280 ELECTRIC CIRCUITS C. General Discussion of Constant-potential Constant- current Transformation 146. In the preceding methods of transformation between ‘ ' constant potential and constant current by reactances, that is, by combinations of inductive and condensive reactances, the constant alternating current is in quadrature with the constant e.m.f. Even in constant-current control by series inductive reactances, the constancy of current is most perfect for light : loads, where the reactance voltage is large and thus the constant- current voltage almost in quadrature, and the constant-current control is impaired in direct proportion to the shift of phase

• of the constant current from quadrature relation. ws ttTITIITTIITITIITII IIT | PTT TT eee TT EE TT TT dh TT aE er | (Se A ot tt arin et . tig ti tt titi tt Pett | pr Waal a. aS POC Cr rae TTS WCE arent | ee DEPT ins .9728ner acne | sc | SEEP AREER Log ttt Tet ee et tt ET Zea eee o t 2 8 4 5 6 7 8 9 © & - KILOVOLTS Fia. 121.

The cause hereof is the storage of energy required to change the character of the flow of energy. That is, the energy supplied at constant potential in the primary circuit, is stored in the react- ances, and returned at constant current, in the secondary circuit.

The storage of the total transformed energy in the reactances allows a determination of the theoretical minimum of reactive power, that is, of inductive and condensive reactances required for constant-potential to constant-current transformation, since the energy supplied in the constant-current circuit must be stored for a quarter period after being received from the constant-po-

tential circuit,

CONSTANT-CURRENT TRANSFORMATION _— 281

Let

p = P(1 + cos 2 @) = Power supplied to the constant-current circuit; thus, neglecting lasses, —_- po = P(1 — cos 2 4) = Power consumed from the constant-potential cir- cuit, and po — p = 2 P cos2 6 ' ' = Power in the reactances. ;

That is, to produce the constant-current power, P, from a single-phase constant-potential circuit, the apparent power, 2 P, must be used in reactances; or, in other words, per kilowatt con- stant-current power produced from a single-phase constant-po- tential circuit, reactances rated at 2 kv.-amp. as a minimum are required, arranged so as to be shifted 45° against the constant- potential and the constant-current circuit.

The reactances used for the constant-potential constant-cur- rent transformation may be divided between inductive and con- densive reactances in any desired proportion.

The additional wattless component of constant-potential power is obviously the difference between the wattless volt-am- peres of the inductive and that of the condensive reactances. That is, if the wattless volt-amperes of reactance is one-half of inductive and one-half of condensive, the resultant wattless volt- amperes of the main circuit is zero, and the constant-potential circuit is non-inductive, at non-inductive load, or consumes cur-

rent proportional to the loads

, If A is the condensive and B the inductive volt-amperes, the resultant wattless volt-amperes is B-A; that is, a lagging watt- less volt-amperes of B-A (or a leading volt-ampere of A-B) exist in the main circuit, in addition to the wattless volt-amperes of the secondary circuit, which reappear in the primary circuit.

147. These theoretical considerations permit the criticism of the different methods of constant-potential to constant-current transformation in regard to what may be called their apparatus economy, that is, the kilovolt-ampere rating of the reactance used, compared with the theoretical minimum rating required.

148. Series inductive reactance, that is, a reactive coil of constant inductive reactance in series with the circuit. This arrangement obviously gives only imperfect constant-current control. Per-

— J

282 ELECTRIC CIRCUITS

mitting a variation of 5 per cent. in the value of the current (that

is, full-load current in 5 per cent. less than no-load current) and , assuming 4 per cent. loss in the reactive coil, a reactance rated at

2.45 kv.-amp. is required per kilowatt constant-current load.

This apparatus operates at 87.9 per cent. economy and 30 per

cent. power-factor.

Assuming 10 per cent. variation in the value of the current, reactance rated at 2.22 kv.-amp. is required per kilowatt constant- current load. This arrangement operates at an economy of 91.8 per cent., and a power-factor of 49.5 per cent.

In the first case, the apparatus economy, that is, the ratio of the theoretical minimum kilovolt-ampere rating of the reactance to the actual rating of the reactance is 88 per cent., and in the last case 92 per cent., thus the objection to this method is not the high rating of the reactance and the economy, but the poor constant- current control, and especially the very low power-factor.

2. Inductive and condensive reactances in resonance condition, the condensive reactance being shunted by the constant-current circuit. In this case, condensive reactance rated at 1 kv.-amp. and inductive reactance rated at 2 kv.-amp. are required per kilo- watt constant-current load, and the main circuit gives a constant wattless lagging apparent power of 1 kv.-amp. Assuming again 4 per cent. loss in the inductive and 2 per cent. loss in the condens- ive reactances, gives a full-load efficiency of 91 per cent. and a power-factor (lagging) of 74 per cent. The apparatus economy by this method is 66.7 per cent.

3. Inductive and condensive reactgnces in resonance condition, the inductive reactance shunted by the constant-current circuit. In this case, as a minimum, per kilowatt constant-current load, condensive reactance rated at 2 kv.-amp. and inductive reactance rated at 1 kv.-amp. is required, and the main circuit gives a con- stant wattless leading apparent power of 1 kv.-amp. The effi- ciency of transformation is at full-load 92.5 per cent., the power- factor (leading) 73 per cent., the apparatus economy 66.7 per cent. .

4, T-connection, that is, two equal inductive reactances in se- ries to the constant-current circuit and shunted midway by an equal condensive reactance. In this case per kilowatt constant- current load, condensive reactance rated at 2 kv.-amp. and in- ductive reactance rated at 2 kv.-amp. are required.

The main circuit is non-inductive at all non-inductive loads, that is, the power-factor is 100 per cent.

CONSTANT-CURRENT TRANSFORMATION — 283 The full-load efficiency is 89.3 per cent. apparatus economy 50 ; per cent. 5. The monocyclic square. In this case a condensive reactance rated at 1 kv.-amp. and inductive reactance rated at 1 kv.-amp. are required per kilowatt constant-current load. The main cir- cuit is non-inductive at all non-inductive loads, that is, the power- . factor is 100 per cent, The full-load efficiency is 94.3 per cent.,, the apparatus economy 100 per cent. . 6. The monocyclic square in combination with a constant- ; potential polyphase system of impressed e.m.f. In this case, per kilowatt constant-current load, condensive reactance rated at 0.5 kv.-amp. and inductive reactance rated at 0.5 kv.-amp. are re- quired. The main circuits are non-inductive at all loads, that is, the power-factor is 100 per cent. The full-load efficiency is over 97 per cent. the apparatus economy 200 per cent. ; 148. In the preceding, the constant-potential to constant-cur- rent transformation with a single-phase system of constant im- pressed e.m.f., has been discussed; and shown that as a minimum in this case, to produce 1 kw. constant-current output, reactances rated at 2 kv.-amp. are required for energy storage. The con- stant current is in quadrature with the main or impressed e.m.f., but can be either leading or lagging. Thus the total range avail- able is from 1 kw. leading, to zero, to 1 kw. lagging. Hence if a constant-quadrature e.m.f. is available by the use of a poly- phase system, the range of constant current can be doubled, that is, reactance rated at 2 kv.-amp. can be made to control the po- tential for 2 kw. constant-current output in the way shown in Fig. 122 for a three-phase, and Fig. 123 for a quarter-phase system of impressed e.m.f. In this case, one transformer feeds a monocyclic square, the other transformer inserts an equal constant e.m.f. in quadrature with the former, which from no-load to half-load is subtractive, from half-load to full-load is additive, that is, at full-load, both . phases are equally loaded; at half-load only one phase is loaded and at no-load one phase transforms energy into the other phase. The monocyclic e.m.f. square in this case, when passing from full-load to no-load, gradually collapses to a straight line at half-load, then overturns and opens again to a square in the opposite direction at no-load. That is, at full-load the trans- formation is from constant potential to constant current, and at '

• |

284 ELECTRIC CIRCUITS no-load the transformation is from constant current to constant potential.

Obviously -with this arrangement the efficiency is greatly increased by the reduction of the losses to one-half, and the con- stant-current control improved.

; CO) CS) te a

i S Es i = CONSTANT OUBRENT = : SINGLE-PHASE Fig 122. .

At the same time, the sensitiveness of the arrangement for dis- tortion of the wave shape, as will be discussed later, is greatly reduced, due to the insertion of a constant-potential e.m.f. into the constant-current circuit.

Obviously the arrangments in Figs. 122 and 123 are not the

: only ones, but many arrangements of inserting a constant-quad-

rature e.m.f. into the monocyclic square or triangle are suitable.

Ino, ino,

oe

| ' oor, |

E &

CONSTANT CORRENT

. GINQLE-PHASE

| Fra. 123.

Different arrangements can also be used of the constant-current

. control, for instance, the inductive and condensive reactances in

resonance condition with their common connection connected

‘ to the center of an autotransformer or transformer, with the

: insertion of the constant-potential quadrature e.m.f. in the latter circuit as shown in Fig. 124, or the T-connection, shown applied to a quarter-phase system in Fig. 125.

i 1 - | | , | / ' CONSTANT-CURRENT TRANSFORMATION — 285 | Constant-potential apparatus and constant-current single- phase circuits can also be operated from the same transformer secondaries in a similar manner, as indicated in Fig. 124 for a three-phase secondary system. In Figs. 122 to 125 the arrangement has been shown as applied to step-down transformers, but in the estimate of the efficiency 2 S, oS oe t ; il SE] Sh zf § a |

= CONSTANT CURRENT - eo SINQLE*PHASE Fig. 124. the losses in these transformers have not been included, since these transformers are obviously not essential but merely for the convenience of separating electrically the constant-current cir- cuit from the high-potential line. It is evident, for instance, in Fig. 124, that the constant-current and constant-potential cir- = Y pr, : S RS He eB é EE = CONSTANT CURRENT ad SINGLE- PHASE Fig. 125. cuits instead of being operated from the three-phase secondaries of the step-down transformers can be operated directly from the three-phase primaries by replacing the central connection of the one transformer by the central connection of the auto- | transformer. |

286 ELECTRIC CIRCUITS D. Problems —_ 149. In the following problems referring to constant-potential to constant-current transformation by reactances; it is recom- mended:

(a) To derive the equation of all the currents and e.m.fs., in complex quantities as well as in absolute terms, while neglecting . the loss of power in the reactances.

(b) To determine the volt-amperes in the different parts of

the circuit, as load, reactances, etc., and therefrom derive the apparatus economy, to find its maximum value, and on which

| condition it depends.

(c) To determine the effect of inductive load on the power of the primary supply circuit, to investigate the phase angle of the primary supply circuit, and the conditions under which it becomes a minimum, or the primary supply becomes non- inductive.

(d) To redetermine the equations of the problem, while con- sidering the power lost in the reactances, and apply these equa- tions to a numerical example, plotting all the interesting values.

7 (e) To investigate the effect of a change of frequency on the equations, more particularly on the constant-current regulation.

(f) To investigate the effect of distortion of wave shape,

: that is, the existence of higher harmonics in the impressed

; e.m.f., and their suppression or reappearance in the secondary

circuit.

| (g) To study the reversibility of the problem, that is, apply

| (a) to (f) to the reversed problem of transformation from constant

current to constant potential.

| Some of the transforming devices between constant potential

and constant current are:

A. Single-phase.

| (a) The resonating circuit, or condensive and inductive reactances, of equal values, in series with each other in the con- stant-potential circuit, and the one reactance shunted by the constant-current circuit. ;

‘ (b) T-connection, as partially discussed in (A).

(c) The monocyelic square, as partially discussed in (B).

(d) The monocyelic triangle: a condensive reactance and an inductive reactance of equal values, in series with each other

across the constant-potential circuit, the constant-current

|

CONSTANT-CURRENT TRANSFORMATION — 287 circuit connecting between the reactance neutral, or the common connection between the two (opposite) reactances, and the neutral of a compensator or autotransformer connected across

_ the constant-potential circuit. Instead of the compensator neutral, the constant-current circuit can be carried back to the neutral of the transformer connected to the constant-potential circuit.

B. Polyphase. _ (a) In the two-phase system the two phases of e.m.fs., eo and jéo, are connected in series with each other, giving the outside terminals, A and B, and the neutral or common con- . nection, C. A condensive reactance and an inductive reactance of equal values, in series with each other and with their neutral or common connection, D, are connected either between A and — B, and the constant-current circuit between C and D, or the reactances are connected between A and C, and the constant- | current circuit between B and D. In either case, several ar- . rangements are possible, of which only a few have a good appara- tus economy. (b) In a three-phase system, a condensive reactance, an induct- ive reactance equal in value to that of the condensive reactance ; and the constant-current circuit, are connected in star connec- tion between the three-phase, constant-potential terminals. Here also two arrangements are possible, of which one only gives good apparatus economy. (c) In a constant-potential three-phase system, each of the | three terminals, A, B, C, connects with a condensive and an | inductive reactance, and all these reactances are of equal value, | and joined together in pairs to three terminals, a, b, c, 90 that each of these terminals, a, b, c, connects an inductive with a condensive reactance. a, b, c, then, are constant-current three- ‘ phase terminals, that is, the three currents at a, b, c,-are constant and independent of the load or the distribution of load, and displaced from each by one-third of a period. This arrange- : ment is especially suitable for rectification of the constant al- ternating-current, to produce constant direct current. 150. Some further problems are: | 1. In a single-phase, constant-current transforming device, as the monocyclic square, the constant current, 7, is in quadrature with the constant impressed e.m.f., é. By inserting a constant- potential e.m.f., #3, into the constant-current circuit, the appa- 4

| | 288 ELECTRIC CIRCUITS ratus economy can be greatly increased, in the maximum can | be doubled; that is, the e.m.f., #; gives constant-power output, and from no-load to half-load, the transformation is from con- stant current to constant potential, that is, a part of the power supply, 3, is transformed into the circuit, of e.m-f., é, that is, the circuit, ¢, receives power. Above half-load the circuit . of é transforms power from constant potential to constant current, into the circuit of e.m.f. Es.

Since 7 is in time quadrature with ¢, with non-inductive secondary load, that is, the secondary terminal voltage, £, in phase with the secondary current, 7, £; should also be in phase with ¢, that is, f; = jes. With inductive secondary load, of phase angle, 6, Z'; should be in phase with £, that is, leading 7 by angle 6, or should be: Hs = jes (1 + ky). :

• It is interesting, therefore, to investigate how the equation of

the constant-potential constant-current devices are changed by the introduction of such an e.m.f., Hs, at non-inductive as well as at inductive load, if #; = jes, or Zs = j(es — je’s), in either case, and also to determine how such an e.m.f., Es, of the proper phase relation, can be derived directly or by trans- formation from a two-phase or three-phase system.

2. If in the constant-potential constant-current transform- ing device one of the reactances is gradually changed, increased or decreased from its proper value, then in either case the regula- tion of the. system is impaired. That is, the ratio of full-load current to no-load current falls off, but at the same time, the no-load current also changes.

With increase of load, the frequency of the system decreases, due to the decreasing speed of the prime mover, if the output of the system is an appreciable part of the rated output. If, therefore, the reactances are adjusted for equality of the frequency of full-load, at the higher frequency of no-load, the inductive reactance is increased, and thereby the no-load current decreased below the value which it would have at constant reactance, and in this manner the increase of current from full-load to no-load is reduced. ;

Such a drop of speed and therefore of frequency, s, can there- fore be found, that the current at full-load, with perfect equality between the reactances, equals the current at no-load, where the reactances are not quite equal. That is, the variation of frequency compensates for the incomplete regulation of the

; ; CONSTANT-CURRENT TRANSFORMATION — 289 : current, caused by the energy loss in the reactances. Further- more, with a given variation of frequency, s, from no-load to full- . load, the reactances can be chosen so as to be slightly unequal at full-load, and more unequal at no-load; the change of current caused hereby compensates for the incomplete current regu- lation, that is, with a given frequency variation, s (within , certain limits), the current regulation can be made perfect from ; no-load to full-load, by the proper degree of inequality of the ; reactances. It is interesting to investigate this, and apply to an example, a, to determine the proper s, for perfect equality of reactance at , full-load; 6, with a given value of s = 0.04, to determine the in- equality of reactance required. Assuming a = 0.03; 6b = 0.01. : 8. If one point of the constant-current circuit, ‘either a terminal or an intermediate point, connects to a point of the constant-potential circuit, either a terminal or some intermediate | point (as inside of a transformer winding), the constant current . is not changed hereby, that is, the regulation of the system is | not impaired, and no current exists in the cross between the two : circuits. The distribution of potential between the reactances, however, may be considerably changed, some reactances re- ceiving a higher, others a lower voltage. It follows herefrom, that a ground on a constant-current system does not act as a ground on the constant-potential system, , but electrically the two systems, although connected with. each other, are essentially independent, just as if separated from each other by a transformer. So, for instance, in the monocyclic square, one side may be short-circuited without change of current in the secondary, but with an increase of current in the other three sides. It is interesting to investigate how far this independence of the circuits extends. . In general, as an example, the following constants may be chosen: In the constant-potential circuit: eo = 6600 volts and t’o = 10 amp. at full-load. In the constant-current circuit: 1 = 7.5 amp., e’ = 7500 volts at full-load. - Or, especially in polyphase systems, e’, respectively, 1’o corresponding to the maximum economy point, I and a = 0.03; b = 0.01. an

' . . 290 ELECTRIC CIRCUITS E. Distortion of Voltage Wave

151. It is of interest to investigate what effect the distortion of the voltage wave, that is, the existence of higher harmonics in the wave of supply voltage, has on the regulation of the con- stant-potential constant-current transformation systems dis-

’ eussed in the preceding. a

Where constant current is produced by inductive reactance

only, higher harmonics in the voltage wave naturally are sup- . pressed the more, the larger the inductive reactance and the higher the order of the harmonic.

An increase of the intensity of the harmonics in the current wave, over that in the voltage wave, and with it an impairment of the constant-current regulation, could thus be expected only with devices using capacity reactance.

As example may be investigated the effect of the distortion of the impressed voltage wave on the 7 connection, and on the monocyclic square.

The symbolic method of treating general alternating waves may be used, as discussed in Chapter X XVII, of ‘Theory and Calculation ‘of Alternating-current Phenomena,” fifth edition, page 379. That is, the voltage wave is represented by

co B= Dp 3 (6's — jnt’n) 1 and the impedance by : . Le Z= r+ jn (Mtn + 29 + =) where nm = order of harmonic. A. T Connection or Resonating Circuit 152. Assuming the same denotation as before, we have, for the nth harmonic:

primary inductive reactance, , Zo = + jnao;

secondary inductive reactance, Z, = + jnz;

condensive reactance, Zz = io.

n

I | CONSTANT-CURRENT TRANSFORMATION — 291 | when neglecting the energy losses in the reactances, load , Z = r(1+jnk) therefore, also for the nth harmonic. G = r(1+jnk) fh | \ B,=E+2Zi 4 | = [r (1+jnk)+ jnz,] I, | and also = —7 77. | E = Jj n T yy hence, . 1 ink j , h = juni tin ) + nui] . 0 . : and . fo=Ith _ to — jn’a, — nr(1 + jnk) i ‘ jxo . , . hence, BE, = Ei + Zolo = {[r (1 + jnk) + jnay] + n[jxo—jn?a. — nr (1 + jnkyi}4 = { — (n? -1)r (1 + jnk) — jnay (n? — 1) + jnzol]; hence, I = 7 jE 0 nto — nxy(n? — 1) + fj (n? — 1) 7r (1 + jnk)

• ao NXy — (n* — 1)[n (a1 + kr) + jr)’ ‘hence, approximately, for higher values of n, _ Eo P=+ n(x, + kr)’ . that is, for larger values of n, { = 0, or the higher harmonics in the current wave disappear. Herefrom, by substituting in the preceding equations, the supply current, fo, the condenser current, /:, their respective e.m.fs., etc., are derived. It is then, in general expression: If 0 Eo = den — jnén') = impressed e.m.f., 1

292 ELECTRIC CIRCUITS T= £1 nay — (n® = 1) [nei + br) + irl =. j(eo — jeo') _ Sinlen — jen’) Xo a (a1 + kr)’ the equation of the secondary current. For instance, let . . Eo = 6600 {1, _ 0.203 - 0.15; + 0.067 _- 0.25 ja} = constant-impressed e.m.f. or, absolute, eo = 6600 Ni + 0.20? + 0.15? + 0.06? + 0.25? = 6600 X 1.062 = 7010 volts, ; and choosing the same values as before, in paragraph 143, . Zo = 880 ohms, az, = 508 ohms, r’ = 930 ohms, k = 0.4; it is, substituting, __ . , 60.03 — 48.8 js — 8.0 js +1.2j7 P= 755+ 508 + 0.4r , or, absolute, . 60.0? + 48.8? + 8.0? + 1.2? = 24 eer Tew Tees Pa yet (608 + 0.41)? | 604,600 = 24 _ 0V4,0W Vio + Gost 04n® hence, at no-load, ' . t= 7.5 X 1.00021 and, at full-load, r = 930, t = 7.5 X 1.00003. . That is, the current wave is as perfect a sine wave as possible, regardless of the distortion of the impressed e.m.f., which, for instance, in the above example, contains a third harmonic of 32 per cent. Or in other words, in the T connection or the resonat- ing circuit, all harmonics of e.m.f. are wiped out in the current wave, and this method indeed offers the best and most conven- ient means of producing perfect sine waves of current from any : shape of e.m.f. waves.

| CONSTANT-CURRENT TRANSFORMATION — 293 | 153. a: Monocyclic Square Assuming the same denotation as before, we have for the nth harmonic: . inductive reactance, ; Zz = + jnto; condensive reactance, Z4=—-j =; load, ‘ ‘ | Z =r(1 + jnk); currents, _fo-f hi ™~ 2 , and I afoot, qs = >) ? e.m.fs., E = Zui + Zo], ZIo= Zid — Za); , ' hence, substituting, we have Bo = — jro(!* — ahs), r(1 + jnk)T = — joo(/ + nfs); thus, = — If, (1 4_ (fh Bo = ~ Stel, - 0) - 15 +9) } nk)t = ~220{7,(h 7(L,4)}- nL + ink) = ~ 3° {To(;- +2) — 1 - 9) }; then, combining, we obtain 1 ; 1) _ , jo ly? (1)? Bo(n +5) +r + Jnky (n— 7) = +757 { (n+ 7)" (n—2) | = + 2 jxol, . and . 1 I= — jo ( + -
| a . . 1 | 2 zo + jr(1 + jnk)(n — =) _ SB t 1) 2 nto + jr(n? — 1)(1 + jnk) and herefrom J, {1, [s, etc.

294 ELECTRIC CIRCUITS ;

Approximately, for higher values of n, and for high loads, r,

_ ike p= ©.

That is, the higher harmonics of current decrease proportion- ally to their order, at heavy loads—that is, large values of r. For light loads, however, or small values of r, and in the extreme case, at no-load, orr = 0, it is

p= — a(n? +0), 2 NLo and, approximately, j= — Een, 2 Zo

That is, the current is increased proportional to the order of the harmonics, or in other words, at no-load, in the monocyclic square, the higher harmonics of impressed e.m.fs. produce increased values of the higher harmonics of current, that is, the wave-shape distortion is increased the more, the higher the harmonics.

mo, In general expression: If ;

Mio c}

Ey = Dlen — jné’n) = impressed e.m.f., 1

l= > jn(n? + 1)(@n — jne’n) ;

72 nxo + jar(n? — 1)(1 + jank) and herefrom [o, 1, [2, etc. For instance, let ‘ Eo = 6600 {11 — 0.203 — 0.25 js — 0.155 + 0.067} = constant-impressed e.m.f., or, absolute, éo = 7010 volts, and, choosing the same values as before, , Zo = 880 ohms, r’ = 930 ohms, k = 0.4; ; it is, substituted, [= 75 — _.25= 24s) 6600 25,740 . 5280 — (9.6 —8j)r 8800 — (48 — 24j5)r

• 19,800 . 12,320 — (134.4 — 48 j)r’

| ne | CONSTANT-CURRENT TRANSFORMATION — 295 herefrom follows, at no-load, r = 0, : { = 7.5 — (3.12 — 2.5 js) — 2.92 + 1.612. That is, at no-load, the secondary current contains excessive higher harmonies, for instance, a third harmonic, ‘ V3.12? + 2.5? = 4.0, or 53.3 per cent. of the fundamental. Absolute, the no-load current is . t= V7.5? + 3.12? + 2.5? + 2.92? + 1.612 = 9.13 amp. At full-load, or r = 930, it is . | I = 7.5 + (2.18 + 1.07 js) + (0.51 + 0.32 7.) — (0.14 + 0.06 jz); that is, at full-load, the harmonics, while still intensified, are less than at no-load, and decrease with their order, n, more rapidly. | The absolute value is t= V7.5? + 2.18% + 1.072 + 0.512 + 0.32?+ 0.142 + 0.062 , = 7.91 amp. Instead of 7.5 amp., the value which the current would have . at all loads if no higher harmonics were present, the higher har- monics of impressed e.m.f. raise the current to 9.13 amp., or by 21.7 per cent. at no-load, and to 7.91 amp., or by 5.5 per cent. at full-load, while the impressed e.m.f. is increased by 6.2 per cent. by its higher harmonics. It follows also that the constant-current regulation of the sys- tem is seriously impaired, and between no-load and full-load the current decreases from 9.13 to 7.91 amp., or by 15.4 per cent., which as a rule is too much for an arc circuit. | 154. It follows herefrom: | While the T connection of transformation from constant poten- tial to constant current suppresses the higher harmonics of im- | pressed e.m.f. and makes the constant current a perfect sine wave, the monocyclic square intensifies the higher harmonics so that the higher harmonics of impressed e.m.f. appear at greatly increased intensity in the constant-current wave. The increase of the higher harmonics is different for the different harmonics and for different loads, and the distortion of wave shape produced hereby is far greater at no-load, and the constant-current regulation of the system is thereby greatly impaired, and at load the dis-

| 296 . ELECTRIC CIRCUITS tortion is.less, and very high harmonics are fairly well sup- . pressed, and the operation of an arc circuit so feasible.

Assuming, then, that in the monocyclic square of constant- potential constant-current transformation, with a distorted wave . of impressed e.m.f., we insert in series to the monocyclic square into the main circuit, Jo, two reactances of opposite sign, which "gre equal to each other for the fundamental frequency, that is, a condensive reactance, Z3 = — , and an inductive

. reactance, Z, = + jnzs. Then for the fundamental, these two reactances together offer-no resultant impedance, but neutralize

each other, and the only drop of voltage produced by them is that due to the small loss of power in them. At the nth harmonic, however, the resultant reactance is . Zit y= +in(n +),

or, approximately, = + jrsn,

, and two such impedances so obstruct the higher harmonics, the more, the higher their order while passing the fundamental sine wave.

Such a pair of equal reactances of opposite sign so can be called . a “‘wave screen.” . Further problems for investigation by the student then are:

1. The investigation of the effect of the distortion of the wave . of impressed e.m.f. on the constant current, with other trans- forming devices, and also the reverse problem, the investigation of the effect of the distortion of the constant-current wave, as ’ caused by an arc, on the system of transformation.
2. What must be the value, 21, of the reactance of a wave screen, to reduce the wave-shape distortion of the secondary current in the monocyclic square to the same percentage as the : distortion of the impressed e.m.f. wave, or to any desired per- centage, or to reduce the variation of the constant current with the load, as due to the wave-shape distortion, below a given percentage? ; 3. Determination of efficiency and regulation in the mono- cyclic square with interposed wave screen, 71, assuming again 3 per cent. loss in the inductances, 1 per cent. loss in the capacities and choosing 2, so as to fill given conditions, regarding wave- shape distortion, or regulation, or efficiency, etc. .

| ; CHAPTER XV | CONSTANT-VOLTAGE SERIES OPERATION 155. Where a considerable number of devices, distributed over a large area, and each consuming 4 small amount of power, are to be operated in the same circuit, low-voltage supply—110 or 220 volts—usually is not feasible, due to the distances, and high- voltage distribution—2300 volts—with individual step-down transformers at the consuming devices, usually is uneconomical, due to the small power consumption of each device. In such a case, series connection of the devices is the most eco- nomical arrangement,’ and therefore commonly used. | Such for instance is the case in lighting the streets of a city, etc. Most of the street lighting has been done by arc lamps operated | on constant-current circuits, and as the universal electric power supply today is at constant voltage, transformation from constant voltage to constant current thus is of importance, and has been . discussed in Chapter XIV. . The constant-current system thus is used in this case: (a) Because by series connection of the consuming devices, as the are lamps in street lighting, it permits the use of a sufficiently high voltage to make the distribution economical. (b) The dropping volt-ampere characteristic of the arc makes it unstable on constant voltage, as further discussed in Chapters II and X, and a constant-current circuit thus is used to secure sta- bility of operation of series arc circuits. The condition (6), the use of constant current, thus applies only where the consuming devices are arcs, and ceases to be pertinent | ’ when the consuming devices are incandescent lamps or other con- } stant-voltage devices. The modern incandescent lamp, however’, is primarily a con- stant-voltage device, that is, at constant-voltage supply, the life of the lamp is greater than at constant-current supply, assuming the same percentage fluctuation from constancy. The reason is: & variation of voltage at the lamp terminals, by p per cent., gives 8 variation of current of about 0.6p per cent., and thus a variation 297

• ~

, | 298 ELECTRIC CIRCUITS of power of about 1.6p per cent., while a variation of current in the lamp, by p per cent., gives a variation of voltage of about nt per cent., and thus a variation of power of about (1 + a)? = 2.67p per cent.

Thus, with the increasing use of incandescent lamps for street illumination, series operation in a constant-voltage circuit be- comes of increasing importance.

If e = rated voltage, « = rated current of lamp or other con- suming device, and e9 = supply voltage, n = “lamps can be op- erated in series on the constant-voltage supply eo. If now one lamp goes out by the filament breaking, all the lamps of the series

circuit would go out, if eo is small; if ¢9 is large, an arc will hold in the lamp or the fixture, and more or less destroy the circuit.

Thus in series connection, especially at higher supply voltage, €0, some shunt protective device is necessary to maintain circuit in case of one of the consuming devices open-circuiting

On constant-current supply, a short-circuiting device, such as a

film cutout, takes care of this. With series connection on con- stant-voltage supply, it is not permissible, however, to short- circuit a disabled consuming device, as this would increase the voltage on the other devices. Thus the shunt protective device in the constant-voltage series system must be such, that in case of one lamp barning out, the shunt consumes such a voltage as to ; maintain the voltage on the other devices the same as before. A film cutout, with another lamp in series, would accomplish this: if a lamp burns out, its shunting film cutout punctures and puts ; the second lamp in circuit. However, in general such arrange- ment is too complicated for use. As practically all such circuits would be alternating-current circuits, and thus alternating currents only need to be considered, . the question arises, whether a reactance shunting each lamp would not give the desired effect. Suppose each lamp, of resist- : ance, r, is shunted by a reactance, x, which is sufficiently large not | to withdraw too much current from the lamp: assuming the cur- rent shunted by z is 20 per cent. of the current in the lamp, or = 5r. With 6.6 amp. in 7, x thus would take 1.32 amp., and the | total, or line current would be: 7 = ~/6.6? + 1.32? = 6.73 amp., | thus only 2 per cent. more than the lamp current. If now a lamp ! | | |

CONSTANT-VOLTAGE SERIES OPERATION 299 burns out, the total current flows through z, instead of 20 per cent. only, and the voltage consumed by = is increased fivefold—assum- ing 7 as constant—this voltage, however, is in quadrature with the current, thus combines vectorially with the voltages of the other ‘ consuming devices, which are practically in phase with the cur- rent, and the question then arises, whether, and under what con- ditions such a reactance shunt would maintain constant voltage on the other consuming devices, or, what amounts to the same, constant current in the series circuit. .

Such a reactance shunting the consuming device could at the same time be used as autotransformer (compensator), to change the current, so that consuming devices of different current re-

: quirements, as lamps of various sizes, could be operated in series on the same circuit, from constant-voltage supply.

166. Let n lamps of voltage, e,, and current, 7,, thus conductance

au 9-3 (1) be connected in series into a circuit of supply voltage, eo = ne (2) and each lamp be shunted by a reactance of susceptance, b.

In each consuming device, comprising lamp and reactance, the

admittance thus is, vectorially, “Yi: =g — jb (3) l if, then, { = current in the series circuit, the voltage consumed by the device comprising lamp and reactance, thus is t I = = 4 in a consuming device, however, in which the lamp is burned out, and only the reactance remains, the admittance is Y: = — jb (5) hence, the voltage, with the entire current, J, passing through the admittance, Ys, . I I = =j-- 6

If, then, of the n series lamps, the fraction, p, is burned out,

leaving n(1 — p) operative lamps, it is:

ns : a | 300 ELECTRIC CIRCUITS ‘ voltage consumed by operative devices: n(l — p)t 1- = voltage consumed by devices with burned-out lamps: . . npE, = inp! thus, total circuit voltage: , . eo = n(l — p) Bi + npEs (7) =nf(iirP4 BP =e] or, K nl(b + jpg ‘ = De 8 eo = b@= 3b) (8) or, absolute, ‘ . = nt Pg , eo m fi + (2) (9) where y =<? + b? = admittance of operative device, absolute, (10) hence, i= —_o _ Pg? (11) maj + (7)

• is the current in the circuit, and the current in the lamps thus is . . _@g: == 12 a y . (12) hence, a - C09 1 = [apg ? : (13) najt + (7) for , p = 0, or all devices operative (“full-load,’’ as we may say), it is ty = = for p = 1, or all lamps out (“‘no-load”’), it is

| CONSTANT-VOLTAGE SERIES OPERATION _ 301 i = — na fi + (2)

(14)

“_ eogb

ony

thus smaller than at full-load. _ As seen from equation (13), the current steadily decreases, from p = Oor full-load, to p = 1 or no-load, and no value of shunted reactance, b, exists, which maintains constant current. With de- creasing load, the current, ¢;, decreases the slower, the higher b is, that is, the more current is shunted by the reactive susceptance, b, and the poorer therefore the power-factor is. : Thus shunted constant reactance can not give constant-voltage regulation.

However, with b = 0.2 g, at no-load the shunted reactance would get five times as much current as at load, and thus have five times as high a voltage at its terminals.

The latter, however, is not feasible, except by making the reactance abnormally large and therefore uneconomical.

In general, long before five times normal voltage is reached, magnetic saturation will have occurred, and the reactance thereby decreased, that is, the susceptance, b, increased, as more fully dis- cussed in Chapter VIII.

This actual condition would correspond to a value, b,, of the shunted susceptance when shunted by the lamp, and a different, higher value, b3, of the shunted susceptance when the lamp is burned out.

The question then arises, whether such values of b; and bz can be found, as to give voltage regulation. The increase of b2 over by

naturally depends on the degree of magnetic saturation in the re- ; actance, that is, on the value of magnetic density chosen, and thus can be made anything, depending on the design.

157. Let then, as heretofore,

; Eo = €) = constant-supply voltage. { = current in series circuit. nm = number of consuming devices (lamps) in series. (15) p = fraction of burned-out lamps. g = conductance of lamp.