16. PHASE CONTROL OF TRANSMISSION LINES 76. If in the receiving circuit of an inductive transmission line the phase relation can be changed, the drop of voltage in the line can be maintained constant at varying loads or even decreased with increasing load; that is, at constant generator voltage the transmission can be compounded for constant voltage at the receiving end, or even over-compounded for a voltage increasing with the load. 1. Compounding of Transmission Lines for Constant Voltage Let r = resistance, x = reactance of the transmission line, CQ = voltage impressed upon the beginning of the line, e = vol- tage received at the end of end line. PHASE CONTROL OF TRANSMISSION LINES 91 Let i = power current in the receiving circuit; that is, P — ei = transmitted power, and ii = reactive current produced in the system for controlling the voltage. i\ shall be considered positive as lagging, negative as leading current. Then the total current, in symbolic representation, is / = i - jii; the line impedance is Z = r + jx, and thus the e.m.f. consumed by the line impedance is Ei = ZI = (r + jx) (i - jii) = ri + jrii + jxi - J2xii; and substituting f — — 1, Ei = (ri + xii) - j (rii - xi). Hence the voltage impressed upon the line Eo = e 4- Ei = (e + ri + xii) - j (rii - xi) ; (1) or, reduced, _ eo = V(e + ri + xii)* + (n\ - xi)*. (2) If in this equation e and eQ are constant, ii, the reactive com- ponent of the current, is given as a function of the power com- ponent current i and thus of the load ei. Hence either eQ and e can be chosen, or one of the e.m.fs. eQ or e and the reactive current ii corresponding to a given power current i. 76. If ii = 0 with i = 0, and e is assumed as given, eQ = e. Thus, _ e = V(e + ri + xi^ + (rii - xi)*; 2 e (ri + xii) + (r2 + x2) (i2 + ii2) = 0. From this equation it follows that ex ± Ve2x2 - 2 eriz2 - i2z* /ox *i - - - — i~ Thus, the reactive current ii must be varied by this equation to maintain constant voltage e = eo irrespective of the load ei. As seen, in this equation, ii must always be negative, that is, the current leading. 92 ELEMENTS OF ELECTRICAL ENGINEERING ii becomes impossible if the term under the square root becomes negative, that is, at the value e2x2 - 2 eriz2 - i2z4 = 0; 1 ,' f-^4 - (4) At this point the power transmitted is This is the maximum power which can be transmitted with- Q (g _ Y\ out drop of voltage in the line, with a power current i = - ^ — . The reactive current corresponding hereto, since the square root becomes zero, is III ,-.....'.; ti = 5[; . . . (6) thus the ratio of reactive to power current, or the tangent of the phase angle of the receiving circuit, is A larger amount of power is transmitted if e0 is chosen > e, a smaller amount of power if eQ < e. In the latter case ii is always leading; in the former case i\ is lagging at no load, becomes zero at some intermediate load, and leading at higher load. 77. If the line impedance Z — r + fa and the received voltage e is given, and the power current ^o at which the reactive current shall be zero, the voltage at the generator end of the line is determined hereby from the equation (2) : eQ = V(e -f ri + xii)2 + (ri{ - xi)2, by substituting i\ = 0, i = z'o, Substituting this value in the general equation (2) : e0 = V(e + ri gives (e + n0)2 + zV = (e + ri + xitf + (rii - xi)2 (9) as equation between i and i\. PHASE CONTROL OF TRANSMISSION LINES 93 If at constant generator voltage e0: at no load, i = 0, e = e0, i\ = i'o, and at the load, (10) i = i o, 6 = BO, i\ = 0 it is, substituted: no load, load io, Thus, (eo + £fc'o) + #V| = (eo + n'o)2 + x2t*o2; or, expanded, iV(r2 + x2) + 2 i'0 xe0 = io2 (r2 + x2) + 2 iQre0. (13) This equation gives i'o as function of io, e0, r, x. If now the reactive current i\ varies as linear function of the power current i, as in case of compounding by rotary converter with shunt and series field, it is Substituting this value in the general equation (eo + n0)2 + *V = (e + ri + a»i)» + (rii - xz)2 gives e as function of i; that is, gives the voltage at the receiving end as function of the load, at constant voltage 60 at the gener- ating end, and e = eo for no load, i = 0, ii = i'o, and e = eQ for the load, i = io, ii = 0. Between i = 0 and i = io, e > eo, and the current is lagging. Above i = io, e < eQ, and the current is leading. By the reaction of the variation of e from eo upon the receiving apparatus producing reactive current z'i, and by magnetic satura- tion in the receiving apparatus, the deviation of e from eo is reduced, that is, the regulation improved. 2. Over-compounding of Transmission Lines 78. The impressed voltage at the generator end of the line was found in the preceding, eo = V(e 4- ri + a»i)a + (rii - xi)2. (2) 94 ELEMENTS OF ELECTRICAL ENGINEERING If the voltage at the end of the line e shall rise proportionally to the power current i, then e = e\ + ai', (15) thus, eo = V[ej. + (a + r) t + a»i]» + (n'i - **)2, and herefrom in the same way as in the preceding we get the characteristic curve of the transmission. If eo — e\t i\ = 0 at no load, and is leading at load. If £o < ei, ii is always leading, the maximum output is less than before. If eo > ei, i\ is lagging at no load, becomes zero at some inter- mediate load, and leading at higher load. The maximum output is greater than at e0 = e\. The greater a, the less is the maximum output at the same GO and €\. The greater eo, the greater is the maximum output at the same e\ and a, but the greater at the same time the lagging current (or less the leading current) at no load. EXAMPLES 79. (1) A constant voltage of e0 is impressed upon a trans- mission line of impedance Z = r + jx = 10 + 20 j. The vol- tage at the receiving end shall be 10,000 at no load as well as at full load of 75 amp. power current. The reactive current in the receiving circuit is raised proportionally to the load, so as to be lagging at no load, zero at full load or 75 amp., and lead- ing beyond this. What voltage e0 has to be impressed upon the line, and what is the voltage e at the receiving end at J£, %, and 1J£ load? Let J = ii — jiz = current, E = e voltage in receiving circuit. The generator voltage is then Eo = e + ZI = e + (r + jx) (ii - ji2) = (e + rii + xiz) — j (n'2 — xii) = (e + 10 n + 20 i2) - j (10 t, - 20 t'O; or, reduced, eQ* = (e + n'j + xizy 4. (n'2 _ ^2. = (e + 10 ii + 20 tj)J + (10 t, - 20 *i)2. When t'i = 75, t, = 0, e = 10,000; PHASE CONTROL OF TRANSMISSION LINES 95 substituting these values, e02 = 10,7502 + 15002 = 117.81 X 106; eo = 10,860 volts is the generator voltage. hence, When ii = 0, e = 10,000, eQ = 10,860, let i2 = i: these values substituted give 117.81 X 106 = (10,000 + 20 i)2 + 100 i2 = 100 X 106 + 400 i X 103 + 500 *2, or, i = 44.525 - 1.25 i2 10~3; this equation is best solved by approximation, and then gives p = 42.3 amp. reactive lagging current at no load. Since eo2 = (e + rii + xi2)2 + (riz - xii)2, it follows that e = Veo2 — (™*2 — xii)2 — (rii + 2^2); or, e = V117.81 + 106 - (10 i* - 20 n)2 - (10 ii + 20 t2). Substituting herein the values of ii and i2 gives e. t'l tz e 0 42.3 10,000 25 28.2 10,038 50 14.1 10,038 75 0 10,000 100 -14.1 9,922 125 -28.2 9,803 80. (2) A constant voltage eQ is impressed upon a trans- mission line of impedance Z = r + jx = 10 + 10 j. The vol- tage at the receiving end shall be 10,000 at no load as well as at full load of 100 amp. power current. At full load the total current shall be in phase with the e.m.f. at the receiving end, and at no Load a lagging current of 50 amp. is permitted. How much additional reactance x0 is to be inserted, what must be the generator voltage e0, and what will be the voltage e at the receiv- 96 ELEMENTS OF ELECTRICAL ENGINEERING ing end at % load and at 1J^ load, if the reactive current varies proportionally with the load? Let XQ = additional reactance inserted in circuit. Let I — i\ — jiz = current. Then e02 = (e + rii + x^)2 + (na - Xiii) 2= (e + 10 ii + x^Y + (10 i2 - where Xi = x + #o = total reactance of circuit between e and eQ. At no load, ii = 0, i2 = 50, e = 10,000; thus, substituting, e02 = (10,000 + 50 zi)2 + 250,000. At full load, ii = 100, ia = 0, e = 10,000; thus, substituting, eo2 = 121 X 106 + 10,000 xj. Combining these gives (10,000 + 50 zi)2 + 250,000 = 121 X 106 + 10,000 a^i2; hence, xi = 66.5 + 40.8 = 107.3 or 25.7; thus XQ = Xi — x = 97.3 or 15.7 ohms additional reactance. Substituting xi = 25.7 gives e02 = (e + 10 ^ + 25.7 i2)2 + (10 i* - 25.7 ii)2, but at full load ii = 100, i2 = 0, e = 10,000, which values substituted give e02 = 121 X 106 + 6.605 X 106 = 127.605 X 106, eo = 11,300, generator voltage. Since e = vV - (10 12 - 25.7 *i)2 - (10 ii + 25.7 i2), it follows that 6 = V127.605 X 106 - (10 it - 25.7 ii)2 - (10 ii + 25.7 t2). Substituting for ii and z*2 gives e. PHASE CONTROL OF TRANSMISSION LINES 97 u ti e 0 50 10,000 50 25 10,105 100 0 10,XKX) 150 -25 9,658 81. (3) In a circuit whose voltage e0 fluctuates by 20 per cent, between 1800 and 2200 volts, a synchronous motor of internal impedance Z0 = r0 + jx0 = 0.5 + 5 j is connected through a reactive coil of impedance Z\ = r\ + jx\ = 0.5 -f- 10 j and run light, as compensator (that is, generator of reactive currents). How will the voltage at the synchronous motor terminals e\, at constant excitation, that is, constant counter e.m.f. e = 2000, vary as function of e$ at no load and at a load of i = 100 amp. power current, and what will be the reactive current in the synchronous motor? Let I = ii — jiz = current in receiving circuit of voltage e\. Of this current 1,—jiz is taken by the synchronous motor of counter e.m.f. 'e, and thus EI = e — Zoji2 = e + X0i2 - jr0i2'} or, reduced, e^= (e + xoit)2 + rjif. In the supply circuit the voltage is Eo = Ei + IZl = e + xoi* - jrQi2 + (ii-jiz) (TI + jxi) = [e + riii + (xQ + xi) iz] — j [(rQ + TI) i2 — ' or, reduced, eo2 = [e + riti + (XQ + xi) itf + [(r0 + fi) i2 Substituting in the equations for e^ and e02 the above values of r0 and XQ: at no load, i\ = 0, we have 6l2 = (e + 5 z2)2 + 0.25 i22 and e02 = (e + at full load, i\ = 100, we have ei2 = (e + 5 i2)2 + 0.25 z22, + = (e + 50 + 15 - 1000)2, 98 ELEMENTS OF ELECTRICAL ENGINEERING and at no load, i\ = 0, substituting e = 2000, we have d2 = (2000 + 5 i2)2 + 0.25 z22, eo2 = (2000 + 15 *2)2 + *22; at full load, i\ = 100, we have ea2 = (2000 + 5 *2)2 + 0.25 *22, e02 = (2050 + 15 z2)2 + fe - 1000)2. Substituting herein e0 = successively 1800, 1900, 2000, 2100, 2200, gives values of i'2, which, substituted in the equation for ei2, give the corresponding values of ei as recorded in the follow- ing table. As seen, in the local circuit controlled by the synchronous compensator, and separated by reactance from the main circuit of fluctuating voltage, the fluctuations of voltage appear in a greatly reduced magnitude only, and could be entirely eliminated by varying the excitation of the synchronous compensator. e = 2000 No load ii = 0 Full load ' 11 - 100 H ei H e\ 1,800 -13.3 1,937 -39 1,810 1,900 - 6.7 1,965 -30.1 1,850 2,000 0 2,000 -22 1,885 2,100 + 6.7 2,035 -13.5 1,935 2,200 + 13.3 2,074 - 6.5 1,970