Electromechanical Systems Exam questions PDF

Preview

Summary

Download Electromechanical Systems Exam questions PDF

Description

ELECTROMECHANICAL SYSTEMS (15ELB003) June 2016

2 Hours

Attempt as much of the paper as you can – 100 marks are available Any University approved calculator is permitted.

1. a)

b)

c)

d)

A coil of 32 turns is rotated at a speed of 1050 revs/min inside an 8-pole arrangement of magnets. If the useful flux/pole produced by the magnets is 2.5 mWb, determine the frequency and the rms value of the emf induced in the coil assuming that the emf is sinusoidal.

[6 marks]

A three-phase 6-pole 50 Hz induction motor operates at 4.5% slip at fullload when its output torque is 150 Nm, determine the speed in revs/min and the output power.

[12 marks]

A small transformer core weighs 4.6 kg. The manufacturer quotes a material density of 8100 kg/m3 and a total loss density of 18630 W/m3 when operating at the transformer's nominal frequency and flux density. Determine the power loss in the core under the nominal conditions.

[4 marks]

A coil of 200 turns is wound uniformly over a plastic ring having mean circumference of 600 mm and a uniform cross-sectional area of 500 mm2. If the current in the coil is 4 A dc and assuming that the flux is confined to the ring, calculate: i.

The coil mmf.

[3 marks]

ii.

The magnetic field strength on the mean flux path.

[3 marks]

iii.

The flux density on the mean flux path.

[3 marks]

iv.

The total flux in the ring.

[3 marks]

v.

The self-inductance of the coil.

[4 marks]

1 of 4

/continued

2.

A single-phase transformer has approximate equivalent circuit parameters rm = 1350 Ω, xm = j850 Ω, R1 = 1.35 Ω, X1 = j2.42 Ω and N1:N2 = 2.5:1. The transformer operates with a constant primary voltage of 110 V.

3.

a)

Determine the no-load primary current, power factor and iron loss.

b)

Estimate the secondary current when the transformer is loaded and operating at its maximum efficiency.

[12 marks]

[9 marks]

The 200-turn coil of the loudspeaker illustrated in Figure Q3 below is of 20 mm mean diameter and has a resistance of 15 Ω. The coil lies in the uniform radial magnetic field of 0.9 T produced by an annular magnet system. The coil is attached to a diaphragm that is supported at its rim by a flexible annular ring of compliance 0.01 m/N. a)

b)

The coil is supplied from a constant voltage supply of 0.6 V, determine the force produced on the diaphragm by the coil and the deflection of the diaphragm in mm (in the x direction).

[9 marks]

The coil is supplied from an alternating voltage supply of 0.6 V rms. Assuming that the supply varies sinusoidally and that the effect of movement can be ignored, determine the peak force produced on the diaphragm by the coil.

[9 marks]

flexible annular ring

conduct or adhesive carrier

S

magnet

x N

S

diaphragm

Figure Q3

2 of 4

/continued

4.

Locked rotor, no-load and load tests were conducted on a small 4-pole woundrotor induction motor. The per/phase (star) parameters, for the approximate equivalent circuit of Figure Q4, were determined from the test results to be: rm = 1050 Ω, xm = j40 Ω, Rs = 0.85 Ω, Rr/ = 1.25 Ω, (Xs + Xr/) = j3.85 Ω The mechanical losses were found to be 370 W at synchronous speed.

Figure Q4 For the load test the motor was supplied at 230 V (line), 50 Hz and the mechanical load increased until a motor line current of 5 A was obtained, at which point the motor input power was 1250 W and its speed 1455 revs/min, determine: a)

The output power.

[17 marks]

b)

The output torque.

[6 marks]

Dr Keith Gregory

3 of 4

/continued

Useful information Magnetic materials and circuits Pe = k e (f Bm )2 W m3 Ph = k hf B mx W m 3 Px = k x (f Bm )3 2 W m3 µ 0 = 4 π ×10 −7 H / m

Transformers

voltage regulation =

no - load voltage − full - load voltage no - load voltage

voltage regulation ≈

I′2 (R1 cos φ2 ± X1 sin φ2 ) V1

η=

n × V2I 2 cos φ2 n × V2 I2 cos φ2 + Pi + n 2 × Pc

Force and torque production

T = 1 I12 2

dM dL1 1 2 dL 2 Nm + 2 I2 + I1I 2 dθ dθ dθ

Induction motors s=

ns − nr Ns − Nr ωs − ωr = = ns Ns ωs

Ir′ =

Pm =

Tm =

Vs

(R s + R′r s)2 + ( X s + X ′r )2 Vs 2

(R s + R′r s)

2

+ ( X s + X ′r )

2

×

R′r (1 − s ) W s

2

×

R′r Nm s ωs

Vs 2

(R s + R′r s)

s tm =

2

+ ( X s + X ′r )

A

Rr′ Rs2

+ ( Xs + X′r )2

4 of 4...