E283_Mid_Term_AY1112_S3

COLLEGE OF ENGINEERING PUTRAJAYA CAMPUS

MID-TERM TEST

SEMESTER III 2011/2012

PROGRAMME : Bachelor of Electrical and Electronics Engineering (Hons) : Bachelor of Electrical Power Engineering (Hons)

SUBJECT CODE : EEEB283

SUBJECT : Electrical Machines and Drives

DATE : 16th March 2012

TIME : 1800 - 2000 (2 hours)

INSTRUCTIONS TO CANDIDATES:

1. This paper contains THREE (3) questions in FIVE (5) pages.2. Answer ALL questions in the answer booklet provided.3. Begin the answer to each question on a new page in the answer booklet.4. Formula list is attached at the end of the question paper.5. Students may use scientific calculator, programmable calculator and geometry

set.

THIS QUESTION PAPER CONSISTS OF 5 PRINTED PAGES INCLUDING THIS COVER PAGE.

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EEEB283, Semester III 2011/2012

QUESTION 1 [35 marks]

Figure 1

Figure 1 shows the specifications and design of an inductor that employs a ferrite core with a relative permeability of 3300. The core has a depth of 20 mm. Two air gaps, g 1

and g2 are inserted in the inductor. Assume that there is no leakage flux and that the fringing flux effect is negligible. The rated current of the inductor is 25 A. (

μ0=4 π ×10−7 H /m)

a) EVALUATE and DRAW the magnetic circuit for Figure 1. Labels used must be described. [8 marks]

b) CALCULATE the equivalent/total reluctance of the magnetic circuit. [10 marks]

c) DESIGN the number of turns N that is required so that the air gap flux densities in air-gap 1 is Bg 1=0.1T , and in air-gap 2 is Bg 2=0.2T . [9 marks]

d) DESCRIBE the magnetic behaviour of a magnetic core and the concept of magnetic core saturation. You should also use diagrams in your descriptions. [8 marks]

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6 mm 6 mm 6 mm

6 mm

6 mm

20 mm

30 mm 30 mm

g1=1mm g2= 1.5 mm

i=25 A

N turns



The undergraduate students of a university have been asked to pre-determine the efficiency and voltage regulation of a 2 kVA, 230 V/115 V, 50 Hz transformer. The results of open circuit and short circuit tests obtained are tabulated in Table 1.

Table 1Test A Test B

V = 230 V V = 13.2 V

I = 0.45 A I = 6 A

P = 30 W P=20.1 W

a) If you were the undergraduate student, DETERMINE the tests that were carried out in A and B and on which side were the tests carried out. Explain your answer. [6 marks]

b) FIND the parameters for the equivalent circuit of the transformer referred to the high-voltage side (primary side) and SKETCH the circuit. [11 marks]

c) DETERMINE the voltage regulation at the rated load and at 0.8 lagging power factor. [10 marks]

d) If only a 0–230V power supply is available, is it possible to determine the transformer efficiency at the rated load and at 0.8 lagging power factor. Justify your answer. [4 marks]

Students of a technical college were asked to conduct open circuit and short circuit tests on the above transformer and they obtained the same results as in Table 1 for the open circuit test. Upon completion of open circuit test, in order to conduct short circuit test, the terminals of the secondary winding were shorted with rated primary voltage.

e) EXPLAIN the worst case scenario that would have happened. [4 marks]

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A three-phase Y-connected 460-V, 30-hp, 50-Hz, four-pole induction motor has the

following parameter values in Ω/phase referred to the stator:

R1=0.917; R2=0.264; X1=2.01; X2=0.557 ; X M=50.32

PRCL=449 W ; Pcore=300W ; PF∧W=400 W ; Pstray=744 W

a) DISCUSS the effects of increasing the speed of an induction motor up to synchronous speed. [7 marks]

b) If the motor is operating at an induced torque of 102 Nm, DETERMINE the

following:

i) The air-gap power PAG [4 marks]

ii) The line current IL [10 marks]

iii) The stator copper loss PSCL [3 marks]

iv) The efficiency of the motor [3 marks]

v) The load torque τ load [3 marks]

-END OF QUESTION PAPER-

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LIST OF FORMULAS

General

S1- = V IA Y connected : V =V T

√3 I=I L

P1- = V IA cos ∆ connected : V =V T I=I L

√3

Q1- = V IA sin 1hp = 746 W

S3- = 3V IA = √3VT IL

P3- = 3V IA cos =√3 VT IL cos

Q3- = 3V IA sin =√3 VT IL sin

Machine efficiency , η=Pout

Pout+Ploss

x 100

Introduction to Machinery Principles

F (mmf) = R F ind=i ( l × B )

F (mmf) = Ni B=A

e ind=( v× B ) ⋅ l

Transformer

V P

V S

=NP

N S

=aI P

I S

=1a

Induction Motor

f r=s f e nsync=120 f e

Pnm=(1−s )nsync

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E283_Mid_Term_AY1112_S3

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