Theory and Calculation of Electrical Apparatus (1917) — part 8 of 21

where the brackets denote that the sum of the product of the corresponding parts of the two quantities is taken.

As discussed in the preceding, the torque of an induction motor, in synchronous watts, equals the power consumed by the primary counter e.m.f.; that is:

D2 1 = P*\

and substituting (10) and (11) this gives:

D2 1 = se o 2 {cost (g cost ± bi sin r) T sin r (b cos r + gf sin r ) }

= seo2 {£l+J_£L^ cos2T± ^l^sin2r} J (12)

and herefrom follows the motor output or power, by multiplying with (1 — s).

The sum of the torques of both motors, or the total torque , is :

2 D t = Di + D 2 = 5c 0 2 { (gi +g) - ( gi - g) cos 2 r} . (13) The difference of the torque of both motors, or the synchronize

ing torque , is :

2 Ds

= se 0 2 (6i — b) sin 2 r,

(14)

where, by (7),

ri — 2 r

0i =

g = »

rrii

m

6i =

SX i

6 =

(15)

m i

m

mi = r i 2 + s 2 xP, m = (r x + 2r) 2 + s 2 x i 2 ,

In these equations primary exciting current and primary impedance are neglected. The primary impedance can be intro- duced in the equations, by substituting (r x + sro) for r x , and (xi + Xo) for Xi y in the expression of m x and m, and in this case only the exciting current is neglected, and the results are suffi- ciently accurate for most purposes, except for values of speed

SYNCHRONIZING INDUCTION MOTORS 163

very close to synchronism, where the motor current is appreciably increased by the exciting current. It is, then:

mi = (n + rso) 2 .+ s 2 ( xx + xo) 2 , j

m = (r i + sr 0 + 2 r) 2 + s 2 (xi + Xo) 2 ‘, j ^

all the other equations remain the same.

From (15) and (16) follows

b\ — b 2 srxi (ri + sr 0 + r)

= : j

2 mm i

(17)

hence, is always positive.

96. (6i — b ) is always positive, that is, the synchronizing torque is positive in the first or lagging motor, and negative in the second or leading motor; that is, the motor which lags in position behind gives more power and thus accelerates, while the motor which is ahead in position gives less power and thus drops back. Hence, the two motor armatures pull each other into step, if thrown together out of phase, just like two alternators.

The synchronizing torque (14) is zero if r = 0, as obvious, as for r = 0 both motors are in step with each other. The syn- chronizing torque also is zero if r = 90°, that is, the two motor armatures are in opposition. The position of opposition is unstable, however, and the motors can not operate in opposition, that is, for r = 90°, or with the one motor secondary short- circuiting the other; in this position, any decrease of r below 90° produces a synchronizing torque which pulls the motors together, to r = 0, or in step. Just as with alternators, there thus exist two positions of zero synchronizing power — with the motors in step, that is, their secondaries in parallel and in phase, and with the motors in opposition, that is, their secondaries in opposition — and the former position is stable, the latter unstable, and the motors thus drop into and retain the former position, that is, operate in step with each other, within the limits of their synchronizing power.

If the starting rheostat is short-circuited, or r = 0, it is, by (15), bi = f), and the synchronizing power vanishes, as is obvious, since in this case the motor secondaries are short-circuited and thus independent of each other in their frequency and speed.

With parallel connection of induction-motor armatures a syn- chronizing power thus is exerted between the motors as long as any appreciable resistance exists in the external circuit, and

164

ELECTRICAL APPARATUS

the motors thus tend to keep in step until the common starting resistance is short-circuited and the motors thereby become inde- pendent, the synchronizing torque vanishes, and the motors can slip against each other without interference by cross-curjrents.

Since the term contains the slip, s, as factor, the syn-

chronizing torque decreases with increasing approach to syn- chronous speed.

Fig. 59. — Synchronizing induction motors: motor torque and synchronizing

torque.

For r = 0, or with the motors in step with each other, it is, by (12), (15), and (16):

se 0 2 (ri + 2 r)

D 2 1 = se 0 2 g =

(IS)

(ri + sr 0 + 2 r) 2 + s 2 (xi + x 0 ) 2 ’

that is, the same value as found for a single motor. (As the resistance r is common to both motors, for each motor it enters as 2 r.)

For r = 90°, or the unstable positions of the motors, it is:

ZV = se 0 2 gi =

seoVi

(19)

(ri + $r 0 ) 2 + s 2 ( Xi + x 0 ) 2 ’ that is, the same value as the motor would give with short-

SYNCHRONIZING INDUCTION MOTORS 165

circuited armature. This is to be expected, as the two motor armatures short-circuit each other.

The synchronizing torque is a maximum for r = 45°, and is, by (14), (15), and (16) :

IX = seo 2 bl -X b . (20)

As instances are shown, in Fig. 59, the motor torque, from equation (18), and the maximum synchronizing torque, from equation (20), for a motor of 5 per cent, drop of speed at full- load and very high overload capacity (a maximum power nearly two and a half times and a maximum torque somewhat over three times the rated value), that is, of low reactance, as can be produced at low frequency, and is desirable for intermittent service, hence of the constants:

Z ± = Z 0 = 1 + j, ■

Y = 0.005 - 0.02 j, e Q = 1000 volts,

for the values of additional resistance inserted into the armatures :

r = 0; 0.75; 2; 4.5,

giving the values:

1

01 = m?

6i

2 8 mi

mi = (1 + s) 2 + 4 s 2 j

9

1 + 2 r m ’

b

sx i m ’

m = (1 + $ + 2 r) 2 + 4 ,s* 2 .

As seen, in this instance the synchronizing torque is higher than the motor torque up to half speed, slightly below the motor torque between half speed and three-quarters speed, but above three-quarters speed rapidly drops, due to the approach to syn- chronism, and becomes zero when the last starting resistance is cut out.

CHAPTER IX

SYNCHRONOUS INDUCTION MOTOR

97. The typical induction motor consists of one or a number of primary circuits acting upon an armature movable thereto, which contains a number of closed secondary circuits, displaced from each other in space so as to offer a resultant closed secondary circuit in any direction and at any position of the armature or secondary, with regards to the primary system. In consequence thereof the induction motor can be considered as a transformer, having to each primary circuit a corresponding secondary cir- cuit — a secondary coil, moving out of the field of the primary coil, being replaced by another secondary coil moving into the field.

In such a motor the torque is zero at synchronism, positive below, and negative above, synchronism.

If, however, the movable armature contains one closed cir- cuit only, it offers a closed secondary circuit only in the direc- tion of the axis of the armature coil, but no secondary circuit at right angles therewith. That is, with the rotation of the arma- ture the secondary circuit, corresponding to a primary circuit, varies from short-circuit at coincidence of the axis of the arma- ture coil with the axis of the primary coil, to open-circuit in quadrature therewith, with the periodicity of the armature speed. That is, the apparent admittance of the primary circuit varies periodically from open-circuit admittance to the short- circuited transformer admittance.

At synchronism such a motor represents an electric circuit of an admittance varying with twice the. periodicity of the primary frequency, since twice per period the axis of the armature coil and that of the primary coil coincide. A varying admittance is obviously dentical in effect with a varying reluctance, which will be discussed in the chapter on reaction machines. That is, the induction motor with one closed armature circuit is, at synchronism, nothing but a reaction machine, and consequently gives zero torque at synchronism if the maxima and minima of the periodically varying admittance coincide with the maximum

SYNCHRONOUS INDUCTION MOTOR

167

and zero values of the primary circuit, but gives a definite torque if they are displaced therefrom. This torque may be positive or negative according to the phase displacement between ad- mittance and primary circuit; that is, the lag or lead of the maximum admittance with regard to the primary maximum. Hence an induction motor with single-armature circuit at syn- chronism acts either as motor or as alternating-current generator according to the relative position of the armature circuit with respect to the primary circuit. Thus it can be called a syn- chronous induction motor or synchronous induction generator, since it is an induction machine giving torque at synchronism.

Power-factor and apparent efficiency of the synchronous in- duction motor as reaction machine are very low. Hence it is of practical application only in cases where a small amount of power is required at synchronous rotation, and continuous current for field excitation is not available.

The current produced in the armature of the synchronous induction motor is of double the frequency impressed upon the primary.

Below and above synchronism the ordinary induction motor, or induction generator, torque is superimposed upon the syn- chronous-induction machine torque. Since with the frequency of slip the relative position of primary and of secondary coil changes, the synchronous-induction machine torque alternates periodically with the frequency of slip. That is, upon the con- stant positive or negative torque below or above synchronism an alternating torque of the frequency of slip is superimposed, and thus the resultant torque pulsating with a positive mean value below, a negative mean value above, synchronism.

When started from rest, a synchronous induction motor will accelerate like an ordinary single-phase induction motor, but not only approach synchronism, as the latter does, but run up to complete synchronism under load. When approaching syn- chronism it makes definite beats with the frequency of slip, which disappear when synchronism is reached.

CHAPTER X

HYSTERESIS MOTOR

98. In a revolving magnetic field, a circular iron disk, or iron cylinder of uniform magnetic reluctance in the direction of the revolving field, is set in rotation, even if subdivided so as to preclude the production of eddy currents. This rotation is due to the effect of hysteresis of the revolving disk or cylinder, and such a motor may thus be called a hysteresis motor.

Let I be the iron disk exposed to a rotating magnetic field or resultant m.m.f. The axis of resultant magnetization in the disk, 7, does not coincide with the axis of the rotating field, but lags behind the latter, thus producing a couple. That is, the component of magnetism in a direction of the rotating disk, /, ahead of the axis of rotating m.m.f., is rising, thus below, and in a direction behind the axis of rotating m.m.f. decreasing, that is, above proportionality with the m.m.f., in consequence of the lag of magnetism in the hysteresis loop, and thus the axis of resultant magnetism in the iron disk, /, does not coincide with the axis of rotating m.m.f., but is shifted backward by an angle, a, which is the angle of hysteretic lead.

The induced magnetism gives with the resultant m.m.f. a mechanical couple :

D = sin a,

where

T = resultant m.m.f.,

<£ = resultant magnetism, a = angle of hysteretic advance of phase, m — a constant.

The apparent or volt-ampere input of the motor is :

P = raSF#.

Thus the apparent torque efficiency:

P

Q = Sln “«

where

Q = volt-ampere input,

HYSTERESIS MOTOR

169

and the power of the motor is :

P = (1 — s) D = (1 — s) sin a,

where

s = slip as fraction of synchronism. The apparent efficiency is :

Since in a magnetic circuit containing an air gap the angle, a, is small, a few degrees only, it follows that the apparent efficiency of the hysteresis motor is low, the motor consequently unsuitable for producing large amounts of mechanical power.

From the equation of torque it follows, however, that at constant impressed e.m.f., or current — that is, constant 2F — the torque is constant and independent of the speed; and there- fore such a motor arrangement is suitable, and occasionally used as alternating-current meter.

For $<0, we have a < 0,

and the apparatus is an hysteresis generator.

99. The same result can be reached from a different point of view. In such a magnetic system, comprising a movable iron disk, I, of uniform magnetic reluctance in a revolving field, the magnetic reluctance — and thus the distribution of magnetism — is obviously independent of the speed, and conse- quently the current and energy expenditure of the impressed m.m.f. independent of the speed also. If, now:

V = volume of iron of the movable part,

(B = magnetic density,

and

r) = coefficient of hysteresis,

the energy expended by hysteresis in th$ movable disk, I, is per cycle:

Wo = Yij® 1 - 6 ,

hence, if / = frequency, the power supplied by the m.m.f. to the rotating iron disk in the hysteretic loop of the m.m.f. is:

Po = /IW- 6 .

At the slip, sf, that is, the speed (1 — s)f, the power expended by hysteresis in the rotating disk is, however:

Pi =

170

ELECTRICAL APPARATUS

Hence, in the transfer from the stationary to the revolving member the magnetic power:

P = Po - Pi = (1 - s) fVr)(& XA] ,

has disappeared, and thus reappears as mechanical work, and the torque is :

D = lihi - ■

that is, independent of the speed.

Since, as seen in “ Theory and Calculation of Alternating-cur- rent Phenomena,” Chapter XII, sin a is the ratio of the energy of the hysteretic loop to the total apparent energy of the mag- netic cycle, it follows that the apparent efficiency of such a motor |.

can never exceed the value (1 — s) sin a, or a fraction of the j

primary hysteretic energy. \

The primary hysteretic energy of an induction motor, as repre- :

sented by its conductance, g, being a part of the loss in the f.

motor, and thus a very small part of its output only, it follows that the output of a hysteresis motor is a small fraction only of the output which the same magnetic structure could give with secondary short-circuited winding, as regular induction motor.

As secondary effect, however, the rotary effort of the magnetic structure as hysteresis motor appears more or less in all induction motors, although usually it is so small as to be neglected.

However, with decreasing size of the motor, the torque of the hysteresis motor decreases at a lesser rate than that of the in- I

duction motor, so that for extremely small motors, the torque as hysteresis motor is comparable with that as induction motor.

If in the hysteresis motor the rotary iron structure has not uniform reluctance in all directions — but is, for instance, bar- shaped or shuttle-shaped — on the hysteresis-motor effect is superimposed the effect of varying magnetic reluctance, which tends to bring the motor to synchronism, and maintain it therein, as shall be more fully investigated under “ Reaction Machine” in Chapter XVI.

100 . In the hysteresis motor, consisting of an iron disk of uniform magnetic reluctance, which revolves in a uniformly rotating magnetic field, below synchronism, the magnetic flux rotates in the armature with the frequency of slip, and the resultant line of magnetic induction in the disk thus lags, in space, behind the synchronously rotating line of resultant m.m.f.

HYSTERESIS MOTOR

171

of the exciting coils, by the angle of hysteretic lead, a, which is constant, and so gives, at constant magnetic flux, that is, con- stant impressed e.m.f., a constant torque and a power propor- tional to the speed.

Above synchronism, the iron disk revolves faster than the rotating field, and the line of resulting magnetization in the disk being behind the line of m.m.f. with regard to the direction of rotation of the magnetism in the disk, therefore is ahead of it in space, that is, the torque and therefore the power reverses at synchronism, and above synchronism the apparatus is an hysteresis generator, that is, changes at synchronism from motor to generator. At synchronism such a disk thus can give me- chanical power as motor, with the line of induction lagging, or give electric power as generator, with the line of induction leading the line of rotation m.m.f.

Electrically, the power transferred between the electric cir- cuit and the rotating disk is represented by the hysteresis loop. Below synchronism the hysteresis loop of the electric circuit has the normal shape, and of its constant power a part, propor- tional to the slip, is consumed in the iron, the other part, pro- portional to the speed, appears as mechanical power. At syn- chronism the hysteresis loop collapses and reverses, and above synchronism the electric supply current so traverses the normal hysteresis loop in reverse direction, representing generation of electric power. The mechanical power consumed by the hysteresis generator then is proportional to the speed, and of this power a part, proportional to the slip above synchronism, is consumed in the iron, the other part is constant and appears as electric power generated by the apparatus in the inverted hysteresis loop.

This apparatus is of interest especially as illustrating the difference between hysteresis and molecular magnetic friction: the hysteresis is the power represented by the loop between magnetic induction and m.m.f. or the electric power in the circuit, and so may be positive or negative, or change from the one to the other, as in the above instance, while molecular mag- netic friction is the power consumed in the magnetic circuit by the reversals of magnetism. Hysteresis, therefore, is an electrical phenomenon, and is a measure of the molecular magnetic fric- tion only if there is no other source or consumption of power in the magnetic circuit.

CHAPTER XI

ROTARY TERMINAL SINGLE-PHASE INDUCTION MOTOR

101 . A single-phase induction motor, giving full torque at starting and at any intermediate speed, by means of leading the supply current into the primary motor winding through brushes moving on a segmental commutator connected to the primary

B

winding, was devised and built by R. Eickemeyer in 1891, and further work thereon done later in Germany, but never was brought into commercial use.

Let, in Fig. 60, P denote the primary stator winding of a single- phase induction motor, S the revolving squirrel-cage secondary winding. The primary winding is arranged as a ring (or drum) winding and connected to a stationary commutator, C. . The single-phase supply current is led into the primary winding, P, through two brushes bearing on the two (electrically) opposite

SINGLE-PHASE INDUCTION MOTOR

173

points of the commutator, C. These brushes, B , are arranged so that they can be revolved.

With the brushes, B, at standstill on the stationary commutator, Cj the rotor, S } has no torque, and the current in the stator, P, is the usual large standstill current of the induction motor. If now the brushes, P, are revolved at synchronous speed,/, in the direc- tion shown by the arrow, the rotor, S, again has no torque, but the stator, P, carries only the small exciting current of the motor, and the electrical conditions in the motor are the same, as would be with stationary brushes, B, at synchronous speed of the rotor, S. If now the brushes, P, are slowed down below synchronism, /, to speed, /i, the rotor, S, begins to turn, in reverse direction, as shown by the arrow, at a speed, / 2 , and a torque corresponding to the slip, s = / - (fi +fs).

Thus, if the load on the motor is such as to require the torque given at the slip, $, this load is started and brought up to full 'speed, / — s, by speeding the brushes, P, up to or near synchronous speed, and then allowing them gradually to come to rest: at brush speed, fi — / — s, the rotor starts, and at decreasing, /i, accelr- ates with the speed /a = / — s —/i, until, when the brushes come to rest: fi = 0, the rotor speed is /2 — f — s.

As seen, the brushes revolve on the commutator only in start- ing and at intermediate speeds, but are stationary at full speed. If the brushes, P, are rotated at oversynchronous speed: fi>f 9 the motor torque is reversed, and the rotor turns in the same direction as the brushes. In general, it is:

fi + /a + s = //

where

fi = brush speed, f% — motor speed,

« = slip required to give the desired torque, / = supply frequency.

102 . An application of this, type of motor for starting larger motors under power, by means of a small auxiliary motor, is shown diagrammatically, in section, in Fig. 61.

Po is the stationary primary or stator, So the revolving squirrel- cage secondary of the power motor. The stator coils of P 0 connect to the segments of the stationary commutator, Co, which receives the single-phase power current through the brushes, P 0 .

174

ELECTRICAL APPARATUS

These brushes, B Q) are carried by the rotating squirrel-cage secondary, Si, of a small auxiliary motor. The primary of this, Pi, is mounted on the power shaft, A , of the main motor, and carries the commutator, Ci, which receives current from the brushes, B x .

These brushes are speeded up to or near synchronism by some means, as hand wheel, H, and gears, G , and then allowed to slow down. Assuming the brushes were rotating in counter-clock- wise direction. Then, while they are slowing down, the (ex- ternal) squirrel-cage rotor, Si, of the auxiliary motor starts and

Fig. Cl. — Rotary terminal sngle-phase induction motor with controlling

motor.

speeds up, in clockwise direction, and while the brushes, Pi, come to rest, Si comes up to full speed, and thereby brings the brushes, P 0 , of the power motor up to speed in clockwise rotation. As soon as B 0 has reached sufficient speed, the power motor gets torque and its rotor, So, starts, in counter-clockwise rotation. As S 0 carries Pi, with increasing speed of S 0 and Pi, Si and with it the brushes, P 0 , slow down, until full speed of the power motor, So, is reached, the brushes, B 0} stand still, and the brushes, B 1} by their friction on the commutator, C 1, revolve together with Ci, Pi and So.

In whichever direction the brushes, B i, are started, in the same direction starts the main motor, So.

SINGLE-PHASE INDUCTION MOTOR

175

If by overload the main motor, So, drops out of step and slows down, the slowing down of Pi starts Si, and with it the brushes, jB 0 , at the proper differential speed, and so carries full torque down to standstill, that is, there is no actual dropping out of the motor, but merely a slowing down by overload.

The disadvantage of this motor type is the sparking at the commutator, by the short-circuiting of primary coils during the passage of the brush from segment to segment. This would require the use of methods of controlling the sparking, such as used in the single-phase commutator motors of the series type, etc. It was the difficulty of controlling the sparking, which side-tracked this type of motor in the early days, and later, with the extensive introduction of polyphase supply, the single-phase motor problem had become less important.

CHAPTER XII

FREQUENCY CONVERTER OR GENERAL ALTERNATING- CURRENT TRANSFORMER

103. In general, an alternating-current transformer consists of a magnetic circuit, interlinked with two electric circuits or sets of electric circuits, the primary circuit, in which power, sup- plied by the impressed voltage, is consumed, and the secondary circuit, in which a corresponding amount of electric power is produced; or in other words, power is transferred through space, by magnetic energy, from primary to secondary circuit. This power finds its mechanical equivalent in a repulsive thrust acting between primary and secondary conductors. Thus, if the secondary is not held rigidly, with regards to the primary, it will be repelled and move. This repulsion is used in the constant-current transformer for regulating the current for constancy independent of the load. In the induction motor, this mechanical force is made use of for doing, the work: the induction motor represents an alternating-current transformer, in which the secondary is mounted movably with regards to the primary, in such a manner that, while set in motion, it still remains in the primary field of force. This requires, that the induction motor field is not constant in one direction, but that a magnetic field exists in every direction, in other words that the magnetic field successively assumes all directions, as a so- called rotating field.

The induction motor and the stationary transformer thus are merely two applications of the same structure, the former using the mechanical thrust, the latter only the electrical power transfer, and both thus are special cases of what may be called the “ general alternating-current transformer,” in which both, power and mechanical motion, are utilized.

The general alternating-current transformer thus consists of a magnetic circuit interlinked with two sets of electric circuits, the primary and the secondary, which are mounted rotatably with regards, to each other. It transforms between primary electrical and secondary electrical power, and also between

1 7 £5

FREQUENCY CONVERTER

177

electrical and mechanical power. As the frequency of the re- volving secondary is the frequency of slip, thus differing from the primary, it follows, that the general alternating-current transformer changes not only voltages and current, but also frequencies, and may therefore be called “frequency converter. ” Obviously, it may also change the number of phases.

Structurally, frequency converter and induction motor must contain an air gap in the magnetic circuit, to permit movability between primary and secondary, and thus they require a higher magnetizing current than the closed magnetic circuit stationary transformer, and this again results in general in a higher self- inductive impedance. Thus, the frequency converter and in- duction motor magnetically represent transformers of high ex- citing admittance and high self-inductive impedance.

104 . The mutual magnetic flux of the transformer is pro- duced by the resultant m.m.f. of both electric circuits. It is determined by the counter e.m.f., the number of turns, and the frequency of the electric circuit, by the equation :

E = /2 7r/ 71 $ 10" 8

where

E = effective e.m.f., f = frequency, n — number of turns,

<I> = maximum magnetic flux.

The m.m.f. producing this flux, or the resultant m.m.f. of primary and secondary circuit, is determined by shape and magnetic characteristic of the material composing the magnetic circuit, and by the magnetic induction. At open secondary circuit, this m.m.f. is the m.m.f. of the primary current, which in this case is called the exciting current, and consists of a power component, the magnetic power current, and a reactive component, the magnetizing current.

In the general alternating-current transformer, where the secondary is movable with regard to the primary, the rate of cutting of the secondary electric circuit with the mutual mag- netic’ flux is different from that of the primary. Thus, the fre- quencies of both circuits are different, and the generated e.m.fs. are not proportional to the number of turns as in the stationary transformer, but to the product of number of turns into frequency.

12

178

ELECTRICAL APPARATUS

105. Let, in a general alternating-current transformer:

s = ratio

secondary

primary

frequency, or “slip”;

thus, if:

/ = primary frequency, or frequency of impressed e.m.f., sf = secondary frequency;

and the e.m.f. generated per secondary turn by the mutual flux has to the e.m.f. generated per primary turn the ratio, s,

s = 0 represents synchronous motion of the secondary;

5 < 0 represents motion above synchronism — driven by external mechanical power, as will be seen;

$ = 1 represents standstill;

5 > 1 represents backward motion of the secondary,

that is, motion against the mechanical force acting between primary and secondary (thus representing driving by external mechanical power).

Let:

n 0 = number of primary turns in series per circuit; n\ = number of secondary turns in series per circuit;

a = — = ratio of turns ; n i

Y - g — jb = primary exciting admittance per circuit;

where:

c i = effective conductance; b = susceptance;

Z o = To --jx o = internal primary self-inductive impedance per circuit,

where:

r 0 = effective resistance of primary circuit; x 0 = self-inductive reactance of primary circuit;

Z u = Vi + -jx i = internal secondary self-inductive im- pedance per circuit at standstill, or for $ = 1,

where;

r i = effective resistance of secondary coil; xi = self-inductive reactance of secondary coil at stand- still, or full frequency, s = 1,

FREQUENCY CONVERTER

179

Since the reactance is proportional to the frequency, at the slip, s, or the secondary frequency, sf, the secondary impedance is:

Zi = r x + jsx i.

_jet the secondary circuit be closed by an external resistance, r, and an external reactance, and denote the latter by x at frequency, /, then at frequency, sf, or slip, $, it will be = sx, and thus:

Z = r + jsx = external secondary impedance. 1

Let:

E o = primary impressed e.m.f. per circuit,

E ' = e.m.f. consumed by primary counter e.m.f.,

Ei = secondary terminal e.m.f.,

‘ E\ = secondary generated e.m.f.,

e = e.m.f. generated per turn by the mutual magnetic flux, at full frequency, /,

10 = primary current,

loo = primary exciting current,

1 1 = secondary current.

It is then:

Secondary generated e.m.f.:

E'i = snie.

Total secondary impedance : *

Zi + Z = (n + r) + js {xi + x ) ; hence, secondary current:

r — 1 ~ ffig

‘ 1 “ Zi + Z ~ (ri + r) + js (xi + x)

1 This applies to the ease where the secondary contains inductive react- ance only; or, rather, that kind of reactance which is proportional to the frequency. In a condenser the reactance is inversely proportional to the frequency, in a synchronous motor under circumstances' independent of the frequency. Thus, in general, we have to set, x = x' + x" + x"\ where x' is that part of the reactance which is proportional to the frequency, x " that part of the reactance independent of the frequency, and x" f that part of the reactance which is inversely proportional to the frequency; and have thus,

at slip, s, or frequency, $/, the external secondary reactance, sx' 4- x ,r -1 —

180

ELECTRICAL APPARATUS

Secondary terminal voltage:

E x = FA i - hZ, = JjZ

f T\ + jsx i ] _ snii(r+jsx)

• ^ 1 ^ + j s ( Xl + jc) j “ (n + r) + is (x x + 2 )

e.m.f . consumed by primary counter e m.f.

E ; = n 0 e;

hence, primary exciting current:

/oo = #'Fo = n 0 e (g - ib).

Component of primary current corresponding to secondary current, / x :

n 0 se .

~~ a 2 { (ri + r) + js (x x -+- x) } ?

hence, total primary current:

h = /oo + To

f 1 1 , g - ib 1 .

Sn ° 6 1 a 2 (n + r) + is (xi + x) s J

Primary impressed e.m.f. :

?o = E'

• IqZq s

a 2

Uq6 j 1 +

r 0 + i^o

Oi + r) + is (xi + x)

• (r 0 + jx 0 ) (f/ - ib)

We get thus, as the

Equations of the General Alternating-current Transformer, of ratio of turns, a; and ratio of frequencies, s; with the e.m.f. generated per turn at full frequency, e, as parameter, the values: Primary impressed e.m.f. :

E 0 = n ? e { 1 + ~ (n +" r y -f jsfxTVx^ + (r ° + jXa) {(J ~ jb) 1 ' Secondary terminal voltage.

T1 ( „ 7*1 + 7SXi

Ei = snxe 1 - 7

l Oi + r) + JS (Xi. + X) Primary current:

€J r + j sx}

(r 1 + r) + js(xi + x)

h

sn^e

I I +y-.ik 1.

a 2 (n + r) + is (zi + as) s I

FREQUENCY CONVERTER

181

Secondary current :

(r* + r) + js (xi + xj

Therefrom, we get: Ratio of currents :

jr • = “ | 1 + ~ (G ~ jb) [(ri + r) + js (*i + a:)] | •

Ratio of e.m.fs.:

• (r 0 + jx o) (ff - jb)

' , , s r 0 + jx o . , , .

Eo _ a\ a 2 (r t + r) + js (x x + x)_ ^ r ° 3X0

Ei s j j n + jsx 1

1 (j"i + r) + js (xi 4- x j

Total apparent primary impedance:

Z t = y-° = ^ } (ri + r) 4- js (xi + x ) }

4 0 o

± n 4 ja?o

« 2 (ri + r) 4- js (xi 4- x)

4- (r 0 4- jx 0 ) (gr — j&)

where:

14- - (g - jb) [(n 4 - r) 4 - js (xi + x)]

o

a? — x f + h --

in general secondary circuit as discussed in footnote, page 179. Substituting in these equations:

S = 1;

gives the

General Equations of the Stationary Alternating-current Transformer Substituting in the equations of the general alternating-current transformer:

Z = 0,

gives the

General Equations of the Induction Motor Substituting:

(r t + r) 2 + $ 2 (xi + x ) 2 = z k 2 ,

182

ELECTRICAL APPARATUS

and separating the real and imaginary quantities :

Eo = n 0 e { [l + 77^7 (To ( r i + r) + sx 0 (xi + x)) + (r a g + xjb ) ]

• j {75^7 ^ sr ° < ' a:i + x ) ~ %( ri + r )) + ( r «£> - **)] : 1

T f rri H- 7* , grT . rs(asi + x) . fell

*• - moe i W + 1 J- j IV + d r

h = 77 { Oi + r) - js (xi + x) | •

Zlc l i

Neglecting the exciting current, or rather considering it as a separate and independent shunt circuit outside of the trans- former, as can approximately be done, and assuming the primary impedance reduced to the secondary circuit as equal to the secondary impedance :

Yo = 0, § = Zi.

Substituting this in the equations of the general transformer we get :

#0 = n 0 e 1 1 + A Dl ( r i + r) + sx 1 ( x y + *)]

[sri (xi + x) - X! (ri + r)] •’

Ei = — 7 ( ( Ti + r) + s 2 x (xi + «)] - js [rxi - arj },

Zfc

h = - 777 { (h + r) - js (xi + x ) },

LLZk

h = 77 Kn +r) -js(xi + x)}.

Z k

106 . The true power is, in symbolic representation:

p = im,

denoting :

gives:

Secondary output of the transformer:

Pi = [-E'1/1] 1 = (— Yr = snt>;

\ Zk /

FREQUENCY CONVERTER

183

Internal loss in secondary circuit:

T> 1 - 0 / ^

P i l = ^iVi = ( ) ri =

\ St /

Tjtal secondary power:

Pi + Px 1 = (r + ri) = sw (r + 7*1) ;

Internal loss in primary circuit:

pi .*2 -22 / snie \ 2

P 0 = r 0 — io 2 ria 2 = ( ) v i = sriw:

\ Zk J

Total electrical output, plus loss:

P 1 = Pi + Pi 1 + Po 1 = (— ) 2 (r + 2 rx) = sir (r + 2 r,) ; Total electrical input of primary:

Po = [Po/o] 1 = s (— ) (r + n + sr x) = w (r + n + sr t ) ; Hence, mechanical output of transformer :

P = Po — P 1 = w (1 — s) (r + ri);

Ratio :

mechanical ou t put P __ 1 — s _ speed

total secondary power Pi + Pi 1 ~ s ~~ slip Thus,

In a general alternating transformer of ratio of turns, a, and ratio of frequencies, $, neglecting exciting current, it is: Electrical input in primary :

p _ sftiV (r + ri + ng) .

° (r a + r) 2 + s 2 (xi + %) 2 ’

Mechanical output:

p = $ (1 - s) n x 2 e 2 ( r + rQ .

(n + r) 2 + s 2 (xi + x) 2 ’

Electrical output of secondary:

Pi =

s 2 ni 2 e 2 r

Oi + r) 2 + s 2 (xi + x) 2 ' Losses in transformer :

2 s 2 ni 2 e 2 ri

Po 1 + PT = P 1 =

(r x + r) 2 + 5 2 (a?i + x) 2

184

ELECTRICAL APPARATUS

Of these quantities, P 1 and Pi are always positive; P o and P can be positive or negative, according to the value of s. Thus the apparatus can either produce mechanical power, acting as a motor, or consume mechanical power; and it can either con- sume electrical power or produce electrical power, as a generator. 107 . At:

5 = 0, synchronism, Po = 0, P = 0, Pi = 0.

At 0 < s < 1, between synchronism and standstill.

Pi, P and P 0 are positive; that is, the apparatus consumes electrical power, P 0 , in the primary, and produces mechanical power, P, and electrical power, Pi + Pi 1 , in the secondary, which is partly, P d, consumed by the internal secondary resistance, partly, Pi, available at the secondary terminals.

In this case:

Pi + Pi 1 _ £ .

P 1 - s’

that is, of the electrical power consumed in the primary circuit, Po, a part P o 1 is consumed by the internal primary resistance, the remainder transmitted to the secondary, and divides between electrical power, Pi + Pi 1 , and mechanical power, P, in the proportion of the slip, or drop below synchronism, s, to the speed: 1—5.

In this range, the apparatus is a motor.

At 5 > 1; or backward driving, P < 0, or negative; that is, the apparatus requires mechanical power for driving.

Then:

Po -Po 1 -Pi 1 <Pi;

that is, the secondary electrical power is produced partly by the primary electrical power, partly by the mechanical power, and the apparatus acts simultaneously as transformer and as alternating^current generator, with the secondary as armature.

The ratio of mechanical input to electrical input is the ratio of speed to synchronism.

In this case, the secondary frequency is higher than the primary.

At:

5 < 0, beyond synchronism,

P < 0; that is, the apparatus has to be driven by mechanical power.

FREQUENCY CONVERTER

185

P o < 0 ; that is, the primary circuit produces electrical power from the mechanical input.

At: (

V 4 - T\

r + n + sri = 0, or, $ = — - ;

the electrical power produced in the primary becomes less than required to cover the losses of power, and P o becomes positive again.

We have thus:

8 <

r + ri

Ti

consumes mechanical and primary electric power; produces secondary electric power.

r + n n

< s < 0

consumes mechanical, and produces electrical power in primary and in secondary circuit.

0 < s < 1

consumes primary electric power, and produces mechanical and secondary electrical power

1 < s

consumes mechanical and primary electrical power; produces secondary electrical power.

108 . As an example, in Fig. 62 are plotted, with the slip, s, as abscissae, the values of:

Secondary electrical output as Curve I.; total internal loss as Curve II.;

mechanical output as Curve III.;

primary electrical output as Curve IV.;

n 16 = 100.0; r = 0.4;

r x = 0.1; x = 0.3;

Xi = 0 . 2 ;

for the values:

186

ELECTRICAL APPARATUS

hence:

Pi = Po 1 + Pi 1 = Po = p =

16.000 s 2 .

1 + s 2 ’

8000 s 2 .

1 + s 2 ’

4000 s (5 + s). 14- s 2 ’

20.000 s(l - s) 1 + s 2

Fig. 62. — Speed-power curves of general alternating- current transformer.

109 . Since the most common practical application of the general alternating-current transformer is that of frequency converter, that is, to change from one frequency to another, either with or without change of the number of phases, the following characteristic curves of this apparatus are of great interest:

1. The regulation curve; that is, the change of secondary terminal voltage as function of the load at constant impressed primary voltage.

FREQUENCY CONVERTER

187

2. The compounding curve; that is, the change of primary impressed voltage required to maintain constant secondary terminal voltage.

In this case the impressed frequency and the speed are con- stant, and consequently the secondary frequency is also constant. Generally the frequency converter is used to change from a low frequency, as 25 cycles, to a higher frequency, as 60 or 62.5 cycles, and is then driven backward, that is, against its torque, by mechanical power. Mostly a synchronous motor is em- ployed, connected to the primary mains, which by overexcitation compensates also for the lagging current of the frequency converter.

Let :