9. VECTOR DIAGRAMS 42. The best way of graphically representing alternating-cur- rent phenomena is by a vector diagram. The most frequently used vector diagram is the crank diagram. In this, sine waves of alternating currents, voltages, etc., are represented as projec- tions of a revolving vector on the horizontal. That is, a vector equal in length to the maximum value of the alternating wave is assumed to revolve at uniform speed so as to make one complete revolution per period, and the projections of this revolving vec- tor upon the horizontal then represent the instantaneous values of the wave. Let, for instance, 01 represent in length the maximum value of current i = I cos (6 — 00). Assume then a vector, 07, to revolve, left-handed or in positive direction, so that it makes a 42 ELEMENTS OF ELECTRICAL ENGINEERING complete revolution during each cycle or period. If then at a certain moment of time this vector stands in position OIi (Fig. 14), the projection, OA^' of Oh on OA represents the instan- taneous value of the current at this moment. At a later moment 07 has moved farther, to 0/2, and the projection, OAZ, of 072 on OA is the instantaneous value. The diagram thus shows the instantaneous condition of the sine waves. Each sine wave FIG. 14. — Crank diagram showing instantaneous values. FIG. 15. — Crank diagram of an e.m.f. and current. reaches the maximum at the moment when its revolving vector, 01, passes the horizontal, and reaches zero when its revolving vector passes the vertical. If Fig. 15 represents the crank diagram of a voltage OE, and a current^O/, and if angle AOE^AOI, this means that the current 01 is behind the voltage OE, passes during the revolu- tion the zero line or line of maximum intensity, OA, later than the voltage; that is, the current lags behind the voltage. In the vector diagram, the first quantity therefore can be put in any position. For_instance, the current 01, in Fig. 15, could be drawn in position 01, Fig. 16. The voltage then being ahead VECTOR DIAGRAMS 43 of the current by angle EOI = 0 would come into the position OE, Fig. 16. This vector diagram then shows graphically, by the projections of the vectors on the horizontal, the instantaneous values of the alternating waves at one moment of time. At any other moment FIG. 16. — Crank diagram. of time, the instantaneous values would be the projections of the vectors on another radius, corresponding to the other time. The angles between the vector representation are the phase differ- ences between the vectors, and the angles each vector makes with the horizontal may be called its phase. The horizontal then FIG. 17. — Vector diagram of two e.m.f.'s acting in the same circuit. would be of phase zero. The phase of the first vector may be chosen at random; all other phases are determined thereby. In this representation, the phase of an alternating wave is given by the time when its maximum value passes the horizontal. 44 ELEMENTS OF ELECTRICAL ENGINEERING Two voltages, e\ and e2, acting in the same circuit, give a resultant voltage e equal to the sum of their instantaneous values. Graphically, voltages ei and e% are represented in intensity and in phase by two revolving vectors, OEi and OEZ, Fig. 17. The instantaneous values are the projections Oei, Oe2 of OEi and OEZ upon the horizontal. Since the sum of the projections of the sides of a parallelogram is equal to thejDrojection of the diagonal,jthe sum of the projec- tions Oei and Oez equals the projection Oe of OE, the diagonal of the parallelogram with OEi and OEZ as sides, and OE is thus the resultant e.m.f . ; that is, graphically alternating sine waves of voltage, current, etc., are combined and resolved by the parallelo- gram or polygon of sine waves. FIG. 18. — Vector diagram. 43. The sine wave of alternating current i = I0 sin 0 is repre- sented by a vector equal in length, 01 0, to the maximum value 70 of the wave, and located so that at time zero 0=0, its projec- tion on the horizontal, is zero, and at times 0 > 0, but < TT, the projection is positive. Thus this vector 0/0 is the negative vertical, as shown in Fig. 18. The voltage consumed by inductance, ez = x!0 cos 0, is repre- sented by a vector OEZ equal in length to x!Q, and located so that at 0 = 0, its projection on the horizontal is a maximum. That is, it is the zero vector OE2 in Fig. 18. Analogously, the counter e.m.f. of self-inductance E'2 is represented by vector OE'Z on the negative horizontal of Fig. 18; the voltage consumed by the resistance r, e\ — e!Q sin 0, is represented by vector OEi equal to r/0, and located on the nega- VECTOR DIAGRAMS 45 tive vertical, and the counter e.m.f. of resistance by vector OE'i on the positive vertical. The counter e.m.f. of impedance: — (r/o sin 0 + x!Q cos 0) - ?Jn sin (ft -\- fi»} sin (6 + 00) then is represented graphically as the resultant, by the parallelo- gram of sine waves of OE\ and OE'2} that is, by a vector OE', equal in length to z!0, and of phase 90 + 00. The voltage consumed by impedance, or the impressed voltage, is represented by the vector OE, equal and opposite in direction to the vector OE' . This vector is the resultant of OEi and OE2 and has the phase 00 — 90, or — (90 — 00), as shown in Fig. 18. An alternating wave is thus determined by the length and direc- tion of its vector. The length is the maximum value, intensity or amplitude of the wave; the direction is the phase of its maximum value, usually called the phase of the wave. 44. As phase of the first quantity considered, as in the above instance the current, any direction can be chosen. The further quantities are determined thereby in direction or phase. The zero vector OA is generally chosen for the most frequently used quantity or reference quantity, as for the current, if a num- ber of e.m.fs. are considered in a circuit of the same current, or for the e.m.f., if a number of currents are produced by the same e.m.f., or for the generated e.m.f. in apparatus such as transform- ers and induction motors, synchronous apparatus, etc. With the current as zero vector, all horizontal components of e.m.f. are power components, all vertical components are reac- tive components. With the e.m.f. as zero vector, all horizontal components of current are power components, all vertical components of current are reactive components. By measurement from the vector diagram numerical values can hardly ever be derived with sufficient accuracy, since the magnitudes of the different quantities used in the same diagram are usually by far too different, and the vector diagram is there- fore useful only as basis for trigonometrical or other calculation, and to give an insight into the mutual relation of the different quantities, and even then great care has to be taken to distinguish between the two equal but opposite vectors, counter e.m.f. and e.m.f. consumed by the counter e.m.f., as explained before. 46 ELEMENTS OF ELECTRICAL ENGINEERING EXAMPLES 45. In a three-phase long-distance transmission line, the vol- tage between lines at the receiving end shall be 5000 at no load, 5500 at full load of 44 amp. power component, and propor- tional at intermediary values of the power component of the current; that is, the voltage at the receiving end shall increase proportional to the load. At three-quarters load the current shall be in phase with the e.m.f. at the receiving end. The generator excitation, however, and thus the (nominal) generated FIG. 19. — Vector diagram of e.m.f. and current in transmission line. Cur- rent leading. e.m.f. of the generator shall be maintained constant at all loads, and the voltage regulation effected by producing lagging or leading currents with a synchronous motor in the receiving cir- cuit. The line has a resistance rx = 7.6 ohms and a reactance Xi = 4.35 ohms per wire, the generator is star connected, the resistance per circuit being r2 = 0.71, and the (synchronous) reactance is x2 = 25 ohms. ^ What must be the wattless or re- active component of the current, and therefore the total current and its phase relation at no load, one-quarter load, one-half load, three-quarters load, and full load, and what will be the terminal voltage of the generator under these conditions? The total resistance of the line and generator is r = TI + r2 = 8.31 ohms; the total reactance, x = Xi + #2 = 29.35 ohms. Let, in the polar diagram, Fig. 19 or 20, OE = E represent the voltage at the receiving end of the line, OIi = I\ the power component of the current corresponding to the load, in phase with OE, and 0/2 = Iz the reactive component of the current in quadrature with OE, shown leading in Fig. 19, lagging in Fig. 20. We then have total current / = 01. VECTOR DIAGRAMS 47 Thus the e.m.f. consumed by resist ance_,j9£'i = rl, is in phase with 7,the e.m.f. consumed by reactance, OEz = xl, is 90 degrees ahead of /, and their resultant is OE3, the e.m.f. consumed by impedance. OW3 combined with 0#, the receiver voltage, gives the genera- tor voltage OE0. FIG. 20. — Vector diagram of e.m.f. and current in transmission line. Cur- rent lagging. Resolving all e.m.fs. and currents into components in phase and in quadrature with the received voltage E, we have Current E.m.f. at receiving end of line, E = E.m.f. consumed by resistance, EI = E.m.f. consumed by reactance, E2 = Thus total e.m.f. or generator voltage, E0 = E + E! + E2 = E + Herein the reactive lagging component of current is assumed as positive, the leading as negative. The generator e.m.f. thus consists of two components, which give the resultant value Phase component Quadrature component /I ~/2 E 0 r/i -r/2 xlz + Z/1 + xI xll - r/ E, = V(E + rh + xI,Y + (xh - r/2)2; substituting numerical values, this becomes + 8.31 At three-quarters load, 5375 + 29.35 72)2 + (29.35 A - 8.31 72)2. E = 3090 volts per circuit, 48 ELEMENTS OF ELECTRICAL ENGINEERING /i = 33, 72 = 0, thus Eo = \/(3090 + 8.31 X 33)2 + (29.35 X 33)2 = 3520 volts per line or 3520 X \/3 = 6100 volts between lines as (nominal) generated e.m.f. of generator. Substituting these values, we have 3520 = \/(E + 8.31 7i + 29.35 72)2 + (8.31 12 - 29.35 /O2. The voltage between the lines at the receiving end shall be: No U M H Full load load load load load Voltage between lines, 5000 5125 5250 5375 5500 Thus, voltage per line -5- \/3, # = 2880 2950 3020 3090 3160 The power components of current per line, II = 0 11 22 33 44 Herefrom we get by substituting in the above equation Reactive component of £°d lotd lo^d ^ ^d current, 72 = -21.6 -16.2 -9.2 0 +9.7 hence, the total current, + /22 = 21.6 19.6 23.9 33.0 45.05 and the power factor, ^ = cos 0 = 0 56.0 92.0 100.0 97.7 the lag of the current, 0 = 90° 61° 23° 0° -11.5° the generator terminal voltage per line is E' = V(E + rj, = V(E + 7.6 A + 4.35 72)2 + (4.35 II - 7.6 72)2 thus: No \i M H. Full load load load load load Per line, E'_ = 2980 3106 3228 3344 3463 Between lines, E' V3 = 5200 5400 5600 5800 6000 Therefore at constant excitation the generator voltage rises with the load, and is approximately proportional thereto.