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

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