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Unformatted text preview: 93. 910. 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 fullload speed of the motor with RaGlj = 175 Q to the fullload 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 P91), 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 P94 and if Rad]. = 175 Q, what is its no—load speed? What is its fullload 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 noload 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 51 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 P91, 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 916. 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 fullload 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
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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 917. 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 torquespeed 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.
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it (1200 r/min) = 915 r/min The resulting torquespeed characteristic is shown below: CumulativelyCOmpounded Dc Motor TorqueSpeed 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
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 Spring '05
 Shaban

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