For instance, in the constant-current transformer, as shown - diagrammatically in Fig. 114, the secondary coils, S, are arranged so that they can move away from the primary coils, P, or in- versely. Primary and secondary currents are proportional to each other, as in any transformer, and the magnetic field between primary and secondary coils, or the magnetic stray field, in which the secondary coils float, is proportional to either current. The magnetic repulsion between primary coils and secondary coils is proportional to the current (or rather its ampere-turns), and to the magnetic stray field, hence is proportional to the square

~ of the current, but independent of the voltage. The secondary

CONSTANT-CURRENT TRANSFORMATION — 251 coils, S, are counter-balanced by a weight, W, which is adjusted so that this weight, W, plus the repulsive thrust between second- ary coils, S, and primary coils, P (which, as seen above, is propor- tional to the square of the current), just balances the weight of the secondary coils. Any increase of secondary current, as, for instance, caused by short-circuiting a part of the secondary load, then increases the repulsion between primary and secondary coils, and the secondary coils move away from the primary; hence more . of the magnetic flux produced by the primary coils passes between primary and secondary, as stray field, or self-inductive flux, less passes through the secondary coils, and therefore the second- ary generated voltage decreases with the separation of the coils,

• and also thereby the secondary current, until it has resumed the same value, and the secondary coil is again at rest, its weight ' balancing counterweight plus repulsion. Inversely, an increase of load, that is, of secondary impedance, decreases the secondary current, so causes the secondary coils to move nearer the primary, and to receive more of the primary flux; that is, generate higher voltage. In this manner, by the mechanical repulsion caused by the cur- rent, the magnetic stray flux, or, in other words, the series induct- ive reactance of the constant-current transformer, varies auto- , matically between a maximum, with the primary and secondary coils at their maximum distance apart, and a minimum with the coils touching each other. Obviously, this automatic action is independent of frequency, impressed voltage, and character of load. ' Jf the two coils P and S in Fig. 114 are wound with the same number of turns and connected in series with each other and with the circuit, Fig. 114 is a constant-current regulator, or a regulating ; : reactance, that is, a reactance which varies with the load so as to maintain constant current. If P is primary and S secondary circuit, Fig. 114 is a constant-current transformer. Assuming then, in the constant-current transformer or regula- tor or other apparatus, a device to vary the series inductive reactance so as to maintain the current constant. Let Eo = eo = constant = impressed e.m.f., Z=r+je, = r (1 + jk) the impedance of the load, and let o = inductive series reactance, as the self-inductive internal reactance of the constant-current transformer. |

252 ELECTRIC CIRCUITS . The current in the circuit then is [= —%, . r + j(to + 2) or, the absolute value, i= ——*%.. r? + (to + 2)?’ and, to maintain the current, 1, constant (i = %o), then requires iy = eo 0 —:::::__|!|!" Vr? + (to +2)?’ or, transposed, . _ eo = : _ . | m= 4) (2) r—gz (11) or, for 2 = kr, = eo? ae | zo = 4 (22)" = 98 — (12) that is, to produce perfectly constant current by means of a variable series inductive reactance, this series reactance must be varied with the load on the circuit, according to equation (11) or (12). For non-inductive load, or z = 0, it is =.|(%)*_ » Zo = \ (2) r | (13) the maximum load, which can be carried, is given by ; %=0 and is e 2=Vi+atarvi+ == (14) 0 As seen from equation (13), the decrease of inductive reactance, %, required to maintain constant current with non-inductive load, is small for small values of resistance, 7, when the r? under the root is negligible. With inductive load, equation (11), the inductive reactance, 2», has still further to be decreased by the inductive reactance of the load, z. Substituting: zoo = £2 | 00 — to as the value of the series inductive reactance at no-load or short- circuit, equations (11), (12), (13) assume the form: . i |

CONSTANT-CURRENT TRANSFORMATION _ 253

General inductive load:

ty = Vina 2, (14)

Induetive load of = = k:

Lo = V x0? — 7? — kr (15) -

Non-inductive load:

Xo = V 200? — 1?. (16)

131. As seen, a constant series inductive reactance gives an approximately constant-current regulation with non-inductive load, but if the load is inductive this regulation is spoiled. Inversely it can be shown, that condensive reactance, that is, a source of leading current in the load, improves the constant- current regulation. .

With a non-inductive load, series condensive reactance exerts the same effect on the current regulation as series inductive re- actance; the equations discussed in the preceding paragraphs re- main the same, except that the sign of 2» is reversed and the cur- rent always leading.

With series condensive reactance, condensive reactance in the load spoils, inductive reactance in the load improves the constant- current regulation. .

That is, in general, a constant series reactance gives approxi- mately constant-current regulation in a non-inductive circuit, and with a reactive load this regulation is impaired if the react- ance of the load is of the same sign as the series reactance, and the regulation is improved if the reactance of the load is of opposite sign as the series reactance.

Since a constant-current load is usually somewhat inductive, it follows that a constant series condensive reactance gives a better constant-current regulation, in the average case of a some- what inductive arc circuit, than the constant series inductive reactance. .

Let Eo = é) = constant = impressed, or supply voltage. Z=r-+ jz = impedance of the load, or the receiver circuit, and : z = kr, . that is, Z =r (1+ jk)

254 ELECTRIC CIRCUITS or, absolute, z=rvV/1 + ke. Let now a constant condensive reactance be inserted in series with this circuit, of the reactance, —z,, then the total impedance of the circuit is Z =r—j (a — kr). (17) The current is

P= Tile. — iy’ (18) or, the absolute value is . Co Vite (ae — bey (09) | the phase angle is tan 0) = — %— (20) and the power-factor is r i 21 ee EF Cae — ry ey for k=0, or non-inductive load, equations (19) and (21) assume the form: . eo r = ———— and cos 0 = —————, | Va Vitae that is, the same as with series inductive reactance. From equation (19) it follows, that with increasing current, ¢, from no-load:

r = 0, hence: ty) = - (22) the current, io, first increases, reaches a maximum, and then decreases again. When decreasing, it once more reaches the value, %o, for the resistance, 71, of the load, which is given by

• €o (7) = —— ———— = —} Vriit+ (z.—kri)? Fe ‘ hence, expanded, 2 kx , NTH (28) | and the maximum value through which 7 passes between r = 0 and r = rj, is given by dr ,

CONSTANT-CURRENT TRANSFORMATION ~ 255 or ohn + (2. — kr)?} = 0 = 27 — 2k(x, — kr); hence, = hte 1, "2 TFR 2 (24) This maximum value is given by substituting (24) in (19), as . = 2Vi +, for =toV1 + k? , . (25) k = 0.4, this value is @? = 1.077 to, that is, the current rises from no-load to a maximum 7.7 per cent. above the no-load value, and then decreases again. | As an example, let €o = 6600 volts impressed e.m.f. and

• 2. = 880 ohm condensive reactance, ze being chosen so as to give | io = = 7.5 amp.; for k = 04, then, i= ___—-6600-— Vr? + (880 — 0.47)” | 008 0p = pee | 0 Vt (680 — 047)” e=2t = 1.077 rn. . . These values of current and power-factor are plotted, with the receiver voltage as abscisse, in Fig. 115.

132. The conclusions from the preceding are that a constant . series reactance, whether condensive or inductive, when inserted in a constant-potential circuit, tends toward a constant-current regulation, at least within a certain range of load. That is, at varying resistance, r, and therefore varying load, the current is approximately constant at light load, and drops off only gradu- ally with increasing load. ;

256 ELECTRIC CIRCUITS | | This constant-current regulation, and the power-factor of the : circuit, are best if the reactance of the receiver circuit is of oppo- site sign to the series reactance, and poorest if of the same sign. That is, series condensive reactance in an inductive circuit, and series inductive reactance in a circuit carrying leading current, Pi ttt tT Ey TT Ty fal Pte ty tt tT eT. | pop tT Tet | | | rt ee ht ttt tt eT MAL IL ei | | | dT dP cpr TNT, el | | | fee] TT TT TING pon | de Sep SSSS000R08F | nm tL ti tt | dep | ttt IZ}

Fie. 115. : give the best regulation; series inductive reactance with an in- ductive, and series condensive reactance with leading current in the circuit, give the poorest regulation.

Since the receiver circuit is usually inductive, to get best regula-

. tion, either a series condensive reactance has to be used, as in Fig.

115, or, if a series inductive reactance

10 1 is used, the current in the receiver cir-

H hy) j , cuit is made leading, as, for instance,

° + E a by shunting the receiver circuit by a

condensive reactance. Fia 116. Assuming, then, as sketched diagram- .

matically in Fig. 116, in a circuit of

constant impressed e.m.f., Zo = eo = constant, a constant in-

. ductive reactance, 2, inserted in series; and the receiver circuit, of impedance,

Z=rt+ je = r(1 + jk) where k = tangent of the angle of lag = =;

CONSTANT-CURRENT TRANSFORMATION 257 let the receiver circuit be shunted by a constant condensive react- _ ance, z-; let then: ; E = potential difference of receiver circuit or the condenser. terminals, { = current in the receiver circuit, or the ‘secondary. current,”’ 7: = current in the condenser, Io = total supply current, or “primary current.” Then fP=I[th (26) and the e.m.f. at receiver circuit is . E=Z] (27) at the condenser, E = — jx) (28) hence, Z i=) z qT (29) | and, in the main circuit, the impressed e.m.f. is . Bo = = E+ jxfo _ (80) | Hence, substituting (26), (27) and (29) in (30), . ZZ | eo = ZI + jzo(T +551) = fe, ae {2 + jxo 22 le | ot Le — 2 . ; | 6) = {z= + joo} » (81) and . gi = Zo 4 jzo If x = 2%, that is, if the shunted condensive reactance equals the series inductive reactance, equations (32) assume the form, oe | f=+ jac in (33) and the absolute value is . eo . = Zo (34)

• that.is, the current, t, is constant, independent of the load and the power-factor. ; 17 | |

258 ELECTRIC CIRCUITS That is, if in a constant-potential circuit, of impressed e.m.f., eo, an inductive reactance, 2», and a condensive reactance, 2-, are connected in series with each other, and if Ze = Xo, (35) that is, the two reactances are in resonance condition with each other, any circuit shunting the capacity reactance is a constant- current circuit, and regardless of the impedance of this circuit, Z =r.+ ja, the current in the circuit is . i= © . Zo 133. Such a combination of two equal reactances of opposite sign can be considered as a transforming device from constant potential to constant current. : Substituting, therefore, (35) in the preceding equation gives: (33) substituted in (29): Current in shunted capacity Z = zi? (36) or, absolute, . . 2€o u1= Zo? (37) and, substituting’ (33) and (36) in (26): primary supply current is Z- jxo Io = re €o (38) or the absolute value is . e to = Ta V0 + Go — 2)* (39) and the power-factor of the supply current is %o—- 2 r tan # = — ——,, cos & = ————————— 40 ET RON Tena O In this case, the higher the inductive reactance, z, of the receiving circuit the lower is the supply current, 7, at the same resistance, 7, and the higher is the power-factor, and if = 2 : . To = — and cos @ = 1 (41) Zo that is, the primary, or supply circuit is non-inductive, and the primary current is in phase with the supply e.mf., and the

| . | | | | CONSTANT-CURRENT TRANSFORMATION — 259 . power-factor is unity, while the secondary or receiver current (33) is 90° in phase behind the primary impressed e.m.f., é. Inserting, therefore, an inductive reactance, 41 = 2% — 2, in series in the receiver circuit of impedance, Z = r + jz, raises the power-factor of the supply current, 7, to unity, and makes this current, %, a minimum. Or, if the inductive reactance, Zo, is inserted in the receiver circuit, thus giving a total imped- ance, Z + jt = r+ j (x + 2) by equation (38), substituting Z + jo instead of Z, gives the primary supply current as Zeo To = Zor (42) or the absolute value as . . 2€o ‘ | lo = Zo (43) and the tangent of the primary phase angle tan Ho = ~ = tan 8, that is, the primary power-factor equals that of the secondary. Hence, as shown diagrammatic- ally in Fig. 117, a combination SBE BERR of two equal inductive reactances te] fe ; in series with each other and with =; ls r | the receiver circuit, and shunted [f° 2 i midway between the inductive re- ig actances by a condensive reactance Fie. 117. equal to the inductive reactance, | transforms constant potential into constant current, and inversely, without any change of power-factor, that is, the primary supply : current has the same power-factor as the secondary current. With an inductive secondary circuit, the primary power- factor can in this case be made unity, by reducing the inductive reactance of the secondary side, by the amount of secondary

• reactance.

134. Shunted condensive reactance, z., and series inductive reactance, 2, therefore transforms from constant potential, @, to constant current, 7, and inversely, if their reactances are equal, z, = 2, and in this case, the main current is leading, with non-inductive load, and the lead of the main current decreases, with increasing inductive reactance, that is, increasing lag, of the \

a | ! 260 ELECTRIC CIRCUITS receiving circuit. The constant secondary current, 1, lags 90° behind the constant primary e.m-f., é9. Inversely, by reversing the signs of zo and 2, in the preceding : equations, that is, exchanging inductive and condensive react- ances, it follows that shunted inductive reactance, 2%, and series condensive reactance, 2z,, if of equal reactance, x = 2, transform constant potential, ¢, into constant current, 7, and inversely. In this case, the main current lags the more the higher the inductive reactance of the receiving circuit, and the constant secondary current, 7, is 90° ahead of the constant primary e.m.f., @.

In general, it follows that, if equal inductive and condensive reactances, % = 2, that is, in resonance conditions, are con- nected in series across a constant-potential circuit of impressed

A A A A

ze si Bo = fe i = r ba = . r = 2 . le z Be Ze Ze B “T B B I 514 pin Iv ° Fig, 118.

e.m.f., é, any circuit connected to the common point between the reactances is a constant-current circuit, and carries the current, + = fo

Zo

Instead of connecting this secondary or constant-current circuit with its other terminal to line, A, so shunting the con- densive reactance with it, and causing the main current to lead (I in Fig. 118), or to line, B, so shunting the inductive reactance with it, and causing the main current to lag (II in Fig. 118), it can be connected to any point intermediate between A and B, - by a autotransformer as in III, Fig.118. If connected to the mid- dle point between A and B, the main current isneither lagging nor leading, that is, is non-inductive, with non-inductive, load, and

with inductive load, has the same power-factor as the load.

The two arrangements, I and II, can also be combined, by connecting the constant-current circuit across, as in IV, Fig. 118, and in this case the two inductive reactances and two conden-

CONSTANT-CURRENT TRANSFORMATION _ 261 sive reactances diagrammatically form a square, with the con- stant potential, eo, as one, the constant current, 7, as the other diagonal, as shown in Fig. 119. This arrangement has been called the monocyclic square. The insertion of an e.m.f. into the constant-current circuit, in such arrangements, obviously, does not exert any effect on the constancy of the secondary current, 7, but merely changes the primary current, 7%, by the amount of power supplied or consumed by the e.m.f. inserted in the secondary circuit. While theoretically the secondary current is absolutely con- stant, at constant primary e.m.f., practically it can not be per- fectly constant, due to the power A lo consumed .in the reactances, but falls off slightly with increase of | iy 1; load, the more, the greater the | Ps yor loss of power in the reactances, | , that is, the lower the efficiency of |, PEAR the transforming device. | Two typical arrangements of | Y oh such constant-current transform- | es gy ing devices are the T-connection : V/41hs or the resonating-circuit, diagram 7 Te Fig. 117, and the monocyclic Fic. 119. ‘square, diagram Fig. 119. From these, a very large number of different combinations of in- ductive and condensive reactances, with addition of autotrans- formers, and of impressed e.m.fs., can be devised to transform from constant potential to constant current and inversely, and by the use of quadrature e.m.fs. taken from a second phase of the polyphase system, the secondary output, for the same amount of reactances, increased. These combinations afford very. convenient and instructive examples for accustoming oneself to the use of the symbolic method in the solution of alternating-current problems. Only two typical cases, the T-connection and the monocyclic square will be more fully discussed. A. T-Connection or Resonating Circuit 185. General—A combination, in a constant-potential circuit, of an inductive and a condensive reactance in series with each | |

262 ELECTRIC CIRCUITS

other in resonance condition, that is, with the condensive react- ance equal to the inductive reactance, gives constant current in a circuit shunting the capacity. This circuit thus can be called the “secondary circuit” of the constant potential constant- current transforming device, while the constant-potential supply circuit may be called the “primary circuit.”

If the total inductive reactance in the constant-current cir- cuit is equal to the condensive reactance, the primary supply current is in phase with the impressed e.m.f.

Let, as shown diagrammatically in Fig. 117,

Zo = value of the inductive and the condensive reactances which are in series with each other. ;

2, = the additional inductive reactance inserted in the constant- current circuit.

Z =r+jz,orz = Vr? + 2? = the absolute value of the im- pedance of the constant-current load. ‘

Assuming now in the constant-current circuit the inductive reactance and the resistance as proportional to each other, as_ for instance is approximately the case in a series arc circuit, in which, by varying the number of lamps and therewith the

load, reactance and resistance change proportionally. Let, then, k= ; = ratio of inductive reactance to resistance of the load, or tangent of the angle of lag of the constant-current circuit. It is then ;

Z =r(1 + jk) and z=rvV/1 $k? (1) let, then, Eo = e = constant = primary impressed e.m.f., or sup- ply voltage, E, = potential difference at condenser terminals, & =secondary e.m.f., or voltage at constant-current circuit, {7o = primary supply current, . q: = condenser current, {7 = secondary current, then, in the secondary or receiver circuit, E=Z] (2) at the condenser terminals

CONSTANT-CURRENT TRANSFORMATION ~— 263 Ei = E+ jul = (Z + jai){ (3) and, also, BE, = — jroli (4) hence, : 4 } haji Thy ©) and the primary current is Z2ti . fo=T+h= (7th) hence, expanded, Z— j(xo — Jo = phate a I (6) . and the primary supply voltage is @ = Bi + j2ofo; . hence, substituting (3) and (6), eo =[(Z + jari)—{Z — j(mo—a)}, or, expanded, . €o = + jxof (7) or, the secondary current is ; = — J p= -® (8) and, substituting (8) in (6) and (5): the primary current is Z —jlte — ; Jo = 2= HA = 2) , (9) the condenser current is p=2the (10) or, the absolute value is -_ £0 1 ro (11) io = Vr + (Zo = Z1 — x)? €o (12) To i= Ce eo (13) Zo tan 6 = = k gives the secondary phase angle (14) and M—-m—-2. . tan 6) = — “—— —— gives the primary phase angle (15)

264 ELECTRIC CIRCUITS This phase angle 6; = 0, that is, the primary supply current is non-inductive, if %M—-%—-2=0, : that is, . 2 = By —2. (16) . The primary supply can in this way be made non-inductive for any desired value of secondary load, by choosing the reactance, 21, according to equation (16). If z = 0, that is, a non-inductive secondary circuit (series in- . candescent lamps for instance), 2, = 20, that is, with a non-in- , a ‘ductive secondary circuit, the primary supply current is always non-inductive, if the secondary reactance, 21, is made equal to the primary reactance, 2o. In this case 2; = %, with an inductive secondary circuit tan 0) = : = tan 6; that is, the primary supply current has the . same phase angle as the secondary load, if all three reactance (two inductive and one condensive reactance) are made equal. In general, z: would probably be chosen so as to make Jo non- inductive at full-load, or at some average load. 136. Example.—A 100-lamp are circuit of 7.5 amp. is to be operated from a 6600-volt constant-potential supply & = 6600 : volts, and 7 = 7.5 amp. . Assuming 75 volts per lamp, including line resistance, gives a maximum secondary voltage, for 100 lamps, of e’ = 7500 volts. Assuming the power-factor of the arc circuit as 93 per cent. lagging, gives : cos 6 = 0.93, or tan 6 = 0.4; . hence, oe k = < = 0.4, and Z = r(1 + 0.4), or 2 = 1.077 r at full-load, if . é& = 7500 volts, , v= 5 = 1000 ohms, hence r’ = 0.93 2’ = 930 ohms, xz’ = 0.4r' = 372 ohms, and . t= Moray = & = 9600 _ 960 ohms. Zo 2 7.5

CONSTANT-CURRENT TRANSFORMATION — 265 — To make the primary current %) non-inductive at full-load, or for z’ = 372 ohms, this requires 2, = % — 2’ = 508 ohms. This gives the equations

• = 7.5 amp., ; e= 7.52 = 8.08 r volts. . fo = yf" + (872 — 0.4r)? x ed 2 a , : ‘

• 7.54] sa) ( _ sam) (gs + 0.423 2200 tan Oy = — 32 — O4r : _ = 04 — 322, . r hence, leading current below full-load, non-inductive at full-load and lagging current at overload.

137. Apparatus Economy.—Denoting by 2’, 1’, x’ the respective full-load values, the volt-ampere output at full-load is : — ay C02! _ cor V1 + ' Q, a 2 Xo? Zo? (17) volt-ampere input, 7: Q, = tote = SE (18) That is, the volt-ampere input is less than the volt-ampere output, since the input is non-inductive, while the output is not. ‘ The power output is = P =i = or (19) = ae) which is equal to the volt-ampere input, since the losses of power in the reactances were neglected in the preceding equations. The volt-amperes at the condenser are : Q’ = 11220; hence, substituting (13), te 2 , 2 2 3 _ @ mite +2)? 2” + (hr + 21) (20) Zo Zo The volt-ampere consumption of the first, or primary inductive reactance, Zo, is Q” = to*Xo ;

266 ELECTRIC CIRCUITS hence, substituting (12), . 2 — 7 — 2 2 _— aoa 2 Q” = r? + (2 = 21) eo? = 7’? + (xo ker 21) e’o? (21) Zo Zo the volt-ampere consumption of the second, or secondary induct- ive reactance, 21, is . QQ” - 24, or . Q’” == eat (22) . Xo? The total volt-ampere rating of the reactances required for the transformation from constant potential to constant current then is Q=9+0" +0" /2 2 U — 2 2 2? (1 TE) + Ber Os, — ee) + Tot: + 221") (23) 0 . and the apparatus economy, or the ratio of volt-amperes output to the volt-ampere rating of the apparatus is f= MQ rto/ 1 + KP Q = -2r’2(1 + k®) + 2 ker’ (2 2 — 20) + (t0® — tori + 2 21?) (24) . . this apparatus economy depends upon the load, 7’, the power- factor or phase angle of the load, k, and the secondary additional inductive reactance, 2.

To determine the effect of the secondary inductive reactance, 2%: The apparatus economy is a maximum for that value of secondary inductive reactance, 21, for which t = 0.

Instead of directly differentiating f, it is preferable to simplify the function f first, by dropping all those factors, terms, etc., which inspection shows do not change the position of the maxi- mum or the minimum value of the function. Thus the numera- tor can be dropped, the denominator made numerator, and its first term dropped, leaving

f' = 2k’ (2a, — to) + (ao? — tots + 2 21) as the simplest function, which has an extreme value for the same value of 21, asf. Then df’ az, = 4 ke’ — Zo + 4a = 0, and m= mie (25)

CONSTANT-CURRENT TRANSFORMATION — 267 substituting (25) in (24), gives __ 8roVith fi= 16 r’? — 8 kr’xo + 7 x0? (26) To determine the effect of the load r’: fi becomes a maximum for that load, r’, which makes fy ar = %

• or, simplified, , 16 r’? — 8 kr’'xo + 7 20? f 1= a . hence af "1 7’ 2 2 ar’ = (32 7’ - 8 kxo) - (16 7 - 8 kr’xo + 7 207) = 0, hence v= zeV7 (27) and, substituting (27) in (26), V1+h = +. 28 fh Vi-k (28) hence, for k = 0: . 1 = —= = 0.378, f: 2 V7 Y= avi = 0.662 20, | m= z = 0.25 2, for k = 0.4: V/1.16° = —.—_— = 0.478 : hi 0d ! . f= zev7 = 0.662 zo Z = V116 avi = 0.712 20 | ; = ; (1 — 0.4+/7) = — 0.016 2 = approximately zero. . At non-inductive load ; k=0 and with non-inductive primary supply, that is, v1 =TZo, |

268 ELECTRIC CIRCUITS by substituting these values in (24), the apparatus economy is f= a (29) = 307° ae) ; which is a maximum for r= x » (30) fo = 4 = 0.25 (31) which is rather low:

That is, non-inductive load and supply circuit do not give very high apparatus economy, but inductive reactance of the load, and phase displacement in the supply circuit, gives far higher appa- ratus economy, that is, more output with the same volt-amperes in reactance.

By inserting in (23), with the quantities, Q’, Q”, and Q’”, coefficients 1, 22, ns, which are proportional respectively to the cost of the reactances per kilovolt-ampere, the expression

mQ’ + niQ”’ + niQ’”’ (32) P then represents the commercial economy, that is, the maximum of this expression, derived by analogous considerations as before, gives the arrangement for minimum cost at given output.

138. Power Losses in Reactances.—

In the preceding equations, the losses of power in the reactances have been neglected. However small these may be, in accurate investigations, they require consideration as to their effect on the

. regulation of the transforming device, and on the efficiency.

Let ;

@ = power-factor of inductive reactance, that is, loss of power, as fraction of total volt-amperes. :

b = power-factor of condensive reactance, that is, loss of power, as fraction of total volt-amperes.

Here a and b are very small quantities, in general b, the loss in the condensive reactance, being far smaller than the loss in the inductive reactance.

Approximately, the inductive reactances are (a + j)zo and (a + j)x1 respectively, and the condensive reactance is (b —j)<o.

Assuming the same denotations as in the preceding paragraphs, receiver circuit

E=Z] (33)

CONSTANT-CURRENT TRANSFORMATION — 269 at condenser terminals - . EF,=E+(@ + pul ; ={Z+ (at full (34) and also By = (b — j)tols (35) hence ; = Z+(at+ fu n= @-Heo ! (36) and forth ~2+6 — j)to + (a + D7 ’ (b — j)xo . _ Z — j(to — 21) + (bao + az1) (6 — j)to © i (37) and the impressed e.m.f. , 6o = Ei + (a + j)xolo; hence, substituting (35) and (37), eo = Zot {Z(a+b) —jao(a—b) +jri(a+b)} + { xoab+2:a(a+b)} I b-j (38) Since a and 6 are very small quantities, their products and squares can be neglected, then €&o = Zo + {Z(a + b) — fee ” b) + jzi(a + b)} I (39) or . a (Sine); T= eth —ima-htimerh}) this can be written 7 jouw 1+ [2+ -j-H +52 @+8)| . 0 Zo hence Jeo ia — jt _4 p=-E{1+ja-j2@ty-F@+H} (a that is, due to the loss of power in the reactances, the secondary . current is less than it would be otherwise, and decreases with increasing load still further.

270 ELECTRIC CIRCUITS Equation (41) can also be written = — (fy _ 7 _ f2te -

: p= -2{f1-2@+0]-i[727@+ 0 - a] } 42) here the imaginary component is very small in the parenthesis, that is, the secondary current remains practically in quadrature with the primary voltage.

; The absolute value is, neglecting terms of secondary order, pa YT }. i= 21 = (a +6) } (43) The primary current is, by equation (37) and (40), Io = Z — j(to — 21) + (bao + a21) eo (44) to + Z(a + b) — jao(a — b) + jui(a + b) ao Z . Eat Z1 _« na l- a) +5) 701 4 4—Fq45)-ja—0)

139. Example.—

Considering the same example as before: a constant-potential circuit of ¢g = 6600 volts supplying a 100-lamp series arc circuit, with 7’ = 7.5 amp., and e’ = 7500 volts at full-load of 93 per cent. power-factor, that is, k = 0.4, and Z = (1—0.4j)r. Assuming now, however, the loss in the inductive reactance as 3 per cent., and in the capacity as 1 per cent., that is, a =0.03 b =0.01, the full-load value of the secondary load impedance is: 2’ = 1000

. ohms, r’ = 930 ohms and 2’ =372 ohms. ‘ To give non-inductive primary supply at full-load, the follow- ing equation must be fulfilled:

X= Xo — 2 = X% — 372.

i From equation (43), the secondary current, at full-load, is a i ae = (a+)} .

; or

7.5 = 9800/1 _ 980 x 0.08);

| ar) Xo ,

hence

| a = 840 ohms, and 2x; = 468 ohms.

|

| CONSTANT-CURRENT TRANSFORMATION 271 Substituting in (42), (43), (44), ; J = — 7.865 | (1 — 0.04 5) + j(0.052 — 0.016 57) | i = 7.86(1 — 0.04 7) e=t = 1077" ; ; ; = 8.46 r(1 — 0.04 7) (sq + 0.027) — j(0.443 — on) Io = 7.86 840 840 _— 0.047r\ , ./0.016r . (1+ gay") + 3 gao7 + 9-002) | and herefrom the power-factor, efficiency, etc. wt TTT TTT Ty yt tt tt ti a ttt ttt ttt ttt tT ttt tt et | woe | | tT Tt | | Tt | | edwerfracton| 1-T | | PT Pett tt eect | TL aT ledlardemlert eet TTT Tt BATA TT tT feet TT ET TE Tt nD ae as eee WV eet ae ee eee | 4 Wet ttt tT tt tT te TT ett tt tt . PlLite Th Tite pep; te Lf lo Fie. 120. In Fig. 120, there are plotted, with the secondary, e.m.f., e, as abscisse, the values: secondary current, 7; primary current, 0; primary power-factor, cos 6, and efficiency. 140. In alternating-current circuits small variations of fre- quency are unavoidable, as for instance, caused by changes of ; load, etc., and the inductive reactance is directly proportional, the condensive reactance inversely proportional to the frequency. Wherever inductive and condensive reactances are used‘in series with each other and of equal or approximately equal reactance, so more or less neutralizing each other, even small changes of frequency may cause very large variations in the result, and in

272 ‘ELECTRIC CIRCUITS | such cases it is therefore necessary to investigate the effect of a

change of frequency on the result: for instance, in a resonating

circuit of very small power loss, a small change of frequency at

constant impressed e.m.f. may change the current over an enor-

mous range.

Since in the preceding, constant-current regulation is produced by inductive and condensive reactances in series with each other, the effect of a variation of frequency requires investigation.

Let, then, the frequency be increased by a small fraction, s.

The inductive reactance thereby changes to 2(1 + 8) and

a(1 +s), and Z = r+ (1 + 8)x respectively, and the conden- 5 Zo . sive reactance to ides

Leaving all the other denotations the same, and neglecting the

loss of power in the reactances, | E =] Bi = {2+ j(1 + 8)zi}f = — jtod 1 1+38’ hence, . . Z+j(1 =f Eto +i + s)zi} | 0 . and (1 Z — j{xo — (1 ay

Ie=T+p, =jh +94 de — OF +8) hes G+ o's)

thus ‘ éo = Ei, + j(1 + 8)xofo =(Z+j(1 + s)ti—(1 + 8)?Z + j(1 + 8) {20 — . (1 + s)*x})7 hence, expanding and dropping terms of higher order, eo= +f] {xo + 8(to — 221 + 4 Zj)—8(38 2, + 25)}, or = — 201 _ 5(y 9% 4 4432)). [= {a s(1-22 +4 ©} (45)

Hence, the current is not greatly affected by a change of frequency. That is, the constant-current regulation of the above-discussed device does not depend, or require, a constancy of frequency beyond that available in ordinary alternating- current circuits.

| 7 ° CONSTANT-CURRENT TRANSFORMATION — 278 | B. Monocyclic Square . 141. General. ’ ‘A combination of four equal reactances, two condensive and two inductive, arranged in a square as shown diagrammatically ; in Fig. 119, page 261, transforms a constant voltage, impressed | upon one diagonal, into a constant current across the other | diagonal, and inversely. ' Let, then, : Eo = eo = constant = primary impressed e.m.f., or supply © ; voltage, E = secondary terminal voltage, FE, = voltage across the condensive reactance, E, = voltage across the inductive reactance, and Io :+ primary supply current, { = secondary current, Zi = current in condensive reactance, ‘7s = current in inductive reactance, these currents and e.m.fs. being assumed in the direction as indicated by the arrows in Fig. 119. Let Zo = condensive and inductive reactances;

• hence, Z: = — jzo= condensive reactance (1) Z; = + j= inductive reactance (2) Then, at the dividing points, fo=UitTs (3) and T=4-h. (4) . hence, n-&5! 6) and n=bfl 6) In the e.m.f. triangles, 6o=Z11 + Za]s (7) 18

274 - ELECTRIC CIRCUITS e and . | B= Zilli — Za) (8) and . E=2] a) substituting (1) and (2) in (7) and (8) gives €o= — jo (fi — J 2) (10) and ZI = — jto (Tis + I) (11) ' and, substituting herein the current, €o = + jrof (12) and , ZI = — jrolo (13) hence, the secondary current is ; : = — 16 . [=-2 (14) the primary current, Z b= (15) the condenser current, _Z+jm ~ rf 1™~ 2 Xo? . (16) and the current in the inductive reactance, _ 4 — jro, | est a7) The secondary terminal voltage is — 4 B= — jer (18)

• 0 , the condenser voltage, Bi = — jr, = — TEE , (9) and the inductive reactance voltage, E2 = + jtol2 = + 1G = i004, (20) Lo The tangent of the primary phase angle is tan 0) = - = tan 0 (21) hence, the absolute value of the secondary current is - = £0 i=] (22)
  ry

CONSTANT-CURRENT TRANSFORMATION = 275 of the primary current, . | ce (23) of the condenser current, , 2 2 i= Vet Get ey € (24) Zo and of the inductive reactance current ig = VOT, (25) The secondary terminal voltage is . e= z €o (26) the condenser voltage, a/r? + (a0 + 2)3 a= VE tea é (27) | and the inductive reactance voltage, 2 _— 2 @: = VO seen eo. (28) 142. From these equations follow the apparent powers, or volt- amperes of the different circuits as: Output, | . _ €o*2 ; : Qo =a = at (29) Input, Qi: = Goto = oe (30) Hence the input is the same as the output. This is obvious, since the losses of power in the reactances are neglected, and it was found (21), that the phase angle or the power-factor of the primary circuit equals that of the secondary circuit. Apparent power of the condensive reactance, 2 2 Qi = et) = mt foto €0%. (31) Inductance, 2 —y)2 Qt = cain = EO os (32) and, therefore, total volt-ampere capacity of the reactances is Q = 2(Q: + Qs) | ;

| ‘ | 276 ELECTRIC CIRCUITS 2 2 2 “ot = €0°} hence | . 2 2 Qa Hee os (33) and, apparatus economy, Q ; : 0 220 . y I= Q "BFa0 G4) henceamaximumfor . z=2 (35) and this maximum is equal to fo = }4, or 50 per cent. (36) ‘ That is, the maximum apparatus economy of the monocyclic , square, as discussed here, is 50 per cent., or in other words, for every kilovolt-ampere output, 2 kv.-amp. in reactances have to be provided. This apparatus economy is higher than that of the 7-connec- tion, in which under the same conditions, that is for 21 = 2, the apparatus economy was only 25 per cent. The commercial, or cost economy would be given by g=—__% __ = maximum (37) 2 (n.Q, + niQ,) where m = price per kilovolt-ampere of condensive reactance, ny = price per kilovolt-ampere of inductive reactance. 143. Example.— Considering the same problem as under A. From a constant impressed e.m.f. ¢9 = 6600 volts, a 100-lamp arc circuit, of 93 per cent. power-factor, is to be operated, requiring ; t = 7.5 amp. Z=r+je =r (1+ jk) where k=7 =04; r hence Z=r(1+04)), and at full-load e’ = 7500 volts. Then, from (22), / z = ? = 880 ohms, z' = © = 1000 ohms; j |

CONSTANT-CURRENT TRANSFORMATION = 277 hence , r’ = 930 ohms, x’ = kr’ = 372 ohms, and, therefore, ¢ = 7.5 amp., = 7.5 t= (.0 880 amp., e = 7.52, and at full-load, or r = 930, when denoting full-load values by prime, =7.5amp., to = 7.93 amp., i’, =6.65amp., ° t's = 4.52 amp., e’ =. = 7500 volts, e’, = 5850 volts, e's = 3980 volts, . \ ° P’ a= P’,, = } 56.25 kv.-amp. P’ a, = 38.9 kv.-amp. P’, = 18.0 kv.-amp. P’,’ = 113.8 kv.-amp. f’ = 0.4943 or 49.43 per cent. that is, practically the maximum. ; 144. Power Loss in Reactances.— In the preceding, as first approximation, the loss of power in | the reactances has been neglected, and so the constancy of current, 7, was perfect, and the output equal to the input. Con- sidering, however, the loss of power in the reactances, it is found that the current, 1, varies slightly, decreasing with increas- ing load, and the input exceeds the output. , Let, then, Z, = (b — jf) Zo = condensive reactance, Zs = (a + j) Zo = inductive reactance, . otherwise retaining the same denotations as in the preceding paragraphs, ; Then, substituting in (7) and (8), . 2=b-Dht+@+Hhs (38) eb -Dh-@+Dh (89) |

278 ELECTRIC CIRCUITS - Assuming a=, + Cs b =¢(—- cr} (40) : —b cq = et} ce = a a (41) Substituting in (38) and (39) = — 9 — Ts) + adi + 22) — (Ti — J) Zi . : nm — ji + Ix) + (01 — P2) — e201 + J), substituting herein from equations (3) and (4) gives gta ta tale | (42) and i t= - al — (+o (43) . and from these two equations with the two variables, J and [ 0 it follows from (43) that Z — Ci1Ly = — 44 To G +e) zt (44) Substituting (44) in (42), transposing, and dropping terms of secondary order, that is, products and squares of c; and ce, gives — — Jeo ‘ee — exe {= Zo {1 + jee a} (45) substituting (45) in (44), and transposing, ( 2— Z2 . Io = Zt + 2jaZ| (46) then, substituting (45) and (46) in (5) and (6), _ &0 Z + jae Ci — Ce . 2e,—C, 00? qh — mal 2 + 2 Xo + jZ 2 2 =| (47) . _ @f[Z1—jto , ater 7 2late, 47) Ta Bp ae + 2 ao) (48) and the absolute value is . p= 11 — ok) 4 (a — eZ)?

• i (1 az) + (cs or)

CONSTANT-CURRENT TRANSFORMATION — 279 ! or, approximately, . me 20(1 _ 6, 2 | i= a(1 1 =), ete. (49) 145. Example.— Considering the same example as before, of a 7.5-amp. 100- lamp arc circuit operated from a 6600-volt constant-potential supply, and assuming again as in paragraph 139: 3 per cent. power-factor of inductive reactance, or a = 0.03. 1 per cent. power-factor of condensive reactance, or b = 0.01. ; It is then, | ¢, = 0.02, co = 0.01, and at full-load, | gt Hig = Lo (1 “1 a) ° or, 7.5 = (1 _ 0.02%); ' Zo Xo . hence, ; . r to = 861, and i = 7.66(1 — 0.0257), and we have, approximately, _, r 0.42 jr Io= 7.66{ <7 + 0.02 +E" }, roo 0.4r | Ty = 3.88 {a5 + 4(1 + Sei) |? r . 0.47 | Ta = 3.83 35; ~ 4(1 — ger) |) @= 2 = 1.077 7. In Fig. 121 are plotted, with the secondary terminal voltage, e, as abscisse, the values of secondary current, 7; primary current, #9; condenser current, 71; inductive reactance current, 72, and efficiency. As seen, with the monocyclic square, the current regulation is closer, and the efficiency higher than with the T connection. This is due to the lesser amount of reactance required with the monocyclic square. The investigation of the effect of a variation of frequency on the current regulation by the monocyclic square, now can be carried out in the analogous manner as in A with the T connection,