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

1400175175261Ra

00Ra

a) Draw the magnetic equivalent circuit.

183832521780518383253035301352550217805001028700122554RRS

0RRS

1181100265430485775217805485775217805

16668753086100018002253086100

2438400189865Rx

Rx

4857751898650029527537465emf

48577528956018383256096000

b) Compute the flux density (B) in the working air gap for x=10 mm.

Magnetic flux = 1000 turns1 x 108turnsWb= 10-5 WbA=pr2= p x (20 x10-3 )2

= 1.257 x 10-3 m2

B= 10-51.257 x 10-3 = 7.96 x 10-3 T

c) Compute the value of the energy stored in Wfld (for x=10 mm).

Wfld = I2R

= 102 x 7.96 x 10-3

= 0.796 W

d) Compute the value of the inductance L (for x=10 mm).

Inductance = magnetic fluxcurrent = 10-5/10 = 10-06 H

e) For a force ffld of 1000 N determine for x=10 mm the current

Force = current x length x B

= 10 x 10 x 10-3 x 7.96 x 10-3

= 7.96 x 10-4 N

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.

Draw the approximate equivalent circuit referred to the primary.

Compute all equivalent circuit parameter values ( eq eq R + jX , Rc, Xm) referred to the primary. W=V1IoCos b 96.4 = 1100 x 1.82 x cos b

Cos b = 0.048152

B = 87.24o

Im=IoSin b = 1.82Sin 87.24 = 1.818 A

Iw = IoSin b = 1.82 Cos 87.24 = 0.0876 A

Xm = V1Im = 11001.818 = 605.1 Rc = 11000.0876 = v1Is = 12557

Req + jXeq = 12577 + j605.1

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

Efficiency = outputinput = 40 000 x 0.840 000+800+320 = 77.8 o/o

d) Compute the voltage regulation at full with a power factor of 0.8 lagging

Per unit voltage regulation = R2cos b + X2sin b = 12557 0.04815 + 605.2 0.999

= 604.62 + 604

= 1208.62 V

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:

The output power Pout of the motor.

Ns = 120 fP = 120 x 50 6 = 1000 rpm

Therefore, s = (Ns N)/Ns = (1000 970)/1000 = 0.03

Output power = (1-s) P2

= (1 0.03) x 3 xI22 x (0.510.03)

= 676037 W

Net Power = Total mechanical Losses

= 676037 1530

= 674508 W

The input line current IL to the motor.

Input current I1 = V1Z1V1 = 415

Z1 = R1 + jX1 = 0.981 + j 1.80

I1 = 415302.05 33 = 116.9 -33o A

c) The efficiency of the motor.

Efficiency = output output+lossesOutput = 674503 W

Losses 1 = 3I12R1 = 3 x 116.92 x 0.981 = 37307 W

Losses 2 = 1530 W

Efficiency = 674503674503+37307+1530 = 94 o/o

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:

The full load efficiency.

Efficiency = outputoutput+losesP = Tw

W = 2pN60 = 2p X 80060 = 83 rads-1

P = 632.6 x 83 = 52505.8 W

Efficiency = 52505.852505.8+1220 = 97 o/o

The full load output torque in Nm.

Eb1 = V - IaRa = 500 (113.6 x 0.295)

= 466.49 V

Where Eb1 emf at no load

Eb2 emf at full load

Taf Torque at full load

Eb2No= fP260 Eb2 = No (fPZ/60A) = V - IafRA = N2 x 0.466 = 500 Iaf x 0.6

Now Taf at full load

Taf = 9.55Eb1Iao/No=9.55 x 466.49 x 113.6/800 = 632.6 Tm

The no load speed in rev/min.

Ns x 0.466 = 500 113.6 x 0.295

Ns = 466.480.466 = 1000 rpm

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:

The input current (magnitude and power factor).

I1 = V1ZeqZeq = 6.7 + j12.9 + 16.5 + j11.4

= 23.2 + j24.3

= 2503033.6 35

= 4.3 -350

Power factor = Cos 35 = 0.819

The air-gap power.

P = I12R1 + I22R2

= 4.32 x 6.7 + 32 x 16.5

= 123.883 + 1487.5

= 272.383 W

The shaft output power.

Output power = 272.383 15

= 257.383 W

The shaft output torque.

P= Tw

T= PwP = 257.383 w= 2pN60 = 2p x 141060 = 147.65 rads-1

T = 257.383147.65 = 1.74 Nm

e) The motor efficiency

Efficiency = outputoutput+losses = 257.383257.383+15 = 94 o/o

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