CHAPTER XV CONSTANT-VOLTAGE SERIES OPERATION 166. Where a considerable number of devices, distributed over a large area, and each consuming a 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 conmionly 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 imiversal 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: (o) Because by series connection of the consuming devices, as the arc lamps in street lighting, it permits the use of a suflBciently high voltage to make the distribution economical. (6) 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 6i constant current, thus applies only where the consuming devices are arcs, and ceases to be pertinent when the consiuning devices are incandescent lamps or other con- stant-voltage devices. The modem 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: a variation of voltage at the lamp terminals, by p per cent., gives a variation of current of about 0.6p per cent., and thus a variation 297 298 ELECTRIC CIRCUITS of power of about l.Qp per cent., while a variation of current in the P lamp, by p per cent., gives a variation of voltage of about jr-^ per cent., and thus a variation of power of about (1 + 7r^)p = 2.67 p 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, i = rated current of lamp or other con- suming device, and eo = supply voltage, n = — lamps can be op- erated in series on the constant-voltage supply eo. If now one lamp goes out by the iSlament breaking, all the lamps of the series circuit would go out, if eo is small; if eo 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, eo, 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 burning 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 x is 20 per cent, of the current in the lamp, or x = 5 r. With 6.6 amp. in r, x thus would take 1.32 amp., and the total, or line current would be: i = V6.6^ + 1.322 = 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 re, instead of 20 per cent, only, and the voltage consumed by x is increased fivefold — assum- ing X 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. 156. Let n lamps of voltage, ei, and current, ii, thus conductance ff = j^ (1) ei be connected in series into a circuit of supply voltage, eo = nei (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) if, then, / = current in the series circuit, the voltage consumed by the device comprising lamp and reactance, thus is in a consuming device, however, in which the lamp is burned out, and only the reactance remains, the admittance is Y2= - jb (5) hence, the voltage, with the entire current, /, passing through the admittance, ¥2^ ^. = f^ = j {■ (6) If, then, of the n series lamps, the fraction, p, is burned out, leaving n(l — p) operative lamps, it is: 300 ELECTRIC CIRCUITS voltage consumed by operative devices: n(l - p)Pi = -JZTjT voltage consumed by devices with burned-out lamps: jnpt npE2 = b thus, total circuit voltage: eo = n(l - p) ^1 + np^2 (7) 1 - P , JP] ="'(^?+ or, or, absolute, g — jb 6 J '• ' big - jb) ^^^ where 2/ = \/9^ + 6^ = admittance of operative device, absolute, (10) hence, i^ ^oy is the current in the circuit, and the current in the lamps thus is hence, 0, or U = Jf (12) . _ eofif for p = is all devices operative (*' full-load," as we may say), it • _ e,g n for p = 1, or all lamps out C'no-load"), it is CONSTANT-VOLTAGE SERIES OPERATION 301 (14) ^ cogb ny thus smaller than at full-load. As seen from equation (13), the current steadily decreases, from p = or full-load, to p = 1 or no-load, and no value of shunted reactance, 6, exists, which maintains constant current. With de- creasing load, the current, f i, decreases the slower, the higher 6 is, that is, the more current is shunted by the reactive susceptance, 6, and the poorer therefore the power-factor is. Thus shunted constant reactance can not give constant-voltage regulation. However, with 6 = 0.2 gf, 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, 6, increased, as more fully dis- cussed in Chapter VIII. This actual condition would correspond to a value, 6i, of the shunted susceptance when shunted by the lamp^ and a different, higher value, 62, of the shunted susceptance when the lamp is burned out. The question then arises, whether such values of 61 and 62 can be found, as to give voltage regulation. The increase of 62 over 61 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. 167. Let then, as heretofore. ^0 — 60 = constant-supply voltage. / = current in series circuit. n = number of consuming devices (lamps) in series. p = fraction of burned-out lamps. g a= conductance of lamp. (15) 302 ELECTRIC CIRCUITS and let 6 1 = shunted susceptance with the lamp in circuit, that is, exciting susceptance of reactor or auto- transformer, and y = \/g^ + bi^ = admittance of complete consuming device. 62 = shunted susceptance with the lamp burned out and let c = — = exciting current as fraction of load ^ current: c < 1. a = ^- = saturation factor of reactor or - auto transformer: o > 1. (16) (17) it is, then: voltage of lamp and reactor: voltage of reactor with lamp burned out: I .7 ?>2 (18) ^2 = — nr = J — jbo thus, with pn lamps burned out, and (1 — p)n lamps burning, total voltage, eo = n(l — p)^i + np ^2 (19) it is (20) = n/ substituting (17), Co = nl g g — jhi "^ 62 J 1 - p , . 11 — jc (21) or. hence, absolute, nl 1 — p(l — oc) + jap g I - JC nt 60 = - V[l-p(l -ac)Y + a^p^ (22) smce. y = gy/l + c^ thus, the current in the series circuit. I = e^y n VU - p(l- ac)Y + d^^ (24) CONSTANT-VOLTAGE SERIES OPERATION 303 158. For, p = 0, or full-load, it is thus. io = ^ (25) n i = ^^ (26) V [1 - p(l - ac)]« + oV The same value of io as at full-load is reached again for the value p = po> where the square root in (24) becomes one, that is, [l-po(l-ac)P+o«po« = 1, hence, 2(1 - ac) , _, P" = a^ + (1 - ac)* ^^^ for, p = 1, or no-load, it is i 7^° (28) The current is a maximum, i = t„, for the value of p = Pm, given by dp ' or, from (26), ^{[l-p(l-ac)P+aV} =0, this gives ^'^ " a2 + (1 - ac)2 (^^ 2' as was to be expected. Substituting (29) into (26), gives as the value of maximum cur- rent z. = io^l + {^^^y (30) and the regulation g, that is, the excess of maximum current over full-load current, as fraction of the latter, thus is • • 5 = V to 304 ELECTRIC CIRCUITS If q is small (31) resolved by the binomial, ^ves As seen, with the shunted susceptance increased by saturation at open circuit, the current and thus lamp voltage are approxi- mately constant over a range of p. That is, with decreasing load, from full-load p = 0, the current i, and proportional thereto the lamp voltage increasea from tg to a maximum value i„, at p ~'n' then decreases again, to io at p = po, and decreases further, to ii at no-load, p = 1. Thus, there exists a regulating range from p= to p a little above po, where the current is approximately constant. Instance : Saturation: a — Excitation: c = Regulation: « - Range: p, - .5 3.1 3.U7 J. 573 1.5 1.5 2.0 0.2 0,3 0.1 0.1030.0670.077 0.510 0.432'0, 345 2.0 0.2 0.044 0.275 2.0 0.3 0.O2O 0.192 2.5 2.5 0.1 0.2 0.0440.020 0.2200.154 2.5 0.3 0.005 0.079 - ■^ *= c=Z :a. ~~- ~~ — . ■^ VI IV <::; ^ ^ ■^ -^ 1 ^ ^~- -^ III "-- ?f= :::; £ - ■~~- ■ t 10 D 1 n Ab illustrations £ Fia. 126. e shown, in Fig. 126, the regulation curres, - from equation (26), for: o - 1.5 - 2.5 - 2.0 -0.1 - 0.3 II III CONSTANT-VOLTAGE SERIES OPERATION 305 169. By the preceding equations, it is possible now to calculate the values of exciting susceptance 61, and saturation 62, required by the shunting reactors to give desired values of regulation with- in a given range. From (32) follows: c = J - V2~q (33) Substituting (33) into (27) gives : 2V2q a = c = Pod + 2q) po(l +2q) -Aq (34) 2V~q From chosen values of q and po, cl and c thus can be calculated, from a and c and the conductance g of the consuming device, 61, 62, i, etc., follow. Instance: n = 100 lamps of f 1 = 6 amp. and ei = 50 volts, are to be oper- ated in series on constant-voltage supply, with negligible line re- sistance and reactance. The regulation shall be within 4 per cent, in a range of 30 per cent. That is, q = 0.04 and po = 0.30. It thus is: 2l = 6 ei = 50 , = 1 = 0.12 n = 100 q = 0.04 po = 0.30 a = 1.75 c = 0.287 61 = 0.0345 62 = 0.0685 From (34) follows: Hence by (17) : and by (16) : by (2): y = 0.1248 eo = 5000 volts 20 ELECTRIC CIRCUITS and by (25) : to = 6.24 amp. thus, by (26) ■ (35) VI - P + 3.31 p' Fig. 127 showa, as curve I, the values ot q = 1, in per cent., that is, the regulation, with p as abscissce. 4 I 1 — ■~~- .^ ^ =: = "^ ■*J ^ - -I ^ ^ ~N N- \ S N \ 1 1 1 i a ». 1 1 "« Fia. 127. 160. In general, the resistance and reactance of the circuit oi line is not negligible, as assumed in the preceding, and the re- actors, especially if used at the same time as autotransformers, contain a leakage reactance, which acts as a series reactance in the circuit, and the lamp circuit of conductance g also may contain a small series reactance. Let then : I'd = line resistance; Xd = line reactance; X = series or leakage reactance per autotransformer or consumption device. The moat convenient way is to represent ro, Xo and x by their equivalent in lamps or reactors. The admittance of each eon- sumption device, comprising lamp and reactor or autotrans- former, is CONSTANT-VOLTAGE SERIES OPERATION 307 Yi = g - jbi = g(l - jc), thus the impedance, 1 1 1 + ic Zi = Yi gd - jc) (7(1 + cr and by (23), _ 1+jc If, then, we add to the resistance tq a part ctq of the reactance, we get an impedance, Z = ro(l+jc), which has the same phase angle as Zi, and thus can be expressed as a multiple of Zi, Z = UlZly where ni = |-^ = royVT+T' (36) thus is the "lamp equivalent" of the line resistance ro plus the part CTo of the reactance. This leaves the reactance, a^i = Xo + n(l — p)x — CTo, and as the reactance of a reactor without lamp is 1 the reactance Xi can be expressed as multiple of Xj, Xi = 712^2 __ !^ where n2 = a;i62 = &2 [iCo + n(l — p)x — cro] (37) thus is the *'lamp equivalent" of the line reactance xo and leakage reactances Xi in burned-out lamps. Thus the addition of the line impedance r© + jxoj and the leakage reactances x, is represented by ni lamps with reactors, and 112 burned-out lamps, or a total of rii + n2 lamps. Thus the circuit can carry rii — {n\ -I- n^ lamps, and its regula- tion curve starts at the point v = — and ends at p = 1 of the complete regulation curve. 308 ELECTRIC CIRCUITS However, in this case, the full-load current, for p = --, would already be slightly higher than in a circuit without line impedance, and all the current values would thus have to be proportionally reduced. Instance: In the case a = 2.0 c = 0.3 given as curve III of Fig. 126 let: n = 100 g = 0.12 ro = 60 xo = 0.5 hence, 6i = q; = 0.036 y = y/g2 + 6j2 =: 0.125 6j = 2 = 0.06, a thus. ni = uy Vl + c^ — 6.54 nj = 62 [xq + n (1— p) X — cfo] = 7.0 ni + n2 = 13.54. Thus, the regulation curve starts at p = — = 0.07 of curve III, Fig. 126, and ends at p = 1 = 0.935 of this curve. For p = 0.07 it is, by equation (26), I = 1.017, thus, all values of curve III, Fig. 126, are reduced by dividing with 1.017, and then plotted from p = 0.07 on, and then give the regulation curve inclusive line resistance shown as curve IV. As seen, the regulation range is reduced, but the regulation greatly improved by the line impedance. This is done essen- tially by the line reactance and leakage reactance, but not by the resistance. 161. Instead of approximating the effect of line impedance and leakage reactance by equivalent lamps and reactors, it can be directly calculated, as follows: CONSTANT'VOLTAGE SERIES OPERATION 309 fo = line resistance Xo = line reactance X = leakage or series reactance per autotransformer (38) the other symbols being the same as (15), (16) and (17). It is then : voltage consumed by line resistance tq: voltage consumed by line reactance XqI voltage consumed by leakage reactances x of the n(l — p) lamp devices : jxn{l - p)/, thus, total circuit voltage: substituting the abbreviation. n 712 = ^ n hs = xg and substituting (17) and (40) into (39), gives ^0 == ~ 1 _ V + JP^ + ^1 + i^2 + i(l - p)hz (40) - 'a [f^^ + '■] +4^"+ --+*.+('- rt».] 1 (41) 9 hence, absolute. eo = jyllr+i + '^ J + [t+§-' + Pa+ '»« + (1 - p)'»3] ' (42) hence, the current, geo 1 = 310 ELECTRIC CIRCUITS for p = (42) and (43) gives the full-load current and voUage, where (12) to = ii I (45) is the full-load line current, for z'l = full-load lamp current. 162. Let, in the instance paragraph 159 and Fig. 126; ro = 50 Xo = 75 X = 0.5 the other constants remaining the same as in paragraph 159, that is: 2l = 6 n = 100 g =0.12 6i = 0.0345 6a = 0.0685 y = 0.1248 a = 1.75 c = 0.287 It is then (40), hi = 0.06 /i2 = 0.09 hz = 0.06 hence, by (45), io = 1.04 X 6 = 6.24 amp. by (44) , eo = 5200 V (0.923 + 0.06) ^ + (0.264 + 0.09 + 0.06)» = 5200 V b.983^ + 4142 = 5200 Vl.l37 = 5200 X 1.066 eo = 5540 volts and by (43), 6.65 I = \/(0.983 - 0.923 p^ + (0.414 + 1.426 p)« CONSTANT-VOLTAGE SERIES OPERATION 311 ^ 6^65 \/l.l37 - 0.634 p + 2.885 p^ . 6.24 ,_, t = . == (46) \/l - 0.558 p + 2.54 p2 Fig. 127 shows, as curve II, the values of ^-sr — 1 from equation (46), that is, the regulation, as modified by line imped- ance and leakage reactance, with p as abscissae. The regulating range, po, of equation (46) is given by 1 - 0.558 po + 2.54 po' = 1, hence, po = 0.22. Thus the regulation range is reduced by the line impedance and leakage reactance, from 30 per cent, to 22 per cent. The maximum value of current, u, occurs at ■ Pn, = '^'= 0.11 and is given by substitution into (46), as, or, q = 0.015. That is, the regulation is improved, by the line and leakage reactance, from g = 4 per cent, to 5 = 1.5 per cent, as seen in Fig. 127. 163. In paragraph 161 and the preceding, the shunted react- ances, 61 and 62, have been assumed as constant and independent of p. However, with the change of p, the wave-shape distortion between current and voltage changes, as with increasing p, more and more saturated reactors are thrown into the circuit and dis- tort the current wave. As 61 is shunted by gf, and carries a small part of the current only, and g is non-inductive, the change of wave shape in 61 will be less, and as 61 carries only a part of the current, the effect of the change of wave shape in 61 thus is practically neg- ligible, so that 61 can be assumed as constant and independent of p. 62, however, carries the total current, at fairly high saturation, and thus exerts a great distorting effect. At and near full-load, with all or nearly all conductances, g, in 312 ELECTRIC CIRCUITS circuit, the entire circuit is practically non-inductive, that is, the current has the same wave shape as the voltage. Assuming a sine wave of impressed voltage, eo, the current, i, at and near full- load thus is practically a sine wave, and the shunting reactance, 62, thus has the value corresponding to a sine wave of current traversing it, that is, the value denoted as "constant-current reactance," Xe, in Chapter VIII. At no-load, with all or nearly all conductances, g, open-circuited, the entire circuit consists of a series of n reactive susceptances, 62. If, then, the impressed voltage, 60, is a sine wave, each susceptance, 62, receives 1/n of the impressed voltage, thus also a sine wave. That is, at and near no-load, the shunted reactance, 62, has the value corresponding to an impressed sine wave of voltage, that is, the value denoted as "constant-potential reactance," Xpy in Chapter VIII. Xc, however, is materially larger than rCp, and the shunting re- actance thus decreases, that is, the shunting susceptance, b2, in- creases from full-load to no-load, or with increasing p. Due to the changing wave-shape distortion, 62 thus is not con- stant, but increases with increasing p, thus can be denoted by 62 = 60(1 + sp) (47) this gives a = :j-^. (48) 1 + sp ^ ^ Substituting (48) into (43) gives, as the equation of current, allowing for the change of wave-shape distortion, i — <"' (49) Assume, in the instance paragraphs 159 and 161, and Fig. 127, that the shunted susceptance, 62, increases from full-load to no- load by 40 per cent. That is, s = 0.4; it is, then, ao a = 1 + 0.4 p Assuming now, that at the end of the regulating range, p = po = 0.22, CONSTANT'VOLTAGE SERIES OPERATION 313 a has the same value as before, a = 1.75, this gives 1.75 = "• 1 + 0.4 X 0.22 a = 1.90 and Substituting now the numerical values in equation (49), gives 6.65 t = \/(0.983 - 0.923 p)« + (0.414 + (a - 0.324)p)« 6^24 V[0.928 - 0.866 pl« + [0.388 + (0.938 a - 0.304) pY (51) Fig. 127 shows, as curve III, the values of ^-^t from equation (51), that is, the regulation as modified by the changing wave shape caused by the saturated reactance. The maximum value of current, im, occurs at p« = ^ = 0.11, and is given by substitution into (50) and (51), as, a = 1.82 -^ = 1 Oil 6.24 ^"^^ that is, q = 0.011 thus, the regulation is still further improved, by changing wave shape, to 1.1 per cent.