13AUG-Work examples.pdf

# 13AUG-Work examples.pdf - 4.10 A lOO-kW 230th generator has...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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, ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern