CHAPTER XIX. SYNCHRONOUS MOTOR. 198. In the chapter on synchronizing alternators we have seen that when an alternator running in synchronism is connected with a system of given E.M.F., the work done by the alternator can be either positive or negative. In the latter case the alternator consumes electrical, and consequently produces mechanical, power ; that is, runs as a synchronous motor, so that the investigation of the synchronous motor is already contained essentially in the equations of parallel-running alternators. Since in the foregoing we have made use mostly of the symbolic method, we may in the following, as an instance of the graphical method, treat the action of the synchronous motor diagrammatically. Let an alternator of the E.M.F., E±, be connected as synchronous motor with a supply circuit of E.M.F., EQ, by a circuit of the impedance Z. If E0 is the E.M.F. impressed upon the motor termi- nals, Z is the impedance of the motor of induced E.M.F., E±. If E0 is the E.M.F. at the generator terminals, Z is the impedance of motor and line, including transformers and other intermediate apparatus. If EQ is the induced E.M.F. of the generator, Z is the sum of the impedances of motor, line, and generator, and thus we have the prob- lem, generator of induced E.M.F. EQ, and motor of induced' E.M.F. El; or, more general, two alternators of induced E.M.Fs., E0, Elf connected together into a circuit of total impedance, Z. Since in this case several E.M.Fs. are acting in circuit 322 ALTERNATING-CURRENT PHENOMENA. with the same current, it is convenient to use the current, /, as zero line OI of the polar diagram. Fig. 188. If I=i= current, and Z = impedance, r = effective resistance, x = effective reactance, and s = Vr2 -f x2 = absolute value of impedance, then the E.M.F. consumed by the resistance is E,, = ri, and in phase with the cur- rent, hence represented by vector OE,, ; and the E.M.F. consumed by the reactance is E2 = xi, and 90° ahead of the current, hence the E.M.F. consumed by the impedance is E = V(£,,)2 + (E2f, or = i Vr2 + x* = is, and ahead of the current by the angle 8, where tan 8 = x / r. We have now acting in circuit the E.M.Fs., E, Elf EQ; or El and E are components of EQ ; that is, EQ is the diagonal of a parallelogram, with El and E as sides. Since the E.M.Fs. Elf Ez, E, are represented in the diagram, Fig. 138, by the vectors OE~lf OE2, OE, to get the parallelogram of £Q, Elt E, we draw arcs of circles around 0 with EQ , and around E with El . Their point of intersection gives the impressed E.M.F., OEQ = EQ, and completing the parallelogram OE EQ E± we get, OE± = E± , the induced E.M.F. of the motor. IOE0 is the difference of phase between current and im- pressed E.M.F., or induced E.M.F. of the generator. IOEi is the difference of phase between current and in- duced E.M.F. of the motor. And the power is the current /times the projection of the E.M.F. upon the current, or the zero line OI. Hence, dropping perpendiculars, E^EJ and E^E^, from EQ and E! upon OI, it is — P0 = iX OE^ = power supplied by induced E.M.F. of gen- erator. PI = / X OE^ = electric power transformed in mechanical power by the motor. P = / x OEl = power consumed in the circuit by effective resistance. SYNCHRONOUS MOTOR. 323 Since the circles drawn with EQ and E± around O and K respectively intersect twice, two diagrams exist. In gen- eral, in one of these diagrams shown in Fig. 138 in drawn Fig. 138. lines, current and E.M.F. are in the same direction, repre- senting mechanical work done by the machine as motor- In the other, shown in dotted lines, current and E.M.F. are in opposite direction, representing mechanical work con- sumed by the machine as generator. Under certain conditions, however, £Q is in the same, E^ in opposite direction, with the current ; that is, both ma- chines are generators. 199. It is seen that in these diagrams the E.M.Fs. are- considered from the point of view of the motor ; that is,. 324 ALTERNATING-CURRENT PHENOMENA. work done as synchronous motor is considered as positive, work done as generator is negative. In the chapter on syn- chronizing generators we took the opposite view, from the generator side. In a single unit-power transmission, that is, one generator supplying one synchronous motor over a line, the E.M.F. consumed by the impedance, E = OE, Figs. 139 to 141, con- sists of three components ; the E.M.F. OE£ — Ez, consumed Fig. 139. by the impedance of the motor, the E.M.F. consumed by the impedance of the line, and the E.M.F. EZ E = E± consumed by the impedance of the generator. Hence, dividing the opposite side of the parallelogram E1E(), in the same way, we have : OEl = E1 = induced E.M.F. of the motor, OEZ = 2?a = E.M.F. at motor terminals or at end of line, OE3 = E3 = E.M.F. at generator terminals, or at beginning of line. OEQ = EQ = induced E.M.F. of generator. SYNCHRONOUS MOTOR. 325 The phase relation of the current with the E.M.Fs. £lt , depends upon the current strength and the E.M.Fs. El and 200. Figs. 139 to 141 show several such diagrams for different values of Elf but the same value of / and EQ. The motor diagram being given in drawn line, the genera- tor diagram in dotted line. Fig. 140. As seen, for small values of E1 the potential drops in the alternator and in the line. For the value of E1 = E0 the potential rises in the generator, drops in the line, and rises again in the motor. For larger values of Ely thfe potential rises in the alternator as well as in the line, so that the highest potential is the induced E.M.F. of the motor, the lowest potential the induced E.M.F. of the gen- erator. 326 ALTERNATING-CURRENT PHENOMENA, It is of interest now to investigate how the values of these quantities change with a change of the constants. Fig. 747. 201. A. — Constant impressed E.M.F. Ev, constant current strength I = i, variable motor excitation Ev (Fig. 142.) If the current is constant, = z; OE, the E.M.F. con- sumed by the impedance, and therefore point E, are con- stant. Since the intensity, but not the phase of EQ is constant, EQ lies on a circle eQ with EQ as radius. From the parallelogram, OE EQ El follows, since E1 EQ parallel and = OE, that El lies on a circle el congruent to the circle eQ, but with Ei} the image of E, as center : OEi = OE. We can construct now the variation of the diagram with the variation of El ; in the parallelogram OE EQ E1 , O and E are fixed, and E0 and El move on the circles £Q. In the first case, El = EQ (Fig. 127), we see that at Fig. 144. very small current, that is very small OE, the current / leads the impressed E.M.F. EQ by an angle EQOf = WQ. This lead decreases with increasing current, becomes zero, and afterwards for larger current, the current lags. Taking now any pair of corresponding points E, EQ, and producing EEQ until it intersects eit in Eif we have ^^ Ei OE — 90°, El = EQ , thus : OE1 = EEQ=OEQ = EQEt ; that is, EE{ = SYNCHRONOUS MOTOR. 331 2EQ. That means the characteristic curve el is the enve- lope of lines EEiy of constant lengths 2EQ, sliding between the legs of the right angle Et OE; hence, it is the sextic hypocyloid osculating circle EQ (Fig. 145) the current cannot equal zero either, but begins at a finite value C±, corresponding to the minimum value of OEQ : // = * — -. At this value however, the alternator E1 is still generator and changes to a motor, its power passing through zero, at the point corresponding to the vertical tangent, onto elf with a very large lead of the impressed E.M.F. against the cur- rent. At H the lead changes to lag. The minimum and maximum value of current in the three conditions are given by : Minimum: Maximum: 1st. 7=0, 7=^. Since tfie current passing over the line at El = O, that is, when the motor stands still, is 70 = EQj z, we see that in such a synchronous motor-plant, when running at syn- chronism, the current can rise far beyond the value it has at standstill of the motor, to twice this value at 1, some- what less at 2, but more at 3. 203. C. EQ = constant, El varied so that the efficiency is a maximum for all currents. • (Fig. 146.) Since we have seen that the output at a given current strength, that is, a given loss, is a maximum, and therefore SYNCHRONOUS MOTOR. 333 the efficiency a maximum, when the current is in phase with the induced E.M.F. EQ of the generator, we have as the locus of EQ the point EQ (Fig. 146), and when E with increasing current varies on ! cos ft,^), (1) thus, — * If f0 = E.M.F. at motor terminals, z = internal impedance of the motor; if eo= terminal voltage of the generator, z = total impedance of line and motor; if t0= E.M.F. of generator, that is, E.M.F. induced in generator armature by its rotation through the magnetic field, z includes the generator impedance also. SYNCHRONOUS MOTOR. 339 The displacement of phase between current i and E.M.F. = z i consumed by the impedance z is : cos (ie) = - sin (/ The parameter <^> has no direct physical meaning, appar- ently. These equations (19) and (20), by giving the values ef el and i as functions of / and the parameter <£ enable us to construct the Power Characteristics of the Synchronous Motor, as the curves relating ev and i, for a given power /, by attributing to <£ all different values. 342 ALTERNATING-CURRENT PHENOMENA. Since the variables v and w in the equation of the circle (16) are quadratic functions of e1 and /', the Power Charac- teristics of the Synchronous Motor are Quartic Curves. They represent the action of the synchronous motor under all conditions of load and excitation, as an element of power transmission even including the line, etc. Before discussing further these Power Characteristics, some special conditions may be considered. 206. A. Maximum Output. Since the expression of el and i [equations (19) and (20)] contain the square root, W02 — 4 rp, it is obvious that the maximum value of / corresponds to the moment where this square root disappears by passing from real to imaginary ; that is, tf _ 4 rp = 0, °r> / = £.. (21) This is the same value which represents the maximum power transmissible by E.M.F., eQ, over a non-inductive line of resistance, r\ or, more generally, the maximum power which can be transmitted over a line of impedance, into any circuit, shunted by a condenser of suitable capacity. Substituting (21) in (19) and (20), we get, and the displacement of phase in the synchronous motor. cor(A,0-^--i tc± z hence, tan fa, /) = -?, (23) SYNCHRONOUS MOTOR. 343 that is, the angle of internal displacement in the synchron- ous motor i§ equal, but opposite to, the angle of displace- ment of line impedance, ('i, 0 = - (', 0, = ~ 2 r, r0; that is, motor E.M.F. > generator E.M.F. In either case, the current in the synchronous motor is leading. 207. B. Running Light, p = 0. When running light, or for / = 0, we get, by substitut- ing in (19) and (20), (26) Obviously this condition cannot well be fulfilled, since p must at least equal the power consumed by friction, etc. ; and thus the true no-load curve merely approaches the curve / = 0, being, however, rounded off, where curve (26) gives sharp corners. Substituting / = 0 into equation (7) gives, after squar- ing and transposing, e* + e<* 4- 3*,-« - 2 ^V - 2 22rV + 2 ra*'V - 2 *2*V = 0. (27) This quartic equation can be resolved into the product of two quadratic equations, 0. | (28) 0. j 344 ALTERNATING-CURRENT PHENOMENA. which are the equations of two ellipses, the one the image of the other, both inclined with their axes. The minimum value of C.E.M.F., eit is ^ = 0 at / = ^2. (29) The minimum value of current, z, is / = 0 at et = e0 . (30) The maximum value of E.M.F., elt is given by Equation (28)', /= e* + 22z2 -e02 - e*y + **z2 (s2 + 8 r2) + 2 j*e* (5 r2 - 22) - 2 / V (32 + 3 ^ = Oi (42) The curve of maximum displacement is shown in dash- dotted lines in Figs. 154 and 155. It passes through the SYNCHRONOUS MOTOR. 349 point of zero current — as singular or nodal point — and through the point of maximum power, where the maximum displacement is zero, and it intersects the curve of zero displacement. 210. E. Constant Counter E.M.F. At constant C.E.M.F., el = constant, If the current at no-load is not a minimum, and is lagging. With increasing load, the lag decreases, reaches a mini- mum, and then increases again, until the motor falls out of step, without ever coming into coincidence of phase. If the current is lagging at no load ; with increasing load the lag decreases, the current comes into coincidence of phase with eQ , then becomes leading, reaches a maximum lead ; then the lead decreases again, the current comes again into coincidence of phase, and becomes lagging, until the motor falls out of step. If eQ <