NATIONAL CERTIFICATE

ELECTROTECHNICS N*

This marking guideline consists of 15 pages.

QUESTION 1

Run the motor on no-load at normal voltage and speed.

Measure the no-load input current ( I ), shunt field current ( ISH ) and determine the armature current. IA = I – ISH

Armature copper loss = Error! Objects cannot be created from editing field codes. . Ra

Shunt field copper loss = ISH.V orError! Objects cannot be created from editing field codes.. RSH

Iron, friction and windage losses = IA.V

Total losses =Error! Objects cannot be created from editing field codes.. RA + ISH.V + IA.V

No-load input to the motor = I .V

(8)

1.2.1

E1 = V – IA1 . RA

= 250 – (60 + 0,35)

= 229 V

E µ N.F But F is constant \ E µ N

T2 = T1 \ IA2 = IA1

(6)

1.2.2

T µ N2 and T µ IA \ N2 µ IA

(4)

[18]

QUESTION 2

2.1

Fundamental

XL1 = 2.p. f.L = 314 x 0,04 = 12,56 W

Z1 = R + jXL1

= 20 + j12,56

= 23,617 Ð32,129º W

Third Harmonic

XL3 = 2.p. f3.L = 942 x 0,04 = 37,68 W

or

XL3 = 3 . XL1 = 3 x 12,56 = 37,68 W

Fifth Harmonic

XL5 = 2.p. f5.L = 1 570 x 0,04 = 62,8 W

or

XL5 = 5 . XL1 = 5 x 12,56 = 62,8 W

Method 1

P = ( IRMS )2 . R

= 3,1182 x 20

= 194,438 W

Method 2

(10)

2.2

(2)

[12]

QUESTION 3

3.1

The secondary winding is open (no-load)

The primary winding is connected to the rated voltage and frequency

The wattmeter reads the iron (core) losses of the transformer

(5)

3.2

3.2.1

(6)

3.2.2

(a)

S = V2 /I2

I2 = S/V2

= 100 x 103/500

= 200 A

Cos q = 0,8 Lagging \ q = 36,87 Sin 36,87 = 0,6

V2 = 500 - (1,712 % x 500)

= 500 - 8,56

= 491,44 V

(6)

(b)

V2 = 500 - (-0,304 % x 500)

= 500 + 1,52

= 501,52 V

(4)

[21]

QUESTION 4

4.1

The distribution factor is the ratio of the actual EMF obtained to the EMF that would be obtained if all the coil sides were concentrated in one slot.

(2)

4.2

n = Number of slots per pole/Phase

Number of slots per pole = n x Phase

= 3 x 3

= 9

a = 180 /Number of slots per pole = 180 /9 = 20

y = 180 - 120 = 60º

kp = Cos y/2 = Cos 60 /2 = 0,866

f = N.p/60 = 1 000 x 3/60 = 50 Hz

Number of slots per pole = Number of slots/Number of poles

Number of slots = Number of slots per pole x Number of poles

= 9 x 6

= 54 slots

Z = Total conductors in series per phase

= (54 x 2) / 3

= 36

EP = EL /Ö3 = 400/Ö3 = 230,94 V

(11)

[13]

QUESTION 5

5.1

(a)

E V

(b)

(c)

In (a) the synchronous motor has been synchronised. E and V are equal in magnitude but in phase opposition.

In (b) the mechanical drive of the motor is removed and E retards by an angle a. E still remains equal to V. ER lets current I flow through the motor.

In (c) the excitation of the motor is strengthened and E increases. An increase in E results in an increase of ER. I is approximately 90 out of phase with ER and leads V by angle q.

(6)

5.2

5.2.1

(4)

5.2.2

Cos q = 0,707 Leading \θ = 45º

Zs = R + jXS

= 1,936 + j 17,424

= 17,531 ë83,66 W

Method 1

Er = I . Zs

= 39,365 ë45 x 17,531ë83,66

= 690,108 ë128,66 V

Ep = Er - Vp

= 690,108 ë128,66 - 1 270,171 ë0

= - 1 701,28 + j538,882

= 1 784,586 ë162,424 V

EL = Ö3 . Ep

= Ö3 x 1 784,586

= 3 090,994 V

Method 2

I.RA = 39,365 x 1,936 = 76,211 V

I.Xs = 39,365 x 17,424 = 685,896 V

Ep = Vp ë-q + IRa ë180 + I.Xs ë-90

= 1 270,171 ë-45 + 76,211ë180 + 685,896 ë-90

= 821,936 - j1 584,043

= 1 784,593 ë-62,576 V

EL = Ö 3 .Ep

= Ö 3 x 1 784,593

= 3 091,006 V

(5)

5.2.3

Method 1

α = 180º - 162,424º = 17,576º Electrical degrees

Method 2

α = 62,576º - 45º = 17,576º Electrical degrees

α = 17,576º / 3 = 5,859º Mechanical degrees

(2)

[17]

QUESTION 6

R2 = S . XO

XO = R2 / S

= 0,5 / 0,1

= 5 W

[9]

QUESTION 7

7.1

At poor power factor the kVA demand is high and Escom charges per kVA demand at improved power factor, kVA is reduced and so the charges.

At poor power factor the current is high and since the line and cable volt drops depend on current, the consumer voltage becomes low.

(2)

7.2

7.2.1

Method 1

q1 = Cos-1 0,8 lagging = -36,87

q3 = Cos-1 0,96 lagging = -16,26

(4)

Method 2 Note: Refer to the phasor diagram in method 1

Q1 = kVAr before power factor correction

Q1 = P x Tan q1

= 1 800 x Tan 36,87

= 1 350 kVAr

Q3 = kVAr after power factor correction

Q3 = P x Tan q3

= 1 800 x Tan 16,26º

= 524,993 kVAr

Q2 = kVAr of the capacitor bank

Q2 = Q1 - Q2

= 1 350 kVAr - 524,993 kVAr

= 825,007 kVAr

or

Q2 = P ( Tan q1 - Tan q3)

= 1 800( Tan 36,87 - Tan 16,26 )

= 825,012 kVAr

(4)

7.2.2

Method 1

(4)

Method 2

(4)

Method 3

IW = IL1 . Cos θ1

= 196,824 x 0,8

= 157,459 A

IL2 = IW ( Tan q1 - Tan q1 )

= 157,459 (Tan 36,97º - Tan 16,26º )

= 72,17 A

(4)

[10]

TOTAL:

100

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