Unit-5. Question Bank

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1 Unit-5 Question Bank Q. 1. A 8 pole lap wound DC generator has 70 slots in its armature with 22 conductors per slot. The ratio of pole are to pole pitch is The diameter of the bore of the pole oe is 0.48m. The length of the pole oe is 0.28m. The air gap flux density is 0.32Wb/m 2 & the generated emf in the armature is 400V. Find the speed of generator. Ans P = 8, o. of slots =70, conductors/slot=22 Total o. of conductors = = 1540 The ratio of pole arc to pole pitch = 0.64 Diameter of bore of pole oe=0.48m & length of pole oe = 0.28m Air gap flux density = 0.32Wb/m 2 emf generated ( g ) =400V Pole arc A = P = 8 (Lap wound) 0.64 Pole pitch D Pole arc = 0.64 pole pitch = = m Area of pole oe (A) = pole arc length of pole oe Wb Z P g = 60 A = 60 8 = = m 2 = 1442rpm Ans.

2 Q. 2. A 6-pole DC unt generator has the following data Field resistance = 120, armature resistance = 0.8 umber of conductors = 350 (Wave connected) Flux per pole= 0.02Wb Load resistance across the terminals = 12, armature rotates at 1000 rpm. Calculate power absorbed by load Ans W R = 120, Ra = 0.8, Z = 350, = 0.02Wb Let the terminal voltage be V t Volts. V I = A and t IL = A V I a =I +I L = Z P g = 350V (A=2 wave wound) 60 A 60 2 g = V t +I a R a = V t or V t = V load current (I L )= A R 12 And power absorbed by load = V t I L = = L W Ans. Q. 3. A separately excited DC generator has terminal voltage 250V with constant field excitation. If the load changes from 200KW to 125KW, find percentage reduction in speed. The armature resistance is and total contact drop at brues is 2V. eglect the armature reaction. The flux & total number of armature conductors remain constant. Ans %

3 Terminal Voltage (V t ) = 250V Ra = Total contact drop at brues = 2V Z1P g1 = 60A g1 1 I a =I L For 200KW load I L1 = A g = V t +I L R a +2 = =264V For 125KW load I L2 = A g2 = V+I L2 R a +2 = g2 = 259.5V g2 2 g1 1 g Percentage reduction speed = 1.704% Ans. 264

4 Q. 4. A 220V DC series motor has an armature resistance of 0.3 and field resistance of 0.2. It runs at a speed of 700 rpm taking a current of 15A. Calculate the resistance to be inserted in series with the armature to reduce the speed to 600rpm. The input current remains constant. Assume that the magnetization characteristics is st. line. Ans V t = 220V, Ra=0.3, R sc =0.2 I=I a1 =I a2 =15A 1 =700rpm, 2 =600rpm b1 = V t I a1 (R a +R sc ) = ( ) b1 =212.5V b2 =V t I a1 (R a +R sc +R) = ( R) b2 = ( R)V 2 b2 1 b2 1 b1 2 b R R = = I I Ans. a1 a2

5 Q. 5. A 220V, 1.5KW, 859rpm, separately excited DC motor has armature resistance of 2.5 and it draws a current of 8A at rated load condition. If the field current & the armature current are fixed at the value of rated speed at rated load, what will be the no-load speed of the motor? Assume losses remain constant between no-load full-load operation. Ans rpm Let b1 be back emf at rated load and b2 at no load I ao =0 (no load) Ia=8A (at rated load) Armature resistance (R a ) = 2.5 Supply Voltage (V t ) = 220V Rated speed () = 859 rpm = 1 (say) b1 = V t I a R a = =200V b2 = V t I ao R a = = 220V Z1P b1 = 60A ZoP bo = 60A b1 1 bo o bo 220 o = b rpm Ans.

6 Q. 6. A 250V, DC unt motor on no-load, runs at a speed of 1000 rpm and takes a current of 5A the armature and unt field resistances are 0.2 and 250 respectively. Calculate the speed when the motor is on-load, and is taking current of 50A. Assume that the armature reaction weakens the field by 3%. Ans rpm Motor on no-load line current, I L =5A Field current I = A I ao =I L I =5 1 = 4A bo =V t -I ao R a = = 249.2V On load I L =50A I a1 =50 1 = 49A b1 =V t I a1 R a = = 240.2V bo o o & b1 1 1 bo o o b1 1 1 armature reaction weakens the field flux by 3% 1 (1 0.03) o = 0.97 o o = 1000 rpm 1 = b1 o o bo = rpm Ans.

7 Q. 7. Determine developed torque and aft torque of 220V, 4-pole series motor with 800 conductors wave connected & supplying a load of 8.2KW by taking 45A from the mains. The flux per pole is 25mWb and its armature circuit resistance is m Ans m Sol P Developed Torque Ta=0.159ZIa For wave connection A=2 A = = m Ans. Armature circuit resistance R = R sc +R a = 0.6 b = V t I a R = = 193V ZP b = A or, 193 = or, = rpm T = Shaft torque, Out put Power= 8.2KW = 8200W 2 T = Out Put Power Or, 2π T = 8200 T = 8200 = m Ans.

8 Q. 8. A 6-Pole, lap wound d.c. generator has 840 armature conductors and flux per pole of 0.018Wb. The generator is run at 1200 rpm. Calculate the emf generated. Ans V P = 6, Z = 840, =0.018Wb, = 1200rpm In lap wound generator, A = P=6 P Z Generated emf, g = 60A V Ans. Q. 9. Calculate the emf generated by a 4-pole wave wound armature with 45 slots, with 18 conductors per slot when driven at 1000 rpm. The flux per pole is 0.02Wb. Ans. 540V P = 4, Z=18 45 = 810, =0.02Wb = 1000rpm, For wave wound generator, A = 2 P Z g = 60A V Ans.

9 Q. 10 A lap-connected 8-pole generator has 500 armature conductors and useful flux of 0.07Wb. Determine the induced emf when it runs at 1000rpm. What must be the speed at which it is to be driven to produce the same emf if it is wave wound? 250rpm Ans V, P = 8, Z =500, =0.07Wb, = 1000 rpm A =P= 8 for lap wound g = P Z V Ans. 60A 608 ow for wave wound Generator, g=583.33, A = 2 = g 60A P Z rpm Ans. Q.11 A lap wound DC generator having 8-poles develops emf of 500V at 400rpm. The armature has 144 slots and each slot contains 6 conductors. Calculate the flux per pole. Ans Wb P = 8, A = 8, g =500V, = 400rpm, Z = 144 6=864 P Z g = 60 A Or, g 60A PZ = Wb Ans.

10 Q.12. Determine the power output of a dc motor armature having 1152 lap-connected conductors carrying 150A and rotating at 300rpm in a 12-pole. The flux/pole is 60mWb. Ans KW Z=1152, P = 12, A = 12, = 300rpm, =60mWb= Wb, I a = 150A 3 P Z Generated mf, g= 345.6V 60A 6012 Power Output = g I a = = Q. 13. A dc unt generator has an armature resistance of 0.25 and the resistance of unt field is 220. While delivering a load current of 50A, it has terminal voltage of 440V. Determine the generated emf. Ans. 453V 440 I = R 220 2A 51.84KW Ans. I a = I L +I = 50+2=52A g = V t +IaRa= 440+( ) = 453V Ans.

11 Q. 14. A 20KW, 220V dc unt generator has an armature resistance of 0.07 and a unt field resistance of 200. Find power developed in the armature when it delivers rated output. 220 I = 1.1 A R 200 Ans W P = V t I L I L = A I a = I L +I = = A g =V t +I a R a = 220+( ) = V Power developed = g I a = = W Ans. Q. 15. A unt Generator has an induced voltage on open circuit of 127V. When the machine is loaded, terminal voltage is 120V. Find the load current if the field resistance is 15 and armature resistance is Ignore armature reaction. 342A Ans. g =127, V t = 120V, Ra=0.02 R=15 g = V t +I a R a I a = g V R a t A 0.02 V 120 I = R 15 8A I L =I a I =350 8= 342A Ans.

12 Q. 16. A d.c. unt generator is supplying load connected bus bar voltage of 220V. It has an armature resistance of and field resistance of 110. Calculate the value of load current and load power when it generates an emf of 230V. Ans. 398A, 87.56KW g =230V V 220 I = R 110 2A g =V t +I a R a V I a = g = =400A Ra Load current= I L =I a I = = 398A Ans. Load Power=V t I L = = 87.56KW AS. Q. 17. A 230V d.c. unt Motor takes 51A at full load. Resistance of armature and field windings are 0.1 and 230. Determine (i) field current (ii) armature current (iii) back emf developed at full load Ans. (i) 1A (ii) 50A (iii) 225V (i) (ii) 230 I= R 230 Field current = Armature current, I a =I L I =51 1= 1A Ans. 50A Ans. (iii) b=v t I a R a = 230 (50 0.1)= 225V Ans.

13 Q. 18. A 250V d.c. unt Machine has line current of 80A. It has armature & field resistance of 0.1 and 125 respectively. Calculate Power developed in armature when running as (a) Generator (b) Motor. Ans. (a) KW (b) 1889KW (a) As a generator I = 250 R 125 2A I a =I L +I = 80+2=82A g =V t +I a R a = =258.2V Power Developed = g I a = = KW Ans. (b) As a Motor I a =I L I=80 2 = 78A b =V t I a R a = 250 (78 0.1) = 242.2V Power developed = b I a = = 18.89KW Ans.

14 Q. 19. A 230V dc unt Motor runs at 800rpm and takes armature current of 50A. Find the resistance to be added to the field circuit to increase speed to 1000 rpm at an armature current of 80A. Assume armature resistance 0.15 and field winding resistance = 250 Ans V 230 I a = 50A, I = 0.92A R 250 I L =I a +I =50.92A b 1 =V t IaRa = = 222.5V ow R is connected to field arnding 2 = 1000rpm I a = 80A, b2 =V t I a R a = = 218V Since b b I ( α I ) I I b1 1 1 b I I 2 = =0.8211A I 2 = ( R R ) R R+( ) = 230

15 230 ( ) R = = Ans. Q. 20. A 200V dc unt Motor running at 1000 rpm takes armature current of 17.5A. It is required to reduce the speed to 600 rpm. What must be the value of resistance to be inserted in the armature circuit if the original armature resistance is 0.4? Take armature current to be constant during this process. Ans I a =17.5 b1 =V t I a R a = 200 ( ) = 193V 1 =1000rpm b2 =V t Ia(R a +R) = (0.4+R) = R 2 = 600rpm b ( =Constant) = b1 1 b = R R 600 R = Ans.

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