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.