ELEN 2001-2015, Assignment 1/3
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, R=20 mm, and d=10 mm. The exciting coil
has N=1000 turns and carries a constant current of I= 10A.
a) Draw the magnetic equivalent circuit.
b) Compute the flux density (B) in the working air gap for
x=10 mm.
c) Compute the value of the energy stored in Wfld (for x=10
mm).
d) Compute the value of the inductance L (for x=10 mm).
e) For a force ffld of 1000 N determine for x=10 mm the current
I=I0 required
N=
1000
turns
N=
1000
turns
R
x
d
µ ? 8 I
I
cylindrical
plunger
working air gap cylindrical shell
µ ? 8
grad
Fig. 1: Cylindrical magnet for 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:
• Open Circuit Test (primary open circuited): VOC= 1100 V, IOC= 1.82 A, POC= 320 W.
• Short Circuit Test (secondary short circuited): VSC= 19.5 V, ISC= 96.4 A, PSC= 800 W.
a) Draw the approximate equivalent circuit referred to the primary.
b) Compute all equivalent circuit parameter values ( eq eq R + jX , Rc, Xm) referred to the primary.
c) Compute efficiency at full load with a power factor of 0.8 lagging.
d) 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 ?, X1= 1.80 ?, and XM= j80 ?. 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 970rev/min:
a) The output power Pout of the motor.
b) The input line current IL to the motor.
c) The efficiency of the motor.
ELEN 2001-2015, Assignment 2/3
Problem 4 (20 MARKS)
A 500 V, 6 pole dc shunt motor has the following name plate details:
RATING 50 kW VOLTAGE 500 V dc
POLES 6 CURRENT 113.6 A
SPEED 800 rev/min CONNECTION SHUNT
The motor parameters are: armature resistance (RA)= 0.295 ?, shunt field resistance (RSHUNT)= 98 ?, friction
& windage losses (PF/W)= 1.22 kW. Determine:
a) The full load efficiency.
b) The full load output torque in Nm.
c) The no load speed in rev/min.
d) The no load input current.
Problem 5 (20 MARKS)
A single phase capacitor start induction motor has the following name plate data:
RATING 0.2 kW PHASES 1
POLES 4 VOLTAGE 250 V
FREQUENCY 50 Hz CURRENT 3.0 A
INSULATION Class B SPEED 1410 rev/min
The approximate equivalent circuit parameters are: r1= 6.7 ?, r2′ = 16.5 ?, Xf= 248 ?, x1= 12.9 ?, and
x2’= 11.4 ?. The friction and windage loss is 15 W.
For normal running condition as a single phase motor with the auxiliary winding open, calculate for a slip of
4%, (not full load slip), determine:
a) The input current (magnitude and power factor).
b) The air-gap power.
c) The shaft output power.
d) The shaft output torque.
e) The motor efficiency.
ELEN 2001-2015, Assignment 3/3
Student Name: Student Number:
ELEN2001- S2 2015, Assignment
Solution Summary Sheet
(To be included with your detailed solution)
Part Problem #1
a) Magnetic equivalent circuit:
b) B =
c) Wfld =
d) L =
e) I0 =
Part Problem #2
a) Approximate equivalent circuit
b) eq eq R + jX = Rc = Xm=
c) ?=
d) %VR=
Part Problem #3
a) Pout=
b) IL=
c) ?=
Part Problem #4
a) ?=
b) Tout=
c) NNL=
d) INL=
Part Problem #5
a) I= pf=
b) Pgap=
c) Pout=
d) Tout=
e) ?=
End of Assignment

ELEN 2001-2015, Assignment 1/3
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, R=20 mm, and d=10 mm. The exciting coil
has N=1000 turns and carries a constant current of I= 10A.
a) Draw the magnetic equivalent circuit.
b) Compute the flux density (B) in the working air gap for
x=10 mm.
c) Compute the value of the energy stored in Wfld (for x=10
mm).
d) Compute the value of the inductance L (for x=10 mm).
e) For a force ffld of 1000 N determine for x=10 mm the current
I=I0 required
N=
1000
turns
N=
1000
turns
R
x
d
µ ? 8 I
I
cylindrical
plunger
working air gap cylindrical shell
µ ? 8
grad
Fig. 1: Cylindrical magnet for 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:
• Open Circuit Test (primary open circuited): VOC= 1100 V, IOC= 1.82 A, POC= 320 W.
• Short Circuit Test (secondary short circuited): VSC= 19.5 V, ISC= 96.4 A, PSC= 800 W.
a) Draw the approximate equivalent circuit referred to the primary.
b) Compute all equivalent circuit parameter values ( eq eq R + jX , Rc, Xm) referred to the primary.
c) Compute efficiency at full load with a power factor of 0.8 lagging.
d) 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 ?, X1= 1.80 ?, and XM= j80 ?. 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 970rev/min:
a) The output power Pout of the motor.
b) The input line current IL to the motor.
c) The efficiency of the motor.
ELEN 2001-2015, Assignment 2/3
Problem 4 (20 MARKS)
A 500 V, 6 pole dc shunt motor has the following name plate details:
RATING 50 kW VOLTAGE 500 V dc
POLES 6 CURRENT 113.6 A
SPEED 800 rev/min CONNECTION SHUNT
The motor parameters are: armature resistance (RA)= 0.295 ?, shunt field resistance (RSHUNT)= 98 ?, friction
& windage losses (PF/W)= 1.22 kW. Determine:
a) The full load efficiency.
b) The full load output torque in Nm.
c) The no load speed in rev/min.
d) The no load input current.
Problem 5 (20 MARKS)
A single phase capacitor start induction motor has the following name plate data:
RATING 0.2 kW PHASES 1
POLES 4 VOLTAGE 250 V
FREQUENCY 50 Hz CURRENT 3.0 A
INSULATION Class B SPEED 1410 rev/min
The approximate equivalent circuit parameters are: r1= 6.7 ?, r2′ = 16.5 ?, Xf= 248 ?, x1= 12.9 ?, and
x2’= 11.4 ?. The friction and windage loss is 15 W.
For normal running condition as a single phase motor with the auxiliary winding open, calculate for a slip of
4%, (not full load slip), determine:
a) The input current (magnitude and power factor).
b) The air-gap power.
c) The shaft output power.
d) The shaft output torque.
e) The motor efficiency.
ELEN 2001-2015, Assignment 3/3
Student Name: Student Number:
ELEN2001- S2 2015, Assignment
Solution Summary Sheet
(To be included with your detailed solution)
Part Problem #1
a) Magnetic equivalent circuit:
b) B =
c) Wfld =
d) L =
e) I0 =
Part Problem #2
a) Approximate equivalent circuit
b) eq eq R + jX = Rc = Xm=
c) ?=
d) %VR=
Part Problem #3
a) Pout=
b) IL=
c) ?=
Part Problem #4
a) ?=
b) Tout=
c) NNL=
d) INL=
Part Problem #5
a) I= pf=
b) Pgap=
c) Pout=
d) Tout=
e) ?=
End of Assignment