12. IMPEDANCE OF TRANSMISSION LINES 54. Let r = resistance; x = 2 irfL = the reactance of a trans- mission line; E0 = the alternating e.m.f. impressed upon the line; I = the line current; E = the e.m.f. at receiving end of the line, and 6 = the angle of lag of current 7 behind e.m.f. E. B < 0 thus denotes leading, 0 > 0 lagging current, and 6 = 0 a non-in- ductive receiver circuit. The capacity of the transmission 0 line shall be considered as negligible. FIG. 27.— Vector diagram ,1 i f , v i. of current and e.m.fs. in a _Assummg the phase of the current transmission line assuming QI = / as zero in the polar diagram, zero capacity. Fig. 27, the e.m.f. E is represented by vector OE, ahead of 07 by angle 0. The e.m.f. consumed by re- sistance r is OEi = Ei = Ir in phase with the current, and the e.m.f. consumed by reactance x is OE% = Ez = Ix, 90 time de- grees ahead of the current; thus the total e.m.f. consumed by the line, or e.m.f. consumed by impedance, is the resultant OES of and O#2, jind is E3 = Iz. Combining OEz and OE gives OEQ, the e.m.f. impressed upon the line. 58 ELEMENTS OF ELECTRICAL ENGINEERING Denoting tan 0i = - the time angle of lag of the line impe- dance, it is, trigonometrically, Since OE02 = OE2 + EEQ2 - 2 OE X EEQ cos ~EEo = OE* = Iz, OEEQ = 180 - 0i + 6, FIG. 28. — Locus of the generator and receiver e.m.fs. in a transmission line with varying load phase angle. E02 = E2 + I2z2 + 2 EIz cos (0! - 6) = (E + Iz)2 - 4 #/z sin2 ^-^, we have and E0 = \I(E -f- Iz)2 — 4 EIz sin2 -^—= — , and the drop of voltage in the line, EQ - E = \ (E + Iz}2 - 4 EIz sin2 -^ E. IMPEDANCE OF TRANSMISSION LINES 59 65. That is, the voltage EQ required at the sending end of a line of resistance r and reactance x, delivering current / at vol- tage E} and the voltage drop in the line, do not depend upon current and line constants only, but depend also upon the angle of phase displacement of the current delivered over the line. If 0 = o, that is, non-inductive receiving circuit, FIG. 29. — Locus of the generator and receiver e.m.fs. in a transmission line with varying load phase angle. E0 = - 4 EIz sin21; that is, less than E + Iz, and thus the line drop is less than Iz. If 0 — 6 1, EQ is a maximum, = E + Iz, and the line drop is the impedance voltage. With decreasing 0, E0 decreases, and becomes = E; that is, no drop of voltage takes place in the line at a certain negative 60 ELEMENTS OF ELECTRICAL ENGINEERING value of 0 which depends not only on z and 0i but on E and 7. Beyond this value of 6, EQ becomes smaller than E; that is, a rise of voltage takes place in the line, due to its reactance. This can be seen best graphically. Choosing the current vector 01 as the horizontal axis, for the same e.m.f. E received, but different phase angles 6, all vectors OE lie on a circle e with 0 as center. Fig. 28. Vector OEz is constant for a given line and given current 7. Since E3EQ = OE = constant, E0 lies on a circle eQ with Es as center and OE = E as radius. To construct the diagram for angle 0, OE is drawn at the angle 0 with 07, and EEQ parallel to ~OE*. The distance E±EQ between the two circles on vector OEo is the drop of voltage (or rise of voltage) in the line. As seen in Fig. 29, EQ is maximum in the direction OE% as OE'0, that is, for 0 = 00, and is less for greater as well, OE" 'o, as smaller angles 6. It is = E in the direction 072"'o, in which case 0 < 0, and minimum in the direction The values of E corresponding to the generator voltages E'Q, E"0, E'"Q, #IV0 are shown by the points E' E" Ef" E™ respectively. The voltages E"Q and Elv$ correspond to a wattless receiver cir- cuit E" and E™. For non-inductive receiver circuit W the generator voltage is OEvo. 56. That is, in an inductive transmission line the drop of voltage is maximum and equal to Iz if the phase angle 0 of the receiving circuit equals the phase angle 00 of the line. The drop of voltage in the line decreases with increasing difference be- tween the phase angles of line and receiving circuit. It becomes zero if the phase angle of the receiving circuit reaches a certain negative value (leading current). In this case no drop of vol- tage takes place in the line. If the current in the receiving cir- cuit leads more than this value a rise of voltage takes place in the line. Thus by varying phase angle 9 of the receiving circuit the drop of voltage in a transmission line with current 7 can be made anything between Iz and a certain negative value. Or inversely the same drop of voltage can be produced for different values of the current 7 by varying the phase angle. / Thus, if means are provided to vary the phase angle of the receiving circuit, by producing lagging and leading currents at will (as can be done by synchronous motors or converters) , the voltage at the receiving circuit can be maintained constant IMPEDANCE OF TRANSMISSION LINES 61 within a certain range irrespective of the load and generator voltage. In Fig. 30 let OE = E, the receiving voltage; /, the power component of the line current; thus OES = Es = Iz, the e.m.f. consumed by the power component of the current in the impe- dance. This e.m.f. consists of the e.m.f consumed by resistance ^Ei and the e.m.f. consumed by reactance OEz- FIG. 30. — Regulation diagram for transmission line. Reactive components of the current are represented in the diagram in the direction OA when lagging and OB when leading. The e.m.f. consumed by these reactive components of the current in the impedance is thus in the direction e1 '3, perpendicular to OEs- Combining OEz and OE gives the e.m.f. OE* which would be required for non-inductive load. If EQ is the generator voltage, EQ lies on a circle eQ with O^o as radius. Thus drawing E^E0 par- allel to e'z gives OEQ, the generator voltage; OE'S = EJZo, the 62 ELEMENTS OF ELECTRICAL ENGINEERING e.m.f. consumed in the impedance by the reactive component of the current; and as proportional thereto, OI' = I', the reactive current required to give at generator voltage E0 and power cur- rent 7 the receiver voltage E. This reactive current 7' lags be- hind E'z by less than 90 and more than zero degrees. 57. In calculating numerical values, we can pro'ceed either trigonometrically as in the preceding, or algebraically by resolv- ing all sine waves into two rectangular components; for instance, a horizontal and a vertical component, in the same way as in mechanics when combining forces. Let the horizontal components be counted positive toward the right, negative toward the left, and the vertical components positive upward, negative downward. Assuming the receiving voltage as zero line or positive hori- zontal line, the power current 7 is the horizontal, the wattless current I' the vertical component of the current. The e.m.f. con- sumed in resistance by the power current 7 is a horizontal com- ponent, and that consumed in resistance by the reactive current /' a vertical component, and the inverse is true of the e.m.f. consumed in reactance. We have thus, as seen from Fig. 30: Horizontal Vertical component component Receiver voltage, E, + E 0 Power current, /, + 7 0 Reactive current, 7', 0 HF 7' E.m.f. consumed in resistance r by the power current, Ir, + Ir 0 E.m.f. consumed in resistance r by the reactive current, 7'r, 0 + 7V E.m.f. consumed in reactance x by the power current, Ix, 0 + Ix E.m.f. consumed in reactance x by the reactive current, I'x, ± I'x 0 Thus, total e.m.f. required, or impressed e.m.f., Eo, E + Ir± I'x + I'r + Ix; hence, combined, Eo = ^/(E + Ir± or, expanded, + 2 E (Ir ± I'x) + (72 + 7'2)*2- IMPEDANCE OF TRANSMISSION LINES 63 From this equation I' can be calculated ; that is, the reactive current found which is required to give E0 and E at energy current 7. The lag of the total current in the receiver circuit behind the receiver voltage is tan 0 = j. The lead of the generator voltage ahead of the receiver voltage is vertical component of EQ horizontal component of EQ B + Ir± 7V and the lag of the total current behind the generator voltage is As seen, by resolving into rectangular components the phase angles are directly determined from these components. The resistance voltage is the same component as the current to which it refers. The reactance voltage is a component 90 time degrees ahead of the current. The same investigation as made here on long-distance trans- mission applies also to distribution lines, reactive coils, trans- formers, or any other apparatus containing resistance and reactance inserted in series into an alternating-current circuit. EXAMPLES 58. (1) An induction motor has 2000 volts impressed upon its terminals; the current and the power-factor, that is, the cosine of the angle of lag, are given as functions of the output in Fig. 31. The induction motor is supplied over a line of resistance r = 2.0 and reactance x = 4.0. (a) How must the generator voltage eQ be varied to maintain constant voltage e = 2000 at the motor terminals, and (b) At constant generator voltage CQ = 2300, how will the voltage at the motor terminals vary? 64 ELEMENTS OF ELECTRICAL ENGINEERING We have / a* — ft —. e = 2000. 63.4°. «= -i l(e + iz)2 — 4 ezz sin2 tan 0i = ~ = 2. cos 0 = power-factor. Taking i from Fig. 31 and substituting, gives (a) the values of e0 for e = 2000, which are recorded in the table, and plotted in Fig. 31. JTPUT .10 20 30 40 50 60 70 80 90 100 110 .120 130 140 150 160 170 180 190 200 FIG. 31. — Characteristics of induction motor and variation of generator e.m.f. necessary to maintain constant the e.m.f. impressed upon the motor. (6) At the terminal voltage of the motor e = 2000, the cur- rent is i, the output P, the generator voltage eQ. Thus at gen- erator voltage e'o = 2300, the terminal voltage of the motor is the current is and the power is 2300 . p, The values of e', i' , P' are recorded in the second part of the table under (6) and plotted in Fig. 32. IMPEDANCE OF TRANSMISSION LINES 65 (a) At e = 2000 Thus, M (6) Hence, at eo = 2300 Output, P = kw. Current, t Lag. Output, Current, »' Voltage, 0 12.0 84.3° 2048 0. 13.45 2240 5 12.6 72.6° 2055 6.25 14.05 2234 10 13.5 62.6° 2060 12.4 15.00 2230 15 14.8 54.6° 2065 18.6 16.4 2220 20 16.3 47.9° 2071 24.4 18.0 2216 30 20.0 37.8° 2084 36.3 22.0 2200 40 25.0 32.8° 2093 48.0 27.5 2198 50 30.0 29.0° 2110 59.5 32.7 2180 69 40.0 26.3° 2146 78.5 42.8 2160 102 60.0 24.5° 2216 110.2 62.6 2080 132 80.0 25.8° 2294 131.0 79.5 1990 160 100.0 28.4° 2382 149.0 96.4 1928 180 120.0 31.8° 2476 156.5 111.5 1860 200 150.0 36.9° 2618 155.0 132.0 1760 7 L IS 30 36 40 50 60 JO 80 90 100 110 120 130 !40 FIG. 32. — Characteristics of induction motor, constant generator e.m.f. 5 66 ELEMENTS OF ELECTRICAL ENGINEERING 69. (2) Over a line of resistance r = 2.0 and reactance x = 6.0 power is supplied to a receiving circuit at a constant voltage of e = 2000. How must the voltage at the beginning of the line, or generator voltage, eQ, be varied if at no load the receiving circuit consumes a reactive current of i* — 20 amp., this reac- tive current decreases with the increase of load, that is, of power current i\, becomes iz = 0 at i\ = 50 amp., and then as leading current increases again at the same rate? REACT CURRE FIG. 33. — Variation of generator e.m.f. necessary to maintain constant receiver voltage if the reactive component of receiver current varies propor- tional to the change of power component of the current. The reactive current, = 20 at = 0 at and can be represented by = 0, = 50, •*2= l - 20 = 20 -0.4 ti; the general equation of the transmission line is £o = V (e H- iir -f izx) = V(2000 + 2n + hence, substituting the value of z*2, (2*2 - e0 = V(2120 - 0.4 n)2 + (40 - 6.8ti)a = V4,496,000 + 46.4 if - 2240 ^. ALTERNATING-CURRENT TRANSFORMER 67 Substituting successive numerical values for ii gives the values recorded in the following table and plotted in Fig. 33. ii eo 0 '2120 20 2114 40 2116 60 2126 80 2148 100 2176 120 2213 140 2256 160 2308 180 2365 200 2430