EEEB413_exam_sem1_1112

MESB443, Semester I 2011/2012

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COLLEGE OF ENGINEERING

PUTRAJAYA CAMPUS

FINAL EXAMINATION

SEMESTER 1 2011 / 2012

PROGRAM : BACHELOR IN MECHANICAL ENGINEERING (Hons.)

PROGRAM

KOD MATAPELAJARAN : MESB443

SUBJECT CODE

MATAPELAJARAN : ELECTRO-MECHANICAL SYSTEMS

SUBJECT

TARIKH : AUGUST 2011

DATE

MASA : 3 HOURS

TIME

Instructions to candidates:

1. Answer ALL questions.

2. Total number of questions in this booklet: FIVE (5)

3. This is a close book examination. Usage of scientific calculator is allowed.

4. All answers and rough work must be done in the examination answer booklet.

5. Useful equations are provided at the last page of this booklet.

DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE INSTRUCTED TO DO SO

THIS QUESTION PAPER CONSISTS OF 13 PRINTED PAGES INCLUDING THIS

COVER PAGE.


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QUESTION 1 [20 marks]

a) Give 2 methods used to control the speed of rotating magnetic field. (2 marks)

b) Explain the advantages of stepper motor. (3 marks)

c) Draw and label the power triangle. (2 marks)

d) Draw and label the B-H curve for ferromagnetic materials inside AC machines.

(3 marks)

e) Explain what is rectifier circuit. Draw the circuit diagram. (4 marks)

f) A 240V shunt dc motor has a nominal armature current of 55A. If the armature

resistance is 0.3Ω, calculate the following:

i) The counter EMF. (2 marks)

ii) The power supplied to the armature. (2 marks)

iii) The mechanical power developed by the motor. (2 marks)

[2+3+2+3+4+2+2+2=20 marks]

End of Question 1


A 208V Y-connected synchronous motor is drawing 40A at unity power factor from a

208V power system. The field current flowing under these conditions is 2.7A. Its

synchronous reactance is 0.8Ω. Assume a linear open-circuit characteristic.

a) Find the torque angle δ. (5 marks)

b) Sketch the phasor diagram. (2 marks)

c) How much field current would be required to make the motor operate at 0.8PF

leading? (6 marks)

d) What is the new torque angle in part (c)? (2 marks)

e) Sketch the new phasor diagram and label it. (2 marks)

f) If rotational losses Prot = 2kW, what is the efficiency at 0.8PF leading? (3 marks)

[5+2+6+2+2+3=20 marks]

End of Question 2


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a) A 15kVA 8000/230V distribution transformer has equivalent impedance referred

to the primary side of 80 + j300Ω. The components of the excitation branch

referred to the primary side are RC = 350 kΩ and XM = 70 kΩ. If the primary

voltage is 7967V and the load impedance is ZL = 3.2 + j1.5Ω, what is the

secondary voltage of the transformer? (6 marks)

b) A 6-pole, 50Hz, 3-phase induction motor has a referred rotor resistance R2 of

0.25Ω per phase and a maximum torque of 10Nm at 875rpm. The stator

impedance is assumed to be negligible. Answer the following:

i) Calculate the torque when the slip is 5%. (5 marks)

ii) Calculate the resistance to be added to the rotor circuit to obtain 60%

of the maximum torque at starting. Justify your answer. (6 marks)

iii) Draw the torque-speed curve of an induction motor and label all the

important points. (3 marks)

[6+5+6+3=20 marks]

End of Question 3


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A 208V three phase power system is shown in Figure 1 above. It consists of an ideal

208V three-phase generator connected through a three-phase transmission line to a load.

The transmission line has an impedance of 0.06 + j0.12Ω per phase, and the load has an

impedance of 24 + j18Ω per phase. For this simple balanced power system, assuming that

the phase angle for one of the VLL,source is 30deg, and the sequence is abc sequence, find:

a) The line current ( LI ). (4 marks)

b) The load’s line-to-line voltage ( LLV ) and phase voltage (VP). (4 marks)

c) The real, reactive and apparent powers consumed by the load. (3 marks)

d) The real, reactive and apparent powers consumed by the line AND supplied

by the generator. (6 marks)

e) Power factor for both the load and the generator. (3 marks)

[4+4+3+6+3=20 marks]

End of Question 4

Figure 1

+ _

+

_

+ _

0.06Ω

0.06Ω

0.06Ω

j0.12Ω

j0.12Ω

j0.12Ω

V

LZ LZ

LZ

LZ = 24+j18

208V

_

+


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Company A is designing a tabletop moving horizontally on a ball screw for its production

line. The motor that drives the tabletop has to be selected to fulfill the requirements above.

4-pole single phase induction motor is thought to be suitable for the system. Based on

the requirements and the catalog provided in Appendix A, select the appropriate motor.

Justify your answer. Note that no-load speed is approximately 1750 rpm and rated speed

is between 1450 rpm to 1550 rpm for a 4-pole motor at 60 Hz. Use safety factor of 2 and

select a pinion shaft, lead wire type motor.

[20 marks]

End of Question 5

Total mass of table and work, m1 = 30 kg

Table speed, V = 15 ± 2 mm/s

External force, FA = 0N

Ball screw is horizontal

Total length of ball screw, LB = 800mm

Ball screw shaft diameter, DB = 20mm

Ball screw pitch, PB = 5mm

Ball screw lead, A = 5mm

Ball screw efficiency, = 0.9

Material of ball screw is iron (density =7.9x103kg/m

3)

Internal frictional coefficient of pilot pressure nut, 0 = 0.3

Friction coefficient of sliding surface, = 0.05

Motor power supply, Single phase 110 V, 60 Hz

Movement time, intermittent operation, 5 hours per day

-Load with repeated starts and stops

-Required load holding

-END OF QUESTION PAPER-

Required specifications and structural specifications


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LIST OF EQUATIONS

22

1 746

120

sync m

sync

F R Ni Hl

lR

A

F Bil

e Blv

P

VP I R

R

hp W

fen

p

n ns

n

1

1

1

3

3

3

cos

sin

3 3

3 cos 3 cos

3 sin 3 sin

3sin

3sin

A

A

A

A T L

A T L

A T L

A

s

A

m s

S V I

P V I

Q V I

S V I V I

P V I V I

Q V I V I

V EP

X

V E

X

1

22 2

122 12

2

1

2

2max max

2 22 22 2

3,

3

2

TH

mTH

m

mind TH

m

sync TH TH

TH

TH

TH TH sync TH TH TH

XV V

X X

RV Xs where R R

X XRR X X

s X X

R Vs

R X X R R X X

I

VZ

EEEB413_exam_sem1_1112

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