ELEN 2001-2015, Assignment 1/3

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