotating through the
    magnetic field. As the sides of the loop cut through the magnetic field, they generate an induced voltage which
    causes a current to flow through the loop, slip rings, brushes, zero-center current meter and load resistor all
    connected in series. The induced voltage that is generated in the loop, and therefore the current that flows,
    depends upon the position of the loop in relation to the magnetic field. Now, lets analyze the action of the loop
    as it rotates through the field.
    ELEMENTARY GENERATOR OPERATION
    Assume that the armature loop is rotating in a clockwise direction and that its initial position is at A
    (zero degrees) of Figure 5.4. In position A, the loop is perpendicular to the magnetic field and the black and
    white conductors of the loop are moving parallel to the magnetic field. If a conductor is moving parallel to a
    magnetic field, it does not cut through any lines of magnetic flux and no voltage is generated in the conductor.
    This applies to the conductors of the loop at the instant they go through position A. No voltage is induced in
    them and therefore no current flows through the circuit. The current meter registers zero.
    As the
    loop rotates from position A to position B, the conductors are cutting through more and more lines of flux until at 90
    degrees (position B) they are cutting through a maximum number. In other words, between zero and 90 degrees, the
    induced voltage in the conductors builds up from zero to a maximum value. Observe that from zero to 90 degrees the
    black conductor cuts down through the field while at the same time the white conductor cuts up through the field. The
    induced voltage in each of the conductors is therefore in series, and the resultant voltage across the brushes (the terminal
    voltage) is the sum of the two induced emfs, or double that of one conductor since the induced voltages are equal to each
    other. The current through the circuit will vary just as the induced emf varies, being zero at zero degrees and rising up
    to a maximum at 90 degrees. The current meter deflects increasingly to the right between positions A and B, indicating
    that the current through the load is flowing in the direction shown. (The reader should verify the polarity of the induced
    voltage and the direction of resulting current by applying either the right hand rule or the cross product relationship.)
    5
    Figure 5.5: How the Elementary Generator Works (cont’d)
    The direction of current flow and polarity of the induced voltage depends upon the direction of the magnetic field and
    the direction of rotation of the armature loop. The waveform shows how the terminal voltage of the elementary
    generator varies from position A to position B. The simple generator drawing on the right is shown rotated by 900 to
    illustrate the relationship between the loop position and the generated waveform.
    As the loop continues rotating from position B (90 degrees) to position C (180 degrees), the
    conductors, which are cutting through a maximum number of lines of flux at position B, cut through fewer
    lines, until at position C they are again moving parallel to the magnetic field and no longer cut through any
    lines of flux. The induced voltage, therefore, will decrease as the loop rotates from 90 to 180 degrees in the
    same manner as it increased from zero to 90 degrees. The current flow will similarly follow the voltage
    variations. The generator action at positions B and C is illustrated in Figure 5.5.
    6
    From zero to 180 degrees the conductors of the loop have been moving in the same direction
    through the magnetic field and, therefore, the polarity of the induced emf has remained the same. As the
    loop starts rotating beyond 180 degrees back towards position A, the direction of motion of the conductors
    through the magnetic field reverses. Now the black conductor cuts up through the field, and the white
    conductor cuts down through the field. As a result, the polarity of the induced voltage and the current flow
    will reverse. From positions C through D back to position A, the current flow will be in the opposite
    direction than that from positions A through C. The generator terminal voltage will also have its polarity
    reversed. The voltage output waveform for the complete revolution of the loop is as shown in Figure 5.6.
    Figure 5.6: How the Elementary Generator Works (cont’d)
    7
    RECTIFICATION BY MECHANICAL COMMUTATION
    A simple AC generator was discussed in the section above. How can this machine be modified to
    produce DC voltage and current which has the property that the polarity of the voltage does not change and
    the direction of current flow is always the same, although it may not be constant? One method for
    producing unidirectional voltage and current flow is shown below in Figure 5.7(a). The scheme is similar
    to the elementary AC generator but the hardware associated with the connection of the rotating wire loop to
    the outside world is modified. Instead of slip rings and brushes, the DC machine has commutator segments
    and brushes. The slip ring of the AC machine has been divided into two halves insulated from each other
    to form a commutator. Referring to Figure 5.7(a) , the right hand brush is always connected to the side of
    the loop which is traveling downward next to the South Pole at the times of maximum induced voltage.
    This means that the voltage at the left hand brush will always be “+” relative to that of the right hand brush.
    A commutator is a rotating switch which reverses the connection between the armature (the rotating loop of
    wire) and the outside world every half turn just at the point in the cycle the polarity would reverse in an
    elementary AC generator. Note that the poles in Figure 5.7 are reversed relative to Figure 5.6. The
    reader should again use the cross product or the right hand rule to verify the direction of current.
    Consider it a homework assignment. The voltage at the left hand brush relative to that at the right hand
    brush is shown in Figure 5.7(b).
    (a) (b)
    Figure 5.7 Mechanical Commutation Produces Unidirectional Output Voltage
    Note that although the voltage across the brushes is unidirectional, it is not very constant. The
    voltage pulsations are undesirable for many applications. How can the voltage generated be made
    smoother? The answer is by including more loops of wire in the armature and correspondingly more
    commutator segments. The armature in Figure 5.8(a) contains two mutually perpendicular loops with the
    corresponding output shown in Figure 5.8(b) and a 3 loop armature machine and output are shown in
    Figure 5.9. The generated DC voltage becomes more constant as more and more loops are included on the
    armature.
    8
    (a) (b)
    Figure 5.8 A DC Generator With Two Loops On Its Armature and Corresponding Output
    (a) (b)
    Figure 5.9 A DC Generator With Three Loops On Its Armature and Corresponding Output
    There are two primary methods of connecting the armature windings together. The turns are
    connected together in parallel in the lap winding method. This provides for greater current capability at a
    lower voltage. The turns are connected in series in the wave winding method providing for higher voltage
    at reduced current. Combinations of these two methods can also be used. The details of these two
    methods are beyond the scope of this course. If you are interested in learning more about either type of
    winding, your instructor can help or direct you to an appropriate reference.
    5.4 THE EQUIVALENT CIRCUIT
    Since the d-c generator is a practical form of the linear generator, the model introduced for the
    latter may be used with only minor modifications. First, because control of the magnetic field strength by
    adjusting current in the field coils is an important feature of the d-c generator, symbols for the field coils
    and their resistance, RF , are included in the model. Second, since the generator is a rotating machine, the
    model will show the armature symbol connected to a rotating shaft which is driven by a prime mover, as a
    constant reminder that energy is being converted from one form to the other. Table 5.2 illustrates how the
    basic relationships for the linear machine are modified to fit the rotating machine.
    9
    V f
    R f
    Prime Mover
    T a
    T d
    Ω
    T f
    R A
    V T
    +
    -
    IA
    +
    -
    φ E = KφΩ
    IF
    Field Armature
    Figure 5.10 Separately Excited DC Generator
    Table 5.2.
    Item Linear Machine Rotating Machine
    Velocity/Speed u m/s 6 rad/s or n rpm
    Current, Armature I amperes I A amperes
    Current, Field ------------- I f amperes
    Magnetic Field B Webers/m2 1 Webers (per pole)
    Voltage, Induced e = BLu newtons E = K16 = k1n volts
    Voltage, Terminal v Volts VT Volts
    Force/Torque (Developed) fd = BLi newtons Td = K1IA N-m
    Force/Torque (Applied) fa newtons Ta newton-meters
    Resistance, Armature r ohms RA ohms
    Mechanical Power Dev. fdu watts Td 6 watts
    Electrical Power Dev. ei watts EIA watts
    Power Equivalence ei = fdu EIA = Td 6
    Machine Constants* ---------------- K,k
    *The constant for a given machine lumps together all the effects of the unchanging quantities
    such as the number of poles, dimensions of the armature core, number of armature coils and
    the method of interconnecting them.
    10
    A few comments are in order for clarifying information given in the table. First, the rotational
    velocity of the machine is stated formally as 6 which is related to tangential velocity of u = radius x 6, but
    tachometers in present use are calibrated in the more practical unit of revolutions per minute, symbol n. Thus
    in the equation
    volts (E=KφΩ=kφn 5-1)
    The two constants K and k differ, but we need not be concerned with their actual numerical values. The
    relationship between the two velocities is
    Ω =2πn /60rad/s This is from (n rev/min)(2Œ rad/rev)(1 min/60sec) (5-2)
    The magnetic field is traditionally represented as flux (1) instead of flux density (B) because of the size of the
    average voltage E is arrived at by considering the total flux encountered by a single revolution of a conductor
    rather than the instantaneous flux densities. Finally, it will be remembered that tangential force applied to a
    rotating cylinder is interpreted as “torque”, which is the product of the tangential force and the radius of the
    cylinder,
    Td = Σfd x (radius) newton-meters (5-3)
    whereΣ fd is the sum of the tangential forces developed by the individual conductors, and the radius is called
    the “torque arm”. Consequently,
    Td=KφIANewton-meters (5-4)
    Where the constant “K” is the same as in E = K1:.
    5.5 THE MAGNETIZATION CURVE
    Magnetic conditions within the generator can be investigated by testing a separately excited generator
    using a voltmeter connected to the output of Figure 5.10, where VT is indicated, and an ammeter in series with
    the field circuit to measure field current. The test procedure is to adjust the speed of the prime mover to the
    rated value as indicated on the generator name plate and then to adjust the variable resistor for increasing
    values of field current (including zero) as read on the ammeter. For each setting of field current, the voltmeter
    indicates the corresponding induced voltage. Data is usually taken for voltages up to 25 to 50 percent above
    the rated value. The data is then plotted on a graph called the “saturation curve” or “magnetization curve”, a
    typical curve is illustrated in Figure 5.11. If for any reason it should be desired to have curves for other
    speeds, they can be obtained without actually running more experimental tests, by using direct proportionality.
    For example if the curve is wanted for 1000 rpm a point whose voltage is 250 x 1000/1200 would be located
    directly below the 250-volt ordinate on the 1200 rpm curve. By calculating a series of points in this manner
    a new curve can be sketched below the 1200 rpm curve. The justification for this procedure is that for any
    given fixed current the flux is the same for any speed and hence induced emf is proportional to speed only.
    We shall next examine the magnetization curve in detail because its shape has a direct bearing on the operation
    of generators and motors.
    11
    50
    100
    150
    200
    250
    300
    0
    0 0.5 1.0 1.5 2.0 2.5
    Field Current in Amps
    Generated Voltage in Volts
    1200 rpm
    Figure 5.11 Magnetization Curve for a Typical DC Generator.
    Now this graph can be interpreted as a plot of air gap flux vs mmf as well as voltage vs current, since
    voltage is proportional to flux when the speed is held constant E=kφn, and the current becomes mmf-perpole
    when multiplied by the number of turns of wire in each field coil. Of course the numerical scales would
    differ, but the point is that with this interpretation we can explain the behavior of the magnetic circuit within
    the machine by referring to the shape of the curve. For this reason the curve has been redrawn with flux and
    mmf coordinates in Figure 5.12.
    Beginning in the lower left corner of the graph it can be seen that a flux exists for zero mmf. This
    should not be surprising since we know from previous study of magnetic phenomena (hysteresis) that residual
    magnetism is left in the steel structure from previous use of the generator. Although this flux is responsible
    for only a small induced voltage, it fulfills an extremely important role in self-excited generators by providing
    the “seed” flux which starts the process of “build-up” to normal flux levels — to be explained in detail later.
    As current (mmf) is increased, it is seen that the curve is straight for an appreciable portion, i.e. flux is directly
    proportional to the mmf. In this region the steel parts of the magnetic circuit account for only a small portion
    of the total mmf, most of the mmf drop occurring in the air gap between the armature core and the pole shoe.
    As current is increased beyond the linear region, rotation of magnetic domains and finally magnetic saturation
    take place in the steel parts resulting in a marked departure from the straight-line relationship of the curve.
    Since this is a gradual process rather than an abrupt one, there is a transition region between the linear part and
    the saturated part, which is referred to as the “knee” of the curve.
    12
    0 MMF, Magnetomotive Force which
    is proportional to Field Current
    FLUX
    0
    LINEAR
    SATURATION
    Figure 5.12 Magnetization curve with modified scales.
    Operating the generator within the linear portion of the curve is recommended for applications where
    terminal voltage is to be varied over a wide range, as in situations where the generator has direct control of a
    single load. On the other hand, if an application requires the voltage not change much once it is set, operation
    in the saturated region is preferred. In the former case the armature is more responsive to control since a given
    change in field current causes a larger change in voltage than in the saturated region.
    5.6 SEPARATELY-EXCITED GENERATORS
    The simplest generator and one which is used for a wide variety of applications is “separately
    excited”. The modifier “separately” is used to distinguish it from another class of generators which are “selfexcited”,
    referring to the manner in which electrical energy is supplied to the field coils. In generators using
    self excitation (to be treated later) the field is connected to the armature; whereas with separate excitation the
    field is connected to a separate source. The obvious advantage of self excitation is the cost savings realized
    by not requiring additional capital investment for a separate source. The advantage of separate excitation
    which offsets the extra cost is the much greater range of control available, including the ability to change
    polarity at will. The principle application for this type of generator is in motor control. The method of wiring
    was indicated in the test for acquiring data for the magnetization curve, but in Figure 5.13 a load resistor has
    been added and the field control potentiometer has been modified to allow reversal of field current. We shall
    next investigate how the generator reacts under load conditions, but first a few words about ratings.
    13
    V f
    R f
    Prime Mover
    T a
    T d
    Ω
    T f
    R A
    VM
    IA
    +
    -
    φ E = KφΩ
    IF R L
    AM
    +
    +
    -
    -
    IL
    Figure 5.13 Separately excited generator.
    The chief parameters for rating rotating machines are power output, voltage, current and speed. These
    ratings assume continuous operation, 24 hours a day and 7 days a week unless otherwise stated. Operation of
    a machine above its current rating for an appreciable length of time could result in overheating and to
    deterioration of insulation or complete burnout. The current rating is the most likely rating to be exceeded
    since it depends directly on the loading of the machine. In the graphs of external characteristics which follow,
    rated (i.e. “full-load) values will be indicated for reference. Operation beyond the rated limits is shown for
    completeness, for the sake of perspective, not as recommended practice.
    The load characteristic for a separately-excited generator is displayed in Figure 5.14. This graph is
    the I-v relationship or terminal characteristic of the generator. The drop in terminal voltage as the current
    increases is caused by the IARAvoltage drop within the armature in accordance with
    VT=E−IARA volts (5-5)
    Just as it was in the linear generator. The voltage regulation of the generator is defined as:
    V.R. = Voltage Regulation = VNL −VFL  /VFL (5-6)
    Suppose it is desired to investigate the effect on the external characteristics of changing the field
    current by readjusting the variable resistor in the field circuit, without changing the load resistance. First, we
    can draw a family of curves for several settings of the field current, as shown in Figure 5.15. The spacing of
    the curves will be equidistant for equal field current increments provided that operation is limited to the linear
    part of magnetization curve, otherwise not. Second, a “load line” is drawn from the origin through the point
    where the coordinates are rated current and voltage. The slope of this line is the resistance of the load RL. The
    load line is the i-v characteristic of RL. The intersection of this load line with the external characteristic of the
    14
    generator is the point at which the generator will operate. For field current IF4 and the load line shown in
    Figure 5.15, rated current and voltage will result. Now it can be seen that increasing the field current above
    the IF4 value results in current and voltage overloads, but reducing it below IF4 affords full control of current
    without overload.
    V FL
    V NL
    V T ,Terminal Voltage
    Rated Voltage
    Rated Current
    Load Current
    IL = IA
    Load Line Slope = VL /IL = RL
    This Load Intersects Machine
    Characteristic at Greater Than
    Rated Load Current Causing
    Overheating
    Figure 5.14 External Characteristic of a Separately-Excited Generator.
    V FL
    V T ,Terminal Voltage
    Rated Voltage
    Rated Current
    Load Current
    IL = IA
    IF1
    IF2
    IF3
    IF4
    IF5
    Figure 5.15 External Characteristics of a Separately-Excited Generator with
    Various Settings of Field Current
    15
    To further demonstrate the versatility of the separately-excited generator, note that moving the
    “slider” of the field control resistance to the left of the centerpoint (Figure 5.13) reverses the polarity of the
    voltage applied to the field coils, which reverses the direction of field current and eventually the magnetic field.
    This extends operation of the generator to negative terminal voltage and negative line current with respect to
    Figures 5.14 and 5.15 or into the third quadrants of these plots.
     EXAMPLE 5-1 
    A separately-excited generator with the following nameplate ratings is delivering energy at rated
    voltage to 8 parallel loads with resistances of 25 ohms each:
    5 Kilowatts 1800 rpm
    125 Volts 40 Amperes
    When the loads are disconnected the voltage rises to 130 volts.
    a. Calculate the voltage regulation.
    b. Calculate the armature resistance.
    c. Calculate the internal torque when supplying the original loads.
    While the loads are still connected, the speed governor malfunctions, reducing the speed to 1400 rpm.
    d. Calculate the new values of IL and V.
    e. Calculate the new value of torque.
    f. To what value will the voltage now rise when the loads are disconnected?
    Solution
    a. V.R. = VNL−VFL/VFL=130 −125/125 =.040
    b. RL=25/8=3.125 O h m s ; IA=IL=V/RL=125/3.125=40 a m p e r e s ,
    RA=E−V/IA=130−125/40=0.125 Ohms
    c. Td=EIA/Ω=130x40/2π 1800 / 60=27.6 N-m.
    d. By proportionality, E2=E1(n2/n1)for a given setting of field current, therefore,
    E=130(1400/1800)=101.1 volts
    IL=IA=E/RA+RL=101.1 /0.125+3.125=31.1 amperes
    V=E−IARA=101.1−31.1x0.125=97.2 volts.
    e. Td=EIA/Ω=101.1x31.1 / 2π 1400 / 60=21.5 N-m.
    f. E = 101.1volts.
    16
    5.7 SELF EXCITED SHUNT GENERATORS
    A large savings in the initial cost and complexity of a DC generator can be realized if the field circuit
    is connected across the armature terminals instead of to a separate source. Moreover, a great amount of control
    over the terminal voltage is still available, with the aid of a “field rheostat” (RC) connected in series with the
    field coils.
    R A
    IA
    +
    -
    E = KφΩ
    φ
    IF
    +
    -
    V T
    R A
    IA
    +
    -
    E = KφΩ
    +
    -
    V T
    IL
    φ
    R F
    R C
    FIELD FIELD
    IF
    V F
    R F
    (a) Separately-excited Generator (b) Self-excited Generator
    Figure 5.16. Comparison of generator connections.
    The term “shunt” is synonymous with “parallel”. Since the field circuit is wired in parallel with the load, as
    far as the armature is concerned, the field circuit is just another load to be supplied with current. From KCL:
    IA=IL+IF (5-7)
    and then using Ohm’s Law:
    IF=VTRF+RC. (5-8)
    There is an interdependence between the field circuit and the armature circuit since the field current supplies
    the magnetic field which produces the armature induced voltage but at the same time the armature supplies
    current to the field current, a feedback situation. This can be expressed in mathematical terms as
    (5-9) E f IF =  
    17
    which describes the magnetization curve, and Equation (5-8) which describes the field circuit. Since Equation
    (5-9) describes a non-linear, non-analytic function it is necessary to resort to a graphical solution of these two
    simultaneous equations.
    Since Equation (5-8) is a linear equation (Ohm’s law) it can be plotted as a straight line on a graph.
    Specifically, we choose to plot it on the same graph as the magnetization curve. Since it is a straight line
    through the origin and has a slope of , one need only pick a single current at random RF+Rc and calculate
    the corresponding voltage to find one point on the line, then drawing a line through that point and the origin
    to complete the job, as illustrated in Figure 5.17. Since the magnetization curve indicates values of E and the
    field resistance line indicates values of VT, the vertical difference between the curves must correspond to the I R A A
    drop in the armature.
    0
    0
    IAR A
    E
    V T
    IF
    Field Resistance Line
    E, VT
    Figure 5.17 Field resistance line drawn on magnetization curve showing
    voltage division in armature.
    Voltage Build-Up
    The build-up of voltage in a self-excited generator can be visualized with the aid of this graph. First
    assume there is no load resistance connected to the terminals of the generator except the shunt field, and since
    the field current is small compared with the rated load current (to keep losses low) it can also be assumed that
    theIARA = IFRA voltage drop is negligible, so VT must equal E in the steady state. Second, assume the prime
    mover has been brought up to rated speed. The small residual magnetic field induces a small voltage in the
    armature which in turn causes a small field current. This field current increases the strength of the magnetic
    field which then causes an increase in induced voltage, which causes an increase in field current, and so on.
    The growth of the magnetic field, with concurrent growth in the magnitude of induced voltage, is thus seen to
    18
    be an automatic, positive feedback situation. Now, what prevents the build-up, the positive feedback cycle,
    from proceeding ad infinitum? The non-linearity of the magnetization curve gives us a clue: saturation. An
    increase in flux for a given increase in field current becomes smaller beyond the knee of the curve, until finally
    an equilibrium condition exists. Then the field current is precisely enough to keep the induced voltage at the
    level required by Ohm’s law to maintain that amount of current in the field circuit.
    Remembering our assumption of negligible drop under no-load conditions, the establishmIARA ent
    of an equilibrium condition must result in E = VT, and the only place on the graph where this is true is at the
    intersection of the resistance line and the magnetization curve. This point can also be interpreted as the
    graphical solution of simultaneous Equations (5-8) and (5-9). We can make use of this knowledge by coupling
    it with the fact that the slope of the field resistance line is equal to the combined field circuit resistances
    RF+RC. To be specific, we can control the voltage at which the intersection occurs by adjusting the field
    rheostat Rc, as illustrated in Figure 5.18. Beginning with the original setting Rc1, which produces a voltage E1
    an increase to RC2 (steeper slope = larger resistance) drops the voltage to E2. Finally, an increase to RC3 drops
    the induced voltage drastically, where build-up is practically nonexistent. This brings up a point: if a selfexcited
    generator fails to build up, what are the usual causes?
    0
    0
    IF
    E, VT R C3 R C2 R C1
    E 1
    E 2
    E 3
    Figure 5.18. Effect of varying field rheostat on no-load voltage build-up.
    19
    Table 5.3. Failure of a Generator to Build Up*
    Symptom Reason Remedy
    1. Voltage low, drops when field
    is disconnected.
    Field rheostat set for too
    much resistance
    Readjust field rheostat for
    lower resistance
    2. Voltage low, rises slightly
    when field is disconnected
    Field mmf is opposing
    residual magnetism
    Switch shunt field
    connections
    3. No voltage No residual magnetism Disconnect shunt field and
    reconnect it to a d-c source
    (“flasher”) for a few
    seconds. Disengage field
    from source and reconnect to
    armature.
    4. No voltage Open circuit Shut down and check circuits
    for continuity
    5. Wrong Polarity Residual magnetism in
    wrong direction
    “Flash” field as described
    above
    *Assuming direction of rotation is correct.
    Load Characteristic
    If a self-excited shunt generator is connected to a load and a test is run for obtaining the load
    characteristics, it is found that the voltage drops more, for a given load current than it does when connected
    for separate excitation, as illustrated in Figure 5.19.
    V NL
    V T ,Terminal Voltage
    IL
    Self-excited
    Separately-excited
    Rated Current
    Figure 5.19 Comparison of External Characteristics For The Same
    Generator Connected Two Ways, With The Same No-load Voltage
    20
    The additional armature current required by the field does not sufficiently increase the IARIA=IL+IF A drop to
    account for the entire additional drop — we must look elsewhere for a reasonable explanation. The answer is found
    in Equation (5-8) where the field current is seen to be directly proportional to the terminal voltage VT. The field current
    does not remain constant as it would in a separately excited generator. A drop in VT causes a drop in IF, which in turn
    results in a decreased induced voltage E. This can be seen graphically in Figure 5.20, where E has dropped from the
    no-load value at the intersection (E1) to the loaded value, E2, with the accompanying change in VT from VT1 to VT2.
    Correspondingly, the drop in VT has reduced field current from IF1 to IF2. In other words, the difference between the two
    curves in Figure 5.19 is accounted for by the difference between E1 and E2 in Figure 5.20.
    0
    0
    IAR A
    E 2 V T2
    IF2
    E, VT
    IF1
    E 1 = VT1
    IF
    Figure 5.20 Effect of loading on location of operating point, in self-excited
    generator. Condition 1 is no-load condition 2 is with load.
    EXAMPLE 5-2
    The generator of example 5-1 is rigged for self-excitation and adjusted for rated voltage and load
    current, with the same load as before. The field current is noted to be 1.85 amperes. When the load is
    disconnected the terminal voltage rises to 138 volts.
    a. Calculate voltage regulation.
    b. Calculate total field resistance.
    c. Calculate generated emf under full-load conditions.
    d. Calculate field current under no-load conditions.
    21
    Solution
    a. V.R. = (138 - 125)/125 = .104
    b. RF+RC=V/IF=125/1.85=67.6 ohms
    c. E=VT+IL+IFRA=125+41.850.125+130.2 volts
    d. IFno−load=VT/RF+RC =138/67.6=2.0 amperes
    5.8 COMPOUND GENERATORS
    The “droop” in the external characteristic of the shunt generator is a disadvantage when it is serving
    a multiplicity of loads, since the voltage at one load location will fluctuate up and down as other loads are
    turned on and off. This effect is rendered all the worse by the resistance of wiring between the generator and
    the loads, which causes additional voltage drop. An ingenious but simple method of overcoming this difficulty
    was thought of in the early days of electrical engineering, consisting of a means of incorporating automatic
    positive feedback proportional to load current — the “series’ field
    The key to offsetting voltage drop is in increasing the induced voltage by adding extra mmf. Although
    this could be done by someone continually adjusting the shunt field rheostat, a better way is to add another
    field winding called a “series field”, as shown in Figure 5.19(a). The series field is wired in series with the
    load and thus carries load current. The physical location of the shunt and series field windings are the same,
    but because the full-load current is typically many times greater than the shunt field current, the series field
    must be wound with much larger wire. Increased wire size is offset, however, by the small number of turns
    needed by the series field to achieve a significant amount of mmf. For example, a 500-turn per-pole shunt field
    with a current of 2 amperes produces an mmf of 1000 ampere-turns, but a series field of only 5 turns per pole
    with a current of 40 amperes produces an mmf of 200 ampere-turns.
    IF R L
    IL
    R C
    SHUNT SERIES
    +
    -
    E
    +
    -
    V T
    V NL
    V T ,Terminal Voltage
    Rated Current
    IL
    Compound
    Separately
    Excited
    Shunt
    (a) Wiring Diagram (b) Comparative External Characteristics
    Figure 5.21. Compound generator
    22
    The external characteristic illustrated in Figure 5.21(b) shows a rising terminal voltage as load current
    increases. This is often more desirable than a horizontal curve because the additional voltage above the noload
    value can be used to cancel the effect of IR drop in the wires leading to the loads, If the voltage rise is
    greater than needed for a given situation, a variable resistor called a “diverter” can be added in parallel with
    the series field to bypass a portion of the load current around the field, thus, lowering the curve. If the curve
    is horizontal, the generator is said to be “flat-compound” because the full-load and no-load voltages are the
    same (voltage regulation is zero). “Overcompound” and “undercompound” generators have characteristic
    curves above or below the flat curve, respectively. Generators are normally furnished by the manufacturer as
    overcompound so the user can adjust the compounding to suit his or her application, by adding a diverter.
    An interesting possibility arises when one considers that the series field might also be connected so
    its mmf opposes the mmf of the shunt field. Under this circumstance the terminal voltage would drop quite
    drastically when a load is connected. Although this defeats the purpose of adding the series field in the
    beginning, there is one application which uses the series field in this way — the d-c arc welding generator.
    The sharply drooping voltage characteristic fits the need in welding for a large voltage to start an arc but once
    the arc begins the voltage needed to sustain it is considerably less.
    Figure 5.22 Physical Location of Shunt And Series Field Windings Shown On One Pole.
    5.9 TACHOMETER GENERATORS
    Aside from its in energy conversion, the d-c generator finds application as a transducer which
    provides an output voltage proportional to shaft speed. This is accomplished quite simply by using a
    permanent magnet to provide a magnetic field of constant strength and coupling its shaft directly to the rotating
    shaft whose speed is to be measured. The armature terminals are then connected to a voltmeter calibrated
    directly in rpm. Another use for tachometer generators is as a link in the feedback path of a servo system, for
    example in a system to maintain motor speed at a steady value in spite of fluctuating load.
    23
    5.10 SUMMARY
     Knowledge of the terminology and construction of d-c machines is essential to understanding their
    operation
     The induced emf and current emf direction in each armature coil undergoes a reversal in traveling the
    distance of one pole pitch. The brush-commutator system serves as a means of rectifying the voltage and
    current so they appear at the brushes as unidirectional.
     The magnetization curve is a plot of induced voltage vs. Shunt field current, determined experimentally by
    separately exciting the generator under no-load conditions. It is useful in explaining and predicting machine
    performance. Curves for speeds other than the test speed can be constructed by direct proportionality.
     The separately-excited generator requires a voltage source to supply power to the field circuit. This allows
    complete control of the generator’s armature voltage from zero up to rated, for either polarity.
     The self-excited generator saves the cost of providing a source for the field by wiring the field in parallel
    with the armature; field current is then supplied by the armature. Initial flux to allow the system to build
    up voltage is supplied by the small residual magnetism in the magnetic circuit. Indefinite build-up of
    voltage is prevented by saturation of the magnetic circuit.
     Reasons for the failure of a self-excited generator to build up voltage are enumerated in Table 5.3.
     Addition of a series field to a self-excited shunt generator, wired in series with the armature*, provides
    additional mmf as the generator load current increases, thus offsetting the voltage “droop” which normally
    occurs with the shunt field only. A generator having both fields, shunt and series, is called a compound
    generator.
    *or the total
    5.11 QUESTIONS
    Q5.1 Describe the function of the commutator in a DC generator.
    Q5.2 What is the advantage of multiple pairs of commutator segments in a DC generator?
    Q5.3 Describe the source of the magnetic field in which the rotor spins for (a) separately excited DC
    generator and (b) a self-excited DC generator.
    Q5.4 What allows the approximation to be made that the no load terminal voltage is about equal to be armture
    induced voltage for a self-excited DC shunt generator?
    Q5.5 What would be the effect of zero residual magnetic field in a self-excited, DC shunt generator upon
    startup? How would you correct this problem?
    Q5.6 What is the major advantage of a compound DC generator?
    24
    5.12 PROBLEMS
    P5.1 Assuming that Figure 5.8 of the **** applies to a particular DC generator, find the open circuit
    voltage of that generator if (a) the field current is 0.5A and the speed is 1800 rpm and (b) the field
    current is 2.0A and the speed is 1000 rpm.
    P52. Repeat Example 5-1 of the **** if 10 parallel loads are supplied, each having a resistance of 20
    Ohms and the generator is rated at 120 Volts, 1800 rpm, 60 Amps and 7.2 kW.
    P5.3 The generator of P5.2 is reconnected for self-excitation and adjusted for rated voltage and load
    current, with the same load as before. The field current is noted to be 1.8 Amps. When the load
    is disconnected, the terminal voltage rises to 140 V.
    a. Calculate the voltage regulation.
    b. Calculate the total field resistance.
    c. Calculate the generated emf under full-load conditions.
    d. Calculate the field current under no-load conditions.
    P5.4 A separately excited DC generator is rated at 125 V, and 1 Amp at 1200 rpm. When the load is
    disconnected, the terminal voltage rises to 130 Volts. Calculate (at rated conditions):
    a. the armature current,
    b. the voltage regulation,
    c. The armature resistance,
    d. And the internal torque when supplying rated load.
    P5.5 If the speed of the generator of P5.4 is increased to 2000 rpm, calculate: (Assume linear operation.)
    a. The load current,
    b. The terminal voltage,
    c. And the mechanical power converted to electrical power.
    P5.6 A self-excited DC shunt generator is rated at 125V, and 1 Amp at 1200 rpm. The field resistance is
    125 Ohms and the armature resistance is 5 Ohms. Calculate:
    a. The field current.
    b. The armature current,
    c. The no-load terminal voltage (the behavior is non-linear, and a graphical solution using the
    curve is required),
    d. The voltage regulation,
    e. And the internal torque when supplying rated load.
    25
    P5.7 A separately excited DC generator rated at 22 kW, 220 V and 100 A at 1200 rpm has the
    magnetization curve shown in Figure 5.8 of the ****.
    a. Given the field resistance of 150 Ohms, estimate the field current if the generator has a no-load
    voltage of 250 Volts.
    b. Determine the armature resistance, RA.
    If the generator is supplying rated voltage and current to the load, calculate:
    c. the developed torque,
    d. the total input power if rotational losses are 0.1 Hp.
    e. the total power out,
    f. the efficiency,
    g. and the % voltage regulation.
    P5.8 A self-excited DC generator rated at 22 kW, 220 V and 100 A at 1200 rpm has the magnetization
    curve shown in Figure 5.8 of the ****. If the no-load voltage is 250 V, calculate:
    a. the field resistance, Rf.
    b. the armature resistance, RA.
    c. the rated load resistance,
    d. and the developed torque.
    If mechanical losses are 0.1 Hp, calculate:
    e. the total power,
    f. the efficiency,
    g. and the % voltage regulation.
    P5.9 If the speed governor for the generator of P5.7 were to fail and the speed increased to 1500 rpm,
    what would the new no-load voltage be?
    P5.10 If the speed governor for the generator of P5.8 were to fail and the speed decreased to 1000 rpm,
    what would be the new no-load voltage?
    P5.11 A separately-excited DC generator is rated at 130 V, and 45 A at 1200 rpm. At rated conditions, the
    mechanical power losses are 250 W. When the load is disconnected, the terminal voltage rises to
    136 V. Sketch and completely label a circuit model for this generator showing the following Ta, Td,
    Tf, 6, Prime mover, E, IL, Rl, V, IA, RA, 1, VF, RF, RC, and IP. At rated conditions calculate:
    a. the power supplied to the load, POUT,
    b. the induced voltage, E,
    c. the percent voltage regulation,
    26
    d. the armature resistance, RA,
    e. the load resistance, Rl,
    f. the angular velocity, 6,
    g. the power converted from mechanical to electrical power, POUT,
    h. the developed torque Td,
    i. the mechanical power in PIN,
    j. and the torque due to friction Tf.
    k. Sketch the power flow diagram for this generator, including numeric values for each power.
    P5.12 A separately-excited DC generator is running at rated conditions of 215 v and 155 A at 1800 rpm.
    It has an armature resistance of 0.12 Ohms and the magnetization curve shown in Figure P5.12
    applies.
    50
    100
    150
    200
    250
    300
    0
    0 1 2 3 4 5
    Field Current in Amps
    Generated Voltage in Volts
    1800 rpm
    Figure P5.12
    a. Sketch a circuit diagram showing all the parameters listed in P5.11 and then determine the value
    of the equivalent load resistance.
    b. Predict the terminal voltage if the load is disconnected.
    c. Determine the amount of field current required at full-load to produce the correct terminal
    voltage.
    27
    d. Predict the terminal voltage for the load found in (a) if the field current is reduced by 20%.
    e. If the field current is restored to its original value and the speed of rotation is slowed to 1000
    rpm, determine the terminal voltage.
    P5.13 A separately-excited DC generator has the magnetization curve shown in Figure P5.13. The rated
    output is 5.5 kW and 210 V at 1200 rpm. The input torque to sustain these rated conditions is 49.5
    N-m.
    50
    150
    200
    250
    300
    350
    0
    0 1 2 3 4 5
    Field Current in Amps
    Generated Voltage in Volts
    1200 rpm
    100
    Figure P5.13
    a. Determine the rated load ILD.
    To produce rated output it is found necessary to set the field current IF, equal to 1.6A. This current
    is supplied from a 200 V DC source. The resistance, RF, in the field winding is 100 Ohms.
    Determine:
    b. The value of the adjustable resistance, RC,
    c. And the equivalent load resistor, RLD, which must be connected to the generator output terminals
    to receive this amount of power.
    d. And the equivalent load resistor, RLD, which must be connected to the generator output terminals
    to receive this amount of power.
    With the machine delivering rated output, the load is disconnected but no other changes are made.
    Study the system carefully and determine:
    28
    e. the no-load terminal voltage, VT,
    f. the value of RA, the armature resistance,
    g. the percent voltage regulation of this machine,
    h. the power converted from mechanical to electrical, PCONV,
    i. the required mechanical input power, PIN,
    j. the friction and windage loss of the machine,
    k. and the efficiency of operation at rated output.
    P5.14 A self-excited, shunt DC generator is rated at 130 V and 95 A at 1200 rpm. The total field
    resistance is 120 Ohms and the armature resistance is 0.13 Ohms. Mechanical losses at rated speed
    are 480 W. Under rated conditions, sketch a complete model of the generator including all
    appropriate labels. Then determine:
    a. the field current,
    b. the armature current,
    c. the induced voltage,
    d. the angular velocity,
    e. the developed torque,
    f. the applied torque,
    g. and the torque due to mechanical losses.
    h. construct a power flow diagram for this generator under rated condition, including the numeric
    value for each power.
    The load is now disconnected. Under unloaded conditions both the induced and terminal voltages
    will increase. If E increases to 145 V, what will be the no load terminal voltage?
    (Note: the IARA voltage drop under unloaded conditions is very small compared with E and VNL.
    Therefore, the assumption that under no load conditions the voltage drop due to the armature
    resistance can be ignored is a good approximation.)
    P5.15 The rated output for a self-excited, DC generator is 5.2 kW. The terminal voltage is 195 V at 1200
    rpm. The following resistances were measured: RA = 1.13 Ohms, and the total field resistance - 116
    Ohms. The friction and windage losses are 250 W. Carefully sketch and label a circuit diagram for
    this generator and then determine:
    a. the terminal voltage and shunt field current for this fully loaded machine,
    b. the value of the generated voltage,
    c. the efficiency of the machine.
    29
    P5.16 A self-excited, DC generator is rated at 22 kW, 220 V and 100 A at 1800 rpm. It has the same
    magnetization curve as shown in Figure P5.12 and its no load voltage is measured at 250 V. At
    rated output, the mechanical losses are 1 Hp. Calculate:
    a. the total field resistance (Hint: plot the field resistance line and make appropriate
    approximations.),
    b. the developed torque, the total input power and the efficiency at rated output.
    P5.17 A DC shunt generator is operating at rated condition, which are: 18 kW, 150 V, and 1500 rpm. The
    total field resistance is 30 Ohms and the armature resistance is 0.08 Ohms. Mechanical losses are
    142 W. Draw an equivalent circuit and a power flow diagram and then calculate:
    a. the field current,
    b. the armature current,
    c. the induced voltage
    d. the power converted,
    e. the angular velocity,
    f. the developed torque,
    g. the total torque due to mechanical losses,
    h. the torque applied to the shaft by the prime mover,
    i. and the input horsepower required to turn the shaft.

  2. #2
    عضو فى جمعية مهندسى الحماية الكهربائية الصورة الرمزية محمد شلهوب
    تاريخ التسجيل
    Jun 2006
    الدولة
    ksa
    المشاركات
    199

    رد: Direct Current Generators

    هل الفائدة من dc generator
    توليد ac current

    or dc current ?

  3. #3
    V.I.P Member الصورة الرمزية dyiaeldeen
    تاريخ التسجيل
    Dec 2006
    الدولة
    مصر
    المشاركات
    1,795

    رد: Direct Current Generators

    شكراً لك اخي الكريم

معلومات الموضوع

الأعضاء الذين يشاهدون هذا الموضوع

الذين يشاهدون الموضوع الآن: 1 (0 من الأعضاء و 1 زائر)

المواضيع المتشابهه

  1. Why High Voltage Direct Current
    بواسطة مهندس محمد خضر الزين في المنتدى قسم وقاية خطوط نقل و توزيع الطاقة الكهربائية Feeders protection
    مشاركات: 0
    آخر مشاركة: 25-03-12, 05:49 PM
  2. current transformer for direct &alternating current
    بواسطة عبدالرحيم موسى في المنتدى قسم هندسة المحولات الكهربية Electrical Transformer
    مشاركات: 7
    آخر مشاركة: 01-12-11, 12:27 PM
  3. High-voltage direct current
    بواسطة مفتاح القماطي في المنتدى قسم هندسة الجهد العالى High voltage engineering
    مشاركات: 3
    آخر مشاركة: 09-11-10, 11:39 PM
  4. Direct Current Generators
    بواسطة Ammar yousif في المنتدى قسم الموضوعات العامة
    مشاركات: 0
    آخر مشاركة: 31-07-06, 09:22 PM

الكلمات الدلالية لهذا الموضوع

المفضلات

المفضلات

ضوابط المشاركة

  • لا تستطيع إضافة مواضيع جديدة
  • لا تستطيع الرد على المواضيع
  • لا تستطيع إرفاق ملفات
  • لا تستطيع تعديل مشاركاتك
  •  
 

 

 

Flag Counter