CHAPTER XVIII POLYPHASE INDUCTION MOTORS 155. The induction motor consists of a magnetic circuit inter- linked with two electric circuits or sets of circuits, the primary and the secondary. It therefore is electromagnetically the same structure as the transformer. The difference is, that in the transformer secondary and primary are stationary, and the electromagnetic induction between the circuits utilized to trans- mit electric power to the secondary, while in the induction motor the secondary is movable with regards to the primary, and the mechanical forces between the primary, and secondary utilized to produce motion. In the general alternating-current trans- former or frequency converter we shall find an apparatus trans- mitting electric as well as mechanical energy, and comprising both, induction motor and transformer, as the two limiting cases. In the induction motor, only the mechanical force be- tween primary and secondary is used, but not the transfer of electrical energy, and thus the secondary circuits are closed upon themselves. Hence the induction motor consists of a magnetic circuit interlinked with two electric circuits or sets of circuits, the primary and the secondary circuit, which are movable with regard to each other. In general a number of primary and a number of secondary circuits are used, angularly displaced around the periphery of the motor, and containing e.m.fs. displaced in phase by the same angle. This multi-circuit arrangement has the object always to retain secondary circuits in inductive rela- tion to primary circuits and vice versa, in spite of their relative motion. The result of the relative motion between primary and secondary is, that the e.m.fs. generated in the secondary or the motor armature are not of the same frequency as the e.m.fs. impressed upon the primary, but of a frequency which is the difference between the impressed frequency and the frequency of rotation, or equal to the "slip," that is, the difference between synchronism and speed (in cycles). 208 POLYPHASE INDUCTION MOTORS 209 Hence, if / = frequency of main or primary e.m.f., 5 = slip as fraction of synchronous speed, sf = frequency of armature or secondary e.m.f., and (1 — s) / = frequency of rotation of armature. In its reaction upon the primary circuit, however, the arma- ture current is of the same frequency as the primary current, since it is carried around mechanically, with a frequency equal to the difference between its own frequency and that of the primary. Or rather, since the reaction of the secondary on the primary must be of primary frequency — whatever the speed of rotation — the secondary frequency is always such as to give at the existing speed of rotation a reaction of primary frequency. 156. Let the primary system consist of po equal circuits, displaced angularly in space by — of a period, that is, — of Vo Po the width of two poles, and excited by po e.m.fs. displaced in phase by — of a period; that is, in other words, let the field Po circuits consist of a symmetrical po-phase system. Analo- gously, let the armature or secondary circuits consist of a sym- metrical pi-phase system. Let no = number of primary turns per circuit or phase; Hi = number of secondary turns per circuit or phase; no a = — ni Pi Since the number of secondary circuits and number of turns of the secondary circuits, in the induction motor — as in the stationary transformer — is entirely unessential, it is preferable to reduce all secondary quantities to the primary system, by the ratio of transformation, a; thus if E'l = secondary e.m.f, per circuit. El = aE'i = secondary e.m.f. per circuit reduced to primary system ; 210 ALTERNATING-CURRENT PHENOMENA if I' I = secondary current per circuit, Ii = -J- = secondary current per circuit reduced to primary system ; if r'l = secondary resistance per circuit, Vi = a-hr'i = secondary resistance per circuit reduced to pri- mary system; if x'l = secondary reactance per circuit, Xi = a^bx'i = secondary reactance per circuit reduced to pri- mary system; if z'l = secondary impedance per circuit, 2i = a^hz'i = secondary impedance per circuit reduced to pri- mary system; that is, the number of secondary circuits and of turns per sec- ondary circuit is assumed the same as in the primary system. In the following discussion, as secondary quantities, the values reduced to the primary system shall be exclusively used, so that, to derive the true secondary values, these quan- tities have to be reduced backward again by the factor a^b ni^pi 157. Let "J> = total maximum flux of the magnetic field per motor pole. We then have E = \/2 TT/io/ ^ 10~^ = effective e.m.f. generated by the magnetic field per primary circuit. Counting the time from the moment where the rising mag- netic flux of mutual induction, (flux interlinked with both electric circuits, primary and secondary), passes through zero, in complex quantities, the magnetic flux is denoted by $ = - i$, and the primary generated e.m.f., E = - e; where e = \/2 xn/$ 10~* may be considered as the "active e.m.f. of the motor," or "counter e.m.f." Since the secondary frequency is sf, the secondary induced e.m.f. (reduced to primary system) is Ei = — se. POLYPHASE INDUCTION MOTORS 211 Let 7o = exciting current, or current through the motor, per primary circuit, when doing no work (at synchronism), and F = g — j6 = primary exciting admittance per circuit = — • We thus have, ge = magnetic power current, ge~ = loss of power by hysteresis (and eddy currents) per primary coil. Hence Poge^ = total loss of power by hysteresis and eddies, as calculated according to Chapter XII. he = magnetizing current, and nobe = effective m.m.f. per primary circuit; hence -wnohe = total effective m.m.f., and — ^nobe = total maximum m.m.f., as resultant of the m.m.fs. V2 of the po-phases, combined by the parallelogram of m.m.fs.^ If (R = reluctance of magnetic circuit per pole, as discussed in Chapter XII, it is V2 Thus, from the hysteretic loss, and the reluctance, the con- stants, g and h and thus the admittance, Y, are derived. Let To = resistance per primary circuit; Xo = reactance per primary circuit; thus, Zo = To -}- jxo = impedance per primary circuit; ri = resistance per secondary circuit reduced to primary system; Xi = reactance per secondary circuit reduced to primary system, at full frequency/; 1 Complete discussion hereof, see Chapter XXXIII. 212 ALTERNATING-CURRENT PHENOMENA hence, sxi = reactance per secondary circuit at slip s, and Zi = ri + jsxi — secondary internal impedance. 158. We now have, Primary generated e.m.f., E = — e. Secondary generated e.m.f.. El = — se. Hence, Secondary current, se ri + jsxi Component of primary current, corresponding thereto, or primary load current, ri-\-jsxi' Primary exciting current, Io= eY = e {g — jb); hence, Total primary current, I = r + h e.m.f. consumed by primary impedance, E, = Zo/ e.m.f. required to overcome the primary generated e.m.f., - E = e; hence. Primary terminal voltage, Eo ^ e + E, = e 1 + s (ro + jxo) - + (ro + jXo) (g - jb) | ri + jsXi We get thus, in an induction motor, at slip s and active e.m.f. e. Primary terminal voltage, TT 111 ^ (^0 + i^") I / , ■ \ r •A^ 1 ?° ^ 'i^ + r,+JSX, + (ro + JXo) (g -Jb)\; POLYPHASE INDUCTION MOTORS 213 Primary current, I = e\ — r^~ + (9 -i^)}; (ri + jsxi ^^ ^ ' ]' or, in complex expression, Primary terminal voltage. £o -e |l+sf^ + ZoF}; Primary current. '.-Al^\ To eliminate e, we divide, and get, Primary current, at slip s, and impressed e.m.f., E^; J s + ZiF ^ Zi + sZo + ZoZiy"""' or, ^ _ s+(ri +isa-i) (g - jb) ^ (ri + isxi) + s (ro + jx^i) + (ro + jxi,) (ri + jsxi) ((/ - jh) ■ °" Neglecting, in the denominator, the small quantity Zi^TixY , it is , s + Z,Y f = z;v^zr s + (ri + jsxi) (g - j6) (ri +jsa;i) +s(ro+ia:o) (s + r,^ + sa^ife) — j (rib — sx^g) Eo, (ri + STo) + is (xi + a^o) or, expanded, [(sfi + sVo) + ri^g + sri (ro? — Xob) + s^a^i (a^ofi' + Xig + rofe)- , i[s- (xq + a^i) + ri^b + sri (x^g + rpb) + s^x^ (xob + a^i^- Ti^g)] {ri-hsror-hs'ixi + xor- Hence, displacement of phase between current and e.m.f., _ s^(^o + Xi) + ri-b + sr^ixog + ro6) + s~Xi{xob + a: 16 - r„g) ° ~ (sri + sVo) +ri2(/ + sr](ro<7-a:o&) +s2a;i(a;og+a;]6i-ro6) Neglecting the exciting current, h, altogether, that is, setting F= 0, We have tan do = „ (ri + sro) — js (xo + X]) . ^(ri + sro)2 + s2(a;o+ a;])^ Eso (ri + sro) +is(a:o + a;])' ri + sro 214 ALTERNATING-CURRENT PHENOMENA 159. In graphic representation, the induction motor diagram appears as follows: — Denoting the magnetism by the vertical vector 04> in Fig. 118, the m.m.f. in ampere-tm-ns per circuit is represented by vector OF, leading the magnetism, O*, by the angle of hysteretic advance a. The e.m.f. generated in the secondary is propor- tional to the slip s, and represented by OEi at the amphtude of 180°. Dividing OEi by a in the proportion of ri -j- sxi, and connecting a with the middle, h, of the lower arc of the circle, OEi, this line intersects the upper arc of the circle at the point, IiTu Thus, Ohri is the e.m.f. consumed by the secondary resistance, and Ohxi equal and parallel to £"1/1^1 is the e.m.f. consumed by the secondary reactance. The angle, EiOhri = 61, is the angle of secondary lag. Fig. 118. The secondary m.m.f., OGi, is in the direction of the vector, OliVi. Completing the parallelogram of m.m.fs. with OF as diagonal and OGi, as one side, gives the primary m.m.f., OG, as other side. The primary current and the e.m.f. consumed by the primary resistance, represented by OIvo, is in line with OG, the e.m.f. consumed by the primary reactance 90° ahead of OG, and represented by OIxo, and their resultant, OIzo, is the e.m.f. consumed by the primary impedance. The e.m.f. gener- ated in the primary circuit is OE', and the e.m.f. required to overcome this counter e.m.f. is OE equal and opposite to OE'. Combining OE with OIzo gives the primary terminal voltage represented by vector OEo, and the angle of primary lag, EoOG - 6*0. POLYPHASE INDUCTION MOTORS 215 160. Thus far the diagram is essentially the same as the diagram of the stationary alternating-current transformer. Re- garding dependence upon the slip of the motor, the locus of the different quantities for different values of the slip, s, is determined thus, Fig. 119. Let Ei = sE'. Assume in opposition to 0$, a point. A, such that OA -^ IiTi = El -^ Iisxi, then ^. IiVxXEi 7iri X sE' ri „, UA = — 7 = — f = — E = constant. llSXi liSXi Xi That is, IiTi lies on a half-circle with OA = — £" as diameter. ^1 That means Gi lies on a half-circle, gi, in Fig. 119 with OC as diameter. In consequence hereof, Go lies on half-circle go with FB equal and parallel to OC as diameter. 216 ALTERNATING-CURRENT PHENOMENA Thus Ito lies on a half-circle with DH as diameter, which circle is perspective to the circle, FB, and Ixo lies on a half- circle with IK as diameter, and Izo on a half-circle with LN as diameter, which circle is derived by the combination of the circles, Itq and Ixq. The primary terminal voltage, Eo, lies thus on a half-circle, eo, equal to the half-circle, Izo, and having to point E the same relative position as the half-circle, Izo, has to point 0. This diagram corresponds to constant intensity of the maxi- mum magnetism, 0$. If the primary impressed voltage, £"0, is kept constant, the circle, eo, of the primary impressed voltage changes to an arc with 0 as center, and all the corresponding points of the other circles have to be reduced in accordance herewith, thus giving as locus of the other quantities curves of higher order which most conveniently are constructed point for point by reduction from the circle of the loci in Fig. 119. Torque and Power 161. The torque developed per pole by an electric motor equals the product of effective magnetism, ^7^, times effective F armature m.m.f., — 7-, times the sine of the angle between both, V2 ^F Z)' = ^ sin i^F). If Hi = number of turns, Ii = current, per circuit, with pi armature circuits, the total maximum current polarization, or m.m.f. of the armature, is pini/i ^' = vr Hence the torque per pole, PlTli^Il . D = ' 7^^ sm (4>/i). 2\/2 U q = the number of poles of the motor, the total torque of the motor is, qpini^Ii . D = ^^ — 7=— sm i^Ii). 2V2 POLYPHASE INDUCTION MOTORS 217 The secondary induced e.ni.f., E), lags 90° behind the inducing magnetism, hence reaches a maximum displaced in space by 90° from the position of maximum magnetization. Thus, if the secondary current, Ii, lags l^ehind its emf., Ei, by angle, 6i, the space displacement ])etween armature current and field magnetism is ^ (/i4>) = 90° + ^1, hence sin ($/i) = cos ^i. We have, however, cos di es 10-1 e = V2 7r7?,i$/10-8, thus, el08 substituting these values in the equation of the torque, it is ^ qpisrie^ 10^ 4 7r/ in' + s^Xi^) ' or, in practical (c.g.s.) units, 4 tt/ (ri2 + s'^xi^) ' is the torque of the induction motor. At the slip, s, the frequency, /, and the number of poles, q, the linear speed at unit radius is q ' hence the output of the motor, P - Dv, or, substituted, „ Pirie-s{\ — s) 2S I, that is, the maximum torque, occurs below standstill, and the torque constantly increases from synchronism down to standstill. It is evident that the position of the maximum torque point, Si, can be varied by varying the resistance of the secondary POLYPHASE INDUCTION MOTORS 221 circuit, or the motor armature. Since the shp of the maxi- mum torque point, St, is directly proportional to the armature resistance, ?'i, it follows that very constant speed and high efficiency brings the maximum torque point near synchronism, and gives small starting torque, while good starting torque means a maximum torque point at low speed; that is, a motor with poor speed regulation and low efficiency. Thus, to combine high efficiency and close speed regulation with large starting torque, the armature resistance has to be varied during the operation of the motor, and the motor started with high armature resistance, and with increasing speed this armature resistance cut out as far as possible. 164. If St = 1, In this case the motor starts with maximum torque, and when overloaded does not drop out of step, but gradually slows down more and more, until it comes to rest. ■ If St > 1, then ri > V^d^ + {xi + Xoy. In this case, the maximum torque point is reached only by driving the motor backward, as counter-torque. As seen above, the maximum torque, Dt, is entirely inde- pendent of the armature resistance, and likewise is the current corresponding thereto, independent of the armature resistance. Only the speed of the motor depends upon the armature resistance. Hence the insertion of resistance into the motor armature does not change the maximum torque, and the current corre- sponding thereto, but merely lowers the speed at which the maximum torque is reached. The effect of resistance inserted into the induction motor is merely to consume the e.m.f., which otherwise would find its mechanical equivalent in an increased speed, analogous to resistance in the armature circuit of a continuous-current shunt motor. Further discussion on the effect of armature resistance is found under "Starting Torque." 222 ALTERNATING-CURRENT PHENOMENA Maximum Power 165. The power of an induction motor is a maximum for that sHp, Sp, where dP ^=«' d f (ri + sro)^ + s2 {xi + a^o)^ or, smce ds [ s (1 — s) expanded, this gives 0; substituted in P, we get the maximum power, P. p,Eo' 2 {(ri + ro) + V'Cri + ny + (a^i + a^o)^} This result has a simple physical meaning: (ri + ro) = r is the total resistance of the motor, primary plus secondary (the latter reduced to the primary), (xi + Xo) is the total reactance, and thus VC^i + ro)^ + {xi + Xoy = z is the total impedance of the motor. Hence P Pi^o^ , ^^ ~2(r + z} is the maximum output of the induction motor, at the slip, ri 5» = ri + 2 The same value has been derived in Chapter X, as the maxi- mum power which can be transmitted into a non-inductive receiver circuit over a line of resistance, r, and impedance, z, or as the maximum output of a generator, or of a stationary transformer. Hence: The maximum output of an induction motor is expressed by the same formula as the maximum output of a generator, or of a stationary transformer, or the maximum output ivhich can he transmitted over an inductive line into a non-inductive receiver circuit. POLYPHASE INDUCTION MOTORS 223 The torque corresponding to the maximum output, Pp, is ^ qVxE,\r, + z) ^^ ^irfzir + z) This is not the maximum torque; but the maximum torque, Dt, takes place at a lower speed, that is, greater slip, since > Vro2 + (xi+a:o)^ ri + V(ri +ro)2 + (xi +^0)2' that is, St > Sp. It is obvious from these equations, that, to reach as large an output as possible, r and z should be as small as possible; that is, the resistances, ri + ro, and the impedances, z, and thus the reactances, Xi + Xo, should be small. Since ri + ro is usually small compared with xi + Xo it follows, that the problem of induction motor design consists in constructing the motor so as to give the minimum possible reactances, Xi + Xq. Starting Torque 166. In the moment of starting an induction motor, the slip is s = 1; hence, starting current, J ^ 1 + (ri-{-jxi) ig -jh) ^ _ (ri -f-jxi) + (ro +ixo) + (ri -{-jxi) + (ro + jXo) {g - jb) "' or, expanded, with the rejection of the last term in the denomi- nator, as insignificant, [(^1 + ro) + ^(ri[ri + ro] + Xifxi + Xo\) + b{roXi - XoVi)]- j ^ i[(xi + Xo) + 6(ri[ri + ro] + Xt[xi + x^]) - g (roXi - XorQ] (ri + ro)2 + (xi + Xo)=^ •■"' and, displacement of phase, or angle of lag, ^^^ Q ^ (xi + xo) + b (ri [ri + ro] + Xi[xi + Xp]) - gir^Xx — rcorQ ('/•i -\- ra) + g (ri [ri + ro] + Xi{X], + Xo]) + 6(roXi - XorO' 224 ALTERNATING-CURRENT PHENOMENA Neglecting the exciting current, g = 0 = b, these equations assume the form, J (ri + ro) - j (xi + xo) Eq (ri + ro)2 + {xi + Xo)" . (ri + ro) + j {xi + Xq) ' or, ehminating imaginary quantities, V(ri + nY + (xi + x,Y z ' and , . Xi + Xq tan do ; ri + ro That means, that in starting the induction motor without additional resistance in the armature circuit — in which case Xi + Xo is large compared with ri + ro, and the total impe- dance, z, small — the motor takes excessive and greatly lagging currents. The starting torque is q-piTiE^"^ Do = 47r/{(ri + ro)=^ + (xi + xo)2 1 qpiEo^ Vi 47r/ Z~' That is, the starting torque is proportional to the armature resistance, and inversely proportional to the square of the total impedance of the motor. It is obvious thus, that, to secure large starting torque, the impedance should be as small, and the armature resistance as large, as possible. The former condition is the condition of large maximum output and good efficiency and speed regula- tion; the latter condition, however, means inefficiency and poor regulation, and thus cannot properly be fulfilled by the internal resistance of the motor, but only by an additional resistance which is short-circuited while the motor is in operation. Since, necessarily, Ti < z, we have, qV,Eo\ ^^' < T^' and since the starting current is, approximately J _ Eq ^ ~ 2 ' POLYPHASE INDUCTION MOTORS 225 we have, Do < I^M. Doo = 1 rEoI would be the theoretical torque developed at 100 per cent, efficiency and power-factor, by e.m.f. Eo, and current /, at synchronous speed. Thus, Do < Doo, and the ratio between the starting torque, Do, and the theo- retical maximum torque, Doo, gives a means to judge the per- fection of a motor regarding its starting torque. This ratio, j^, exceeds 0.9 in the best motors. -L'OO E Substituting / = — in the equation of starting torque, it avssumes the form, 47r/ Since = synchronous speed, it is: The starting torque of the inductioji motor is equal to the resistance loss in the motor armature, divided by the synchronous speed. The armature resistance which gives maximum starting torque is dDo or smce, dri _ qpiE,^ rj_ 47r/ (ri-Hro)2+ {x. + XoY dri I ri 1 ' expanded, this gives, ri = Vro" + {xi -\- XoY, the same value as derived in the paragraph on "maximum torque." Thus, adding to the internal armature resistance, r'l, in start- ing the additional resistance, r"i = Vro' + {x, ^ Xo)' - r'l, 15 226 AL TERN A TING-C URREN T PHENOMENA makes the motor start with maximum torque, while with increasing speed the torque constantly decreases, and reaches zero at synchronism. Under these conditions, the induction motor behaves similarly to the continuous-current series motor, varying in speed with the load, the difference being, however, that the induction motor approaches a definite speed at no-load, while with the series motor the speed indefinitely increases with decreasing load. 10 20 30 40 50 70 90 100?. 1 q: . 0'=s 20 H.P. THREE-PHASE INDUCTION MOTOR 110 VOLTS, 900 REVOLUTIONS, 60 CYCLES Y = .l-.4j r,=.02/.045/.18 / .75 UJ Z,=.02+.085j ^ ,^ k in 28 26 24 22 20 18 16 14 12 10 g ~rt ba^ ^ k ^^ y < \ r" \ 1 ^, p< N ^ \y \ / \ \ \i\ y - / ■^ ^\ V 6 4 2 / "^ \ A \ / 1 1 ^ 0 10 20 30 40 50 60 70 80 SPEED, PER CENT SYNCHRONISM 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 AMPERES Fig. 120. 90 The additional armature resistance, r"i, required to give a certain starting torque, is found from the equation of starting torque: Denoting the internal armature resistance by r'l, the total armature resistance is ri = r'l -f r"i, and thus, r'l + r'\ Do hence. qjhEl 4x/ (r'i + r", + ro)2-i-(a:i + Xo)2' r"i = - r'l - ro 4- 8-^ ^^fWwY - f^-(-^+-o)^^ POLYPHASE INDUCTION MOTORS 227 This gives two values, one above, the other below, the maxi- mum torque point. Choosing the positive sign of the root, we get a larger armature resistance, a small current in starting, but the torque constantly decreases with the speed. Choosing the negative sign, we get a smaller resistance, a large starting current, and with increasing speed the torque first increases, reaches a maximum, and then decreases again toward synchronism. These two points correspond to the two points of the speed- torque curve of the induction motor, in Fig. 120, giving the desired torque, Do. The smaller value of r'\ gives fairly good speed regulation, and thus in small motors, where the comparatively large start- ing current is no objection, the permanent armature resistance may be chosen to represent this value. The larger value of r'\ allows to start with minimum current, but requires cutting out of the resistance after the start, to secure speed regulation and efficiency. 167. Approximately, the torque of the induction motor at any slip, s: qpiTiEo^s D = 47r/}(r, + .sro)2 + s2(a:i + a;o)2} can be expressed in a simple and so convenient form as function of the ynaximum torque: Dt = qp\Eo- 8 wfiro + Vro2 + (xr+ x^} or of the starting torque: s — 1: r) qpiriEo^ " 47r/{(ri + ro)2 + (a:i+Xo)=^}* Dividing D by Dt we have in + sro)2 -h §2 {x, + xo^ " Since ro, the primary resistance, is small compared with X ==^ Xi -]- Xq, 228 ALTERNATING-CURRENT PHENOMENA the total self-inductive reactance of the motor, it can be neg- lected under the square root, and the equation so gives: ^ ^ 2 ris (ro -\rx) ^ or, still more approximately: and the starting torque, for: s — 1: hence, dividing, _ 2r,x or, if ri is small compared with x, that is, in a motor of low- resistance armature: From the equation: .ST" D = -^^^Z)o, it follows that for small values of s, or near synchronism: by neglecting s^x^ compared with ri^: For low values of speed, or high values of s, it follows, by neglecting ri^ compared with s^x^: that is, approximately, near synchronism, the torque is directly proportional to the slip, and inversely proportional to the armature resistance, that is, proportional to the ratio — —7 r~ ; near standstill, the torque is inversely pro- armature resistance portional to the slip, but directly proportional to the armature resistance, and so is increased by increasing the armature resist- ance in a motor of low-armature resistance. POLYPHASE INDUCTION MOTORS 229 Synchronism 168. At synchronism, s = 0, we have, L = EAg-jh); or. P = 0, Z) = 0; that is, power and torque are zero. Hence, the induction motor can never reach complete synchronism, but must shp sufficiently to give the torque consumed by friction. Running near Synchronism 169. When running near synchronism, at a slip, s, above the maximum output point, where s is small, from 0.01 to 0.05 at full-load, the equations can be simplified by neglecting terms with s, as of higher order. We then have, current, g + ri (g - jh) 1 = iio, or, eliminating imaginary quantities, angle of lag. -^ + &)'+6^^o; ^^2 x. + x, ^ ^^^ s2 (xi + To) + ri'^b " r^ tan do = , o = r "' sri + ri^g s + r^g PiEoh P = D = or, inversely. qpiEp^s ' ~ PiEo'' ^ 4.7rfr^D that is, Near synchronism, the slip, s, of an induction motor, or its drop 230 ALTERNATING-CURRENT PHENOMENA in speed, is proportional to the armature resistance, ri, and to the power, P, or torque, D. EXAMPLE 170. As an example are shown, in Fig. 120, characteristic curves of a 20-hp. three-phase induction motor, of 900 revolutions synchronous speed, 8 poles, frequency of 60 cycles. in 20 H.P. THREE PHASE INDUCTION MOTOR O a. CURRENT DIAGRAM o z 82 SO 28 26 24 22 20 18 16 14 12 10 8 6 4 2 n Y=.1-.4j Zo=.03 -(-.OQj Z,= .02 + .085J in u. /^ ^ N o 7 / / TC )RQl. JE \ o ^^ / y >v \ \ UJ \ C1.UJ >^ / \\ < S 100 90 ""■*"- ■-:: ■'-/, ~7- /_ SPE ED ^ ^ / y ^' --- -~.^ ^•4( V; ""~-- --. \ / ^>^1 ^s-v^ ^s^ 70 / / "- -.. A / ^~v -^^ 50 40 30 / \ 1. 1 / \ / 1 1 1 20 10 1 1 50 100 150 200 AMPERES Fig. 121. 250 300 350 The impressed e.m.f. is 110 volts between lines, and the motor star connected, hence the e.m.f. impressed per circuit: ~% = 63.5; or^o The constants of the motor are: 63.5. POLYPHASE INDUCTION MOTORS 231 Primary admittance, Y = 0.1 — OAj. Primary impedance, Z = 0.03 + 0.09 j. Secondary impedance, Zi = 0.02 + 0.085 j. In Fig. 120 is shown, with the speed in per cent, of synchronism, as abscissas, the torque in kilogram-meters as ordinates in drawn Hnes, for the values of armature resistance: Ti = 0.02 : short-circuit of armature, full speed. ri = 0.045: 0.025 ohms additional resistance. Ti = 0.18 : 0.16 ohms additional, maximum starting torque. Ti = 0.75 : 0.73 ohms additional, same starting torque as n = 0.045. / / \ s 28 20 H.P. THREE-PHASE / \ 24 INDUCTION MOTOR / \ E 20 110 VOLTS, 900 REVOLUTIONS Rr> rvri pc io' ^:> y ' 0 16 SPEED DIAGRAM Y =.1 -4j Zo=.03 +.09J Zi=.045 + .085j — -' ^ ' \ 5 12 pt ,'' \ \ a- 8 ^V h- 4 Speed , Percent of Synchronism -100-80 -60_;4a--'20-''"0 20 4 0 c 0 SO 1 X) 150 1 ^ )0 250 3( K) 2[o lis 16 14 1I2 l!o 3 6 4 .2 C . 2-t. 4 - 6 -.8 Sli 5S 1 -8 --- — ■ > -12 • 1 t< ^ — -— \ L16 '/ H 1 \ -20 / / = 0 / / Bac kw: rd 1 Jota ion -0 c c -28 / Ab ove Syn chro nist 1 (/) -32 / _. if . Fig. 122. On the same figure is shown the current per line, in dotted lines, with the verticals or torque as abscissas, and the hori- zontals or amperes as ordinates. To the same current always corresponds the same torque, no matter what the speed may be. On Fig. 121 is shown, with the current input per line as abscissas, the torque in kilogram-meters and the output in horse- power as ordinates in drawn lines, and the speed and the mag- netism, in per cent, of their synchronous values, as ordinates in dotted lines, for the armature resistance, ri = 0.02, or short- circuit. 232 ALTERNATING-CURRENT PHENOMENA In Fig. 122 is shown, with the speed, in per cent, of synchro- nism, as abscissas, the torque in drawn hne, and the output in dotted hne, for the value of armature resistance ri = 0.045, for the whole range of speed from 120 per cent, backward speed to 200 per cent, beyond synchronism, showing the two maxima, the motor maximum at s = 0.25, and the generator maximum at s = — 0.25. 171. As seen in the preceding, the induction motor is charac- terized by the three complex imaginary constants, Ya = fl'o — J&o, the primary exciting admittance, Zq = Tq -\- jxo, the primary self-inductive impedance, and Zi = Ti -\- jxi, the secondary self-inductive impedance, reduced to the primary by the ratio of secondary to primary turns. From these constants and the impressed e.m.f., eo, the motor can be calculated as follows: Let, e = counter e.m.f. of motor, that is, e.m.f. generated in the primary by the mutual magnetic flux. At the slip, s, the e.m.f. generated in the secondary circuit is se. Thus the secondary current. where h = rTTT^i = ^ («i - i«^), sri s~Xi tti = — r-j — , — ; and 05 The primary exciting current is, /oo ^ eVo = e (go — jbo); thus, the total primary current, /o = /i + /oo = e (bi — jbi), where, &i = «! + go, and 62 = ^2 + bo. The e.m.f. consumed by the primary impedance is, E^ = hZo = e (ro -r jxo) (6i - jbi); the primary counter e.m.f. is e, thus the primary impressed e.m.f., POLYPHASE INDUCTION MOTORS 233 Eo = e -h E^ = e (ci - jct), where, Ci = 1 + robi + 2:062 and C2 = ro62 — a;o6i, or, the absolute value is, hence, Substituting this value gives. Secondary current, Primary current, 61 — .762 /o = eo /67 Impressed e.m.f., Eo = eo Vci^ + Ca^ Thus torque, in synchronous watts (that is, the watts output which the torque would produce at synchronous speed), D - [eI,Y eo^ai hence, the torque in absolute units, _ D _ eo^ai 2irf (Ci^ + C22) 27r/ where/ = frequency. The power output is torque times speed, thus: P. = 0(1 -,)='"''■■''-/'• ^ ^ ^ Ci^ + c-r The power input is, Po = [£^0/0] = [E,hY - 3 [EoUV = Po' +jPo^ _ ep^ (biCi + 62^2) _ . eo^ (bgCi — 61C2) 234 ALTERNATING-CURRENT PHENOMENA The volt-ampere input, Par, = eoh 3 INDUCTIO^ MOTOR LOAD CURVES oof: a: E^ 1- z 2 = =.l+ 3j ^ r = .o 1— . J .^~ 8000 a. / / / / 5000 100 SPEED / / / 90 / /^ / ^ ^ 4000 80 / ^^"^ / / -~--^ V 70 '¥ ^ J / / ,•< / / '"X K fiOOO 60 1 ! / /- / < .f «0 // / / y y' X / 2000 40 1 i 1 1 1 ! / / ^ 30 //' / ^ 1000 30 1 ^ 10 i/ 10 w S( 00 SI 00 40 aWER BO OUTPl H T DO 60 JO 70 DO Fig. 123. hence, the efficiency is, Pi_ ^ ai (1 - s) Po' &lCl + 62C2' the power-factor, Po^ hiCi + &2C2 ^«0 V(6x2 + fe22)(Ci2 + C22)^ POLYPHASE INDUCTION MOTORS 235 the apparent efficiency, Pi tti (1 — s) the torque efficiency/ ai D Po^ biCi + 62C2 r- DUC ION- ) CI MOT JRVI s :s % ^' -.1- -.Jj '~ .ul- -.Ij « 1- V- < z 0 100 s DO ^ 180 0 cu 'REN ' N ,' .''^ '1 160 8000 80 70 > 5^ :x / z' 140 7009 CO BQii 3^ cvtji r >< >^ r \ 120 C009 50 . ' ■vi^ lOt, ..-^ ^ l_t> 'f^\ oj^ /' -^ s \, \ 100 J009 40 ISh >^;- / \ \ I 80 4009 -80- Sjy -^ s=r. "" >t , t '0^-' \ \ 60 3000 20 ^p >KBE )T__ — i*-'— \ A 40 2009 "To' — _.- \\ 20 1000 1.0 1 . . ( LIP. 8 1 1 0 Fig. 124. and the apparent torque efficiency,^ ^ Ox ^-0 ~ V(6i2 + 62^) (ci^ + ct^) 172. Most instructive in showing the behavior of an induction motor are the load curves and the speed curves. The load curves are curves giving, with the power output as abscissas, the current input, speed, torque, power-factor, effi- ciency, and apparent efficiency, as ordinates. The speed curves give, with the speed as abscissas, the torque, 1 That is the ratio of actual torque to torque which would be produced, if there were no losses of energy in the motor, at the same power input. ^ That is the ratio of actual torque to torque which would be produced if there were neither losses of energy nor phase displacement in the motor, at the same volt-ampere input. 236 ALTERNATING-CURRENT PHENOMENA current input, power-factor, torque efficiency, and apparent torque efficiency, as ordinates. The load curves characterize the motor especially at its normal running speeds near synchronism, while the speed curves characterize it over the whole range of speed. In Fig. 123 are shown the load curves, and in Fig. 124 the speed curves of a motor having the constants: Fo = 0.01 — 0.1 j; Zo - 0.1 + 0.3 i; and Zj = 0.1 + 0.3 j.