UNIVERSITY OF BOLTON WESTERN … power input = ωT e = EI a E = KФN OR E = V – I a R a P e = V 1...

OCD82

UNIVERSITY OF BOLTON

WESTERN INTERNATIONAL COLLEGE FZE

BENG(HONS) ELECTRICAL & ELECTRONIC ENGINEERING

SEMESTER TWO EXAMINATION 2015/2016

ELECTRICAL MACHINES AND POWER DRIVES

MODULE NO: EEE6001

Date: Saturday 21 May 2016 Time: 10:00 – 1:00 INSTRUCTIONS TO CANDIDATES: There are Five questions on this paper. Answer any FOUR questions. All questions carry equal marks.

of 9 Western International College FZE BEng (Hons) Electrical and Electronic Engineering Semester Two Examination 2015/2016 Electrical Machines and Power Drives Module No. EEE6001

Question 1

a) Various quantities which affect the magnitude of induced e.m.f in a transformer

are:

Ф = flux

Фm = maximum value of flux

N1 = number of primary winding turns

N2 = number of secondary winding turns

f = frequency of supply voltage

E1 = R.M.S. value of the primary induced e.m.f.

E2 = R.M.S. value of the secondary induced e.m.f.

Develop expressions for E1 and E2

(5 marks)

b)

Figure 1

For the magnetic circuit of above Figure 1, find the flux density and flux in each of the

other limbs and the central limbs. Assume Ur for iron of the core to be

i) ∞

(8 marks) ii) 4500

(12 marks) Leakage and fringing flux are neglected.

Total 25 marks Please turn the page


Question 2

a) A DC generator is connected to a 220 V dc mains as shown in Figure 2. The

current delivered by the generator to the mains is 100 A. The armature resistance

is 0.1 Ω. The generator is driven at a speed of 400 rpm. Calculate

Figure 2

i) the induced emf (1 mark)

ii) the electromagnetic torque

(1 mark)

iii) the mechanical power input to the armature neglecting iron, windage and friction losses

(2 marks)

iv) power input and output of the armature when speed drops to 350 rpm

(5 marks)

v) State whether the machine is generating or motoring. Assume constant flux. (4 marks)

Question 2 continued over the page…


Question 2 continued… b) A transformer on no-load takes 4.5 A at a power factor of 0.25 lagging when

connected to a 230 V, 50 Hz supply. The number of turns of the primary winding is

250. Draw the vector diagram and Calculate

i) The magnetizing current

(2 marks)

ii) The core loss

(2 marks)

iii) The maximum value of the flux in the core.

(4 marks)

iv) Vector diagram

(4 marks)

Total 25 marks

Question 3

a) Explain the circuit arrangement and conditions for the following operation modes of

a DC motor in variable speed application.

i) Motoring (4 marks)

ii) Regenerative braking

(4 marks) iii) Dynamic breaking

(4 marks) iv) Plugging

(4 marks)

Question 3 continued over the page…


Question 3 continued…

b) The speed of a separately excited dc motor is controlled by a full-wave converter

shown in Figure 3. The field current is also controlled by a full-wave converter and the

field current is set to the maximum possible value. The AC supply voltage to the

armature and field converters is one phase, 440 V, 60 Hz. The armature resistance Ra

is 0.25 Ω., the field circuit resistance is Rf = 175 Ω, and the motor voltage constant is

Kv = 1.4 V/A rad/s. The armature current corresponding to the load demand is Ia = 45

A. The viscous friction and no-load losses are negligible. The inductance of the

armature and field circuits is sufficient to make the armature and field currents

continuous and ripple free. If the delay angle of the armature current is αa = 600 and

the armature current is Ia = 45 A, Determine

i) the torque developed by the motor Td

(3 marks) ii) the speed ω

(3 marks) iii) the input Power Factor of the drive.

(3 marks)

Figure 3 Total 25 marks

Please turn the page


Question 4 a) A 220 V dc shunt motor takes an armature current of 25 A on a particular load. The

armature circuit resistance is 0.4 Ω. Find the resistance required in series with the

armature to reduce the speed by 50% if

i) the load torque is constant and

(5 marks) ii) the load torque is proportional to the square of the speed.

(5 marks)

b) A 3-phase, squirrel cage induction motor takes a starting current of 6 times the full-

load current. Find the starting torque as a percentage of full-load torque if the motor

is started

i) Direct-on-line

(8 marks) ii) through a star-delta starter; the full-load slip of the motor being 4%

. (2 marks)

(iii) Also evaluate the results. (5 marks)

Total 25 marks



Question 5 a) When a three-phase supply is given to a synchronous motor with rotor excited, no

steady starting torque is developed. As such a synchronous motor is not self-

starting,

i) Explain the three starting methods of synchronous motors.

(6 marks) ii) Evaluate advantages of synchronous motors over other types of motors.

(4 marks) b) A three phase bridge rectifier has a purely resistive load of R. The rectifier delivers

Idc = 90 A at an output voltage of 530 V DC and the source frequency is 50 Hz.

Analyse the required voltages and currents. (5 marks)

Determine the following performance parameters:

i) The efficiency (1 mark)

ii) The Form Factor (FF) (1 mark)

iii) The Ripple Factor (RF) (1 mark)

iv) The Transformer Utilization Factor (TUF) (3 mark)

v) The Peak Inverse Voltage (PIV) (2 mark)

vi) The Peak Current through Diode (2 mark)

Total 25 marks

END OF QUESTIONS



EQUATION SHEET

F = Ni

Ф = F/R

R g=ℓg/µ0A

β = Ф/A

R c = ℓc/µ0µrAc

Te = EIa/ω Mechanical power input = ωTe = EIa

E = KФN OR E = V – IaRa

Pe = V1I0 cos Ф0 Eg = KvωIa OR Eg = KvωIf Td = KtIfIa

Va = 2Vm/π x cos αa for 0 <= αa <= π PF = Pi / VI OR PF = (2√2/π) x cos αa Vm = Vdc/1.654 Vdc = (3√3/π) x Vm

Vrms = (3/2 + 9√3/4π)1/2 x Vm

Pdc = VdcIdc

Pdc = VrmsIrms

η = Pdc/Pac FF= Vrms/Vdc

RF= Vac/Vdc = √(FF2 – 1) Vs = Vm/√2 = 0.707Vm


Is = 0.7804Im

VA = 3VsIs

TUF= Pdc/VsIs Vdc = (3√3/π)x Vm

PIV = √3Vm

Average current through each diode Id = (Im x2π) sin (π/6)

END OF EQUATION SHEET

UNIVERSITY OF BOLTON WESTERN … power input = ωT e = EI a E = KФN OR E = V – I a R a P e = V 1...

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