ELECTROMECHANICAL ENERGY CONVERSION (EE 413/ECE 412)
THE DYNAMO-PRINCIPLES AND CONSTRUCTION
Dynamo – is a rotating electrical machine in which the energy transformation takes place. There are two general types of dynamo, namely:
a. generator- mechanical energy is converted to electric energy b. motor – electric energy is converted to mechanical energy Note: Generators and motors are fundamentally similar in construction - and this is particularly true of dc machines - they differ only in the way they are used.
DC Generator and Motor Principles The fundamental principles governing generator action and motor action were originally discovered by Michael Faraday in 1831. Briefly summarized, these basic principles may be stated as follows: Generator action – involving the development of voltage, may result (a) by moving a conductor in such a manner that it cuts across magnetic lines of force (dc generator), (b) by moving magnetic lines of force in such a manner that they cut across a conductor (ac generator/alternator), and (c) by changing the number of lines of force that link with a wire or coil of wire (transformer).
General Voltage Equation for DirectCurrent Generator *Remembering that the generated voltage depends
upon the rate at which flux is cut and that 1 volt results from cutting 108 lines of force per second.
The ff. analysis will lead to a very useful fundamental equation: 1) Each one of Z conductors cuts Φ x P line of force per revolution, where Φ is the flux supplied by each of the poles P.
General Voltage Equation for Direct-Current Generator 2) Assuming a parallel armature paths, the number of conductors per path will therefore be Z/a. 3) If the speed of the armature is represented by rpm, the speed in revolutions per second is rpm/60. 4) If (Φ x P) is multiplied by rpm/60, the product would represent the flux cut by each conductor per second.
General Voltage Equation for Direct-Current Generator 5) Since 1 volt is generated for every 108 lines cut per second, multiplying the product in (4), i.e., (Φ x P x rpm/60) by 10-8 would give the voltage generated in each conductor. 6) Finally, multiplying (Φ x P x rpm/60 x 10-8) by (Z/a) would yield the total generated voltage, Eg. Thus the fundamental voltage equation becomes,
General Voltage Equation for Direct-Current Generator Eg
Z Φ P = x 10 − 8 60 a
Where Eg = total generated voltage Φ = flux per pole, Maxwell P = number of poles, an even number N = speed of the armature, rpm Z = total number of conductors effectively used to add to resulting voltage a = number of armature paths connected in parallel (determined by type of armature winding)
Direction of a Generated Voltage The direction of the generated voltage in a conductor, or more correctly in a coil of wire, as it is rotated to cut the lines of force produced by the electromagnets in a generator, will depend upon two factors only: 1) the direction of the flux determined by the magnet polarity 2) the direction of motion of the conductor
Direction of a Generated Voltage Right-Hand Rule – used to determine the direction of the generated voltage Thumb – represents the direction of the motion of the conductor Forefinger – represents the direction of the flux (from north pole to south pole) Middle finger – represents the direction of the generated voltage Dot – generated voltage in the conductor will be toward the observer Cross – generated voltage in the conductor will be away from the observer
Direction of a Generated Voltage magnetic lines of force
N
S
Fig.1 two-pole generator
The Elementary Alternating-current Generator The dc generator is fundamentally an ac generator because, internally, in the armature conductors, the current reverses periodically as the wires move to cut lines of force successively under the north and south poles. The frequency, f in cycles per second (cps), of the alternating current is proportional to both the speed in revolutions per second, rpm/60, and the number of pairs of poles, P/2.
The Elementary Alternating-current Generator P rpm ( P )(rpm ) f = x = 2 60 120 Where: f = frequency, cps (Hz) P = no. of poles rpm = speed of revolution
The Elementary Alternating-current Generator The ff. diagrams illustrate graphically how the number of cycles per revolution is affected by the number of poles:
N S 1 cycle
Fig. 2 Two poles- 1 cycle per revolution
The Elementary Alternating-current Generator
N
N
S
S 2 cycles
Fig.3 Four poles - two cycles per revolution
Commutation in DC Generator From the foregoing discussion, it should be clear that the generated voltage, as well as the current, in dc armature winding is alternating. It is true, of course, that nothing can be done in the modern generator to develop an internal dc emf; what can be done, however, is to rectify the internal alternating current so that the brush voltage – the external voltage – is direct current. The mechanism for doing this consists of the commutator (in its simplest form it may be represented by a split ring) and its brushes. (See figure)
Commutation in DC Generator
Figure 4. Commutator and Brushes
Commutation in DC Generator Brushes are located so that they touch two segments exactly on top and bottom. Each conductor is permanently connected to a segment (or a semi ring). The split ring rotates with the rotating coil. Brushes and poles are stationary.
Generation of Unidirectional Current a x
N
S
S
b
+
y
x
-
a
N
y
b
I
Load
Load
(a)
(b)
When the plane of the coil is vertical, it will be short-circuited by the brushes, thus, the generated voltage is zero. When conductor a is moving downward (clockwise rotation) and cutting the flux under a north pole, semi ring x will be negative; at the same time, conductor b will be moving upward and cutting flux under a south pole, thus making semi ring y positive. The brush touching semi ring y will therefore be positive, while the other brush will be negative; the current through the load will be from left to right.
Generation of Unidirectional Current b
y
N
S
S
a
b
+x
y
-
N
x a I
Load
Load
(c)
(d)
During the next half revolution, conductor a will change places with conductor b under the poles, and this exchange will cause the generated voltages in the two conductors to reverse their direction. However, when this happens, the semi rings, to which they are connected, automatically change places under the stationary brushes.
Generation of Unidirectional Current a
x
N
S y
It follows therefore that the polarity of the brushes does not change. Hence, the current through the load will always be from left to right.
b
Load (e)
It is true that the magnitude of the current will change as the conductors a and b occupy different positions under the poles, but there will be no reversal of current through the load (see figure).
(b)
(a)
(d)
(c)
Fig. 5
(e)
Generation of Unidirectional Current resultant
coil A
coil B
When several coils are joined together properly so that their combined effect acts are additively, the result is not only increased voltage, but also voltage pulsations that are not so violent; in other words, the voltage wave becomes smoother as the number of coils are increased (see figure).
Fig. 6 Two coils in series
Strictly speaking, a dc generator does not deliver a pure direct current, as does a storage battery, for example, but approaches such a current very closely as the number of coils and commutator segments are increased.
Motor action – involving the development of force, results when a current-bearing conductor is placed in a magnetic field so that it is not parallel to the direction of the lines of force.
Force and Torque Developed by Direct-current Motors Force
S
N
Fig. 7 Fields produced by main poles and by current-carrying conductors
S
N Force
Fig. 8 Resultant field and force produced by magnet poles and current-carrying conductors
The first important point to be made in connection with the study of motor is this: if a current-bearing wire is in nonuniform magnetic field so that the flux density on one side of the conductor is greater than that of the other side, the conductor will experience a force action in a direction away from the higher density to the lower density.
Force and Torque Developed by Direct-current Motors In the actual dc motor the nonuniform flux distribution results from the interaction of two magnetic fields, one being the field produced by the stationary main poles and the other field created by a large number of current-carrying conductors on the armature core. Secondly, the force action exerted by a current-carrying conductor placed in a magnetic field depends upon: 1) the strength of the main field 2) the value of the current through the conductor
Force and Torque Developed by Direct-current Motors Experiment has shown that a force of 1 dyne will
be exerted upon a conductor 1 cm long carrying a current of 10 amp when placed under a pole area of which is 1 cm2 and producing one line of force. This leads to the equation:
F=
β Il 10
, dynes
Force and Torque Developed by Direct-current Motors If the units of F, β, and ℓ are specified in more practical terms, that is, pounds, lines per in2 and inches respectively, the equation becomes:
F=
β Il 11 ,300 ,000
Where: β = flux density I = current in the conductor ℓ= length of the conductor
, lb
Commutation in DC Motors It should be clear here that the function of the commutator and the brushes in a dc motor is to act as an inverter, that is, to change the direct current to alternating current, because the current in the armature conductors must be alternating if rotation in the same direction is to continue.
Arrangement of Generator and Motor Parts For the purposes of description, the electric generators and motors may be divided into two sections, namely: the stationary part (stator) - The most important function of the stator is to serve as the seat of the magnetic flux that must be made to enter the armature core. The field generally consists of a cylindrical yoke or frame to which is bolted a set of electromagnets.
Arrangement of Generator and Motor Parts Field-Pole cores – built up of a stack of steel laminations, about 0.025 in. thick per lamination, having good magnetic qualities; rivets are driven through the holes in the sheets to fasten together a stack of such laminations equal to the axial length of the armature core. The shape of the assembled core is such that the smaller cross section is provided for the field winding or windings, while the spread-out portion called the pole shoe permits the flux to spread out over a wider area where the flux enters the armature core.
Arrangement of Generator and Motor Parts Field windings - Each of the main pole cores may have one of three types of the field-winding construction depending upon whether the machine is to be operated as shunt, series, or compound dynamo: (1) shunt winding – has a comparatively large number of turns of fine wires; its resistance is therefore high enough so that it may be connected directly across the armature voltage or to a separate source of emf of about the same order of magnitude. (2) series winding – has relatively few turns of heavy wire and is connected in such a way that high values of current usually pass through it; its resistance is extremely low so that even when carrying normal load current, its voltage drop will be small.
Arrangement of Generator and Motor Parts (3) compound winding – a combination of the shunt and series field. The series coil is wound over the shunt coil; this is good general practice because the series field, carrying high values of current, is kept cool more readily when placed on the outside. the rotating part (rotor) - which is the real source of the electric (generator) power or the mechanical (motor) power, is built up of laminated steel core, slotted to receive the insulated copper armature winding. The number of slots is carefully selected in conjunction with the number of commutator segments, on the basis of good design.
Arrangement of Generator and Motor Parts
Arrangement of Generator and Motor Parts commutator – built-up group of hard-drawn copper bars, wedgeshaped in section when viewed on end, and having V-shaped grooves at each end. Together with the stationary brushes that ride over its rotating surface, it is assigned the duty of changing an internally generated alternating current to an external direct current in the generator and of changing an externally applied direct current to an internal alternating current in the motor.
Arrangement of Generator and Motor Parts
armature winding – virtually the heart of the dynamo; it is where the voltage is generated in the generator or where torque is developed in the motor. The armature-coil ends are soldered to the commutator.
Armature Windings Function: It is where the electric power originates in the generator and where the torque is developed in the motor. Types of Armature winding: The two types of armature winding used on modern dc machines are designated lap and wave. They may be distinguished from each other in two general ways: 1) from the standpoint of construction they differ only by the manner in which the coil ends are connected to the commutator bars.
Armature Windings Lap coil
Wave Coil
Coil end
Coil end
nearly 360 Electrical degrees
Adjacent Commutator Segments
(a) Simplex-Lap Connections
(b) Simplex-Wave Connections
Fig. 9 Sketches showing how the coil ends are connected to the commutator in lap- and wave-wound armature windings
Armature Windings Recognizing this simple construction difference, it should, therefore, be clear that: (a) a lap winding is one in which the coil ends are connected to the commutator segments that are near one another and; (b) a wave winding is one in which the coil ends are connected to commutator segments that are some distance from one another- nearly 360 electrical degrees apart.
Armature Windings 2) from the standpoint of an electrical circuit they differ in the number of parallel paths between the positive and the negative brushes. Like for example, simplex-lap windings have as many parallel paths as main poles, while simplex-wave windings have two parallel paths regardless of the number of poles.
Armature Winding Parameters Coil Pitch (or Span) - refers to the distance between the two sides of the individual coils, measured in terms of the number of slots. It is determined in exactly the same way for all windings, whether lap or wave. The fundamental rule that fixes the coil pitch in any given machine is: the distance between the two sides of the coil must be equal (or very nearly so) to the distance between two adjacent poles.
Armature Winding Parameters
S Y S= − k P Where YS = coil pitch, slots S = total number of armature slots P = number of main poles k = any part of S/P that is subtracted to make YS an integer
Armature Winding Parameters Example: Calculate the coil pitches and indicate the slots which the first coils should be placed for the following armature windings: (a) 28 slots, 4 poles; (b) 39 slots, 4 poles.
28 −0=7 a) YS = 4 39 3 b) YS = − =9 4 4
Slots 1 and 8 Slots 1 and 10
Armature Winding Parameters Commutator Pitch- refers to the distance on the commutator between the two ends of a coil element, measured in terms of commutator segments. Its value is determined in a different way for lap and wave windings. For Lap: YC is equal merely to the degree of multiplicity – the plex – of the lap winding. Thus, YC equals 1,2,3,4 etc., for simplex-, duplex-, triple-, quadruplex-, etc., lap windings, respectively.
Armature Winding Parameters For Wave:
C±m Y c= P 2
Where: YC = commutator pitch C = total number of commutator segments P = number of poles m= multiplicity Note: Use +m if winding is progressive. Use –m if winding is retrogressive.
Armature Winding Parameters Example: Calculate the commutator pitches for the following pole and commutator segment combinations: (a) 6 poles, 34 segments; (b) 8 poles, 63 segments.
34 − 1 = 11 a) YC = 3 63 + 1 b) YC = = 16 4
Tracing, 1-12-23-34 Tracing, 1-17-33-49-2
Armature Winding Parameters number of parallel paths, a- number of groups of coils in series connection and connected in parallel between the “+” and “-“ brushes. When the current passes through any armature winding, it always divides into an even number of parallel paths. For Lap:
For Wave:
For Frog-leg:
a L = mP
aW = 2m
aFL = 2 P
Armature Winding Parameters reentrancy – All dc armatures have closed-circuit windings; this implies that they may be traced completely from any point through all or part of the winding, and such tracing will always lead back to the starting point. For Lap: The degree of reentrancy is that number which is the highest common factor between the number of commutator segments and the commutator pitch.
Armature Winding Parameters YC For Wave: Reentrancy = m, if is integer. m YC is NOT an integer. Reentrancy = 1, if m multiplicity, m (simplex, duplex, triplex, etc.) – is the number of segments progressed or retrogressed by the end terminal of a set of successive coils in series from its starting point in tracing around the armature.
Armature Winding Parameters Example: Determine the commutator pitch for a four-pole simplexwave-wound armature having 21 segments. Also list the commutator segments in the proper order as the coils are traced through the entire winding from segment 1 until it closes.
Solution:
21 ± 1 YC = = 10 or 11 2
Armature Winding Parameters Using YC = 10 , the succession of commutator segments is as follows: 1-11-21-10-20-9-19-8-18-7-17-6-16-5-15-4-14-313-2-12 then reentering segment 1 Using YC = 11 , the succession of commutator segments is as follows: 1-12-2-13-3-14-4-15-5-16-6-17-7-18-8-19-9-20-1021-11 then reentering segment 1
Armature Winding Parameters number of brushes, b For Lap: b = P For Wave: b = 2 or P width of each brush = m segments Note: Brushes are positioned to short circuit conductors in neutral position. total number of coils = number of slots
Armature Winding Parameters Pole pitch (or span)- is the number of slots spanned by two successive, opposite polarity poles (i.e., pair of N & S poles).
S For Lap and Wave: Y P= P Where: S = number of slots P = number of poles Note: If YP= integer, then armature winding is a full-pitch winding. If YP is not an integer, then the armature winding is a fractionalpitch winding.
Armature Winding Parameters Example: Given: Duplex, progressive lap winding, S = 8, C= 8, P =2 Required: Draw a complete winding diagram (assume dynamo to be a generator rotating clockwise) Solution: a) YS = (8/2) – 0 = 4 slots b) YC = 2 segments c) a = (2)(2) = 4 parallel paths d) Reentrancy = HCF of 8 & 2 = 2 (doubly reentrant) e) m = 2 f) b = 2 g) width of each brush = 2 segments h) total no. of coils = 8 coils (single-element coil) i) YP = 8/2 = 4 slots (full-pitch)
Armature Winding Parameters h g
a
2 1
f
8 8
7
2
6
3 5
7
3
1
6
4
4
b
5
e
c d
Armature with More Segments than Slots Modern armatures are generally constructed with more commutator segments than slots for the following reasons: 1) As the number of segments is increased, the voltage between those that are adjacent to each other decreases. For a given terminal voltage, therefore, this also decreases the number of turns of wire in the coil or coils connected to adjacent segments. The result is that, from the performance standpoint, commutation is improved.
Armature with More Segments than Slots (2) As the number of core slots is reduced, the teeth become mechanically stronger, and this results in less damage to laminations and coils when these are handled in manufacture. (3) Assuming that a comparatively large number of segments has been selected for good commutation, the choice of an armature with one-half, one-third, one-fourth, etc., as many slots means that fewer coils will be constructed; this reduces the manufacturing cost.
Multi-Element Winding When there are n times as many segments as slots, each complete coil must have n coil elements. Thus, if the ratio of segments to slots is 2, 3, 4, etc., the individual coils will have 2, 3, 4, etc., elements. a) Number of active elements = number of commutator segments, C Number of coils = C; if C=S (single-element coil) Number of active elements = C; if C is not equal to S (multi-element coil)
Multi-Element Winding b) Number of elements per coil =
C +K S
Where: K = a decimal number to be added to C / S to round it off to the nearest integer Note: If C / S is an integer, then no dummy element is present If C / S is not an integer, then a dummy element is present c) Total number of elements = number of coils X number of elements per coil = C + dummy
Multi-Element Winding
d) Number of element-sides (conductors) per slot = 2 X number of elements per coil
Multi-Element Winding Example: Given: Simplex- lap, S = 12 slots, C = 24 segments, P = 4 poles Required: Draw a complete winding diagram.
Solution: a) YS = (12/4) – 0= 3 slots b) YC = 1 c) a = (1)(4)= 4 parallel paths d) Reentrancy = HCF of 24 & 1 = 1 (singly-reentrant) e) m=1 f) b = 4
Multi-Element Winding g) width of each brush = 1 segment h) Number of elements per coil = 24/12= 2 elements per coil i) Number of active elements = 24 elements j) Number of conductors per slot = (2)(2)= 4 conductors per slot
Multi-Element Winding 2-element coil
2
3
1 12
11
23 22 21 20 19 18
24 1
2 3
+
4 5 6
5
7
17 16
4
+
15 14 13 12
8 9 10 11
6
10 7 9
8
Dead, or Dummy, Elements in Armature Winding When the ratio of segments to slots (C/S) is not a whole number, it will always be found that there is one complete element of a multi-element coil that cannot be used electrically; there are not sufficient segments for exactly two ends of one element. The unconnected element is called a dead or dummy element. It serves only to keep the revolving structure balance mechanically.
Dead, or Dummy, Elements in Armature Winding
Equalizer Connections for Lap Windings The voltages generated in the various paths of lapwound armature are rarely the same. This situation arises in the practical machine because the air gaps under all the poles are not always alike, due to some degree of misalignment, and because the reluctances of the several iron magnetic circuits are unequal. As a result of such voltage inequalities, circulating currents flow in the armature winding and tends to heat the armature to temperatures well above those caused by the normal load current.
Equalizer Connections for Lap Windings Moreover, these undesirable currents pass across the brush contacts as they circulate from one path to another, and this produces an unusual amount of arcing and burning at the commutator; in fact, if the situation becomes serious, a flashover between positive and negative brushes is likely to occur, a situation that represents a direct short circuit across the supply lines. To overcome the detrimental effects resulting from the circulating currents, it is customary to use equalizer connections in all lap-wound armatures.
Equalizer Connections for Lap Windings Equalizer connections – these are low-resistance copper wires that connect between points on the armature winding that are 360 electrical degrees apart; they are placed on the armature outside the influence of the magnetic field. They are nonpotential-generating wires and carry equalizing currents only.
Equalizer Connections for Lap Windings Functions: They relieve the brushes of the circulating current load by causing the latter to be bypassed. They create a magnetic effect that actually reduces the flux under those poles where there is too much magnetism and increases the flux under those poles where there is too little magnetism.
Equalizer Connections for Lap Windings A
a S
Equalizer
N
c' a'
C N
S
b c S
N b'
B
Fig. 11 Sketch illustrating one equalizer connection in a 6-pole machine
Equalizer Connections for Lap Windings Since equalizers must connect points that are exactly 360 electrical degrees apart, it follows that the total number of coils in an armature winding must be divisible by half the number of poles. Thus for 100% equalization,
C no. of equalizers = P 2 Where: C = number of commutator segments P = number of poles
Frog-leg Winding Another construction of the armature winding, which combines the advantages of both lap and wave types and which is used on machines manufactured by the AllisChalmers Manufacturing company, is called a frog-leg winding. The term frog-leg is used to indicate the similarity between this type of coil and the legs of a frog. In this discussion, it must be noted that the real purpose of this type of winding is to eliminate the equalizer connections and yet to retain their advantages. The wave portion of the frog-leg winding, acting together with the lap portion, serves to replace the equalizers, but acts, in addition, as a current-carrying winding.
Frog-leg Winding It is thus possible to obtain 100% equalization of the winding and also to make the maximum use of all copper placed on the armature. Going back to figure 11, it discloses the fact that, theoretically, points A, B, and C are at the same potential. Since this is so, it is at once evident that points a and a’, b and b’, and c and c’ are also at the same potential because these points are connected to the equalizer and are themselves outside the influence of the magnetic field.
Frog-leg Winding A
a
Equalizer
D S
N
c' a'
C N
S
E
b c S
N
F b'
B
Fig. 12 Sketch illustrating a sort of lap-wave winding
It is quite possible, without affecting the winding in any way, to connect points a’ and b by connection E; points b’ and c by connection F; and points c’ and a by connection D. Figure 12 indicates the change suggested here.
Frog-leg Winding By carefully looking at figure 12, you will observe that it really represents a sort of combination lap-wave winding. The wave winding was introduced when the second set of connections was made, i.e., E, F and D. Suppose that the wire representing each of the single-turn coil of Fig.12 is slit in half lengthwise from each of the commutator segments up to the points a, a’, b, b’, c and c’. Electrically, no change has taken place from such an imaginary slitting process. Furthermore, the equalizer can now be omitted for the reason that any wave elements, such as E, and the succeeding lap element, such as B, connect two points on the commutator exactly two pole pitches apart.
Frog-leg Winding A
a
In addition, the net voltage theoretically generated in elements E and B is zero. Elements E and B together therefore have the two important characteristic properties C that must be possessed by an equalizer connection and may serve in replace of the removed connections. This combination lapwave coil has been appropriately named frog-leg coil by the engineers of the Allis-Chalmers Manufacturing Company to be known as the frogleg winding. The connections are as given in Figure 13.
Lap Coil
D S
N
c' a'
Wave Coil N
S
b c S
N
F b'
B
Fig. 13 Frog-Leg winding
E
Frog-leg Winding In practice, the lap portion of the frog-leg winding is always simplex, so that it is necessary to give the wave portion a multiplicity equal to P/2. Thus for frog-leg winding, the number of parallel paths is given by
aFL = a L + aW
aFL = mP + 2m aFL
P = (1)P + 2 = 2 P 2
Frog-leg Winding Example: Determine the coil and commutator pitches for a 24-slot, 48-segment, 6-pole frog-leg armature winding.
Solution:
24 −0=4 For both lap and wave sections: YS = 6 Yc = 1 Lap portion will be simplex:
Wave portion must be triplex:
48 − 3 Yc = = 15 3
DIRECT CURRENT GENERATOR CHARACTERISTICS
Types of DC Generators According to the type of the main field winding used a) Series Generator - it uses only the series field winding
Types of DC Generators b) Shunt Generator - it uses only the shunt field winding
c) Compound Generator - it uses both the series and the field windings
Types of DC Generators RSE
RSH
According to the Source of Excitation for its field windings a) Self-Excited DC Generator – the field windings are excited by current supplied by its own armature.
Types of DC Generators b) Separately- Excited DC Generator – the field windings are excited by current supplied by a separate source.
Types of DC Generators c) Dual Excited DC Generator – the source of excitation for the field windings is both the armature and a separate source. This applies to compound generators.
Types of DC Generators Types of Self-Excited Compound Generators According to the connection of the field windings with respect to the armature a) Short-shunt Compound Generator - the shunt field is directly connected across the armature. b) Long-shunt Compound Generator – the shunt field is connected in parallel across the armature through the series field.
Types of DC Generators
Types of DC Generators According to the direction of the magnetic field produced by the series field and the shunt field windings a) Cumulative compound – the direction of magnetic fields for series and shunt are the same. b) Differential compound – the direction of the magnetic fields for series and shunt are opposite.
Types of DC Generators According to the relative magnitude of the output terminal voltage at no-load and full-load (for cumulative compound) a) Flat Compounded – the no-load voltage is equal to the full-load voltage b) Under Compounded – the no-load voltage is greater than the full-load voltage c) Over compounded – the no-load voltage is less than the full-load voltage
No-Load Characteristics of DC Generators When a shunt or compound generator operates without load- that is, when it is driven by a prime mover, is properly excited, and has none of the load switches closed – a voltage will appear at the terminals that are normally connected to the electrical devices. This generated voltage will depend, for a given machine, upon two factors: (1) the speed of rotation (2) the flux
No-Load Characteristics of DC Generators
No-Load Characteristics of DC Generators If the flux is kept constant while the speed is increased or decreased, the voltage will rise or fall, respectively, in direct proportion to the change in speed. 300 250
Eg
200 150 100 50 0 0
500
1000
1500
Speed (rpm)
2000
2500
No-Load Characteristics of DC Generators Similarly, if the speed is held constant while the flux (not the field current) is varied, the voltage will change in direct proportion to the change in magnetism. However, to show that the generated voltage is directly proportional to the flux is much more difficult because magnetism measurements are not made as readily as are those of amperes and volts. This determination is not particularly important from a practical point of view, because it is more desirable to know how the no-load generated voltage is affected by changes in field current.
No-Load Characteristics of DC Generators This relationship is not a direct one for all changes in excitation, because magnetic saturation sets in after the field current is increased beyond a certain value. To show the relationship between the generated voltage and the field current, a so-called saturation curve (sometimes called magnetization curve) can be plotted. Note particularly that the curve is virtually a straight line up to the so-called “knee”; this is true because, in this region, the iron portions of the magnetic circuit are unsaturated and require a comparatively low percent of the total mmf.
No-Load Characteristics of DC Generators 400 350
accelerating voltage
300 knee
Eg
250
build-up voltage
200 150 100 excitation line
50 0 0
0.5
1
1.5
2
2.5
3
3.5
Field Current
Saturation Curve for dc generator operating at constant speed
No-Load Characteristics of DC Generators With increasing values of flux density the iron saturates, the magnetic permeability drops, and a greater percent, of the field ampere-turns are required for the iron. It should also be observed that the initial voltage is not zero at zero field current; its value Er, usually low, is due to residual magnetism.
No-Load Characteristics of DC Generators Significance of the Saturation (Magnetization) Curve Such a curve emphasizes the extremely important fact that the generated voltage is directly proportional to the flux and not the field current. Curve such as this is extremely important for the purpose of analyzing, predicting, and comparing the operating performance of the various types of generator.
Building Up the Voltage of a SelfExcited Shunt Generator To build up means to rise from its residual voltage, Er, to its normal operating value. Requirements for Build-Up: The machine must develop a small voltage resulting from residual magnetism. The voltage of a self-excited shunt generator will not rise much above an extremely low residual value if the residual flux is insufficient; generators that are expected to operate at voltages up to 250 V should have residual values of flux so that 4 to 10 residual volts are developed.
Building Up the Voltage of a SelfExcited Shunt Generator The total field resistance must be lower than the so-called critical resistance. A generator will fail to build up if the slope of excitation line is 180 ohms 600 150 ohms about equal to or greater than 500 125 ohms 400 the straight-line portion of the 110 ohms 100 ohms 300 magnetization; in fact, a 200 generator will not build up if 100 the total field resistance is 0 0 0.5 1 1.5 2 2.5 3 3.5 4 greater than the so-called Field Current critical value, the latter being defined as the resistance below which machine will build up and above which it will not. Eg
700
Building Up the Voltage of a SelfExcited Shunt Generator The speed of the armature must be above the socalled critical speed.
Eg
A generator will, in fact, fail to 400 build up if, for a given field 2000 rpm 350 1800 rpm 300 resistance, the speed is below the 1600 rpm 250 1400 rpm so-called critical speed, the latter 1200 rpm 200 1000 rpm 150 being defined as the speed above 100 which build-up will occur and 50 0 below which it will not. The critical 0 0.5 1 1.5 2 2.5 3 3.5 speed may be determined Field Current experimentally if, starting from rest, the armature speed is gradually increased; the critical speed will be indicated by a sudden rapid rise in voltage.
Building Up the Voltage of a SelfExcited Shunt Generator There must be a proper relation between the direction of rotation and the connections of the field to the armature terminals. A generator will not build up if the initial field current, at the instant the field switch is closed is in such a direction that the residual flux is opposed; under this condition the machine will build down, not up. This means that there must be a definite relation between the direction of rotation and the connections of the field terminals with respect to the armature terminals.
Building Up the Voltage of a SelfExcited Shunt Generator Thus, if a generator fails to build up, and other conditions have been fulfilled, the difficulty may be corrected: (a) by reversing the direction of rotation or (b) by interchanging the field terminals with respect to the armature terminals. However, if rotation is reversed, the electrical polarity of the brushes will change.
Behavior of a Shunt Generator under Load After a self-excited shunt generator builds up to a required voltage, a no-load voltage, it is ready to supply power to a number of electrical loads up to, and a little above, its rated capacity. One of the most important characteristics of any generator is its behavior with regard to the terminal voltage when the load current is increased. In the shunt type of generator, the voltage always falls down as more current is delivered to the load. There are three reasons for this:
Behavior of a Shunt Generator under Load As more current is delivered by the armature, the voltage drop in the armature IARA increases, thus making a lower emf available at the load terminals. When the armature terminal voltage falls, the field winding suffers a corresponding reduction in current, which, in turn, reduces the flux; the latter further reduces the generated emf.
Behavior of a Shunt Generator under Load When the armature winding carries increasing values of load current, the armature core becomes an electromagnet, apart from the effect of the main poles; this electromagnetic action of the armature reacts with the main field flux further to reduce the flux, the result being that the generated emf suffers an additional drop.
Behavior of a Shunt Generator under Load VNL IARA VFL
Rated output Characteristic load vs. output curve of self-excited shunt generator
In this analysis, it should be clearly understood that the generated voltage, which depends upon the flux is always greater than the terminal voltage by exactly the amount of voltage drop in the armature circuit. This leads to the equation
Vt = Eg – IARA Vt = kΦ - IARA
Compound Generator Behavior under Load – Cumulative The addition of the series field connected to aid the shunt field has the important fundamental purpose of creating additional values of flux with increasing load currents so that the armature will generate greater voltages and thus compensate for the normal tendency of the shunt machine to lose terminal voltage. The behavior of a cumulative compound generator will depend upon the degree of compounding of the said machine, i.e., whether a given generator is flat-, over-, or undercompounded.
Compound Generator Behavior under Load – Cumulative This degree of compounding is determined primarily by the number of series-field ampere-turns with respect to the shunt-field ampere turns. If the series field will produce a sufficient amount of ampere-turns to permit the generated voltage to increase by an amount that is exactly equal to the armature voltage drop when the armature current changes from zero to IAFL, then VFL can be made equal to VNL; under such condition the machine is said to be flat-compounded.
Compound Generator Behavior under Load – Cumulative If, on the other hand, the series field has an overcompensating effect so that EG increases to a greater extent between no load to full load than the armature resistance voltage drop, then VFL will exceed VNL; under this condition the machine is said to be overcompounded. Finally, if the full-load generated voltage is more than the no-load value by an amount that is somewhat less than the armature-resistance drop, the external characteristic may droop; under this condition the generator is said to be undercompounded.
Compound Generator Behavior under Load – Cumulative
under flat
V
FL
over Rated output Characteristic Curves for Cumulative Compound Generator
Compound Generator Behavior under Load – Cumulative Degree of Compounding Adjustment It is customary in manufacturing practice to equip compound generators with sufficient series-field turns so that they will operate considerably overcompounded. Then, by connecting a very lowresistance shunt across the series field, the no load voltage may be brought up to almost any desired value to meet individual demands. It is therefore possible to modify an overcompound generator so that it will be flat- or under- compounded.
Compound Generator Behavior under Load – Cumulative The effect of the series shunt is to by-pass or diverts a portion of the normal load current from the fluxproducing series-field winding, under which the condition the degree of compounding is lessened. The so-called diverter is located where it will have no magnetic influence; moreover, its ohmic value, compared with that of the series field, will determine how much current is diverted. When the diverter resistance is extremely large, the diverted current will be small and the external characteristic will be that of an overcompounded generator.
Compound Generator Behavior under Load – Cumulative On the other hand, if the resistance of the diverter approaches that of a short circuit, practically the load current will be diverted around the series field and the external characteristic will resemble that of a shunt generator. Since the series-field resistance RSE and the diverter resistance RD are in parallel, the total line current IL will divide so that ISE and ID are related to each other by an inverse ratio of the respective resistances.
Compound Generator Behavior under Load – Cumulative Thus
I SE RD = I D RSE
Since
I L = I SE + I D
It follows that
I SE
RD = IL x RD + RSE
Series-Generator Behavior under Load Since the armature, series field, and load are all connected in series, any current that is delivered to the load must, among all other things, simultaneously serve to perform the following functions: It must develop useful energy to the load. It must provide the necessary excitation for the series field so that a voltage is generated in the armature. It must create demagnetizing armaturereaction effect.
Series-Generator Behavior under Load When the load is zero (on open circuit), the current is zero; under this condition the series field ampere-turns will be zero and the generated voltage will be the residual value Er. If the circuit is closed through a load resistance, a current I will flow, in which event the series field will create additional flux and thereby cause a higher voltage to be generated; at the same time the armature will develop a demagnetizing action, and a voltage drop will occur in the armature and series field resistances.
Series-Generator Behavior under Load Therefore, the voltage that will appear at the series generator terminals will be stabilized at some value that is a function of the net generated voltage (due to the net flux) and the I(RA + RSE) voltage drop. The terminal emf Vt will, obviously, rise with the load current so long as the overall voltage increases more rapidly that those factors which tend to reduce it. However, for considerable loads, the iron portions of the magnetic circuit becomes highly saturated under which condition the subtractive effects exceed the slowly rising generated emf; the terminal voltage begins to drop.
Series-Generator Behavior under Load Thus, as the load current increases, the external characteristic of a series generator rises rapidly from its initial Er value during the initial stages, then tapers off to a maximum, and finally drops to zero. Because of the varying nature of the terminal voltage with respect to load, the series generator has few practical applications, and then only when Vt vs. I curve is advantageous to the installation. The series generator is sometimes used in a dc system for voltage-boosting purposes or to minimize leakage currents in grounded dc systems so that electrolytic action in underground structures may be reduced.
Series-Generator Behavior under Load
Magnetization Curve
1600 1200 800 Field Ampere Turns
200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 400
Generated Voltage
External Characteristic
0
10 20 30 40 50 60 Load Amperes
70 80
90
Graphical method to determine the external characteristic of a series generator
VOLTAGE REGULATION Voltage Regulation – is a measure of the extent to which the voltage of a generator changes as the load is gradually lowered from its rated value to zero load. The foregoing may be expressed in percent form as follows:
V −V % Voltage Regulation = x 100% V L
FL
FL
Example: The following data were obtained for the magnetization curve of a 4-pole interpole shunt generator, each field coil of which has 1000 turns. If
E
If
E
If
E
0
6
0.8
160
1.56
260
0.1
20
1.0
200
1.92
280
0.4
80
1.14
220
2.40
300
0.6
120
1.32
240
3.04
320
a) Draw the magnetization curve.
b) If the total shunt field resistance (including the field rheostat) is 125 Ω, determine the voltage to which the machine will build up as a self-excited generator. c) Determine the full-load voltage and the percent regulation of the generator if the full-load armature resistance voltage drop is 30 V.
Solution: 400 350 300 250 200
Build-up voltage = 300 volts
150 100 50 0 0
0.5
1
1.5
2
2.5
3
3.5
400 350
IAFLRA= 30 V
300 250 200
VFL = 252 V
150 100 50 0 0
0.5
1
1.5
2
2.5
3
3.5
V −V x 100% %V .R . = V L
FL
FL
300 − 252 = x 100% 252 = 19.05%
Example: A 20-kW 250-volt short shunt compound generator has a series field whose resistance is 0.022 Ω and each of whose four coils has 6 ½ turns. If a diverter having a resistance of 0.058 Ω is connected across the series field, calculate the series-field ampere-turns per pole at full load.
Solution:
20 kW I = = 80 A 250V 0.058 I = (80 A) = 58 A 0.058 + 0.022 FL
SE
( I )
SE
1 = (58 A) 6 turns 2 = 377 Ampere − turns per pole
Magnetic Action of Armature (Armature Reaction) Armature Reaction – is produced by the load current in the armature conductors that results in a magnetic field whose direction is displaced 90 electrical degrees with respect to the main field. It depends upon and directly proportional to the load current.
Magnetic Action of Armature (Armature Reaction) Effects of Armature Reaction Field Distortion. One very important thing to note in the analysis of the reaction of the armature with respect to the main field flux is to recognize the fact that the current in the armature winding created a field of its own and that this field is superimposed on the main field. Note particularly that this armature flux is in quadrature with the main field flux. The two fields then react with each other and the resulting magnetic action of the armature then tends to distort and alter the direction of the uniformly distributed main field and create a slight demagnetizing effect. The resultant field has a new direction obliquely downward.
Magnetic Action of Armature (Armature Reaction) It seriously affects commutation because sparking will occur at the brushes. Since the resultant field is badly twisted out of shape and is directed obliquely downward, the magnetic neutral is shifted in the direction of rotation so that it will always be at right angles to the resultant field. Such displacement affects commutation, because sparking will occur at the brushes unless they are shifted to locations that reduce sparking.
Magnetic Action of Armature (Armature Reaction) It reduces the generated voltage. To understand why this is so, it is necessary to recognize the fact that the armature flux weakens the field on one half of each pole and strengthens the field on the other half. If the decrease is the same as the increase, the magnitude of the resultant flux would remain unchanged. But, this is not the case because the decrease is usually greater than the increase because of magnetic saturation; the net result is a reduction in flux (demagnetization). In most practical cases, the reduction in flux may be from 1 to 4 percent between the no-load and full-load values.
Magnetic Action of Armature (Armature Reaction) The extent to which the cross-magnetizing armature reaction affects the main field may be determined by the following analysis: The armature has Z/P conductors under each pole and each one carries IA/a amperes. Therefore, the total number of ampere-conductors per pole will be ZIA/aP.
Magnetic Action of Armature (Armature Reaction) However, only those conductors that are directly under the pole faces have a measurable magnetizing or demagnetizing effect upon the main poles, because the conductors between the pole tips, in the interpolar zones, act upon high-reluctance magnetic circuits. If τ represents the ratio pole arc to pole pitch, the maximum effective ampere-conductors per pole will be τZIA/aP. And since two conductors are the equivalent of one turn, it follows that one half of each pole is effectively magnetized and the other half effectively demagnetized.
Magnetic Action of Armature (Armature Reaction) Cross − magnetizin g amp − turns per pole =
τZIA 2aP
Example The lap-wound armature of a 6-pole dc generator has a total of 378 conductors and carries 800 A at full load. If the pole arc is 6.75 in. and the armature is 20 in., calculate the maximum cross-magnetizing ampere-turns per pole that has a magnetizing or demagnetizing effect on each pole tip.
Magnetic Action of Armature (Armature Reaction) Solution:
6.75 τ= = 0.645 20 π 6
(0.645)(378)(800) Cross − magnetizin g I/pole = (2)(6)(6) = 2710 amp. - turns
Some Corrective Methods to Counteract the Effects of Armature Reaction Brush Shifting Using a Chamfered-pole Design Using Pole-Core Laminations with One Pole Tip Using Interpoles and Compensating Windings
Brush Shifting When the armature of a dc machine (without interpoles) carries current, the magnetic and mechanical neutrals do not coincide. This affects commutation because sparking will occur at the brushes unless they occupy positions that short-circuit coil sides in the neutral zone; the brushes must therefore, be shifted to locations that reduce sparking. Moreover, the brushes must be shifted back and forth continually as the load changes because the effect of armature reaction depends upon the value of the armature current; or they must located in some compromise position that represents the best average load.
Brush Shifting However, when this is done a certain Mechanical number of armature ampere turns Neutral A tend to demagnetize the main field New Magnetic apart from the demagnetizing action Neutral that results from field distortion. Figure a illustrates the S demagnetizing action of the N armature when the brushes are not yet shifted to a new position. Only those conductors that are directly under the pole faces have a A considerable demagnetizing effect a) Without Brush Shifting upon the main poles since those conductors in the interpolar zones act upon high-reluctance magnetic circuit.
Brush Shifting However, when the brushes are shifted to an angle AO, the A A directions of the currents in c a the conductors in the AOregion will therefore change. Now then, if diametrically S N opposite conductors are paired in an angle 2AO, it is seen that their effect, as b d ampere turns, is to b) With Brush Shifting demagnetize the main field. In other words, a brush shift of AO means that the number of conductors in 4AO, that is, a to c and b to d, are involved in a direct demagnetizing action of two main poles. Mechanical Neutral
New Magnetic Neutral
Brush Shifting Thus,
1 4 A ZIA Demagnetiz ing amp − conductors / pole = X X 2 360 a AZIA Demagnetiz ing I / pole = 360a
Brush Shifting Mechanical Neutral
A New Magnetic Neutral
c
a
S
N b
The remaining armature conductors will obviously produce a cross-magnetizing action where the conductors a to d and c to b are involved in creating quadrature field. In a two-pole machine there will be [(360 – 4A)/360] x Z crossmagnetizing conductors, while in a P-pole machine their number will be [(360 – 2PA)/360] x Z.
Brush Shifting It follows, therefore, that in a P-pole machine the 360 − 2 PA ZIA Total cross − magnetizin g I / pole = x 360 2 aP 360 − 2 PA = x ZIA 720 aP
Brush Shifting Example The armature has a commutator whose diameter is 15 in. If the brushes are shifted 1.25 in. in the direction of rotation for the purpose of improving commutation, calculate (a) the demagnetizing ampere-turns per pole; (b) the total crossmagnetizing ampere-turns per pole.
Brush Shifting Solution:
1.25 A= x 360 = 9.55 π x 15 9.55 x 378 x 800 Demagnetiz ing I / pole = 360 x 6 = 1335 AT
Brush Shifting 360 − 2 x 6 x 9.55 Total cross − magnetizin g I / pole = x 378 x 800 720 x 6 x 6 = 2865 AT
Note: Brush shift does not alter the direction of the fluxdensity distribution. It can only improve commutation, although, as indicated above, it is accompanied by an additional undesirable demagnetizing influence.
Chamfered Pole Shoe and Pole-Core Laminations with One Pole Tip One method that partly counteracts distortion and demagnetization involves a pole-shoe construction that increases the reluctance between the pole tips and the surface of the armature core; this reduces the flux produced by the armature mmf. Two designs that employ this idea are shown here. In one of these the rounded surface of the pole shoe is not concentric with the circular armature core, i.e., the pole shoe is chamfered. A second scheme uses pole-core laminations with one pole tip; in assembling the laminations, the pole tips are alternated from one side to the other, so that the cross-sectional area of the iron is one-half as much under the pole tips as under the center section.
Chamfered Pole Shoe and Pole-Core Laminations with One Pole Tip
Short air gap
Armature-core surface
Armature-core surface
Wide air gap
(a) Chamfered-pole design
(b) Lamination with one pole tip
Interpoles (Commutating Poles) Interpoles are narrow poles placed exactly halfway between the main poles, centering on the mechanical neutral planes. The exciting windings for these poles are always permanently connected in series with the armature winding because the interpoles must produce fluxes in their air gaps that are proportional to the armature current. Such a relationship can exist only when the iron portions of the magnetic circuits are unsaturated, which means that the interpoles must be operated below the knee of the magnetization curve.
Interpoles (Commutating Poles) a
IA
b
S
S
N
a'
b'
S
N
N IA
(a) Without Interpoles
(b) With Interpoles
Interpoles (Commutating Poles) Note: 1) The polarities of the interpoles in a generator are the same as that of the succeeding main poles in the direction of rotation, whereas in a motor, they must be the same as the preceding main poles in the direction of rotation. 2) The interpoles have no effect upon the armature mmf that distorts the main field. Field distortion is still present and flashover can occur if abnormally heavy loads are suddenly applied.
Interpoles (Commutating Poles) Interpole Ampere-turns Assuming the same number of interpoles as main poles, each one must be provided with sufficient ampere-turns to accomplish three things simultaneously. These are: It must oppose the total cross-magnetizing ampereturns in its own commutating zone; It must inject flux into the armature to nullify the effect of self-induction and thus overcome the reluctance of its own air gap; and
Interpoles (Commutating Poles) It must overcome the reluctance of its iron magnetic-flux paths. To oppose the total cross-magnetizing ampere-turns per pole, the interpole must develop, first of all, ZIA/2aP ampturns. Secondly, the interpole must send ΦC maxwells across the air gap, whose equivalent length, δe in inches, is slightly larger than the actual air gap because of the slotted armature. But
φc =
β gi
6.45
x Agi
Interpoles (Commutating Poles)
And
mmf 0.4π x ( I )gi φc = = ℜ ( δe x 2.54 ) / Agi
Where: βgi = flux density in the air gap, lines/in2 Agi = air-gap area, cm2 (NI)gi = interpole amp-turns for the air gap
Interpoles (Commutating Poles) Equating both values of Φc
β gi x Agi 6.45
0.4π x ( I )gi x Agi = δe x 2.54
From which
( I )gi = 0.313 β gi x δe
Interpoles (Commutating Poles) The third item, requiring that the interpole have sufficient ampere-turns to overcome the reluctance of the iron portions of the magnetic circuit, is usually estimated as being about 0.4 to 0.8 times the air-gap ampere-turns. Taking an average value of about 0.6(NI)gi for the iron, the total interpole ampere-turns is practically,
ZIA ( I )i = + ( 0.313 β gi x δe ) + ( 0.6 x 0.313 β gi x δe ) 2 aP
Interpoles (Commutating Poles) And
ZIA ( I )i = + 0.5 β giδe 2 aP
Interpoles (Commutating Poles) Example Calculate the number of ampere-turns required by an interpole to overcome the reluctance of the air-gap for a flux density of 14 800 lines/in2 if the equivalent length of the air gap is 0.28 in.
Solution (NI)gi = 0.313 x 14800 x 0.28 = 13000 amp-turns
Interpoles (Commutating Poles) Example The armature of a 6-pole machine has a wave winding with a total of 328 conductors and carries a current of 280 A at full load. If the air-gap flux density under each interpole is 12500 lines/in2 and the equivalent air-gap length is 0.24 in., calculate (a) the number of amp-turns required by each of the six interpoles; (b) the number of turns on each interpole.
Interpoles (Commutating Poles) Solution 1 328 x 280 + ( 12500 x 0.24 ) = 5320 amp − turns ( I )i = 2 2 x 6 5320 = = 19 turns per pole 280
Compensating Windings Compensating windings are used for the purpose of neutralizing the effect of armature reaction in the S zones outside the influence of the N interpoles and particularly to maintain a uniform flux distribution under the faces of the main poles. They are special windings placed in slots or holes in the pole faces and carry, as do the interpole windings, the total armature current.
Compensating Windings In the simple two-pole sketch, one set of connectors would join the upper three pole-face conductors on the north pole to the upper three conductors on the south pole, while similar connections would be made for the lower six conductors of the north and south poles; the two sets would then be joined in series and connected in series with the armature winding. The current directions in these conductors are opposite to those in the wires of the armature winding directly below in order to neutralize only that portion of the armature cross-magnetizing ampere-turns that lie directly under the pole faces. It must, therefore, for 100% compensation, always build up an mmf that is equal to the armature mmf per pole face;
Compensating Windings This is τZIA/2aP amp-turns per pole. But the compensating winding current is the total armature current IA. It follows, therefore, that
τZIA
CIA = 2 aP 2
where C is the number of conductors in each pole face. Hence
C=
τZ
aP
Compensating Windings Example A 3000-kW, 600-V. 16-pole generator has a lapwound armature with a total 3250 conductors. If the pole faces cover 63% of the entire circumference, calculate (a) the current in the compensating winding; (b) the number of conductors in each pole face of the compensating winding. (Neglect the shunt-field current.)
Compensating Windings Solution
3000 kW IA = = 5000 A 600 V
0.63 x 3250 C= = 8 conductors 16 x 16
Compensating Windings Note: The use of compensating windings together with properly designed interpole windings in dc machine will provide sparkless commutation and eliminate the possibility of flashover, at least insofar as armature reaction is concerned. These desirable operating characteristics come about because the resultant field is absolutely uniform and the interpolar zones are supplied with the necessary flux to combat the voltage of self-inductance; moreover, such conditions prevail at all values of armature current because the neutralizing effects are caused by the very current that is initially responsible for the difficulty.
Need for Operation of Generators in Parallel Power plants will generally be found to have several small generators rather than large single units capable of taking care of the maximum peak loads. This is true of both dc and ac stations. The several units can then be operated singly or in various parallel combinations, on the basis of the actual load demand. Such practice is considered extremely desirable from the standpoint of: Efficiency Continuity of service Maintenance and Repair problems Addition to plant capacity as the service demands change
Need for Operation of Generators in Parallel When generators are operated in parallel, they function together to supply power to a common load; moreover under ideal conditions, 1) the combined rating of the several machines is approximately equal to the total load 2) each generator assumes its proportionate share of the total load on the basis of its rating in comparison with those of the others.
Need for Operation of Generators in Parallel Conditions for Parallel Operation of Generators The generators must have identical external characteristics, i.e., the voltage changes of all machines must be exactly the same for equal changes in per cent change of load. The generators must have the same polarity. The generators must have the same terminal voltage.
Need for Operation of Generators in Parallel Operation of Shunt Generators in Parallel It was previously shown that shunt generators have “drooping” voltage-vs.-load characteristics, i.e., that the voltage drops as the load increases. If two shunt generators have identical external characteristics, then the two machines will divide the total load in proportion to their relative capacities.
Need for Operation of Generators in Parallel However, if the external characteristics are not similar, then if both have been adjusted for the same voltage at rated load currents, generator A will always deliver a larger percent of its rated capacity than will generator B of its rated capacity. A circuit diagram illustrating how the connections should be made for operating two shunt generators in parallel is shown here.
Need for Operation of Generators in Parallel - Bus To Load
To Load
+ Bus
IA Load Ammeter Main Switch
IB
SA
A
DPDT Switch
Load Ammeter
SB Main Switch
V
B
Voltmeter FB
FA
Shunt Field Field Rheostat
Field Rheostat
Wiring connections for the operation of two shunt generators in parallel.
Assume that generator A is connected to the bus bars through switch SA and that it carries a load. As the load increases, it will ultimately become necessary to (1) connect a larger generator than A in parallel with the latter, after which the smaller machine, when gradually unloaded, is disconnected from the line, or (2) connect another generator in parallel with A and have two machines operate jointly to supply the total load
Need for Operation of Generators in Parallel The procedure for accomplishing this is as follows: Generator B is brought up to speed by its prime mover. Field switch FB is closed, whereupon the voltage will build up. With the DPDT switch closed to the left, the bus voltage (generator A in this case) is observed. This switch is then closed to the right, and the voltage of B is adjusted by means of its field rheostat until it equals the voltage of the line. It is important that the polarity of the incoming generator be exactly the same as that of the line polarity, i.e., plus to plus and minus to minus. Caution! If this is not the case, a serious short circuit will occur when switch SB is closed.
Need for Operation of Generators in Parallel After proper adjustments are made, quickly close switch SB. This places generator B in parallel with generator A. Generator A will still be supplying the entire load, while generator B will be running idle. It said to be “floating”. To shift the load from A to B, it is necessary merely to manipulate both field rheostats simultaneously, cutting in resistance in the field of A and at the same time cutting out resistance in the field B. Any degree of load shifting can be readily accomplished in this way; in fact, the entire load can be shifted to B and the main switch SA opened to remove A from the line.
Need for Operation of Generators in Parallel While the load is shifted from one generator to the other, load ammeters IA and IB should be carefully watched to make sure that overloading does not occur. It is also important that the field rheostats are not manipulated beyond the point where A is carrying no load because, if they are, generator A will attempt to operate as a motor and thus drive its prime mover.
Need for Operation of Generators in Parallel Operation of Compound Generators in Parallel When two compound generators are to be operated in parallel, it is necessary to use essentially the same wiring as that employed for shunt machines, except that an equalizer connection must be added. If the latter connection is not used, the two generators will not operate satisfactorily in parallel.
Need for Operation of Generators in Parallel Equalizer – is a very low-resistance copper wire that joins together identical ends of the series fields not otherwise electrically connected. The figure on the left is a simple schematic diagram showing how this is done for two compound generators. Note particularly that the two series fields are permanently connected in in parallel, a condition that results in a division of the total current so that the ratio of the two series-field currents (ISE)A and (ISE)B are inversely proportional to their respective resistances, that is, ISEA RSEB
ISEB
=
RSEA
Need for Operation of Generators in Parallel If two identical overcompounded generators are operating in parallel without an equalizer and delivering a total current IT, the machine may thus be said to be in unstable equilibrium because any slight mechanical or electrical disturbance will immediately initiate a series of reactions that will cause complete instability and the resulting opening of protective circuit breakers. If, for illustrative purposes, it is assumed that the speed of generator B increases momentarily, its generated emf will increase; this is the initial-cause of instability since B immediately delivers a slightly greater share of the total load while A supplies a correspondingly lower value of current than before, the terminal voltage and the total current remaining substantially constant.
Need for Operation of Generators in Parallel Remembering that both machines have rising voltage vs. load characteristics, the increased current through the series field of B will cause its flux to increase and raise the generated voltage further; conversely, the small decrease of the current through the series field of A will reduce its flux and lower the generated voltage. The result of these incremental changes is to make B take a still larger part of the load, while A is losing more of its proportion. This process of load transfer having once been started continues until generator B delivers the entire load while that of A falls to zero.
Need for Operation of Generators in Parallel If this is permitted to continue, the current in A will reverse and generator A will be operated as a differential-compound motor to drive its prime mover; under this condition, A, taking its power from B, imposes a still greater load on B. The final stage- a most serious one- occurs when the series-field flux completely nullifies the shunt field flux so that the counter emf of A falls to zero; machine A then becomes a virtual short circuit on machine B. The method for connecting one compound generator in parallel with another already supplying load current follows essentially the same procedure as for shunt generators. Load transfer, or disconnecting one machine from the line, may be accomplished in a similar manner.
Need for Operation of Generators in Parallel Two Cases in the Analysis of Generators in Parallel CASE I: External Characteristics are given (VNL, VFL, kW, voltage regulation, etc.) By ratio and proportion: VNLA VNLB new VBUS inititial VBUS
∆V VLB − VFLB = ∆IB IFLB
VFLB VFLA
∆I A IBNEW
IB
IA
∆V VLA − VFLA = ∆I A IFLA
IFLA
IFLB
Need for Operation of Generators in Parallel But
V L − VFL %V .R = x 100% VFL
Therefore,
∆V %VRA x VFLA = ∆IA 100 x IFLA
VBUS EW = VBUS IITIAL ± ∆V
∆V %VRB x VFLB = ∆I B 100 x IFLB
IB EW = IB ± ∆IB IA EW = IA ± ∆IA
Note: +∆V if there is a decrease in the load current -∆V if there is an increase in the load current +∆I if there is an increase in the load current -∆I if there is a decrease in the load current
Need for Operation of Generators in Parallel CASE II: Internal Characteristics are given (generated voltage, armature equivalent resistance, shunt-field resistance, series field resistance, interpole-winding resistance, etc.)
Note: Use network analysis.
Need for Operation of Generators in Parallel PROBLEMS 1. Two shunt generators A and B, with ratings of 250 and 400 kW, respectively, and having identical straight-line voltage vs. per cent kilowatt-output external characteristic, are connected in parallel. If the no-load voltage is 260 V and the full-load voltage is 240 V, calculate (a) the kW output of each machine and the total kW load when the terminal voltage is 245; (b) the kW output of each machine and the terminal voltage when the total output is 575 kW.
Need for Operation of Generators in Parallel 2. Two shunt generators, A and B, are connected in parallel to deliver a common load. Generator A has a no-load voltage of 240 V and a voltage of 220 V when it delivers 120 A. Generator B has a no-load voltage of 235 V and a voltage of 220 V when it delivers the same current as A. Assuming straight-line external characteristics for both machines, calculate (a) the total line voltage and total kW load when generator B is “floating”; (b) the load delivered by each machine and the total load when the terminal emf is 225 V.
DIRECT CURRENT MOTOR CHARACTERISTICS
Operating Differences between Motors and Generators DC GENERATOR
DC MOTOR
a) When a generator is in a) When a motor is in operation, operation, it is driven by a it is “fed” by an electric mechanical machine such as current from an electrical an engine, a water turbine, or supply; the motor current then even an electric motor; the produces two stationary rotation through a magnetic magnetic fields, one by the field generates a voltage, field poles and the other by which, in turn, is capable of the rotating armature, which producing a current in an reacts with each other to electric circuit. develop torque, which, in turn, produces mechanical rotation.
Operating Differences between Motors and Generators DC GENERATOR
DC MOTOR
b) The load on a generator b) The load on a motor constitutes those electrical constitutes the force that devices that convert electrical tends to oppose rotation and energy into other forms of is called a countertorque; such energy; loads such as electric loads may be fan blades, lighting, electric furnaces, pumps, grinders, boring mills, electrical welding, electric crushers, excavators, motors, electric battery elevators, turntables, churns, charging, etc. drills, food mixers and a host of other commonly used machines.
Operating Differences between Motors and Generators DC GENERATOR
DC MOTOR
c) The voltage of a generator c) The speed of rotation tends to tends to change when the load change as the load varies. changes. d) The voltage of a generator can d) The speed of rotation can be always be adjusted by doing changed by varying either or either or both of two things: both of two things: (1) the (1) changing the speed and strength of the magnetic field (2) changing the strength of and (2) the voltage impressed the magnetic field. across the armature terminals.
Operating Differences between Motors and Generators DC GENERATOR e)
DC MOTOR
Generators can be, and e) Motors usually operate as a frequently are, operated in single independent unit to parallel with others to supply a drive their individual loads, common load; infrequently, although in special they may be connected in applications they may be series for the same purpose. connected in parallel or in series for the purpose of performing particular jobs at varying speeds; example of series-and parallel-connected motors are electric excavators and traction equipment.
Operating Differences between Motors and Generators DC GENERATOR
DC MOTOR
f) Generators are always started f) Motors may or may not have a without electrical loads; the mechanical load when they procedure is to bring them up are started; as a matter of to speed, adjust the voltage, practical significance, it is and then close the main quite customary for a motor to switch that permits the start a load that is often equal machine to deliver current. or greater than the rated name-plate value.
Note: Structurally, dc generators and motors are identical, except for minor differences that may permit them to function in accordance with known practical requirements.
Classification of Direct-Current Motors There are also three general types of motor, namely, series, shunt and compound, all of which are widely used in many applications. Each type of motor has very definite operating characteristics that differ markedly from those of the other two, so that it is important to know the load requirements before a proper selection is made. For the purpose of classification, it is convenient, therefore, to indicate how a motor behaves between no load and full load by using such terms as constant speed and variable speed.
Classification of Direct-Current Motors Constant-speed type – if a change from no mechanical load to full load causes the speed to drop approximately 8% or less; shunt motors fall into this classification.
Variable-speed type – motors in which the speed changes by greater values than 8%; series and compound motors behave in this manner. Whenever the speed of a motor can be controlled by an operator who makes manual adjustment, it is said to be of the adjustable-speed type. It is possible, therefore, to have a constant-speed-adjustable-speed motor or a variablespeed-adjustable-speed motor.
Counter Electromotive Force (Counter EMF) With the armature rotating as a result of motor action, the armature conductors continually cut through the resultant stationary magnetic field, and because of such flux cutting, voltages are generated in the very same conductors that experience force action. This can only mean that when a motor is operating, it is simultaneously acting as a generator. Obviously, motor action is stronger than generator action, for the direction of the flow of current in the armature winding is fixed by the polarity of the source supply. The generated voltage does, however, oppose the impressed emf and, in this respect, serves to limit the current in the armature winding to a value just sufficient to take care of the power requirements of the motor. Since the generated voltage opposes the flow of current, it is called a counter electromotive force (counter emf).
Counter Electromotive Force (Counter EMF) Clearly, this counter emf can never be equal to, and must be less than, the voltage impressed across the armature terminals, because the direction in which the current flows determines first the direction of rotation and thus the direction of the counter emf. This can only mean that the armature current is controlled and limited by the counter emf. Therefore, by Ohm’s Law,
V A − EC IA = RA '
Where: IA = armature current VA = impressed voltage across armature winding EC = counter emf RA’ = equivalent armature resistance
Counter Electromotive Force (Counter EMF) Since the counter emf is a generated voltage, it depends, for a given machine, upon two factors: 1. the flux per pole, Φ 2. the speed of rotation in rpm, N Hence,
E C = kφ V A − kφ IA = RA '
Counter Electromotive Force (Counter EMF) As a matter of practical importance, it should be stated that the counter emf developed in the armature of a motor is usually between 80 and 95 per cent of the voltage impressed across the armature terminals; the higher percentages usually apply to the larger motors, while the lower percentages apply to those near or in the fractionalhorsepower ranges. It is also significant that motors in which EC is a high percentage of the armature terminal voltage VA will operate most efficiently, while those in which EC is small compared with VA will have low efficiency. This may be seen from the following analysis:
EC = V A − I A RA '
EC I A = V A I A −
2 I A RA
Counter Electromotive Force (Counter EMF) Note that ECIA is the power in watts developed by the armature because it is equal to the power in watts supplied to the armature, VAIA minus the copper loss in the armature IA2RA. Therefore, for a given load current IA, it should be clear that a motor will develop the greatest power when the counter emf is a maximum.
Counter Electromotive Force (Counter EMF) Example: A 115-volt shunt motor has an armature whose resistance is 0.22 Ω. Assuming a voltage drop across the brush contacts of 2 volts, what armature current will flow when the counter emf is 108 V?
Counter Electromotive Force (Counter EMF) Solution
V A − VB − I A RA − EC = 0 V A − V B − E C 115 − 2 − 108 IA = = = 22.7 A RA 0.22
Counter Electromotive Force (Counter EMF) Example: A compound motor operates at a speed of 1520 rpm when the voltage impressed across the armature terminals is 230 V. If the flux per pole is 620 000 maxwells and the armature resistance is 0.43 Ω, calculate: (a) the counter emf and (b) the armature current. Assume a value of k = 2.2 x 10-7 and a brush drop of 2 volts.
Counter Electromotive Force (Counter EMF)
Solution (a) E C = kφ = 2.2 x10 −7 (620000)(1520) = 207.5 V V A − V B − E C 230 − 2 − 207.5 (b) I A = = 47.7 A = RA ' 0.43
Starting a DC Motor At the instant a dc motor is started, the counter emf is zero because the armature is not revolving. It should be understood that at the instant of starting, the armature current would be extremely high unless some resistance, called starter, were added to offset the lack of EC. As the speed increases, the starter may be cut out gradually because EC rises; finally, when the motor has attained normal speed, the starter is completely cut out of the armature circuit.
Starting a DC Motor Manual Starters for Shunt and Compound Motors There are two standard types of starter for shunt and compound motors, namely, three-point and four-point. Three-point Starter – It has three terminals labeled L, F, and A that are connected respectively to one line terminal, one-shunt field terminal, and one armature terminal.
Starting a DC Motor Manual Starters for Shunt and Compound Motors
Note: Three-point starters are not completely satisfactory when used with speed-controlled motors.
Starting a DC Motor Manual Starters for Shunt and Compound Motors Four-point Starter Of particular importance, compared with the internal connections of a three-point starter, it should be noted that one change has been made. The holding coil has been removed from the shunt-field circuit and, in series with a current limiting resistor, has been placed in a separate circuit in parallel with the armature and the shunt field. With this arrangement, the holding coil current is independent of any field-rheostat changes and thus overcomes the objection of the three-point starter.
Starting a DC Motor Manual Starters for Shunt and Compound Motors
Starting a DC Motor Whenever a starter, whose duty is to accelerate a motor from rest to normal speed, is equipped with a means of governing, in some predetermined manner, the electric power delivered to the apparatus to which it is connected, it is called an electric controller. The basic functions of a controller are acceleration, retardation, line closing, reversing, braking, protection, and others.
Starting a DC Motor Automatic Starters for Shunt and Compound Motors The use of starters that will perform the function of accelerating motors automatically, although somewhat more expensive, is preferable to the manual types. There are several reasons for such preference: Automatic starters are reliable and, when properly adjusted, will bring motors up to speed without the blowing of fuses of the opening of circuit breakers under all conditions of loading.
Starting a DC Motor Automatic Starters for Shunt and Compound Motors Push-button stations for starting and stopping may be conveniently located for remote-control operation. The starting resistors may be cut out at a desired time rate so that acceleration may be uniform and in accordance with the demands of the load. To conserve power, a motor is more likely to be stopped when it is idle by simply pressing a button.
Starting a DC Motor Automatic Starters for Shunt and Compound Motors Electric braking facilities may be readily provided so that a motor may be brought to rest quickly and smoothly. Overload and temperature protection of equipment is usually incorporated.
Starting a DC Motor Automatic Starters for Shunt and Compound Motors Three Types of Automatic Starters Counter EMF (Speed-Limit) Starter – In this scheme, a number of relays are connected across the armature where the counter emf increases as the motor accelerates, and the former are adjusted to pick up at predetermined values of voltage.
Starting a DC Motor Automatic Starters for Shunt and Compound Motors OL - overload - normally open contacts - normally closed contacts
Relays – 1AX, 2AX, 3AX Contactors – M, 1A, 2A, 3A
Shunt Field
M
OL
1AX
R1
R2
R3
1A
2A
3A
2AX M 3AX 1AX
1A
2AX
2A
3AX
3A
Start
OL
Stop M1
Wiring Diagram of a Counter-EMF Starter connected to a Shunt Motor
Starting a DC Motor Automatic Starters for Shunt and Compound Motors Time-Limit Acceleration Starter – In this type, a group of relays is timed to operate at preset intervals, by means of devices that function mechanically, pneumatically, or electrically.
Starting a DC Motor Automatic Starters for Shunt and Compound Motors
Starting a DC Motor Automatic Starters for Shunt and Compound Motors Current-Limit Acceleration Starter – In this type, the relays are designed so that they are sensitive to current changes in the armature circuit.
Starting a DC Motor Automatic Starters for Shunt and Compound Motors
Starting a DC Motor Example: The armature of a 220-volt shunt motor has a resistance of 0.18 Ω. If the armature current is not to exceed 76 A, calculate: (a) the resistance that must be inserted in series with the armature at the instant of starting; (b) the value to which this resistance can be reduced when the armature accelerates until EC is 168 volts; (c) the armature resistance at the instant of starting if no resistance is inserted in the armature circuit. Assume a 2-volt drop at the brushes.
Starting a DC Motor Solution a)
V A − VB − EC IA = R A '+ R starter R starter
V A − V B − VC 230 − 2 − 0 = − RA ' = − 0.18 = 2.82 Ω IA 76
b) Rstarter = V A − VB − VC − RA ' = 230 − 2 − 168 − 0.18 = 0.61 Ω IA
230 − 2 c) I A = 0.18 = 1267 A.
76
Starting a DC Motor Example: A 20-hp 220-volt 540-rpm shunt motor has an armature resistance of 0.12 Ω and a field resistance of 52.4 Ω. If the resistance of a starter is 1.93 Ω, what line current does the motor take at the instant of starting? Assume a motor efficiency of 88% and a brush drop of 3 volts.
Starting a DC Motor Solution At the instant of starting, EC = 0:
I Astarting I Line starting
220 − 3 − 0 = = 105.85 A 1.93 + 0.12 = I SH + I Astarting 220 = + 105.85 = 110.05 A 52.4
Loading a Motor- Effect upon Speed and Armature Current In practice, an electric motor generally receives its electrical power at substantially constant voltage. It then converts electrical power into mechanical power, doing so by developing torque as it rotates its mechanical load. If the mechanical load on the motor changes, either torque or the speed, or both, must change. When a load is applied to a motor, the natural tendency of the latter is to slow down because the opposition to motion (i.e., the countertorque) is increased. Under this condition, the counter emf decreases, for the reason that EC is proportional to the speed.
Loading a Motor- Effect upon Speed and Armature Current The reduction in the speed immediately results in an increase in the armature current. Obviously, this increase in armature current must be exactly that required by the motor to drive the increased load because any increase in mechanical
driving power must be met by a corresponding increase in electrical power input to the armature. Thus it is seen that loading a motor always results in two changes: (1) a reduction in speed and (2) an increase in armature current.
Loading a Motor- Effect upon Speed and Armature Current The speed at which a motor operates when it is driving its rated load, its so-called rated horsepower, is called the normal
speed. On the other hand, reducing the load on a motor causes its armature to take less current while it speeds up. In both shunt and compound motors, the no-load speed is very definite and stable; these types of motor do not attempt to operate at excessive unsafe speeds when running idle. If the mechanical load is completely removed from a shunt motor, it will operate at a speed only slightly higher than the normal speed; this will generally be between 2 and 8 percent higher than the normal speed.
Loading a Motor- Effect upon Speed and Armature Current Doing the same thing to a compound motor will result in a rise in speed of about 10 to 25 percent. On the other hand, the series motor does attempt to race, or operate at very high speed when the load is removed. This fact is well recognized in practice that a series motor is always geared or coupled to its load so that a countertorque will always exist; it is never belted to the load, because the accidental “throwing” of the belt will instantly result in a dangerous racing motor, a motor that is said to “run away”.
Torque Characteristics of DC Motors The power developed by a motor must be sufficient to drive the mechanical load and take care of its own mechanical (friction and windage) losses. This is equal to
Pd = EC I A But
Therefore
ZφP EC = x 10 − 8 60 a ZφP Pd = x 10 − 8 x I A 60 a
Torque Characteristics of DC Motors Also
2πT T hp = = 33000 5250
Where T = lb-ft so that
T Pd = x 746 5250
Equating the two values of Pd
T ZφP −8 x 746 x 10 x I A = 5250 60 a
Torque Characteristics of DC Motors
T
ZP 5250 xφ x I A = − 8 60 x 746 x 10 a
Torque Characteristics of DC Motors
T=
k
xφ x I A
Torque Characteristics of DC Motors
T=
k
xφ x I A
Hence, the torque developed depends upon two factors: (1) the flux created by the main poles (2) the current flowing in the armature winding
Torque Characteristics of DC Motors The foregoing analysis leads to the following general conclusions concerning the manner in which the torque developed by a motor varies with the load, and therefore the armature current: The torque of a shunt motor varies directly with the armature current IA, since the shunt field current is practically constant for all conditions of loading and is fixed only by the field resistance and the impressed voltage; this implies that the shunt-field flux remains substantially the same for all values of load current.
T = k 1I A
Torque Characteristics of DC Motors The torque developed by a series motor varies with changes in both armature current and flux because the series-field ampere-turns, which influence the magnitude of φ, are directly proportional to the load current. At light loads, when the iron on the magnetic circuit
is not saturated, φ is directly proportional to IA.
T = k (k 1I A )I A T = k2 I
2 A
Torque Characteristics of DC Motors At heavy loads, when the magnetic-circuit iron is
saturated, the flux will change very little or not at all with variations in IA.
T = k1 I A The torque vs. load characteristic of a compound motor, where the series-field and shunt-field ampere turns aid each other (cumulative compound), is a composite of the shunt and series motors, and the extent to which the curve departs from that exhibited by the shunt machine depends upon the strength of the series field with respect to the shunt field.
Torque Characteristics of DC Motors a dR
Co m
po un d
Sh (c un t u Se ri e m u la s tiv e)
Rated Torque
ng
e
rlExamination of the curves indicates oa e v O that between no load and full load the shunt motor develops the greatest torque, while the series motor develops the least; the torque develop byFullthe compound falls load between these two. At overloads, the overload torque of a series motor is considerably higher than that developed by a shunt motor; the compound motor again falls Rated between the other two. I A
Armature Current Characteristic torque vs. armature-current curves for three types of motor
Torque Characteristics of DC Motors Example: (a) Calculate the torque in pound-feet developed by a dc motor, given the following particulars: Poles = 4; Z = 828; Φ = 1.93 x 105 maxwells; IA = 40 A; winding = wave. (b) What will be the horsepower of the motor when operating at a speed of 1750 rpm?
Torque Characteristics of DC Motors Solution ZP 5250 (a) T = xφ x IA 8 60 x 746 x 10 a 828 x 4 5250 5 (40) 1 . 93 x 10 = 8 60 x 746 x 10 2
(
= 15 lb − ft
)
Torque Characteristics of DC Motors Solution
2πT (2π )(1750)(15) (b) hp = = 33000 33000 = 5 horsepower
Torque Characteristics of DC Motors Example: A series motor develops 62 lb-ft of torque when the current is 48 A. Calculate the torque if the load increases so that the motor takes 56 A.
Solution
T=
2 k2 I A
T2 = T1
2 I A2 2 I A1
2
T2 56 = 2 62 48 T 2= 84.4 lb − ft
Torque Characteristics of DC Motors Example: A compound motor develops a torque of 271 lb-ft when it is operating at 1200 rpm, under which condition its armature current is 215 A. What will be the torque and hp of the motor if the load is increased so that it slows down to 1120 rpm in which case IA changes to 238 A and the total flux increases by 8%?
Torque Characteristics of DC Motors Solution
T = kφt I A T2 φt 2 I A2 = T1 φt1 I A1 hp2
(1.08φt1 )( 238) T2 = 271 φt1 ( 215) T2 = 324 lb − ft
( 2π )(1120)(324) =
33000 = 69.1 horsepower
Starting Torque and Overload Capacity of Motors The ability of a motor to start a load, overcoming the static friction and inertia of heavy moving parts of its own rotor and the application to which it is connected, is frequently an important requirement in certain installations; this is usually referred to as the percent starting torque or percent “breakaway” torque, based upon the rated full-load torque. Shunt motors develop moderate starting and maximum torques, about 300 and 275 percent, respectively, operate within a speed variation of about 10%, and should be applied to such loads as line shafts, fans, blowers, centrifugal pumps, metal and woodworking machines, elevators, conveyors, laundry washing machines and vacuum cleaners.
Starting Torque and Overload Capacity of Motors Series motors develop extremely high values of starting and maximum torques, about 500 and 425 percent, respectively, operate over a considerable range of speed, tend to race at very light loads, and should be applied to such loads as cranes, traction machines, coal and ore bridges, bucket and mine hoists, gates, car dumpers, turntables, and car retarders. Compound motors have medium starting and maximum torques, about 450 and 350 percent, respectively, operate within a speed variation up to 30%, and should be applied in such applications as plunger pumps, shears, crushers, rotary and flat-bed presses, rolling mills, punch presses, geared elevators, hoists, pressure blowers, compressors, circular saws, bending rolls, and hydroextractors.
Speed Characteristics of DC Motors It was previously pointed out that the speed of a shunt motor rises about 2 to 8% when the rated load is completely removed the speed of a compound motor rises approximately 10 to 25% when the rated load is completely removed the speed of a series motor rises very rapidly when the load is removed and must, therefore, always drive some load if it is to prevented from racing dangerously, i.e., “running away.”
Speed Characteristics of DC Motors
V A − I A RA ' = kφ
Speed Characteristics of DC Motors Example: The armature of a 230-V shunt motor has a resistance of 0.30 Ω and takes 50 A when driving its rated load at 1500 rpm. At what speed will the motor operate if the load is completely removed, i.e., when it is running idle, a condition under which the armature current drops to 5 A? Assume that the brush drops at full load and no load are 2 V and 1 V, respectively.
Speed Characteristics of DC Motors Solution
( 230 − 2) − (0.30 x 50) 1500 = kφ ( 230 − 1) − (0.30 x 5) 2 = kφ
227.5
2 kφ 227.5 = = 213 1500 213 kφ 2 = 1602.11 rpm
Speed Characteristics of DC Motors Example: A 220-V long-shunt compound motor has an armature resistance of 0.27 Ω and a series-field resistance of 0.05 Ω. The full-load speed is 1400 rpm when the armature current is 75 A. At what speed will the motor operate at no load if the armature current drops to 5 A and the flux is reduced to 90% of its full-load value? Assume brush drops the same as in the previous example.
Speed Characteristics of DC Motors Solution
(220 − 2) − 75(0.27 + 0.05) 1400 = kφ FL
2
( 220 − 1) − 5(0.27 + 0.05 ) = k (0.90φ FL )
2 = 1400
217.4
k (0.9φ FL ) = 194 kφ FL
2 = 1743.18 rpm
217.4
0.9 194
Speed Regulation of DC Motors Speed Regulation – the natural or inherent change in speed of a shunt or compound motor between full load and no load
L − FL % speed regulation = x 100 FL
Speed Regulation of DC Motors Example: The name-plate speed of a 25-hp shunt motor is 1150 rpm. If the motor speed rises to 1210 rpm when the load is removed, calculate the per cent regulation.
Solution L − FL %speed regulation = x 100% FL 1210 − 1150 = x 100% 1150 = 5.22%
Speed Regulation of DC Motors Example: The following information is given in connection with a 230-volt shunt motor: line current at a rated speed of 1200 rpm = 20 A, line current at no load = 5 A, armature resistance = 0.3 Ω, shunt-field resistance = 115 Ω, brush drops at rated load and no load equal 2 volts and 1 volt, respectively. Calculate the per cent speed regulation of the motor.
Speed Regulation of DC Motors Solution L
230 − 1 − ( 5 − 2 )( 0.3 ) = kφ
FL =
1230 − 1200 %S .R = x 100% 1200 = 2.5%
230 − 2 − ( 20 − 2 )( 0.3 ) = 1200 kφ
[ L 230 − 1 − ( 5 − 2 )0.3] / kφ = 1200 [230 − 2 − ( 20 − 2 )0.3] / kφ L = 1230 rpm
Speed Control of DC Motors Speed Control – is the physical/manual adjustment of the speed of a motor by an operator or automatic control equipment. Whenever the speed of a motor can be controlled by an operator who makes manual adjustment, it is said to be of the adjustable-speed type. There are two general methods to control the speed of a motor: by varying the voltage across the armature by varying the flux These methods are based on the principle that the speed is
directly proportional to the counter emf and inversely proportional to flux.
Speed Control of DC Motors Speed control can readily be done with dc motors in one or more of three different ways: by inserting a field rheostat in the shunt-field circuit of a shunt or compound motor (Field-Resistance Control) – in this method, the speed increases as resistance is cut in by the field rheostat; this is true because the speed rises as the flux is reduced.
V A− I A R A ' $ = k φ var
Speed Control of DC Motors by inserting a resistance in the armature circuit of a shunt, compound, or series motor (Armature-Resistance Control) – in this method, the speed decreases as resistance is inserted in the armature circuit; the greater the value of the inserted resistance, the lower the speed.
V A− I A (R A '+ R var ) $ = kφ
Speed Control of DC Motors by varying the voltage across the armature circuit of a shunt or compound motor while, at the same time, maintaining constant the voltage across the shunt field (Armature-Voltage Control) – in this method, it is necessary to have two sources of direct current for the controlled motor. The shunt field must be connected to a constant potential supply so that flux of unvarying intensity is created, while the armature is permanently connected to the armature terminals of a special generator whose voltage can be varied. This method permits speed variations above and below normal.
Speed Control of DC Motors $ =
V
−I ARA ' kφ
A var
Armature Voltage Control
WARD LEONARD METHOD OF SPEED CONTROL
Speed Control of DC Motors This variable-voltage control system (Ward Leonard Method of Control) has many important applications when extremely wide speed ranges, often as much as 10:1, are desired. It is frequently found on electric excavators, on freight-handling ships, and in blooming and paper mills and for the operation of passenger elevators in tall buildings. The chief disadvantage of the Ward Leonard system is its high first cost and its low overall efficiency. To offset these disadvantages, it must be said that it provides excellent stepless speed control for a motor which must have a very wide range of speed.
Reversing the Direction of DC Motors There are two general methods for reversing the direction of rotation of a dc motor: changing the direction of current flow through the armature changing the direction of current flow through the field circuit The direction of rotation of a dc motor cannot be reversed by interchanging the connections to the staring switch, because this reverses the current flow through both the armature and the field. To reverse the direction of rotation of a compound motor, it is necessary to reverse the current flow through the armature winding only or through both the series and shunt fields.
Efficiency of Dynamos A dynamo is a machine that converts energy from one form to another. When this conversion takes place at a uniform rate, that is, when the energy received by the machine per unit time and the energy delivered by the machine in the same unit of time are constant, then it is proper to say that a dynamo converts power from one form to another. The power received by a dynamo is called its input; the power delivered by the dynamo is called its output. The power input to a dynamo is always more than its power output.
Efficiency of Dynamos The difference between the power input to a machine (in watts) and its power output (in watts) is called the power loss because it is unavailable to drive mechanical load in a motor or to supply electrical power in a generator. This power loss always produces heating in the dynamo; therefore, the greater the power loss, as a percentage of the input, the hotter will the machine tend to become. If this loss should reach an excessive value, the temperature rise might be high enough to cause failure.
Efficiency of Dynamos POWER LOSSES IN DYNAMOS STRAY-LOAD LOSS ROTATIONAL LOSSES (STRAY-POWER LOSSES)
ELECTRICAL LOSSES
copper losses
bearing and friction brush friction wind friction (windage) Hysteresis eddy current loss
brush contact loss
Hysteresis Loss The hysteresis loss takes place in the revolving armature core because the magnetic polarity in the iron changes in step with changing positions of the magnetic material under various poles. When an armature-core tooth is passing under a north pole, its polarity will be south; the iron particles are then oriented with their north ends pointing toward the shaft center. When this same tooth moves under a south pole, its polarity will be north and the iron particles will then be directed away from the shaft center. The rapid “jerking” around of the tiny magnetic molecules in the armature-core iron as it revolves rapidly causes a sort of magnetic particle friction and produces heating.
Hysteresis Loss In the modern dynamo, it depends upon the flux density in the armature-core iron, the speed of rotation, the quality of the magnetic iron, and the weight of the iron.
Ph = k h fβ w 1.6
Ph = k h ' $β w
For usual grades of iron used,
kh = 6.2 x 10-10
1.6
Hysteresis loss is unaffected by whether or not the core is laminated.
Eddy-Current Loss As the armature core revolves, voltages are generated in the iron exactly as they are in the copper wires. These voltages are objectionable, however, because they create a flow of current in the iron core in “eddies”. These eddy currents result because the generated voltages in the iron near the outside surface are greater than those closer to the center of the shaft because of the higher speed; the difference in potential then causes currents to flow in the iron. Eddy currents may be minimized by slicing or laminating the armature core and then coating each lamination with a highresistance varnish.
Eddy-Current Loss Eddy-current loss depends upon the core flux density, the speed of rotation, the thickness of the laminations and the volume of iron; it is independent of the quality of the magnetic iron.
Pe = ke f t β V 2 2
2
Pe = ke ' $ t β V 2 2
2
ke depends upon the resistivity of the iron and the dimensions employed for other factors.
Copper Losses Copper losses always occur when there is a current flow through the various copper circuits. The various copper losses are (1) the armature-winding, (2) the shunt field, (3) the series field, (4) the interpole field, and (5) the compensating-winding field. The shunt field loss is the only one that remains nearly constant (except for minor line-voltage changes). Since the other fields are always connected in the armature circuit, their losses are almost proportional to the square of the load
Stray-load Loss It results from such factors as: the distortion of the flux because of armature reaction lack of uniform division of the current in the armature winding through the various paths and through the individual conductors of large cross-sectional area short-circuit currents in the coils undergoing commutation The indeterminate nature of the stray-load loss makes it necessary to assign it a reasonable value arbitrarily; it is usually assumed to be 1% of the output of the machine when the rating is about 150 kW (200 hp) or more; for the smaller ratings, the stray-load loss is generally neglected when efficiency calculations are made, without much loss of accuracy.
Efficiency of DC Generators The efficiency of a dc generator is the ratio of the electrical power output to the mechanical input, converted to watts.
watts output %η = ×100% watts input Since watts input = watts output + watts losses,
watts output %η = ×100% watts output + watts losses
Efficiency of DC Generators The two methods for determining the efficiency of a generator are: by directly measuring the total power output and the total power input (direct efficiency test) – this involves an actual test upon the generator in which electrical instruments measure the output, while a calibrated motor drives the machine under test. by making certain necessary tests from which various power losses are determined (conventional method) - this involves actual test upon the generator to determine the resistances of the armature, interpole winding, series field winding, compensating winding, and shunt field winding after which the various copper losses are determined by calculations. A test is also performed to measure the rotational loss.
Efficiency of DC Motors The efficiency of a dc motor is the ratio of the mechanical power output, converted to watts to the electrical input.
watts output %η = ×100% watts output + watts losses The conventional efficiency of a motor may be determined in exactly the same way as for a generator, with one exception. In making the test for the rotational loss, the impressed voltage across the armature must equal the terminal emf minus the brush contact and armature-resistance drops at full load.
Stray-Power Loss Measurement Either of two procedures may be followed to determine the straypower loss: No-load Test – the machine, whether generator or motor, is operated free of any load as a shunt motor at rated speed and with a voltage across the armature circuit equal to the normal generated emf. For Generator: EA = V + Vb + IaRa’ Note: The drops being full load values
For Motor: EA = V - Vb - IaRa’
Stray-Power Loss Measurement
SPL = E A I A − I RA ' 2 A
Where: SPL = stray-power loss EA = supply voltage during the test IA = armature current during the test RA’ = armature equivalent resistance
Stray-Power Loss Measurement Separate-Motor Test – the machine whose SPL are to be determined is operated as a separately excited generator being coupled to and driven at rated speed by a comparatively small shunt motor (the size be about the same order of magnitude as the losses to be measured and its normal speed nearly the same as the tested machine). Two sets of data must be taken, namely: (1) With the driving motor running free (2) With the machines coupled together
Stray-Power Loss Measurement
SPL = ( E A I A − I RA ' ) coupled − ( E A I A − I RA ' )uncoupled 2 A
2 A
Maximum Efficiency If a test is performed upon a dynamo, or calculation are made, to determine its performance, it will be found that the efficiency increases with increasing values of load, reaches a maximum and then proceeds to drop. Condition for Maximum Efficiency
Constant Losses = Copper Losses
Maximum Efficiency
%η max
watts output = watts output + 2constant losses
Prony Brake Test This test is used to determine the output horsepower of the motor. In this test, a hollow cast-steel pulley is placed on the shaft extension of the motor and a set of brake shoes, fitted to a lever that acts on a scale, is made to ride on the flat cylindrical surface of the brake drum.
Prony Brake Test
Prony Brake Test
Prony Brake Test
Prony Brake Test
Prony Brake Test
2π$T Hp = k Where: T = (scale reading – dead weight) x length of arm N = speed in rpm Hp = horsepower output k = 33 000 if T is in lb-ft = 44 760 if T is in N-m
Sample Problems 1) The hysteresis and eddy current losses of a dc machine running at 1000 rpm are 250 W and 100 W respectively. If the flux remains constant, at what speed will the total iron loss be halved. 2) In a dc generator, the iron losses at 1000 rpm are 10 kW at a given field current. At a speed of 750 rpm and at the same field current, the total iron losses become 6 kW. Determine the iron losses at 500 rpm.
Sample Problems 3) The following information is given in connection with a 10 kW 250-V flat-compound generator: Rsh = 125 Ω, Ra = 0.4 Ω, Rse = 0.05 Ω, SPL = 452 watts, Vb (assumed constant) = 3 volts. Calculate the efficiency (a) at full load; (b) at 50% load 4) Using the given data of Prob. 3, calculate (a) the power output of the generator when the efficiency is maximum; (b) the maximum efficiency
Sample Problems 5) A 10-kW 220-V 1400-rpm shunt generator is operated at rated speed as a motor. The armature takes 2.95 A from a 232-volt source. The total armature resistance including brushes is 0.26 Ω, and the shunt field resistance is 146.5 Ω. Determine (a) the SPL; (b) the full load efficiency when the machine is operating as a generator. 4)
The following information is given in connection with a prony-brake test upon a motor: F = 37 lb, tare weight of brake = 3.5 lb, L = 3 ft, rpm = 1160. Calculate the horsepower output under this condition.
