dc Motor Drives 63 Example 5.1 A 200 V, 10.5 A, 2000 rpm shunt motor has the armature and field resistances of 0.5 and 400 Q respettively. It drives a load whose torque is constant at rated motor torque. Calculate motor speed if the source voltage drops to 175 V, — Solution If flux at 200 V, is @, then flux at 175 V. Be 62= HS x 91 = 0875 4, ' Zz, Since the load torque is constant we 2 Ge had = had Re w* kr Rada” Ey, =Vi—[uRy=200-105x05=195V Wr To Ey) =Vy—IgR,=175-114x05=16093V 4% tal E« gN enter oh Lom oe Ey” Nz 5 EB. aot y ML ~ 169.3 1 = ai Na= BX XN = Oe x Tapas X 2000 = 1984.5 rpm Exampte 5.2 A 220 V de series motor runs at 1000 rpm (clockwise) and takes an armature current of 100 A when driving a load with a constant torque. Resistances of the armature and field windings are 0.05 Q each. Find the magnitude and direction of motor speed and armature current if the motor terminal voltage is reversed and the number of turns in field winding is reduced to 80%. Assume linear magnetic circuit. Solution When the number of turns is reduced to 80% then the value of flux for the same field (or armature) current will also be reduced to 80%. ~ YO T= Kila = KE Ty = Kebalar= Ke eae Since T= ) KID, = KL 0812, * - Woy, fer DOSSince or $1.3 als of Electrical Drives ta 100 =-— = =-1118A Ia=— 9g” Jos current has a negative sign because the supply voltage has been reversed, E, = 220 — 100(0.05 + 0.05) = 210 V E, = - [220 - 111.8(0.05 + 0.04)] = — 209.94 V E=KygN = E,= Kila, qo\* 4 Ey = KX 0.81q.Ny ash Ent ee cal Neel 209.98) 100 EF “Ta “08 = 210 * Sire xO 1117.7 rpm \ Compound Motor ra 4Examp Le 5.3 A 220 V, 200 A, 800 rpm de Separately excited motor has an armature resistance of 0.06 Q. The Motor armature is fed from a variable voltage source with an internal resistance of 0.04 Q. Calculate internal voitage of the variable voltage source when the motor is Operating in regenerative braking at 80% of the rated motor torque and 600 pm. Sn Solution p ww Since torque is proportional to the armature Current, motor armature current when regenerating ——— I = 0.8 x 200 = 160 A x E, = 220 - 200 x 0.06 = 208 V ¢ yf , N, on a ce 4 - =—2 f= OW = ) Bx= 5 Ei = gpg X 208 = 156 V 0 & Internal voltage of the variable voltage source = 156- 160(0.06 + 0.04) = 140 V. Clee ap OtEXampLe 5.4 The series motor of Example 5.2 is operated under dynamic braking at twice the rated torque and 800 rpm. Calculate the value of braking current and resistor. Assume linear magnetic circuit. Solution Since T= Kyl; Ts = Kel, Ty = Ty, (T/T, = 100 V2 = KelaiNy; Ez = KelagN2 — faz) No ~ 141.4 ,. 800 = Ea = TX RX Er = To6- Fogg (220- 100 x 0.1) = 237.55 V Now Ey = I,(Rg +01) or 237.55 = 141.4 (Rg +0.1) or Rg = 1.582 EXAMPLE 5.5 A dc shunt motor has the armature resistance of 0.04 Q and the field winding resistance of 10 Q. Motor is coupled to an overhauling load with a torque of 400 N-m. Following magnetisatiot curve was measured at 600 rpm:de Motor Drives 71 Field current, A 25 5 75 10 1215 eel S een) 75eee 20 beni 22-oiee eS. Back emf, V 25 50 73.5 90 1025 110 116 121 125 129 Motor is braked by self-excited dynamic braking with a braking resistance of 1 Q. At what speed motor will hold the load? Solution Motor equivalent circuit when working under self-excited dynamic braking is shown in Fig. ESS. Ret, 7 Lei (S2)1,= > OF = 1, (El) raat E Re 7 v3 Re = —___iE BE K=O = QO x 2160 = Ox €2) ~ T= Kl, 3) =e From magnetisation characteristics and Eqs. (E.1)-(E.3) following relationships are obtained: OE I 5 225 20 115 Ko 2.053 1.99 1.926 1.846 T 564.6 492.4 423.7 355.4 From K,@ and T vs /; plots, for a braking torque of 400 N-m, [; = 19.13 A and K,@ = 1.898. Now for I; = 19.13 A. I= Wp= 11 x 19.13 = 210.43 E=V41,R,= [Re + Ry = 19.13 x 10 + 210.43 x 0.04 = 199.72 V ; —-E_, 60 _ 199.72, 30 _ Speed im rpm = 95 x 31 = Pag: X 32 = 1005 rpm EXAMPLE 5.6 Calculate the value of Rg when motor of Example 5.5 is required to hold overhauling load at 1200 rpm. we , "ye" Solution « For a given /;, E « speed. Hence the magnetization characteristic at 1200 rpm will be: IA 2.5 5 75 10 125 15 17.5 20 22.5 25 EV 50 100 «147 180 205.220 232 242-250-258 Power deveoped Py = TX @m = 400 x 1200 20 = 160007 watts Also Py=El or 424%72. Fundamentals of Electrical Drives 160007 x 0.04 _ 2010.6 Pa oo EK -1,R, 2 E- a RHE- E ey Also V= Rly = 10K, tes From Fg. (E4), and E vs J; relation at 1200 rpm (as given above), V vs Ir relation is obtaineg ,, 4 25 22.5 20 f 241.96 233.7 ks v 250.2 The value V must simultaneously satisfy this relationship and Eq. (E.5). The intersection gy Vvs J; curves based on Eq. (E.5) and relation (E.6) gives V=250V and y=25A_ for which E = 258 V _ E=V _ 258 = 250 _ 999 Wa aR = ogg = 200 Now a 0 lee Be Ay =A Ig = Ia - Ip = 200 - 25 = 175A 250 _ Rp i3g = 1.429. Examete 5.7 The magnetisation characteristic of a de series motor when running at 500 rpm is given by Current, A 20 30 40 50 60 70 80 emf, V 215 310 381 437 485 519 550 Total resistance of the armature and field windings is 0.5 ohm. When connected for dynamic braking against an overhauling load of 500 N-m, motor speed is to be maintained at 600 rpm. What resistance must be connected across the motor terminals? Solution K.o= Oe and T=K,l, From these relations and the data of the magnetization characteristics: 1, 20 30 40 50 60 70 80 ko 4.1 5.92 7.277 8.347 9.264 9.913 10.505 T 82.1 177.63 291 417.3 555.8 693.9 840.4 These K.9 and T vs /, relations are plotted in Fig. E.5.7. For a torque of 500 N-m, J, = 56.A and K.9 = 8.9. For a speed of 600 rpm Egdc Motor Drives 73 Ty A Fig. E.5.7 E=K,9o, = 8.9 x 900 OO x 2 = 559.2V Now (R, + Rp) = £ 22 = 9.986 and Rp = 9.986 - 0.5 = 9.486 Q74 Fundamentals of Electrical Drives Plugging © - 0 T (b) Series (a) Separately excited Fig. 5.10 Plugging speed-torque curves e Yd Motoring Motoring Plugging Plugging (b) Series (a) Separately excited Fig. 5.11 Counter-torque braking Plugging gives fast braking due to high average torque, even with one section of braking resistance Ry. Since torque is not zero at zero speed, when used for stopping a load, the supply must be disconnected when close to zero speed. Centrifugal switches are employed to disconnect the supply. Plugging is highly inefficient because in addition to the generated power, the power ie supplied by the source is also wasted in resistances. supplied by the source is also wasted in resistances. Exampte 5.8 A 220 V, 970 rpm, 100 A de separately excited motor has an armature resistance of 0.05 Q. Itis braked by plagging from an initial speed of 1000 rpra. Calculate (a) resistance to be placed in armature circuit to limit braking current to twice the full load value (b) braking torque, arid (c) torque when the speed has fallen to zero. 4 \ x Solution \o ae At 970 rpm E = 220 -0.05 x 100= 215 V V s a+Drives 74 Fundamentals of Electrical 40m Motoring — 0 T (b) Series (a) Separately excited Fig. .10 Plugging speed-torque curves On, 4p Motoring Motoring Plugging Plugging (b) Series (a) Separately excited Fig. 5.11 Counter-torque braking Plugging gives fast braking due to high average torque, even with one section of braking resistance Rg. Since torque is not zero at zero speed, when used for stopping a load, the supply nust be disconnected when close to zero speed. Centrifugal switches are employed to disconnect the supply. Plugging is highly inefficient because in addition to the generated power, the power supplied by the source is also wasted in resistances. Exampte 5.8 ‘4.220 V, 970 rpm, 100 A de separately excited motor has an armature resistance of 0.05 Q. Itis braked by plagging from an initial speed of 1000 rpma. Calculate (a) resistance to be placed in armature circuit to limit braking current to twice the full load value (b) braking torque, and (c) torque when the speed has fallen to zero. | ~~ Solution At 970 rpm E = 220 -0.05 x 100 = 215 VAt 1000 rpm. = jm x 215 = 221.65 V (a) For plugging operation Ry+ R= EtV a21.es +220 =2219 “J a Rpg = 2.21 - 0.05 = i ohms EX, _ 221.65 x 200 oo) On 1000 x 27/60 (c) At zero speed E=0 = 423.3 N-m Peavy 20 e N= RR = Dar = 9554 L € ye dc Motor D en _ de 99.557 As T& ly @w T= 423.3 x B95’ 210.7 N-m ge =. . a x JG —AMPLE 5.11 220 V, 500 A, 600 rpm separately excited motor has armature and field resistance of 0.02 and 10 ©. respectively. The load torque is given by the expression 7, = 2000 - 2V, N-m, where V90 Fundamentals of Electrical Drives _— is the speed in rpm. Speeds below the rated are obtained by armature voltage cont Tol and. above the rated are obtained by field control. Se, (i) Calculate motor terminal voltage and armature current when the speed is 450 q (ii) Calculate field ee oe ee eae current when the speed is 759 om Solution (i) At 450 rpm, 7, = 2000 - 2 x 450 = 1100 N-m At rated operation E, = 220 - 500 x 0.02 = 210 V Fila _ 210 x 500_ _ 7 Rated torque = >" = gag a7 765 = 1671 N-m 1100 bs For a torque of 1100 N-m, [2 = 1200 x 500 = 329 A ee 450 Al 1. = = é 1450 rpm. Ey = 33h x 210 = 157.5 V = Ey + [iaR, = 157.5 + 329 x 0.02 = 164 V Aw * (ii) At 750 rpm T,, = 2000 - 2 x 750 = 500 N-m At this operating point, let the flux and armature current be @’ and If respectively. Then K.9'1, = 500 rd ty From rated operation / w | = 2107 | KO1= SoOx Tare = 3342 Further at 750 rpm, 0, = 230 x 29 = 78.54 and V = K.0'0', + LR, yp : a 20m + E.R, or 220 = 78.54K.g +0.02/, Ne \ Substituting from Eq. (i) A 220 = 78.54 x 500 + 0.021, or 0.02172 — 2201; + 39270 = 0 This equation has solutions 181.5 A and 21647 A. Ignoring the unfeasible value gives [= 181.5 aah wr Nt Ne From Eq. (i) Ko'= FRO. = 2.755 Y Keg" 2.155 . Field voltage = 220 x °F = 220 x 2355 = 1813 V Gi :de Motor Drives 91 Exawpie $.12 A 2-pole separately excited de motor has the ratings of 220 V, 100 A and 750 rpm. Resistance of the armature is 0.1 Q. The motor has two field coils which are normally connected in parallel Iris used to drive a load whose torque is expressed as 7, = 500 - 0.3N, N-m where N is the motor speed in rpm. Speeds below and above rated are obtained by armature voltage control and by connecting the two field windings in series respectively, (i) Calculate the motor armature current and speed when the armature voltage is reduced to 110 V. (u) Calculate the motor speed and current when field coils are connected in series. Solution ‘At rated operation, E, = 220 - 100 x 0.1 = 210 V Bed Q, x 2n= 250 E, _ 250 Kgs K= 5 = 350 = 2.674 (i) Let the motor speed and current be N> and /,2, respectively. Ey = K@y = 2.674 x Maxon = 0.28N> Ve Ey + lyk, or 110 = 0.28N) + 0.1/3 a) Since T=T, Kl, = 500 - 0.3 or 2.67414) = 500 - 0.3N> or 500 = 0.3N2 + 2.674),3 2) Simultaneous solution of Eq. (1) and (2) gives Ig = 148.9 A and N2 = 339.7 rpm (ii) When field coils are connected in series k= 2814 «1337 2 If armature current and speeds are [3 and N3 2 Ey = 1.337N;x e = 0.14N; VeEyt lake or 220 = 0.14N3 + 0.13 3)92 Fundamentals of Electrical Drives Since T=T, 1.3371,3 = 500 - 0.3N3 or 500 = 0.3N3 + 1.337143 Simultaneous solution of Eqs. (3) and (4) yields lag = 25.48 A and N= 1553.2 rpmExam 5.13 ‘A 200 V, 875 rpm, 150 A separately excited de motor has an armatu fed from a single phase fully-controlled rectifier with an ac sourc Assuming continuous conduction, calculate re resistance of 0.06 Q. It is ¢ voltage of 220 V, 50 Hz. fed motor torque and 750 rpm. 2 angle for rated motor torque and (500) rpm. (iii) motor speed for a = 160° and rated torque. Solution 3° W iw At rated operation E = 200 - 150 x 0.06 = 191 V “ G 750 = = 163.7V (i) E at 750 rpm, E 375 x 191 = 16: V, = E + IR, = 163.7 + 150 x 0.06 = 172.7 V omens Now 2a cos & = Vy or 2% 2207 ong @=172.7 or cos @= 0.872 or a= 29.3° (ii) At -500 rpm = 52% x91 = -109V Since V,=E + LR V, =— 109 + 150 x 0.06 = — 100 V Now cas cos @ = V, ue axa? cos @ = -100 or cosa@=-0.5 or a@=120°de Motor Drives 103 (iii) At a= 160° 2Vn 2x 220 Vy Fee cos op = 2H 220V? 6 t69e = _186,V z sine VZE+hR, — 186 = E + 150 x 0.06 or E=-195V =19 Speed = = X 875 = -893.2 rpm EXAMPLE 5.14 If armature circuit inductance of motor of the drive of Example 5.13 be 0.85 mH, calculate the motor torque for (i) or = 60° and speed = 400 rpm. ‘Now external inductance of 2 mH is added to the armature circuit to reduce the region of discontinuous conduction. Calculate the torque for (ji) a= 120° and speed = - 400 rpm (iii) a= 120° and speed = — 600 rpm Solution = tan-! 22% 50 x 0.85 x 10"? @ = tan 0.06 or g= tan! 4.45 = 77.34° and cot $= 0.2247 Z= {RF+(OL,)? = (0.06? + (2nx 50 x 0.85 x 10-5)? = 0.2737 a 191. 60 _ For motor, = Bs SE = 2.084 Rim _ 0.06 X 22012 _,, « ZK = UaT3T x 2084 = 327 From Eq, (5.82) fa Yi cote One = sin (a — o[ ser = ° yf 1+ 0.494 = 32.73 sin (60° - 77.34 [eae | = +288 rad/see or +275 rpm Since motor speed is greater than @p., the drive is operating under discontinuous conduction. a104 Fundamentals of Electrical Drives 400 =87.3V ag x 191 At 400 rpm From Eq. (5.78) ‘ean-o-€+[£- inca nore 03 22042 ne 220N2 on (B- 77.34°) - 873 4 [Ba — 220N7 sin (60° — 77.34° a7 0.06 * | 0.06 ~ Oara7 Sn (0 “ | x exp [-(»- wu] cot nae] =0 or 1136.74 sin (B - 77.34°) — 1455 + 237686792477 = 9 or sin (B — 77.34°) + 2.09¢°°72478 = 128, Trial solution of this equation gives B = 230°. From (5.80) —_ Vm (Cos & - cos B) + (1+ a - B)E t Ps 4 220-12 (cos 60° — cus 230°) + [x Oe rs o?7o oo = 8 T = Kl, = 2.084 x 512 = 1067 N-m -3 ii) = tan 2X 50% 3.85 10" = tan“ 14.92 = 86.17° cot $ = 0.067, 87.3V Z= Ri + (L,)? = 0.8974 RVm _ 0.06 x 22042 ZR ~ 0.8974 x 2.084 = 9-982 From Eq, (5.82) ye = 9.982 sin (120° — sar 7 e =~ 52.94 rad/see or ~ 505.5 pm Since motor speed, ~ 400 rpm, is higher than the speed on boundary between continuous and discontinuous conduction, the drive is Operating under discontinuous conduction. ae a ennen RnR TEREESETEAEAETTTAT EETFrom Eq. (5.78) de Motor Drives 105 220 _ =873 sin (B ~ 86.172) - =87:3 87.3 20/2, 3 ‘08074 0.06 + ae ‘Gao7e Sin (120° - 86.17°) x ¢7010678p0.067%1208/180 346.7 sin (B ~ 86.17°) + 1455 ~ 1896¢-9%78 = 9 or sin (B — 86.17°) + 4.2 ~ 5.46876" 997 = 9 Solution of this equation by trial and error yields B= 281° From (5.80) 2202 (cos 120° - cos 281°) + (z+ ae + ay 873) z =-77.63V Ve Vj-E_ —77.63 — (-87.3) Ry T = KI, = 2.084 x 161.2 = 335.9 N-m =1612A (iii) Since the motor speed (- 600 rpm) is less than the critical speed (~505.5 rpm), drive operates in continuous conduction. Now 2m gg = 2X 22002 a5 129° = -99 V nu a B= 280 y 191 = -131V 875 Y=E _ -99- (131) 0.06 =5333A T = KI, = 2.084 x 533.3 = 1111.5 N-m EXxamee 5.15 Motor of drive of Example 5.13 has armature circuit inductance of 2.85 mH. Calculate motor Speed for: (i) @= 120° and T = 1200 N-m (ii) @ = 120° and T = 300 N-m Solution From part (ii) of Example 5.14 for c= 120° One = — 59.94 rad/sec or ~ 505.5 rpm At the critical speed —— #106 Fundamentals of Electrical Drives 10.34V E 25055 x 191 = 875 ‘As at this point (critical speed), the conductiot ax moth cos 120° = -99 ‘on is continuous een 0.06 Torque at the critical speed = 189 x 2.084 = 393.9 N-m (i) Since torque of 1200 N-m is greater than T,, drive is operating in continuous conduction, Now 99 (as above) -1_ 10 12 b= yqge 75709 B= Vp~ 1pRy =~ 99 ~ 576.9 X 0.06 = — 133.6V Speed = ws x 875 = - 612 1pm i) As the torque of 300 N-m sles than Te, drive is operating in discontinuous conduction From Eq. (581) EL Vq(cosa-cosB) __—- R7 RB-a)—-K(B-@) 0 From Eq. (5.78) Von Yasin (B- 9) ~sin(ae— ge} = fuer) Substituting from Eq. (1) Vo F Msin (B- 6) — sin (a — p)e"P-@"*] = | Yalcos a - cos 8) ___xP [ R,(p-a) weal! — e-P-ancot oy a From example 5.14, Z = 0.8974, cot @ = 0.067, K = 2.084 22042 ~ 0.8974 Yn Z = 346.7, 9 = 86.17° »dc Motor Drives 107 substituting values of various Parameters in Eq. (2) 467s (B ~ 86.17°) ~ sin (120° ~ 86.17°) exp (iz x 1x 0.067) x coo] _ [eae 120° - cos 3) mx 300 0.06( 3 — 12077180) ~ 2.084(8 — 12077180) x [ ~ exp (ta iw 7™* 0.067) x coor] 346.7[sin (6 — 86.17°) — 0.641¢-.967A) _ | 5185.4(0.5 + cos B) + 452.24 ~ (2.094 — B) (2.094 — B) or h = 0.14¢70678 } This is a nonlinear algebraic equation. Trial and error solutions give B = 233.492°. From Eq. (1) Vn (cos & — cos B) mR,T rr 3 a) KB-ay ° 233.492° — 120° B- a= PRI x w= 1.98 Substitution of various values in Eq. (3) 220 -¥2 (cos 120° ~ cos 233.4929) _ x 0.06 X 300 _j yy E= 1.98 4 x 1.98 Speed = $73 x 1.2 = 5.5 pm eee nat ON DOOTILTILCD CONTROL OFEXampLe 5.16 A 220 V, 960 rpm, 12.8 A separately excited de motor has armature circuit Fesistance an, inductance of 2 ohm and 150 mH, respectively. It is fed from a single-phase half-controtieg Tectifier with an ac source voltage of 230 V, 50 Hz. Calculate (i) Motor torque for @ = 60° and speed = 600 rpm. (ii) Motor speed for a = 60° and T = 20 N-m. Solution First it should be ascertained whether motor operates in continuous or discontinuous conduction The critical speed, Separating continuous conduction from discontinuous, is given by RV [sin g.e-®%0_ sin (or— gye-Reoe Ope = [sesamiae rer) (5.92) } -3 Now 6 = tan! 2250 X50 x10" = tan“! 23.562 = 87.57° cot @ = 0.04244, e-7°* = 0.8752, e-%°? = 0.9565 Z= \27 447.12? = 47.17 At rated operation: E = 220 - 12.8 x 2 = 194.4 ka == 1944 “Om — (960/60) x 27 1.9337 Rm _ 2x 23002 _ KZ = 19337 xa7.a7 = 1132 Substituting in Eq. (5.92) 1 - 0.8752 = 77.76 rad/sec or 742.54 rpm Since motor speed of 600 rpm is less than the critical speed, drive is operating under continue conduction, for which sin 87.57° x 0.9565 — sin (60° — 87.57°) x 0.8752 Ome = 7.132} ——————_ rrr 230-2 nu V,=“™ +¢08 a) = (1 + cos 60°) = 155.3 = 000 = At 600 rpm E = oem x 194.4 = 121.5 V -E 155.3 - 121.5 Ra } = 1.9337 (85) = 32.68 N-m r= Ki,=K(dc Motor Drives \11 (ii) Motor back emf for the critical speed 752.54 rpm = 742.54 54 Fe 60 V,-E, he (A) = 1.9337( 155.3 = 150.37 - ; x 194.4 = 150.37 Since motor torque of 20 N-m is higher than the critical torque T,, drive is operating in continuous conduction. Now a =-2_-_20__ N= R= Togg7 = 10344 E = V4~ IgRy = 155.3 - 10.34 x2 = 134.6 V Speed = 1 S 5.960 = 664.8 rpmde Motor Drives 113 From Eqs. (5.7), (5.8), (5.79) and (5,99) Vn Om = Tap (1 + cos or) 7 (5.100) v, vs curve has same nature as shown in Fig. 5.31(a). Consequently, drive operates only in quadrant 1 Exannle 5.17 4.220 V, 1500 rpm, 50 A separately excited motor with armature resistance of 0.5 Q, is fed from a 3-phase fully-controlled rectifier. Available ac source has a line voltage of 440 V, 50 Hz. A star- delta connected transformer is used to feed the armature so that motor terminal voltage equals rated voltage When converter firing angle is zero, (1) Calculate transformer turns ratio. ii) Determine the value of firing angle when: (a) motor is running at 1200 rpm and rated torque: (b) when motor is running at — 800 rpm and twice the rated torque. Assume continuous conduction. Solution For 3-phase fully-controlled rectifier from Eq, (5.97) az. 3° cosa@ For rated motor terminal voltage a = 0° 220 _ V, 3 cos0° = 230.4. V rms converter input voltage between lines = 230.4//2 = 162.9 V For star-delta transformer connection, ratio of turns between phase windings of primary and _ 440/43 _ secondary = STOW = 1.559. (ii) (a) At 1500 rpm E=220-0.5 x 50=195V 2 = 1200 = At 1200 rpm E = 4200 x 195 = 156V V, = E+ 1,R, = 156 + 50x 0.5 = 181 V Since v= 2 Vy COS Of 181 _ cos a= Ty = 0.8227 or a= 34.65° (b) At ~ 800 rpm x 195 = -104 V114 Fundamentals of Electrical Drives V,= E+ LR, =—- 104 + 100 x 0.5 =~ s4y From Eq. (i) coa.2 a =54 * Vn 3 * 230.4 = ~ 9.2454 @= 104,20° or\ A Examp.e 5.19 source has a voltage of 230 V. Assuming continuous conduction. 230 V. 960 rpm and 200 A separately excited de motor has an armature resistance of 0.02 0. The motor is fed from a chopper which provides both motoring and braking operations. The () Calculate duty ratio of chopper for motoring operation at rated torque and 350 rpm. (11) Calculate duty ratio of chopper for braking operation at rated torque and 350 rpm. ible motor (ii) If maximum duty ratio of chopper is limited to 0.95 and maximum permiss current is twice the rated, calculate maximum permissible motor speed obtainable without field weakening and power fed to the source. (iv) If motor field is also controlled in (iii), calculate field current as a fraction of its rated value for a speed of 1200 rpm. Solution At rated operation E = 230 - (200 x 0.02) = 226 V = 350 5.996 = E at 350 rpm = 960 * 226 = 82.4 V Motor terminal voltage V, = E + IyR, = 82.4 + (200 x 0.02) = 86.4 V (i) 5 5 = 804 _ Duty ratio 5 = S84 = 0.376 Gi) V, = E-L,R, = 82.4 - (200 x 0.02) = 78.4 V 784 pa O= 739 = 974_ de Motor Drives 127 ~ aximum available V, = 0.95 x 230 = 218.5 v i j E= Vy + IaRy = 218.5 + (200 x2 x 0.02) = 226.5 V yeximm permissible motor speed = 2265 x 960 = 962 rpm sssaming lossless chopper, power fed into the source Valy = 218.5 x 400 = 87.4 kW iy) Asin (iti) E= 226.5 Vv for which at rated field current speed = 960 rpm. Assuming linear magnetic circuit, E will be inversely proportional to field current. Field current as a ratio of its rated value = 960/1200 = 0.8. 3 Exawrve 5.20 Motor of Example 5.19 is now operated in dynamic braking with chopper control with a braking resistance of 2 Q. (i) Calculate duty ratio of chopper for a motor speed of 600 rpm and braking torque of twice the rated value. (ii) What will be the motor speed for a duty ratio of 0.6 and motor torque equal to twice its rated torque? Solution 600 (i) E at 600 rpm = 920 x 296 = 141.25 V ay) at IPM = 960 * ye ork E Ck y Ree=(1-5)Re=-R, wu : “we eo C 141.25 4 vw) (1 ~ 8) x2 = 4435 _ 0102 o 5 = 0.83 ee? Gi) E=I,{(1 - 8)Ry + Ral = 400{(1 - 0.6) x 2 + 0.02] = 328 V 328 S| =o = peed 226 x 960 = 1393.3 rpmAe Exampte 5.22 uu Motor of Example 5.21 is now controlled in regenerative braking by a chopper with a source voltage of 220 V. (i) Calculate motor speed for a duty ratio of 0.5 and motor braking torque equal to the rated motor torque. (i) Calculate maximum allowable motor speed for a maximum permissible current of 70 A and maximum permissible duty ratio of 0.95. (in) What resistance must be inserted in armature circuit for the drive to run at 1000 rpm without exceeding armature current beyond 70 A? The duty ratio of chopper has a range from 0.05 to 0.95. (iv) To what extent the number of turns in field winding should be reduced to run the motor at 1000 rpm without exceeding the armature current beyond 70 A. Solution (i) At rated motor torque, I, = 70 A. Now Ey = 8V + IyRy = 0.5 x 220 + 70 x 0.12 = 118.4 V From magnetization characteristic, for J, = 70 A and N = 600 rpm, E = 202 V Required motor speed Nj = sd x 600 = 351.7 rpm130 Fundamentals of Electrical Drives (ii) At 70 A and Spux = 0-95 E; = 0.95 600 rpm, E = 202 V x 220 + 70 x 0. 12 =2174V (from the magnetization curve) For I, = 70 A and N Er oy 2174 = Required speed Np === X N= 2 x 600 = 645.7 rpm iii) For J, = 70 A and speed Ns = 1000 rpm = 1000 x 202 = seer which gives the resistance to be inserted, R = (iv) It is assumed that even after changing field turns R, = 0.122 Eq = (0.95 x 220) + (70 x 0.12) = 217.4. V This is the back emf developed by machine at 1000 rpm. At 600 rpm = 600, = = O00 x 217.4 = 130.44 Fraction to which the number of turns in the field are reduced Et _ 130.44 . = & = 1304 - 0.646 ») E202 2 dy “Ny Exampte 5.23 ay Motor of Example 5.21 is now controlled in dynamic braking. Available chopper provides variation in duty ratio from 0.1 to 0.9. (i) Calculate braking resistor so that maximum braking speed at the armature current of 70 A will be 800 rpm. (ii) Also calculate the maximum available motor torque for a speed of 87 rpm with braking resistance as calculated in (i). Solution (i) From magnetization curve, motor back emf at 800 rpm and J, = 70 A £, = 800 x 202 = 269.33 V 00 Effective value of braking resistance E 9.33, Rov = 7 ~ Ram GG~ ~ 0.12 = 3.73.2dc Motor Drives 131 For a given value of Rg, maximum value of Rye is obtained at minimum value of duty ratio Opin» Thus C1 ~ Spine = Roe or (1 ~ 0.1)Rg = 3.73 which gives Rg = 4.14 Q (ii) For a given speed, torque will be maximum when duty ratio is maximum . Total armature circuit resistance at maximum duty ratio Spay R= Rp(1 - Smax) + Ry = 4.14(1 - 0.9) + 0.12 = 0.534. 2 Now E=Rl, @ Equation (i) must be satisfied for a speed of 87 rpm. Trying various values of /, and the value of corresponding E (at 87 rpm = 26.68 V) obtained from magnetization characteristic, ave a approximate solution of J, = 50 A. y P g PP! a go L a \a _ 184 « y AtSOA, K.O= ape oaTeD 7298 ro 4 i % T = Kel = 2.928 x 50 = 146.4 N-m pa © vo