CHAPTER XXXV BALANCED SYMMETRICAL POLYPHASE SYSTEMS 303. In most applications of polyphase systems the system is a balanced symmetrical system, or as nearly balanced as possible. That is, it consists of n equal e.m.fs. displaced in phase from each other by - period, and producing equal currents of equal phase displacement against their e.m.fs. In such systems, each e.m.f. and its current can be considered separately as constituting a single-phase system, that is, the polyphase system can be resolved into n equal single-phase systems, each of which consists of one conductor of the polyphase system, with zero impedance as return circuit. Hereby the investigation of the polyphase system resolves itself into that of its constituent single-phase system. So, for instance, the polyphase system shown in Fig. 208, at balanced load, can be considered as consisting of the equal single- phase systems :0— 1;0 — 2;0 — 3; . . . 0 — n, each of which consists of one conductor, 1, 2, 3, . . . n, and the return conductor, 0. Since the sum of all the currents equals 0, there is no current in conductor 0, that is, no voltage is consumed in this conductor; this is equivalent to assuming this conductor as of zero impedance. This common return conductor, 0, since it carries no current, can be omitted, as is usually the case. With star connection of an apparatus into a polyphase system, as in Fig. 200, the impedance of the equivalent single-phase system is the impedance of one conductor or circuit; if, however, the appa- ratus is ring connected, as shown diagrammatically in Fig. 201, the impedance of the ring-connected part of the circuit has to be reduced to star connection, in the usual manner of reducing a circuit to another circuit of different voltage, by the ratio _ ring voltage star voltage' or, as these voltages are usually called in a three-phase systeni, delta voltage c = Y voltage 448 BALANCED SYMMETRICAL POLYPHASE SYSTEMS 449 That is, all ring voltages are divided, all ring currents multiplied with c; all ring impedances are divided, all ring admittances multiplied with the square of the ratio, c^. For instance, if in a three-phase induction motor with delta- connected circuits, the impedance of each circuit is Z = r -{- jx, and the voltage impressed upon the circuit terminals E, and the motor is supplied over a line of impedance, per line wire, ■^0 = ''o ~r JXo, the motor impedance, reduced to star connection, or Y impe- dance, is Z' = r ^2 = 3 (^ +J^), and the impressed voltage, reduced to Y circuit, and the total impedance of the equivalent single-phase circuit is therefore Zo + Z' = (ro + jxo) + I (^ + i^)- Inversely, however, where this appears more convenient, all quantities may be reduced to ring or delta connection, or one of the ring connections considered as equivalent single-phase circuit, of impedance Z + c^Zo = (r + jx) + 3(ro + jXo). Since the line impedances, line currents and the voltages con- sumed in the lines of a polyphase system are star, or (in a three- phase system) Y quantities, it usually is more convenient to reduce all quantities to Y connection, and use one of the F-cir- cuits as the equivalent single-phase circuit. 304. As an example may be considered the calculation of a long-distance transmission line, delivering 10,000 kw., three-phase power at 60 cycles, 80,000 volts and 90 per cent, power-factor at 100 miles from the generating station, with approximately 10 per cent, loss of power in the transmission line, and with the line conductors arranged in a triangle 6 ft. distant from each other. 29 450 ALTERNATING-CURRENT PHENOMENA 10,000 kw. total power delivered gives 3,333 kw. per line or single-phase branch (F power). 3,333 kw. at 90 per cent, power-factor gives 3,700 kv.-amp. 80,000 volts between the lines gives 80,000 -^ \^ = 46,100 volts from line to neutral, or per single-phase circuit. 3,700 kv.-amp. per circuit, at 46,100 volts, gives 80 amp. per line. 10 per cent, loss gives 333 kw. loss per line, and at 80 amp., this gives a resistance per line, 333,000 ^ 802 = 52 ohms, or, 0.52 ohms per mile. The nearest standard size of wire is No. 0 B. & S., which has a resistance of 0.52 ohms, and a weight of 1680 lb. per mile. Choosing this size of wire so requires for the 300 miles of line conductor, 300 X 1680 = 500,000 lb. of copper. At 0.52 ohms per mile, the resistance per transmission line or circuit of 100 miles length is, r — 52 ohms. The inductance of wire No. 0, with d = 0.325 in. diameter, and 6 ft. = 72 in. distance from the return conductor, is calculated from the formula of line inductance^ as, 2.3 mil-henrys per mile; hence, per circuit, L - 0.23 henry, and herefrom the reactance, X = 2TrfL = 88 ohms. The capacity of the transmission line may be calculated directly, or more conveniently it may be derived from the inductance. If C is the capacity of the circuit, of which the inductance is L, then is the fundamental frequency of oscillation, or natural period, that is, the frequency which makes the length, I, of the line a quarter-wave length. Since the velocity of propagation of the electric field is the ve- ^ "Theoretical Elements of Electrical Engineering." BALANCED SYMMETRICAL POLYPHASE SYSTEMS 451 locity of light, v, with a wave-length, 4 I, the number of waves per second, or frequency of oscillation of the line, is f - — and herefrom then follows: hence, for V I Vlc I = 100 miles, V = 186,000 miles per second, L = 0.23 henry, C = 1.26 mf. and the capacity susceptance, 6 = 2 tt/C = 475 X 10-«. Representing, as approximation, the line capacity by a con- denser shunted across the middle of the line We have, impedance of half the line, Z = ^ +j| = 26 + 44johms. Choosing the voltage at the receiving end as zero vector, e = 46,100 volts, at 90 per cent, power-factor and therefore 43.6 per cent, induc- tance factor, the current is represented by 7 = 80 (0.9 - 0.436 j) =72-35 j. ^ Or. ii fi = permeability, k = dielectric constant of the medium sur- rounding the conductor, it is hence, V [^ I = \W or. C = (4) 452 ALTERNATING-CURRENT PHENOMENA This gives: Voltage at receiver circuit, e = 46,100 volts; current in receiver circuit, Z = 72 — 35 j amp. ; impedance voltage of half the line, ZI = 3410 + 2260 j volts. Hence, the condenser voltage, Ei = e -\- ZI = 49,510 + 2260 j volts; and the condenser current, + jbEi = — 1.1 + 23.8 j amp.; hence, the total, or generator current, 7o = / + jbEi = 70.9 — 11.2 j amp. The impedance voltage of the other half of the line, ZIo — 2330 - 2830 i volts; hence, the generator voltage, Eo = Ei -\- ZIo — 51,840 + 5090 j volts; and the phase angle of the generator current, 11.2 tan 01= ;^^ = 0.158; 0i = 9.0° The phase angle of the generator voltage, tan 02 = - 5Y^ = - 0.098; 02 = - 5.6°; the lag of the generator current, 00 = 0i — 02 = 14.6°; hence the power-factor at the generator, cos 0o = 96.7 per cent. And the power output, 3 [/, eY = 10,000 kw.; the power input, 3 [7o, £"0]^ = 11,190 kw.; the efficiency = 89.35 per cent.; the volt-ampere output, 3 ie = 11,110 T nm hence, for a three-phase system, 1 = 5 ' 6 n and for a quarter-phase system, with two coils in quadrature, n V2 In the investigation of the armature reaction of synchronous machines. Chapter XXII, the armature reaction of an 7«-phase machine is, by §271, Worn/ . 454 ALTERNATING-CURRENT PHENOMENA where m = number of phases, no = number of turns per phase, effective, that is, allow- ing for the spread of turns over an arc of the periph- ery in machines of distributed winding, I = current per phase, and when, in Chapter XX, the armature reaction is given by nl, the number of effective turns, n, is, accordingly, for a polyphase alternator, m hence, in a three-phase machine, n = —7= = 1.5 no v2; ^/2 in a quadrature-phase machine, n = no \/2- 306. When replacing a balanced symmetrical polyphase system by its constituent single-phase systems, it must be considered, that the constants of the constituent single-phase circuit may not be the same which this circuit would have as independent single- phase circuit. If the branches of the polyphase circuit, which constitute the equivalent single-phase circuits, are electrically or magnetic- ally interlinked, the constants, as admittance, impedance, etc., of the equivalent single-phase circuit often are different from those of the same circuit on single-phase supply, and the poly- phase values then must be used in the equivalent single-phase circuits which replace the polyphase system. This is the case in induction machines, in the armatures of synchronous machines, etc., where the phases are in mutual in- duction with each other. Let, in a star or Y-connected three-phase induction motor: Y = g -jb be the exciting admittance and e the impressed voltage per three- phase Y circuit or constituent single-phase circuit. BALANCED SYMMETRICAL POLYPHASE SYSTEMS 455 The exciting current per circuit then is: I = eY or, absolute: i = ey if n = number of turns per circuit, f = ni = effective value of the m.m.f. per phase, and F= 1.5 X V^ ni = resultant m.m.f. of all three phases. F then produces in the magnetic circuit the flux $, which con- sumes the impressed voltage e. Assuming now, that instead of impressing three three-phase voltage e on the three constituent single-phase circuits of the motor, we impress only a single-phase voltage e on one of the three circuits. The current in this circuit then must produce the same flux , and have the same maximum m.m.f. F, as was given by the re- sultant of all three phases. With n turns, that means, the current ii under the single- phase e.m.f. e is given by: V2 nil and since we had, under the same voltage e and flux $, three- phase: F - 1.5 \/2m it follows: ii = 1.5 i That is, with a single-phase voltage, e, the current, ii, and thus the admittance, Fi, of the circuit, is 1.5 times the current, i, and thus the admittance, Y, which is produced in the same circuit by the three-phase voltage: Yi = 1.5 Y or: Y =^V3Yi That is: If we measure the admittance of one of the motor circuits by single-phase supply voltage, this is not the admittance of this circuit as constituent single-phase circuit of the three-phase motor, but The admittance of the constituent or equivalent single-phase 456 ALTERNATING-CURRENT PHENOMENA circuit of a three-phase induction motor is two-thirds of the ad- mittance of this same circuit as independent single-phase circuit. We can look at this in a different way: As the three-phase circuits combine to a resultant which is 1,5 times the m.m.f. of each circuit, each circuit requires only two-thirds of the m.m.f., and thus two-thirds of the exciting admittance, as equivalent single-phase circuit of a three-phase motor, which it would require, if as independent single-phase circuit it had to produce the entire m.m.f. 307. The same applies to the self-inductive reactance: as the self-inductive or leakage flux, which consumes the reactance voltage, is produced by the resultant of the currents of all three phases, and this resultant is 1.5 times the maximum of one phase, each phase produces only two-thirds, that is, the impedance current of each phase of the motor on three-phase voltage supply is only two-thirds that of the same circuit at the same voltage of single-phase supply, and the impedance thus is ^ = 1.5 times. That is: The effective admittance of the equivalent or constituent sin- gle-phase circuit of a three-phase induction machine is two-thirds of the admittance, and the effective impedance is 1.5 times the impedance of this circuit as independent single-phase circuit. The same applies to synchronous machines: The three-phase synchronous reactance per armature circuit, that is, the synchronous reactance of this armature circuit as equivalent single-phase circuit of the three-phase system, is 1.5 times the single-phase synchronous reactance of the same armature circuit, that is, synchronous reactance of this circuit as single-phase machine. In dealing with the constituent single-phase circuits of a three- phase system, the proper "three-phase" values of the constants of the equivalent circuit must be used.