1. Thirty-six dry cells are connected in four parallel groups of nine cells in series per group. Ifthe voltage and current rating of each cell is 1,45 volts and 4 amp respectively,

The number of poles has been chosen such that the motor mechanical envelop has been met and has high efficiency. Since the motor is used for position control applications, the motor should have minimal cogging torque.In order to achieve this; fractional slot configuration has been chosen. Certain pole slot combinations are preferred. Identify the motor speed – this should be given in revs per minute (rpm) and will typically follow the following for a 50Hz motor: 4a. Rpm between 2800 and 3000 – 2 pole motor 4b. Rpm between 1350 and 1500 – 4 pole motor 4c. Rpm between 850 and 1000 – 6 pole motor.

(a) calculate the voltage, current and power rating of the entire combination.

(b) Recalculate the problem for nine parallel groups of four cells in series per group.

Rqr’d:

(a)Voltage, Current and Power Rating

(b)Voltage, Current and Power Rating

Sol’n:

(a)V= (1.45V)(9)(b) V= (1.45V)(4)

V= 13.05VV= 5.8V

I= (4A)(4)I= (4A)(9)

I=16AI= 36A

P= V IP= (5.8V)(36A)

= (13.05V) (16A)P= 208.8 Watts

P= 208.8 Watts

2.Calculate the average voltage generated in a moving conductor if it cuts 2.5 x 10⁶ maxwells in (a) 1/40 seconds and (b) 1/80 seconds.

Rqr’d:

E_av

Sol’n:

E_av=Ө / t x 10⁸

(a)E_av = 2.5 x 10⁶ / (1/40) x 10⁸

= 1 Volt

(b) E_av= 2.5 x 10⁶ / (1/80)10⁸

= 2 Volts

3.A six-pole DC-generator has an armature winding with 504 conductors connected in six parallel paths. Calculate the generated voltage in this machine if each pole produces 1.65 x 10⁶ maxwells and the armature speed is 1800 rpm.

Rqr’d:

E_g

Sol’n:

E_g= (P Ө R Z / 60 A ) x 10^-8

E_g = ((6(1.65 x 10⁶)(1800)(504)) /

60(6)) x 10^-8

= 249.48V ~ 250 V

4.The armature winding of the generator of Prob. 3 is modified so that it has two parallel paths instead of six. At what speed should the machine be driven if is to develop the same voltage as before the change, assuming all other conditions to remain unchanged?

Rqr’d:

R = Armature speed

Sol’n:

E_g= (P Ө R Z / 60 A ) x 10^-8

R= ((250)(60)(2) x 10⁸) / (6(1.65 x 10⁶)(500))

R= 601.25 rpm

5.Calculate the voltage generated by a four-pole DC machine given the following particulars: number of slots in armature = 55; number of conductors per slot = 4; flux per pole = = 2.62 x 106 maxwells; speed =1200 rpm; number of parallel paths in armature=2.

Rqr’d:

E_g

Sol’n:

E_g= (P Ө R Z / 60 A ) x 10^-8

E_g = (4(2.62 x 10⁶)(1200)(55)) x 10^-8 / (60(2))

= 230V

6.Ifthe armature winding in prob.5 had four parallel paths, what should have been the generated voltage?

Rqr’d:

E_g

Sol’n:

E_g= (P Ө R Z / 60 A ) x 10^-8

E_g = (4(2.62 x 10⁶)(1200)(55))

x 10^-8 / (60(4))

E_g = 115V

7. A four-pole machine generates 250 volts when operated at 1,500 rpm. If the flux per pole is 1.85 x 10⁶ maxwells, the number of armature slots is 45, and the armature winding has two parallel paths, calculate (ɑ) the total number of armature conductors; (b) the number of conductors in each slot.

Req’d: ɑ.) no. of amurture conductors

b.) no. condutors in each slots

Sol’n:

Eg = (PORƵ / 60ɑ) x 10⁸ Let N = no.of conductors

Ƶ = slots (v)

(a) Ƶ = ((250) (60) (2) ) /(4 (1.89 x 10⁶) (1,500) (10^-8))

Ƶ = 270.27

(b)

N = 270.27 / 45

N = 6 conductors

8.How many total conductors and conductors per slot would be necessary in the armature of Prob. 7 if the winding had four parallel paths?

Given:

ɑ = 4 R = 1500 rpm

P = 4 ϕ = 1.85 x 10⁶

Eg = 250v

Reqd: Ƶ and no. of conductors or slot

Soln:

Eg = (PORƵ/60ɑ)x 10^-8

(a) 250 =( ((4) (1.85 x 10⁶) (1500) Ƶ) / 60 (4) ) x 10^-8

60,000 = 1.11 x 10¹⁰Ƶx 10^-8 /240

Ƶ = 540.54 slots

For A 5kw Dc Motor The Number Of Slots Per Pole Should Be Paid

(b) no. of conductors = 540.54 / 45

= 12.01

9. The speed of the generators of Prob. 7 is decreased to 1,350 rpm. (ɑ) What will be the generated voltage if the flux per pole is maintained at the same value, i.e., 1.85 x 10⁶ maxwells? (b)To what value of flux per pole should the excitation be adjusted if the generated voltage is to remain the same, i.e., 250 volts?

Sol’n: ɑ.) Eg = PORƵ x 10^-8/ 60ɑ

=(4)(1.85x10⁶)(1350)(270) x 10^-8 / 60(2)

= 224.78v

b.) 250 = 4 (ϕ)(1350)(270) x 10^-8 / 60(2)

ϕ = 2.056 x 10⁶ maxwell

10. What is the frequency of the alternating voltage generated in the armature conductors of (ɑ) a six-pole 900-rpm machine? (b.) an eigth-pole 750-rpm machine? (c.) a 10-pole 500-rpm machine?

Given:

ɑ.) P = 6 poles b.) P = 8 poles c.) P = 10

R = 900 rpm R = 750 rpm R = 500 rpm

Read: F=? Read: F=? Read: F=?

Soln: Soln: Soln:

F=PR /120F=PR/120F=PR/120

F= 6(900) / 120 F= 8(750) / 120 F= 10(500) / 120

F= 45hƶ

F= 50hƶF= 41.67 hƶ

11. How many poles are therenin a generator in which the armature frequency is 30 cps when operating at a speed of 600 rpm?

Give: Read: P=?

F= 30 cps

R= 600 rpm

Soln:

F=PR / 120

30= P(600) / 120

P= 6 poles

12.At what speed is an armature rotating in a 12-pole machine if the frequency in the armature conductors is 50 cps?

Given: Read: R=?

P= 12 poles

F= 50 cps

Soln:

F= PR / 120

R= f(180) / P

= 50(120) / 12

R= 500 rpm

13. Calculate the force exerted by each conductor, 6 inch long on the armature of the dc motor when it carries the current of 90 amp and lies in a field the density of which 52, 000 lines per square inch.

Req'd:

P=?

Soln:

F= (PxR)/120

P= (30x120)/600

P= 6 poles

14. What torque will the conductor of problem 13 develop if it lies on an armature the diameter of which is 9 inch?

Given: Req'd:

For A 5kw Dc Motor The Number Of Slots Per Pole Should Bet

D=9in T=?

F=2.485 lbs

Soln:

T= Fxr

r=D/2

r=9in/2=4.5in

r=4.5in x (1ft/12in) = 0.375ft

T= (0.375ft) x (2.485lbs)

T=0.93ft-lb.

15. The armature of a dc motor has 31 slots with 16 conductors in each slots. Only 68 percent of the conductors lie directly under the pole faces, where the flux density is uniform at 46,000 lines per square inch. If the armature-core diameter is 7.25 inch, its length 4.25 inch and the current in each conductor is 25 amp, calculate : (a) the force exerted by the armature tending to produce rotation; (b) the torque in pound-feet.

Given:

Slots=31 }68%

Conductors=16 }68%

B= 46000lines/square inches

D=7.5in

L=4.25in

I=25amp

Req'd:

a) F (lb)

b) T (lb-ft)

Soln:

a)

F= (B x I x L)/11.3x10⁶

F= (31x16x0.68x46000x25x4.25)/ 11.3x10⁶

F= 146lbs.

b)

T= F x r

T= (146lbs) x (7.25in/2) x (1ft/12in)

T= 44.10 lb-ft

16. Using the data of prob.15 calculate the torque that will develop if the flux density is reduce by 5 percent while the current is increase to 40 amp.

Given:

B= 46000-[46000x (0.05)] = 43700

I=40amp

D=7.5in

L=4.25in

For A 5kw Dc Motor The Number Of Slots Per Pole Should Be Lowered

Slots=31 }68%

Conductors=16 }68%

Req'd:

T=?

Soln:

F= (B x I x L)/11.3x10⁶

F= (31x16x0.68x43700x40x4.25)/ 11.3x10⁶

F=221.74lbs.

T= F x r

T=221.74lbs x (7.25in/2) x (1ft/12in)

T=67lb-ft

17. What total current must the armature of a dc motor carry, given the ff. information: armature slots = 72; conductor per slot = 6; pole arcs cover 70.5 percent of circumference; flux density = 58,000 lines per square inch; armature core length = 8 inch; armature core diameter = 21 inch; no. of armature paths in parallel = 6; torque = 1050 lb-ft.

Given:

P=6 }70.5%

Slot=72

B=58000

D=21in

T=1050lb-ft

Req'd:

I=?

Soln:

r= [(21in/2) x (1ft/12in)]/2

r= 0.875ft

F= T/r

F= 1050lb-ft/0.875ft

F=1200lbs

I= [(1200) x (11.3x10⁶)]/[(58000) x (6) x (0.705) x (72) x (8)]

I=95.96amp

For A 5kw Dc Motor The Number Of Slots Per Pole Should Be Replaced

18. What must be the total armature winding current in prob.17 if the torque increases to 1200 lb-ft while the flux density drops by 4%.

Given:

T=1200

B=58000-[58000 x (0.04)] =55680

r= 0.875

P=6 }70.5%

Slot=72

Soln:

F= T/r

F= 12000lb-ft/0.875ft

F=1371.43lbs

I= [(1371.43) x (11.3x10⁶)]/[(55680) x (6) x (0.705) x (72) x (8)]

I=114.23amp