DCM Solutions

DCM Solutions - 9-3. 9-10. If the motor is operating at...

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Unformatted text preview: 9-3. 9-10. If the motor is operating at full load and if its variable resistance Rad]. is increased to 250 Q, what is the new speed of the motor? Compare the full-load speed of the motor with RaGlj = 175 Q to the full-load speed with Rad]. = 250 9. (Assume no armature reaction, as in the previous problem.) SOLUTION If Radj is set to 250 Q, the field current is now 1F _Vz_ flL_MZQ6g6A =Radj+RF =250§2+100$f 3259 Since the motor is still at full load, E A is still 218.3 V. From the magnetization curve (Figure P9-1), the new field current I F would produce a voltage E M of 247 V at a speed no of 1200 r/min. Therefore, 218.3 V 247 V A n=———-— : E 0 A0 (1200 r/min) = 1061 r/min Note that Rad]. has increased, and as a result the speed of the motor n increased. If the motor is connected cumulatively compounded as shown in Figure P9-4 and if Rad]. = 175 Q, what is its no—load speed? What is its full-load speed? What is its speed regulation? Calculate and plot the torque- speed characteristic for this motor. (Neglect armature effects in this problem.) a = Cumulativer compounded I = Differentially compounded VT=240V Note: Figure P9—4 shows incorrect values for RA + Rs and RF in the first printing of this book. The correct values are given in the text, but shown incorrectly on 77;? re. This ill cted at thsond rt 7 SOLUTION At no-load conditions, E A = VT 2 240 V. The field current is given by VF 240 V _ 240 V I :——-———:——————-———_ ” Radj+RF 17so+100§2 2759 = 0.873 A From Figure P9‘1, this field current would produce an internal generated voltage E Ab of 271 V at a speed no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be 5-1 EAo "0 n= —" no: w (1200r/min)21063r/min E 271v At full load conditions, the armature current is ——VT—=55A — 0.87A = 54.13A IA=1L—1F=1L—R +R adj F ~ The internal generated voltage E A is EA =VT —IA (RA +RS) = 240 V — (54.13 A)(0.44 Q) =216.2 V The equivalent field current is I; = I, + NSE IA :0.873 A+—2—7“Hi N F 2700 turns (54.13 A): 1.41 A From Figure P9-1, this field current would produce an internal generated voltage E Ao of 290 V at a speed no of 1200 r/min. Therefore, n = i no = 2162 V (1200 r/min) = 895 r/min EAo 290 V The speed regulation is SR = ".11 — "r1 X100% : meotx) 218.8% no 895 r/nnn 9-16. The motor described above is connected in shunt. (a) What is the no—load speed of this motor when Radj = 120 (2? (b) What is its full-load speed? (c) Under no—load conditions, what range of possible speeds can be achieved by adjusting Rad}. ? (a) If Rod). = 120 Q, the total field resistance is 320 Q, and the resulting field current is VT 240 v F=—————=———-——=0.75A RF+Rodj 2009 + 1209 I This field current would produce a voltage E Ao of 256 V at a speed of no = 1200 r/min. The actual E A is 240 V, so the actual speed will be n=§A—n =Eflfl (1200r/min)=1125 r/min E U 256V A0 ’l E E E S § i i i i i g (b) Atfull load, IA=IL—IF =100A — 0.75A = 99.25 A,and EA = V, —IARA = 240 v — (99.25 A)(0.14 9): 226.1 v Therefore, the speed at full load will be E A 226.1 V ’12—” = E 0 256V A0 (1200 r/min) = 1060 r/min (c) If Rodj is maximum at no—load conditions, the total resistance is 500 Q, and VT 240 v =———=————-——=0.48A RF+Rodj 2009 + 3009 [F This field current would produce a voltage E Ao of 200 V at a speed of no = 1200 r/min. The actual E A is 240 V, so the actual speed will be n = in = M (1200 r/min) = 1440 r/min E 0 200V A0 If Radj is minimum at no—load conditions, the total resistance is 200 Q, and V, 240 v :——=———-——=1.2A RF+Radj 2009 + 09 IF This field current would produce a voltage E Ao of 287 V at a speed of no = 1200 r/min. The actual E A is 240 V, so the actual speed will be 240V n = o (1200 r/min) = 1004 r/min E 287 v 9-17. This machine is now connected as a cumulatively compounded dc motor with Rad]. = 120 9. V (a) What is the full—load speed of this motor? I (b) Plot the torque-speed characteristic for this motor. (c) What is its speed regulation? SOLUTION (a) At full load, IA =1L —1F =100 A — 0.75 A = 99.25 A, and EA =VT "IA'(RA +Rs)= 240 V — (99.25 A)(0.14 o+0.05 £2): 221.1 V The actual field current will be ‘ VT 240 V [F =———=————=O.75 A RF +Radj 200 Q + 120 Q and the effective field current will be I; = 1F + NSE 1A = 0.75 A+m(99.25 A): 1.54 A NF 1500 turns This field current would produce a voltage E A0 of 290 V at a speed of no = 1200 r/min. The actual E A is 240 V, so the actual speed at full load will be EA 221.1 V n =———n : EA” 0 290 V, i. it a i it (1200 r/min) = 915 r/min The resulting torque-speed characteristic is shown below: Cumulatively-COmpounded Dc Motor Torque-Speed Characteristic 1600 r—‘ -—1¥ —'—-‘ r— r—-—-V ! F ! 1500.. .............. ................ ................ ............... . ................ ................ .............. ..: ............. .............. ..: .............. “.1 3 = - a . ' : ‘ 1100 1000 - i 1 0 20 40 60 80 100 120 140 160 180 200 "ind (N'm) 900 L—-" 5 -—i —I—5 i (c) The no—load speed of this machine is the same as the no—load speed of the corresponding shunt dc motor with Rad]. = 120 9, which is 1125 r/min. The speed regulation of this motor is thus SR = X 100% = ——————>< 100% = 23.0% nfl 915 r/min nnl — nfl 1125 r/min - 915 r/min l i ...
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DCM Solutions - 9-3. 9-10. If the motor is operating at...

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