Theory and Calculation of Electric Circuits — part 14 of 15

. . | 302 ELECTRIC CIRCUITS | and let bi = shunted susceptance with the lamp in circuit, . that is, exciting susceptance of reactor or auto- transformer, and (16) y = V9? + b;* = admittance of complete consuming device. by = shunted susceptance with the lamp burned out and let c= br _ exciting current as fraction of load 9 current: c < 1. . (17) a= g = saturation factor of reactor or 2 autotransformer: a > 1. it is, then: voltage of lamp and reactor: I ar eT a8) voltage of reactor with lamp burned out: I _f EE, = —jb, 7 by (19) thus, with pn lamps burned out, and (1—p)n lamps burning, it is total voltage, éo = n(1—p)E. + np EB * (20) =en{{itP 472 = nl 5 jo, +9 ba substituting (17), i _ nljl—p . 6 = {45 + ia (21) or, ni 1 — p(1 — ac) + jap, 6 = — g 1 —je hence, absolute, . c= 7 ViL= pa = ac) + att (22) since, y=gV1lt+e thus, the current in the series circuit,

• Coy EE mJi—pd-ajp rap =

CONSTANT-VOLTAGE SERIES OPERATION — 303 158. For, p = 0, or full-load, it is ip = SH (25) thus, _ « to SS ———— a (26) : Vv [1 — p(1 — ac)]? + ap? ‘The same value of 7 as at full-load is reached again for the ~~ value p = po, where the square root in (24) becomes one, that is, . [1 —po(1 —ac)]?+a po* = 1, hence, . , 2(1 — ac) Pom Gt (1 — act (27) for, p = 1, or no-load, it is . to i=——— . 28 av 1 + c? (28) The current is a maximum, ¢ = 7,, for the valueof p = pn, given by di . dp 0, or, from (26), d dp ill — p(l — ac)]* + a%pt} = 0, . this gives 1 — ac Pm = oF + (1 — acy (28) = Po 2? as was to be expected. Substituting (29) into (26), gives as the value of maximum cur- rent . . 1 — ac? ln = to 1 + (—*) (30) and the regulation q, that is, the excess of maximum current over full-load current, as fraction of the latter, thus is _ tm — to q= to 1-— 2 =yi+(A)-1 — 6

; 304 ELECTRIC CIRCUITS . If q is small (31) resolved by the binomial, gives . l/l — 2 , 7= 3( a “) (32)

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 increases from % to a maximum value t,, at p = > then decreases again, to ip at p = po, and decreases further, to 11 at no-load, p = 1.

Thus, there exists a regulating range from p= 0 to p= a little above po, where the current is approximately constant.

. Instance:

; Saturation: a = |1.5 /1.5 3 2.0 |2.0 {2.0 re 2.5 |2.5 Excitation: c = |0.1 (0.2 |0.3 0.1 (0.2 |0.3 0.1 |0.2 (0.3 Regulation: g = |0.147/0.103 0.067|0.07710.044 0.020!0. 044'0.020/0. 005 Range: po = |0.573/0.510/0. 482,0. 34510. 275)/0. 192/0.220/0. 154/0.079 tt tt yg || | tt it pee | PAT at | | PARA PP eA Tt TTT tad pitt bv Se Tt TT Teal PT tT tT TTT PANEAATEOT TET EAST Tal SO SSS tel pT TT ETT TT TT TT PSSST teal tt te EEE T TTT etry eee ET ET TT PT TTT tT tet ye tt ET ETE ET TE PET TTT T TTT ett et te EET Tt bt obiaotiot tpl ol ol wm] » | » | ma |

Fia. 126. As illustrations are shown, in Fig. 126, the regulation curves, a from equation (26), for: . a=1.5 c= 0.2 Curve I = 2.5 = 0.1 II = 2.0 = 0.3 Ill

CONSTANT-VOLTAGE SERIES OPERATION 305 159. By the preceding equations, it is possible now to calculate the values of exciting susceptance b,, and saturation be, required by the shunting reactors to give desired values of regulation with- in a given range. , . From (82) follows: wai V%q (33) Substituting (33) into (27) gives: . 2/2 q | a, a= . poll + 2q) | (34) | ¢ = PUL +29) — 49. | 2V4 | From chosen values of g and po, a and c thus can be calculated, ‘ ' from a and c and the conductance g of the consuming device, b:, : bz, 2, etc., follow. ; . Instance: n = 100 lamps of 7, = 6 amp. and e, = 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 percent. That is,g = 0.04 and po = 0.30. , It thus is: 11 = 6 ée: = 50 = 4012 g= 5 = 0. n = 100 g = 0.04 . po = 0.30 From (34) follows: a = 1.75 c = 0.287 Hence by (17): b: = 0.0345 bs = 0.0685 and by (16): , y = 0.1248 by (2): éo = 5000 volts 20

306 ELECTRIC CIRCUITS and by (25): to = 6.24 amp. thus, by (26): i- 6.24 . (35) . V1 — p+3.31 p? ;

Fig. 127 shows, as curve I, the values of gq = Z — 1, in per cent., that is, the regulation, with p as-abscisse.

Pit dey eet tT tT TE TE TET TE TTT PET TT tet te TE ET EE ET TT Pt Peer Tn] | PA EE Lass | | RT TT Pi Tet tT Ty dt tT | PASE TT SERN SNE PET et tT tT tT tT tT tT TT tT PUA TAN Pt ttt tpt et ey tt tT tT Td PN UNS pppoe | fot pf No BREEN

ee eee Pitt tit ttt te ETT TT TT TT Pitt tt? tT tt tt i-Fre tt tt Pel eth brennan Og! |

Fie. 127.

160. In general, the resistance and reactance of the circuit or 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:

ro = line resistance;

2% = line reactance;

x = series or leakage reactance per autotransformer or consumption device.

The most convenient way is to represent ro, 29 and x by their equivalent in lamps or reactors. The admittance of each con- sumption device, comprising lamp and reactor or autotrans- former, is

|

‘-CONSTANT-VOLTAGE SERIES OPERATION 307 . Yi =g — jb = g(l — je), | thus the impedance, Z<t=-—l tie "Yi gQ = je) g(l + c*)’ and by (23), . Z, = itje , yV1 + cf If, then, we add to the resistance ro a part cro of the reactance, we get an impedance; Z=r(1+je), ; which has the same phase angle as Z;, and thus can be expressed as a multiple of Z,, . Z= mZi, where | n= z = ryV1 + c? ; (36) | | . thus is the “lamp equivalent” of the line resistance ro plus the part cro of the reactance. This leaves the reactance, : “1 = Zo + n(l — p)z — cro, . | and as the reactance of a reactor without lamp is _1 . : t= by the reactance x; can be expressed as multiple of 23, | Dy = Nata | _ 7 : _ be where Ne = Libs = ba (xo + n(l - p)x - Cro] (37) thus is the “lamp equivalent”’ of the line reactance x» and leakage reactances x, in burned-out lamps. . Thus the addition of the line impedance 7» + jaro, and the leakage reactances z, is represented by m; lamps with reactors, and nz burned-out lamps, or a total of n, + nz lamps. Thus the circuit can carry n; — (m; + m2) lamps, and its regula- tion curve starts at the point p = a and ends at p = 1 — met 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 value,s would thus have to be proportionally reduced. Instance: . ; In the case = a = 2.0 c = 03 given as curve III of Fig. 126 let: n = 100 , g =0.12 : ro = 50 ; ‘ to = 0.5 hence, bi = cg = 0.036 y =Vg? + b,? = 0.125 . = g = : bs a 0.06, thus, ; m= ry V1 +c? = 6.54 ne = be [to + n (1—p) & — cro] = 7.0 mi + me = 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),

• = 1.017, to : 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 Let. To = line resistance %o = line reactance (38) x = leakage or series reactance per . autotransformer the other symbols being the same as (15), (16) and (17). It is then: voltage consumed by line resistance ro: rol voltage consumed by line reactance 79: jxol voltage consumed by leakage reactances 2 of the n(1 — p) lamp devices: jen(l — p)f, | thus, total circuit voltage:
, _ 7{(l—p)n , .pn . a | 9 = {SO + 7 + ro + jz + 501 — p)nz} (39) | substituting the abbreviation, ; | hy = 709 | ‘on | hy = “8 (40) hs = 79g and substituting (17) and (40) into (39), gives ° Inj1 - p,. . . € = yiim% + jpa + hy + jhe + j(1 — p)hs} in(fl-p [U—p)e _ . = ella t a] te pie + pa + ha + (1 — pha] } (40) hence, absolute, in [fl.- >? 2 T(1— p)c _ 2 eo = eS SE tha] +[ Be + pat hat (1 p)ha] (42) hence, the current, j=— eee | | —p * pape _ *(43) ; n [3 +4] +[ Ide + pa +h: + (1 pyhs|

310 ELECTRIC CIRCUITS for p = Q (42) and (43) gives the full-load current and voltage, : ton 1 3 c 2 eo = ON lie + ha| + let he + ha| (44) where (12) . . .y to= = 45 o= tro (45) is the full-load line current, for +; = full-load lamp current. 162. Let, in the instance paragraph 159 and Fig. 126; To = 50 Zo = 75 . z=05 the other constants remaining the same as in paragraph 159, that is: ty = 6 . ; n = 100 g = 0.12 b, = 0.0345 bs = 0.0685 y = 0.1248 : a@ = 1.75 c = 0.287 It is then (40), h, = 0.06 he = 0.09 ‘ hs = 0.06 hence, by (45), to = 1.04 X 6 = 6.24 amp. by (44), é9 = 5200 V (0.923 + 0.06)? + (0.264 + 0.09 + 0.06)* = 5200 V0.983? + 414? = 5200 V1.137

• = 5200 X 1.066 €o9 = 5540 volts and by (43), i= 6.65 V (0.983 — 0.923 p)? + (0.414 + 1.426 p)?

CONSTAN T-VOLTAGE SERIES OPERATION 311 _ 6.65 /1.137 — 0.634 p + 2.885 p? . 6.24 i= — 46 V1 — 0.558 p + 2.54 p? ; (46) Fig. 127 shows, as curve II, the values of ao — 1 from equation (46), that is, the regulation, as modified by line imped- ance and leakage reactance, with p as abscisse. 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, 7,,, occurs at = Po | Pu = 5 0.11 and is given by substitution into (46), as, . Tm ‘ : . 594 = 1.015, . - or, q = 0.015. That is, the regulation is improved, by the line and leakage reactance, from g = 4 per cent. to g = 1.5 per cent. as seen in Fig. 127. 163. In paragraph 161 and the preceding, the shunted react- ances, b; and b2, 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 so and more saturated reactors are thrown into the circuit and dis- tort the current wave. ; As b; is shunted by g, and carries a small part of the current | only, and g is non-inductive, the change of wave shape in b; will be less, and as b; carries only a part of the current, the . effect of the change of wave shape in 6; thus is practically neg- ligible, so that b: can be assumed as constant and independent of p. bz, 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 1

. 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, 7, at and near full- load thus is practically a sine wave, and the shunting reactance, bs, thus has the value corresponding to a sine wave of current traversing it, that is, the value denoted as “constant-current , reactance,”’ x,, 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, be. If, then, the impressed voltage, éo, is a sine wave, each susceptance, be, receives 1/n of the impressed voltage, thus also a sine wave. That is, at and near no-load, the shunted reactance, be, has the value corresponding to an impressed sine wave of voltage, that . is, the value denoted as “constant-potential reactance,” x», ‘in Chapter VIII. : 2, however, is materially larger than z,, and the shunting re- actance thus decreases, that is, the shunting susceptance, be, in- , creases from full-load to no-load, or with increasing p. Due to the changing wave-shape distortion, b: thus is not con- . ' stant, but increases with increasing p, thus can be denoted by be = bo(1 + sp) (47) this gives ay 775 + =p . (48) Substituting (48) into (43) gives, as the equation of current, allowing for the change of wave-shape distortion, t= 5 2 7d mY 3 9) —?P — pie Pao _ 7 ny [a ta] [Soe pret pyha|* Assume, in the instance paragraphs 159 and 161, and Fig. 127, that the shunted susceptance, bs, increases from full-load to no- load by 40 per cent. That is, , s = 0.4; it is, then, a=, 1+04p 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 , ao | “15 = TF 04 x OR : a = 1.90 | and , 19 . a= 1+04p (50) Substituting now the numerical values in equation (49), gives _ 6.65 (0.983 — 0.923 p)? + (0.414 + (a — 0.324)p)? (51)

• 6.24 [0.928 — 0.866 p]? + [0.388 + (0.938 a — 0.304) p]? . Fig. 127 shows, as curve III, the values of 54 from equation (51), that is, the regulation as modified by the changing wave shape caused by the saturated reactance. The maximum value of current, i,, occurs at Pa = & = 0.11, and is given by substitution into (50) and (51), as, , a = 1.82 . im : $247 1.011 that is, . ; q = 0.011 thus, the regulation is still further improved, by changing wave shape, to 1.1 per cent.

| CHAPTER XVI LOAD BALANCE OF POLYPHASE SYSTEMS 163. The total flow of power of a balanced symmetrical poly- phase system is constant. That is, the sum of the instantaneous values of power of all the phases is constant throughout the cycle. In the single-phasesystem, however, or ina polyphase system with unbalanced load, that is, a system in which the different phases are unequally loaded, the total flow of power is pulsating, with , double frequency. To balance an unbalanced polyphase system . thus requires a storage of energy, hence can not be done by any method of connection or transformation. Thus mechanical momentum acts as energy-storing device in the use as phase bal- . aneer, of the induction or thesynchronousmachine. Electrically, energy is stored by inductance and by capacity. The question then arises, whether by the use of a reactor, or a condenser, con- nected to a suitable phase of the system, an unequally loaded polyphase system can be balanced, so as to give constant power during the cycle. In interlinked polyphase circuits, such as the three-phase sys- tem, with unbalanced load carried over lines of appreciable im- pedance, the voltages of the three phases become unequal. This ‘makes voltage regulation more complicated than in a balanced system. A great unbalancing of the load, such as produced by operating a heavy single-phase load, as a single-phase railway or electric furnace, greatly reduces the power capacity of lines, trans- __- formers and generators. Unbalanced load on the generators causes a pulsating armature reaction: at single-phase load, the armature reaction pulsates between more than twice the average value, and a small reversed value, between F(cos a + 1) and F(cos a — 1), where cos a is the power-factor of the single-phase - load. Especially in alternators of very high armature reaction, as modern steam-turbine alternators, a pulsation of the armature reaction is very objectionable. It causes a pulsation of the field flux, leading to excessive eddy-current losses and consequent re- duction of the output. The use of a squirrel-cage winding in the 314

LOAD BALANCE OF POLYPHASE SYSTEMS 315

___ field pole faces of the single-phase alternator reduces the pulsation

of the field flux, but also increases the momentary short-circuit stresses. ;

Thus, it is of interest to study the question of balancing unbal- anced polyphase circuits by stationary energy-storing devices, as reactor or condenser.

164. Let a voltage,

e = E cos ¢ (1) ; be impressed upon a non-inductive load, giving the current i=Icos¢ (2) The power then is p = ei = EI cos*¢ = 2 1 + cos 2 4) =Q+Qcos2¢ (3) where 1 E. . Q=>5 (4) that is, in a non-inductive single-phase circuit, the power consists of a constant component,

• EI Q=5° and an alternating component, EI Q= oO cos 2 ¢, of twice the frequency of the supply voltage, and a maximum value equal to that of the constant component. The instantane- ous power thus pulsates between zero and 2 Q, by equation (3). If the circuit is inductive, of lag angle a, the current is , t = I cos (¢ — a) (5) and the instantaneous power thus, p = EI cos ¢ cos (¢ — a) , = FIT cos a + 008 (2 4 — a) | = P + Qcos (29 — a), thus consists of a constant component, P = = cos a = Qos a (7)

316 ELECTRIC CIRCUITS and an alternating component, ; Q cos (2 ¢ — a); it thus pulsates between a small negative and a large positive value, P — Q and P + Q. If the circuit is completely inductive, that is, the current lags 90° or 5 behind the voltage, the current is : . w t= cos(¢ - 5) (8) and the instantaneous power thus, p= EI cos ¢ cos (¢ — 5) EI .. = “> sin 24 — = —™)1 Q cos (2 o 5) (9) Thus, the power comprises only an alternating component, but no continuous component; in other words, no power is consumed, but the power surges or alternates between +Q and —Q, that is, power is stored and then again returned to the circuit. If the circuit is closed by a capacity, C, the current leads the impressed voltage by » thus is ° Tv i = I cos ( + 5) (10) and the instantaneous power thus, : ® | p = EI cos ¢ cos($ + 5) vs . =Q cos (2 o+ 5) (11) - thus, comprises only an alternating component, surging be- tween —Q and +Q, with double frequency. The power consumed by a condenser, equation (11), is opposite in sign and thus in direction, from that consumed by a reactor (9), . ™ 7). Q c05(24 + 5) @ cos(2¢ — 5) 165. If a number of voltages, e; = E; cos (¢ _- ¥:) (12) ; 1 “Engineering Mathematics,” Chapter III, paragraphs 66 to 75.

LOAD BALANCE OF POLYPHASE SYSTEMS 317 of a polyphase system, produce currents, : t; = I; 0038 (¢ — ¥5 — a) (13) | the instantaneous power of each voltage e; is Di = ests ; . = Qi {cos a; + cos (2 ¢ — 24; — a)} (14) | and the total instantaneous power of the system thus is p = Zps = 2Q; cos a; + ZQ; cos (2 @¢—-2%- a) = P + Qcos (2 ¢ — a) (15) where P = 2Q; cos a; (16) ‘is the total effective power of the system, and Q = 2Q: cos (2 ¢ — 271 — a) (17) is the total- resultant alternating component of power, or the resultant power pulsation of the system.

Thus, the power of the polyphase system pulsates, with double : frequency, between P — Q and P + Q.

In this case, P may be greater than Q, and frequently is, and the

power thus pulsates between two positive values, while in the single-phase circuit (6) it pulsated between positive and negative value. ;

It thus can be seen, that in any system, polyphase or single- phase, with any kind of load, the total instantaneous power of the system can be expressed,

p = P + Qeos (2 $ — a) (18) where P is the constant component of power, and Q the amplitude of the double-frequency alternating component of power, and Q may be larger or smaller than P.

It must be noted, that Q is not the total reactive power of the system—which would have to be considered, for instance, in power-factor compensation etc.—but Q is the vector resultant of the reactive powers of the individual circuits, while the total reactive power of the system is the algebraic sum of the individual reactive powers (see “Theory and Calculation of Alternating- current Phenomena,” Chapter XVI).

Thus, for instance, in a system of balanced load, even if the

_ load is reactive, Q = 0. Thus, Q is the unbalanced reactive

318 ELECTRIC CIRCUITS , power of the system, and does not include the reactive power, which is balanced between the phases and thereby gives zero as vector resultant.

166. The expression of the power of a polyphase system of gen- eral unbalanced load is by (15)

p =P + Qcos (2 $ — a) (19) this also is the expression of power of the single-phase load of lag angle a, of the impressed voltage and current,

e = Ecos¢ i = I cos ($ — a) (20) where, from (20), P=Qsine _ EI (21) Q-F while in the general case (19) P and Q may have any values. Suppose now we select from the polyphase system a voltage, , e’ = E’ cos ($ — B) (22) and load it with an inductive load of zero power-factor, | i = I’ cos($ — 6 - 3) (28) Ud that is, we connect a reactor of = a into the phase e’. The power of (22) (23) then is p’ = Q cos(2 ¢ — 26 — 5) (24) where E gy == (25) and the total power of the system, comprising (19) and (25), thus is Po =ptp' = P + Qcos (2 — a) + Q' cos (24 — 26 — ) . and this would become constant, and the double-frequency term ‘ eliminated, that is, the system would be balanced, if Q’ and 8 are chosen so that Q cos (2 — a) +Q' cos(26-28-7)=0 (26)

LOAD BALANCE OF POLYPHASE SYSTEMS 319 hence, y=Q (27) 2¢—-28-5=2¢-a-—" or, = 247 | 2 7a 2 | r=" (20) E’ E” . z= T = 20 (30) thus, a vT = veals-(G+9)] an is the voltage, which, impressed upon a reactor of reactance, E’? . T= 36 (30) balances the power, ‘p =P + Qcos (2 ¢ — a) (24) of an unbalanced polyphase system. That is, ’. _ (& us e’ = E’ cos [¢ (3 + *)| (31) impressed upon the reactance, z, gives the current, 7 = 2Q (2437 i = Fp 00s [ - (5 + 7) | (32) and thus the power, = _(*47 (243% p= Qeos[-(5 +3) ]eos[o-(5+ 7) | = — Qcos (2 ¢ — a) (33) and this reactive power, p’, added to the unbalanced polyphase power, p, gives the balanced power, . p=ptp =P 167. Comparing (31) with (20) or (24), it follows: The unbalanced load of a single-phase voltage, e= E cos 9,

320 ELECTRIC CIRCUITS of lag angle, a, or in general, the unbalanced load of a polyphase system with the resultant instantaneous power of lag angle, a, p=P+ Qcos (2 ¢ — a)

can be balanced by a wattless reactive load, p’, having the same volt-amperes, Q’, as the alternating component, Q, of the unbal- anced load, and having a phase of voltage lagging by

a T

at4 or by 45° plus half the lag angle, a, of the unbalanced load or un-

an balanced single-phase current.

Just as the unbalanced polyphase load, p, (24) may be single- phase load on one phase, or the vector resultant of the loads on different phases, so the wattless reactive compensating volt-

_ amperes (33) may be due to a single reactor connected into the compensating voltage, e’, (31), or may be the vector resultant of several voltages, e’1, loaded by reactances, 21, so that their vector resultant is p’ (33). , ,

If a capacity is used for energy storage in balancing unbalanced load (24); the compensating voltage (22),

e' = E’cos(— 8), impressed upon the capacity gives the reactive leading current, : ve 7) _ us , to v= cos (¢ 6+5) (34) hence the compensating reactive power, p’ = E'T' cos (29-26 +5) (35) and therefrom, by the same reasoning as before, a ,3r . Beaty (36) = E'[ cos o- (5 + 5)] 2 4 (37) ‘= 2Q _(%47 . i = Fy cos - (5 + 3)

That is, when using a capacity for balancing the load, the com-

pensating voltage, e’, has the phase, a ,3r gta’

LOAD BALANCE OF POLYPHASE SYSTEMS 321 or, what is the same as regards to the power expression, a_T 2° 4’ thus lags by half the phase angle, a, minus 45° (or plus 135°). 168. As instance may be considered a quarter-phase system with one phase loaded. Let e, = E cos e=E cos (¢ _ 5) (38) be the two phase-voltages of the quarter-phase system. Let the first phase, e:, be loaded by a current lagging by phase angle, a, 1, = I cos (¢ — a) (39) while the second phase, és, is not loaded. The power then is Pp = et EI = “g {cos a + cos(2 ¢ — a)} (40) and is compensated or balanced by a reactance connected to a compensating phase, e’ = E" cos ($— 8) (41) and consuming the reactive current, ° Tv - i’ = I’ cos($ — 6 ¥ 5) (42) where the — 5 represents inductive reactance, the -+- 5 capacity reactance. . The compensating reactive power then is p' = ei . E’I’ _* and this becomes equal to EI — “y cos (2¢— a), for : E’'l’ = EI B= 5 + i (44) 21

322 ELECTRIC CIRCUITS and the compensating circuit thus is ¢ = E’cos(¢- 5 F 4) 2° 4 ; (45) ce _ &@ oT = T'cos(¢— 5 F7) it is, then, p’ = et’ = E'T’ cos (2 ¢ — a ¥ x) = —EI cos (2 ¢ — a) hence, - Po= p+ p' ~ Fl og 25% for a = 0, or non-inductive load, it is e = Ef cos (¢ + i) E’ r = ryeitae ot cos(¢ - 5) |, hence, if we choose, E' = E~/2, hence, . I I’ =) V2 it is = ey + es (46) ° that is, connecting the two phases in series, gives the compensat- ing voltage for non-inductive load. Or:

““Non-inductive single-phase load, on one phase of a quarter- phase system, can be balanced by connecting a reactance across both phases in series, of such value as to consume a current equal to the single-phase load current divided by +/2, that is, having the same volt-ampere as the single-phase load.”

169. In the general case of inductive load of power-factor, a, the compensating voltage (45) can be written,

to. BY aur in(2 + ™)si eé=E {cos (5 + 7) cos o+ sin (5 + 3) sin o} =f ar oat 7 B'{cos (5 + Z) eos ¢ + 0s (F ¥ 7) oos(# — 5) },

| | : LOAD BALANCE OF POLYPHASE SYSTEMS 323 or, choosing, | E’=#8#, thus, I =i, it is, by (38), e’ = aye: + Gees where a Tv a = cos (Z + 7) (47) : a anon Ge) and the upper sign applies to the reactor, the lower to the con- denser as compensating circuit. The current then is a _ 273), = Tes (¢ - £ $=) (48) The compensating voltage e’ thus can be produced by connect- ing a transformer of ratio a; into the first phase, e;, a transformer of ratio, a3, into the second phase, es, and connecting their second- aries in series across a reactor or condenser of suitable reactance. The current, 7’, in the compensating circuit consumes a current, ax’, in the first phase, e:, and a current, a7’, in the second phase, és. As the latter phase has no load, the total current in the second phase is | . ‘ ; is = asi’ = I 008(% $2) cos (6 — $ + 32) the total current in the first phase is 41° = 4, + ay?’ . = _ aye _a#,3n = T{cos (# — a) + 008 (5 + 5) oos (#- 5 # F)] = 1{ cos(¢ — a) + 0.5 cos ( F 5) + 0.5 cos (¢ — a F r)} ! - _ 7 = 0.5 1 {cos ( — a) + cos(¢ + 5) } a T a T | = 1{c00(5 ¥ Z) 08 (6 - $F 9)}) hence has the same value as i, but differs from it by 5 or 90° in phase, thus has to its voltage, e:, the same phase relation as 7s . |

324 ELECTRIC CIRCUITS has to its voltage, es. That is, the system is balanced in load, in phase and in armature reaction. In the unbalanced single-phase load, the power-factor is @; = COS a in the balanced load, the power-factor is ar = c06(5 + j) thus, is materially reduced for a reactor as compensator, +43 it is in general increased for a condenser as compensator, -7 170. Instead of varying the phase angle of the compensating voltage, e’, with varying phase angle, a, of the single-phase load, compensation can be produced by compensating voltages of constant-phase angle, utilizing two such voltages and varying the proportions of their reactive currents, with changes of a. Thus, if t, = I cos (¢ — a), is the load on phase, é: = Ecos ¢, and the second phase Tv is not loaded, thus giving the unbalanced power, p = AL eos a + 008 (2 6 — a)} (49) as compensating voltage may be used, the voltage of both phases connected in series, @e=@é + ee = EV/2 cos (¢ - 1) (50) and the voltage of the second phase, is @ = E cos (¢ - 5)" (51) Let, then, Yo 7 _ 3x t’ = I’ cos ( rl ) , be the reactive current of the compensating phase, 6, and ; t's = I',cos (¢ — =),

LOAD BALANCE OF POLYPHASE SYSTEMS 325 = — I’; cos ¢ the reactive current of the compensating phase, és. The powers of the two compensating circuits then are ~ p = ew’ = av) cos (2 ¢ — x) and p's = ex's EI, = — “77 000(26 - 5) (58) and the condition of compensation thus is EI _ El'V2 EI’, 7 = 008 (2 — a) = Y= cos 2. + = cos (2.6 — 5) (54) or, resolved, _

(I cos a — I’+/2) cos 2 ¢ + (I sin a — I's) sin2 ¢ = 0, and as this must be an identity, the individual coefficients must vanish, that is,

I'= I cos a v2 . (55) I’, = [sin a = I cos (a — 5) thus, the compensating voltages and currents, which balance the single-phase load, é: = Ecos¢ ti: = I cos (¢ — a) (56) are € = €1 + ee TT “iim (6-9 ~ BV cos #— 4 (67) , _ cosa ( *) v= cos|¢ — | and is é: = E cos (¢ - 5) i's = — Ic0s(a — 7) cos ¢ (58) = I sin a cos ¢

326 ELECTRIC CIRCUITS As seen, this means loading the second phase with a reactor giving the same volt-amperes, , _&I., ets = =y sin a, as the unbalanced single-phase load (56), and thereby balancing the reactive component of load, and then balancing the energy component of the load by the compensating voltage e, + é:, as given by (46). If the single-phase load is connected across both phases of the quarter-phase machine in series, e=at ez T = Ev? cos (¢ + i) (59) . f = V2 cos(¢ + 4 «) in the same manner the conditions of compensation can be de- rived, and give the compensating circuit, a ¢ = B’cos(# — 5) ao: . (60) 7? _~f 4 i = T’eos(s — 5 - 5) where E'l’ = EI. For non-inductive load, a= 0, this gives é= 1, that is, one of the two phases is compensating phase for the re- sultant. 171. As further instance may be considered the balancing of single-phase load on one phase of a three-phase system. Let é: = Ecos 2a er = E co (# — =) (61) ‘ 4n éeg=E cos (¢ - 3) be the three voltages between the three lines and the neutral.

LOAD BALANCE OF POLYPHASE SYSTEMS 327 The voltage from line 1 to line 2, then, is €;? = @; — €2 T = E+/3 cos (¢ + 4) (62) and if a = lag of current behind the voltage, the current produced by voltage, e, is i = Ic0s(¢ +% — a), thus the power, EIV3 . and this is balanced by the compensating voltage and current, as discussed before, e = EV8 cos(6 — (2 + 5)) a, 8 (64) i = Teos(# — (F + 35) it is p' = ea’ EIvV3 - Qn =—3 cos (2.6 — a — ) , EIV3 w = — —}“c0s(26-a + 3) (65) thus, Po=pt+p' ~ FIV oon thus balanced. The balancing voltage (64), a : s é= EV3 cos(¢ -3 - 4): lags behind the load voltage, e (62), by a is 2 + 4’ or by half the lag angle of the load, plus 45°. If a= 6.

328 ELECTRIC CIRCUITS or 30° lag, it is ; ¢ = EV3 cos(¢ - =) (66) thus the compensating voltage, e’, is displaced in phase from the load voltage, ¢12 (62), by 60°, if the lag angle of the load is 30°, and in this case, the second phase of the three-phase system thus can be used as compensating voltage, C13 = @1 — €3 s = EV3 cos ( _- 5) =e,

In the general case, for any lag angle, «, the compensating vol- tage (64) can be produced by the combination of the two-phase voltages, €; and és, as

é= 101 — Gee3 similar as was discussed in the quarter-phase system.

The second phase, ¢:s, as compensating voltage, loaded by a re-

actor, balances the load of phase angle, a = B’ or 30°. For other

. angles of lag, either another phase angle of the balancing voltage is necessary, or, if using the same balancing voltage, the balance is incomplete.

Let thus: the load

ei: = EV3 cos (¢ + 5): i = Icos(¢ +5 — a), be balanced by reactive load on the second phase, €13 = EV/3 cos(# — =), _ 2x i = Toos(# — 3), it is: power of the load, p= C193 ; = FEW | coo + coe (26-+5 ~ a) |; balancing power, p' = €132' EIV3 rs = — =F coe(26+ §) |

; LOAD BALANCE OF POLYPHASE SYSTEMS 329 thus, total power, Po=pt+p’ = Bra cos a +008 (264% _ a) ~ cos (24+ *)f = FIV eos + sin (5 _ 8) cos (20-2+7)|; and . T _@G-9) q COS a is the ratio of the remaining alternating component of power, to the constant power, and may be called the coefficient of unbalancing.

CHAPTER XVII CIRCUITS WITH DISTRIBUTED LEAKAGE

172. If an uninsulated electric circuit is immersed in a high- resistance conducting medium, such as water, the current does not remain entirely in the “circuit,” but more or less leaks through the surrounding medium. The current, then, is not the same throughout the entire circuit, but varies from point to point: the currents at two points of the circuit differ from each other by the current which leaks from the circuit between these two points.

Such circuits with distributed leakage are the rail return circuit of electric railways; the lead armors of cables laid directly in the

; ground; water and gas pipes, etc. With lead-armored cables in

ducts, with railway return circuits where the rails are supported above the ground by sleepers, as in interurban roads, the leakage may be localized at frequently recurring points; the breaks in the ducts, the sleepers supporting the rails, etc., but even then an assumption of distributed leakage probably best represents the conditions. The same applies to low-voltage distributing sys- tems, telephone and telegraph lines, etc.

The current in the conductor with distributed leakage may be the result of a voltage impressed upon a circuit of which the leaky conductor is a part, as is the case with the rail return of electric railways, or occurs when a cable conductor grounds on the cable armor, and the current thereby returns over the armor; or it may be induced in the leaky conductor, as in the lead armor of a single-conductor cable traversed by an alternating current; or it may enter the conductor as leakage current, as is the case in cable armors, gas and water pipes, etc., in those cases where they pick up stray railway return currents, etc.

When dealing with direct-current circuits, the inductance and the capacity of the conductor do not come into consideration except in the transients of current change, and in stationary con- ditions such a circuit thus is one of distributed series resistance and shunted conductance.

| Inductance also is absent with the current induced in the cable armor by an alternating current traversing the cable conductor, 330

CIRCUITS WITH DISTRIBUTED LEAKAGE 331 and with all low- and medium-voltage conductors, with the com- mercial frequencies of alternating currents, the capacity effects are so small as to be negligible.

In high-voltage conductors, such as transmission lines, etc., in general, capacity and inductance require consideration as well as resistance and shunted conductance. This general case is fully discussed in ‘Theory and Calculation of Transient Electric Phe- nomena and Oscillations,” and in ‘Electric Discharges, Waves and Impulses,’’ more particularly in the fourth section of the former book.

173. Let, then, in a conductor having uniformly distributed leakage, or in that conductor section, in which the leakage can be considered as approximately uniformly distributed,

r = resistance per unit length of conductor (series resistance),

g = leakage conductance per unit length of conductor (shunted

conductance), and assume, at first, that no e.m.f. is induced in this conductor.

The voltage, de, consumed in any line element, dl, of this con- ductor, then is that consumed by the current, i, in the series resistance of the line element, rdl, thus:

de = irdl. (1)

The current, di, consumed in any line element, di, that is, the difference of current between the two ends of this line element, then, is the current which leaks from the conductor in this line element, through the leakage conductance, gdl, thus:

di = egdl. (2)

Differentiating (2) and substituting into (1) gives

ad . ae = 79. (3) This equation is integrated by (see “Engineering Mathematics,” Chapter IT) t= Ae, (4) Substituting (4) into (3) gives aAe = rgAe—# hence, a= tv7g, and thus, the current, <= Aye—virol + Agetvial (5)

332 ELECTRIC CIRCUITS where A, and A; are determined by the terminal conditions, as integration constants. . Substituting (5) into (2) gives as the voltage, e= vi Axe vrat — Art vit} (6)

174. (a) If the conductor is of infinite length, that is, of such great length, that the current which reaches the end is negligible compared with the current entering the conductor, it is

t= 0 forl = o. . This gives A, = 0, hence, t= Ae-vrot = r —yvrgl e 4 Ae~ v9 (7)

Vi

=al-¢

. g

That is:

A leaky conductor of infinite length, that is, of such great length that practically no current penetrates to its end, of series resist- ance, r,and shunted conductance, g, per unit length, has an effect- ive resistance,

ir = ,/- 8 To m ( )

It is interesting to note, that a change.of r or g changes the effective resistance, ro, and thus the current flowing into the con- ductor at constant impressed voltage, or the voltage consumed at constant-current input, much less than the change of r or g.

(b) If the conductor is open at the end I = lo, it is

*+=0 forl = kh, hence, substituted into (5) 0= Aje7 vial + Aget vrgle and, putting A = Ajevr0o = — Aget viol, it is t= Afetvra lod — —— vr9 (lo-D} r to — (9) e= g4 {e v7 Qo-D + 7 via (lo—D }

CIRCUITS WITH DISTRIBUTED LEAKAGE 3338 (c) If the conductor is grounded at the end 1 = |h, it is e=0 forl = lh, hence, substituted into (6), O = Aye—vroe — Agetvrale , and, putting A= Aje7Vrole = Agetviote it is Sc ; = Ales ye—n + e~ Vra(e—} ; (10) e= V5 Afetvrae—) - e7Vra(e—0} (d) If the circuit, at 1 = lo, is closed by a resistance, R, it is = R forl = by hence, substituting (5) and (6), gives Aje via — Aget viol = R Aye V7! + Agetvrate Vi g hence, Vi7* As = Aye-2vr0to NG, r vf +R Thus, r— ft i = Afe-vit — __VY ,- 2e-Dvin} R+ vt r- ft e= Vt Afe-v7at 4. EF ¢- @ho- via} Ng R+ vi 175. Substituting, fo = vf (8) as the ‘effective resistance of the leaky conductor of infinite length,”

. 334 ELECTRIC CIRCUITS and a= V7 (12) as the “attenuation constant” of the leaky conductor, it is . R— ro = -a _ & 770 |~o(2—-2) t= Afe Rin o—2)} R — 1o (2-2) (18) = -al —a(2lo— € = rmAfe"? + Rite € oD}

These equations (13) can be written in various different forms. They are interesting in showing in a direct-current circuit features which usually are considered as characteristic of wave trans- mission, that is, of alternating-current circuits with distributed capacity.

The first term of equations (13) may be considered as the out-

‘ flowing components of current and voltage respectively, the sec- ond terms as the reflected components, and at the end of the circuit of distributed leakage, reflection would be considered as occurring at the resistance, R.

If R>ro, the second term is positive, that is, partial reflection of current occurs, while the return voltage adds itself to the in- coming voltage. If R = , the reflection of current is complete.

If R<ro, the second term is negative, that is, partial reflection of voltage occurs, while the return current adds itself to the incoming current. If R = 0, the reflection of voltage is complete.

If R = ro, the second term vanishes, and equations (13) be- come those of (7), of an infinitely long conductor. That is:

A resistance, 2, equal to the effective resistance, ro = V5 of the

infinitely long conductor of distributed resistance and shunted conductance, as terminal of a finite conductor of this character passes current and voltage without reflection. A higher resist- ance partially reflects the current and increases the voltage, and a lower resistance partially reflects the voltage and increases the current. Infinite resistance gives complete reflection of current and doubles the voltage, while zero resistance gives complete re- _ flection of voltage and doubles the current.

The term, ro = Vo thus takes in direct-current circuits the same position as the “surge impedance” Je or Ve in alternating-cur-

i rent circuits. |