Electromagnetic and Electromechanical Energy Conversion (ELEN 2001)

Assignment, S2 2015

(Please submit both solution summary sheet on page 3 and your detailed solution)

Due: Friday October 23, 2015, before 14:00, Assignment Office

Problem 1 (20 Marks)

The cylindrical iron-clad solenoid magnet shown in Figure 1 has a plunger which can move a relatively short

distance X developing a large force ffld. The plunger is guided so that it can move in vertical direction only.

The radial air gap between the shell and the plunger is grad = 1 mm, R220 mm, and (1le mm. The exciting coil

has N=1000 turns and carries a constant current of I = 1oA.

a) Draw the magnetic equivalent circuit. workmg a” gap I cy’mdr’ca’ She”

b) Compute the flux density (B) in the working air gap for I

x=10 mm. N: N=

1000 1000

c) Compute the value of the energy stored in Wfld (for x=10 turns X , turns

mm). X . R

(1) Compute the value of the inductance L (for x=10 mm). _1> I.

e) For a force ffld of 1000 N determine for X=10 mm the current ‘_ grlad

I =10 required I

cylindrical ll T) 00

plunger

Fig. I: Cylindrical magnetfor Problem 1

Problem 2 (20 Marks)

The open circuit and short circuit test results for a 40 kVA single phase transformer that steps up the voltage

from 415 V to 1100 V are:

0 Open Circuit Test (primary open circuited): VoC= 1100 V, 1oC: 1.82 A, Poe: 320 W.

0 Short Circuit Test (secondary short circuited): VSC= 19.5 V, I so: 96.4 A, PSC= 800 W.

a) Draw the approximate equivalent circuit referred to the primary.

b) Compute all equivalent circuit parameter values (Req + jX eq, R C, X m) referred to the primary.

c) Compute efficiency at full load with a power factor of 0.8 lagging.

(1) Compute the voltage regulation at full with a power factor of 0.8 lagging.

Problem 3 (20 Marks)

A three phase induction motor with “Design B” characteristics has the motor name plate details are as follows:

RATING 22 kW VOLTAGE 415 V

POLES 6 CURRENT 39.8 A

SPEED 965 rev/min CONNECTION DELTA

FREQUENCY 50 Hz NEMA DESIGN B

The friction-windage loss plus the core loss Pf/w + Pcore= 1.53 kW may be considered constant over the speed

range of the motor and the motor parameters are: R1= 0.981 (2, X]: 1.80 (2, and X M= j80 Q. When operating at

a speed of 970 rev/min (not full load) the load torque is 190.5 Nm and the input power factor is 0.906 lagging.

Calculate for a speed of 97OreV/min:

a) The output power PM of the motor.

b) The input line current I L to the motor.

c) The efficiency of the motor.

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