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**Unformatted text preview: **4.10. A lOO-kW, 230th generator has R, = 0.05 Q and RI = 57.5 Q. If the generator operates at
rated voltage, calculate the induced voltage at (a) full—load and (b) half full-load. Neglect brush- Q contact drop. See Fig. 4-19; 1/: 230/575 = 4 A. 57.5 n Fig. 4-19 Fig 4-20
4.11. -
3 :1;;V,_2i5§;‘:zshp:rtl—shpnt compound generator has the following data' R = 0 06 Q R — 0 04
, f— . acuatetheinduceda’ . a "I , Ie— .
2 v as the tow! bmSh-comact drop. ”(ﬁe voltage at rated load and termmal voltage. Take See Fig. 4-20. 4.24. A lO-hp, 230-V shunt motor takes a full-load line current of 40 A. The armature and ﬁeld
resistances are 0.25 0 and 230 0, respectively. The total brush—contact drop is 2 V and the core
and friction losses are .380 W. Calculate the efﬁciency of the motor. Assume that stray—load loss is 1% of output. 4.25. A 230-V shunt motor delivers 30 hp at the shaft at 1120 rpm. If the motor has an efﬁciency of 87% at this load, determine (a) the total input power and (b) the line current. (c) If the torque lost due
to friction and windage is 7% of the shaft torque, calculate the developed torque. Q 11 4.26. A lO-kW, 250-V shunt generator, having an armature resistance of 0.1 Q and a ﬁeld resistance of 250 Q, delivers full-load at rated voltage and 800 rpm. The machine is now run as a motor while
taking 10 kW at 250 V. What is the speed of the motor? Neglect brush-contact drop. 84 . DC MACHINES [CHAR 4 4.9. MM rotates‘atrl-80.0wrpr‘n in a uniform l. l-T ﬁeld. Determine the maximum value of the v
in the coil. N”"“‘“~~M..., » AKlRO—turn square coil of side 200 mm is mounted on a cylinder 200 mm in diameter. The cylinder oltage induced From (4.2), Mn, .. Em = BNAw = (l.l)(lO)(O.200)2(27r x 1sow/6&5"‘”=“"”82:94«.v-,_‘_nh A lOO—kW, 230-V, shunt generator has R, = 0.05 Q and Rf = 57.5 (2. If the generator operates at rated voltage, calculate the induced voltage at (a) full-load and (b) half full—load. Neglect brush-
contact drop. See Fig. 4—19; If: 230/575 = 4 A. 3
(a) [L = M = 434.8 A
230 1.. = 1L + If = 434.8 + 4 = 438.8 A
1012“ = (438.8)(0.05) = 22 v E=V+IaRa=230+22=252V (b) [L = 217.4 A
In = 217.4 + 4 = 221.4 A
rare“ = 11 V E=230+ll=241V 57.5 .0 Fig. 4—20 A 50~kW, 250-V short—shunt compound generator has the following data: Ra = 0.06 D, R“z = 0.04 O, and Rf: 125 Q. Calculate the induced armature voltage at rated load and terminal voltage. Take
2 V as the total brush-contact drop. See Fig. 4-20. CHAP. 4] DC MACHINES 85 3
[L = M = 200 A
250
[LRR = (200)(0.04) = 8 V
Vf=250 +8 =258V €4,209 I = 200 + 2.06 = 202.06 A a 10R. = (202.06)(0.06) = 12.12 V E=250 +12.12 +8 +2 =272.12V Q 4.12. Repeat Problem 4.11 for a long-shunt compound connection (Fig. 4-21). 1L=200A
1f=ﬂ=2A
125 Ia=200+2=202A
14R“ + R“) = 202(0.06 + 0.04) = 20.2 V E = 250 + 20.2 + 2 = 272.2 V Fig. 421 Fig. 4-22 4.13. The generator of Problem 4.10 has 4 poles, is lap wound with 326 armature conductors, and runs
at 650 rpm on full—load. If the bore of the machine is 42 cm (in diameter), its axial length is 28 cm,
and each pole subtends an angle of 60°, determine the airgap ﬂux density. A portion of the machine is illustrated in Fig. 4—22. From Problem 4.10, E = 252 = .9513. 5. hence q) = 71.35 111%
60 a The pole surface area is A = r01 = (0.21)(7t/3)(O.28) = 0.0616 m1 -3
Hence B=§=w=lJ6T 0.0616 90 DC MACHINES [CHAR 4 ““31me —load = 2370 + (100x13) = 3170 At per pole This mmf generates an emf of z247 V (from Fig. 4—23). Since 71 is proportional to E/q) or E/mmf,
712 = 1800 32 3373 = 1296 rpm
218 3170 4524/ A -hp, 230—V shunt motor takes a full—load line current of 40 A. The armature and ﬁeld
resistances are 0.25 0 and 230 Q, respectively. The total brush—contact drop is 2 V and the core
and friction losses are 380 W. Calculate the efﬁciency of the motor. Assume that stray—load loss
is 1% of output. input = (40)(230) = 9200 W
. _ 230 _
ﬁeld—resrstance loss - _ (230) - 230 W
230
armature-resistance loss = (40 — 1)2(0.25) = 380 W
core loss and friction loss = 380 W
brush—contact loss = (2X39) = 78 W
stray-load loss = 10 X 746 = 78 W 100
total losses = 1143 W
power output = 9200 _ 1143 = 8057 W
efﬁciency = “8057 = 87.6%
9200 4425. 1 A 230—V shunt motor delivers 30 hp at the shaft at 1120 rpm. If the motor has an efﬁciency of 87%
I“ at this load, determine (a) the total input power and (b) the line current. (c) If the torque lost due
L/ to friction and windage is 7% of the shaft torque, calculate the developed torque. output = (3 0)(746) a in ut ower = _ = 25.72 kW
( ) p p efﬁciency 0.87
(b) input current = W = 15.729 = 111.8 A
mput voltage 230
(c) output torque = output power - (30X746) = 190.8 N - m angular velocity " (21: x 1120)/60 developed torque = (l.07)(190.8) = 204.2 N - m Q . 4] DC MACHINES 91 A 10-kW, 250—V shunt generator, having an armature resistance of 0.1 Q and a ﬁeld resistance of
250 Q, delivers full-load at rated voltage and 800 rpm. The machine is now run as a motor While
taking 10 kW at 250 V. What is the speed of the motor? Neglect brush—contact drop. As a generator: 3
I=_25_O=1A IL=_1_(_)X_I9_=4OA
f 250 250 In = 40 + 1 = 41 A [HRH = (41)(0.1) = 4.1 V Eg = 250 + 4.1 = 254.1 V As a motor: 3
1L=M=40A g=3§9=1A
250 250 1 = 40 — 1 = 39 A [HRH = (39)(0.1) = 3.9 v a Em = 250 — 3.9 = 246.1 V Now E E .
"_ = __ or nm = .1: n = 312 (800) = 774.8 rpm
n 138 Eg g 254.1 4.27. Figure 4-24 depicts the Ward-Leonard system for controlling the speed of the motor M. The
generator ﬁeld voltage, vfg, is the input and the motor speed, com, is the output. Obtain an expression
for the transfer function for the system, assuming idealized machines. The load on the motor is
given by J03"l + bcom, and the generator runs at constant angular velocity cog. Fig. 4-24 From Fig. 4-24, (4.16), and (4.19), the equations of motion are: di _ - f _
vﬂz ' R/g’rg + Lrg 7; or Vrg ‘ (ng + Lfgs) Ifg
eg = kgmgifg = Ri + kmlﬁnrom or kgmglﬂg = R1 + [9,11];an T : mIfmi=Jcb +bm or kII=(b+Js)Qm m m In ﬁn Hence, ...

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