ELECTRICAL TECHNOLOGY: POWER SYSTEMS
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMOARANDUM
MAY/JUNE 2019
INSTRUCTIONS TO THE MARKERS
QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY (GENERIC)
1.1 An accident is an unplanned, uncontrolled event ✓ caused by unsafe acts and or unsafe conditions ✓ resulting in a personal injury, illness or the death of an employee. ✓
An accident means an accident arising out of and in the course of an employee’s employment and resulting in personal injury, illness or the death of the employee.
Note: The definition in the textbook is flawed but will be accepted. (3)
1.2 Move in an orderly manner. ✓
Follow the evacuation route as displayed in your workshop. ✓
Move to the designated assembly point in a calm and orderly manner. (2)
1.3 To prevent oneself form being shocked, ✓ therefore precautions must be taken to isolate oneself when rendering assistance.
To prevent injury. (1)
1.4 A third degree burn occurs when all layers of skin have been burnt, ✓ causing permanent skin damage affecting fat, muscle and even bone. ✓
A serious burn deserves 1 mark (2)
1.5 A person who intentionally and recklessly interferes with equipment will cause damages ✓ and this will render equipment unsafe and compromise the safety of the user. ✓ (2)
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QUESTION 2: RLC CIRCUITS (GENERIC)
2.1
2.1.1 Capacitive reactance is the opposition ✓ to an alternating current by the reactive component of a capacitor ✓ in an ac circuit. (2)
2.1.2 Inductive reactance is the opposition ✓ to an alternating current by the reactive component of an inductor ✓ in an ac circuit.
If only the formula is given by the learner = 1 mark
If the formula is accompanied by the correct explanation = 2 marks. (2)
2.2
2.2.1 XL x 2 x π x f x L
= 2 x π x 60 x 44 x10-3
= 16,59 Ω ✓ (3)
2.2.2 XC = 1
2 x π x f x C
= 1 ✓
2 x π x 60 x120 x10-6
= 22,1 Ω (3)
2.2.3 Z √R2 + (XC - XL)2
= √252 + (22.11- 16.59)2
= 25,6 Ω ✓ (3)
2.3
2.3.1 IC = VS
VC
= 220
60
= 3,67 A ✓ (3)
2.3.2 IX = IL - IC
= 6 - 3,67
= 2,33 A ✓ (3)
2.3.3 The phase angle is lagging ✓ because IL is greater than IC.✓ (2)
2.4 2.4.1 At resonance frequency XL = XC
XC = 50,27 Ω (1)
2.4.2 C = 1
XC x 2π x f
= 1
50,27 x 2π x1000
= 3,17 x 10-6
= 3,17 μF (3) ✓
2.4.3 The value of the current will be halved as the circuit resistance is inversely proportional to the applied voltage. ✓
The value of the current will be halved if the resistance is doubled = 1 mark (2)
2.5
2.5.1 fr = 1
2π √L x C
= 1
2π √80 x 10-3 x 33 x 10-6
= 97,95 Hz (3)
2.5.2 I = VS (Z = R at resonance)
Z
= 120
30
= 4 A (3) ✓
2.5.3 VL = I x XL
= 4 x 49,24
= 196,94 V (3)
2.5.4 The voltage across XL is VL = IXL=. ✓
During resonance current is at a maximum. ✓
As a result VL would be higher due to the current being at a maximum. ✓ However because VL = VC the reactive voltage is zero ✓and effectively this voltage increase does not affect the supply voltage.
XL is greater than R, therefore VL will be greater than VR during resonance because the current in a series circuit is common through all components leading to the voltage across the inductor being greater than the supply voltage. (4)
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QUESTION 3: THREE-PHASE AC GENERATION
3.1
3.1.1 The power factor meter is used to measure the power factor.✓ (1)
3.1.2 The Kilowatt-hour meter is used to measure the amount of electrical power consumed ✓ by the load over a certain period of time. ✓ (2)
3.2 Generation✓
Transmission✓
Distribution✓
3.3
(5)
3 marks for phases
1 mark for degrees
1 mark for rotation
3.4 Excessive current will be drawn by the system. ✓ More losses will occur in the form of heat ✓and the efficiency of the system will be reduced.✓ (3)
3.5
3.5.1 Current in each phase
IPH = V
R
= 380
470
=0.81A (3)
3.5.2 Line current
IL = √3 x IPH
=√3 x 0.81
= 1.40A (3)
3.5.3 Total power
PT = P1 + P2
= 500.65 + 421
= 921.65W (3)
3.6 Single phase system appliances are cheaper than three-phase appliances. ✓
Single phase voltage is lower than three-phase and therefore safer. (1)
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QUESTION 4: THREE-PHASE TRANSFORMERS
4.1
4.1.1 Copper losses ✓
Iron losses ✓
Dielectric losses
Stray losses (2)
4.1.2 Insufficient circulating air for cooling the transformer. ✓
Insufficient oil in which the transformer is immersed.✓
Constant overloading.
Internal faults (2)
4.2
4.3
4.3.1 It ensures that the:
4.3.2 To ensure that the transformer is
4.4
4.4.1 Secondary line current
S = √3 x IL x VL
IL = S
√3 x VL
= 10000
√3 x 500
=11.55A(3)✓
4.4.2 Transformer ratio
V ph = VL
√3
= 6000
√3
=3.46kV
VL = Vph
Vph(p) = Np
Vph(s) Ns
= 3460
500
= 7:1
OR
= 6,92 : 1 (6)✓
4.4.3 Input power
Pin = √3 x VL x IL x Cosθ Pin = S x Cosθ
= √3 x 500 x 11.55 x 0.97 = 10000 x 0.97
= 9.70 kW =9.70kW (3)
4.4.4 η = Pin - Ploss x 100
Pin
= 9700 - 80 x 80
9700
= 99.17%(3)✓
4.5 The kVA rating in the primary windings and the secondary windings of a transformer is identical. The secondary line current in question 4.4.1 will be higher than the primary line current, ✓ because the voltage in the secondary windings is less ✓ than voltage in the primary winding. ✓ (3)
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QUESTION 5: THREE-PHASE MOTORS AND STARTERS
5.1 The speed of the rotating magnetic field ✓ in the stator windings. ✓ (2)
5.2
5.2.1 To ensure:
5.2.2 To ensure that there is no electrical connection between:
5.3
5.4
5.4.1 MC3 (N/C)✓ and MC2 (N/C) ✓ (2)
5.4.2 MC1 (N/O1) is connected in parallel with the start button so that when start button is pressed and released, ✓ current could flow in the circuit through it thus keeping the starter on. ✓
To latch/retain the circuit. (2)
5.4.3
5.5
5.5.1 Synchronous speed
The total of 18 poles = 6 poles per phase = 3 pole pairs per phase (p = 3) ✓
NS = 60 x f
p
= 60 x 50
3
= 1000 rpm (4)
5.5.2 Percentage slip
= NS - Nr x 100
NS
= 1000 - 955 x 100
1000
= 4.5%(3)✓
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QUESTION 6: PROGRAMMABLE LOGIC CONTROLLERS (PLC)
6.1 Relays requires regular maintenance and repair. ✓
Uses much more energy. ✓
It requires a lot of space to be wired. ✓
It has lower response time.
When a single relay develops faults in a panel, it takes much time to locate and fix which might delay production. (3)
6.2
6.2.1 The function of the central processing unit is to execute the instructions as per the inputs ✓ and provide the outputs. ✓ (2)
6.2.2 The function of the modem is to isolate the processor against damage ✓ from random high current or voltage spikes ✓ and to pass information signals in and out of the processor (2)
6.3
6.3.1 Supply lines to the PLC should be installed with either a fuse or a circuit breaker to prevent ✓ excessive over-current. ✓ (2)
6.3.2 Wirings and connections should be checked before connecting the supply to the PLC to ensure that wiring is correctly connected, ✓ to prevent damage ✓ to the unit. (2)
6.4 An ON-delay timer contact would not operate ✓ until a pre-set delay time has passed,✓ after it has first been energised.
An OFF-delay timer contact will immediately operate ✓and remain in this active state once energized. Its contacts will only deactivate and open after the pre-set time has passed,✓ after it has been de-energised. (4)
6.5
6.5.1 An ON delay timer is applied in the control circuit to prevent the light from being switched on ✓before the pre-set time lapses after the start button has been pressed. ✓ (2)
6.5.2 When the start button is pressed, coil Y will be energised✓ closing latching circuit Y and contactor Y.✓
The timer will start timing until the pre-set time has lapsed.
Contactor T will close ✓and the lamp will be switched “ON”✓
The Lamp will remain switched “ON” until the Stop button is pressed. ✓ (5)
6.6
6.6.1 An analogue input is represented by FIGURE 6.6 A,✓ because it varies freely within a certain range ✓ (2)
6.6.2 A digital input is represented by FIGURE 6.6 B, ✓because it varies with definite steps which is digital in nature. ✓ (2)
6.7
6.7.1 Exclusive OR-gate ✓ (1)
6.7.2
6.8
6.8.1 Light sensor is applied in:
Cell phones: ✓✓ The device uses the ambient light sensor to automatically control the brightness of the screen in situation where the intensity of light is high or low.
To sense the light intensity level. (2)
6.8.2 Temperature sensor is applied in:
Motor cars radiator: ✓✓ If the temperature of the water circulating in the engine exceed a preset value, the sensor will warn the driver by flicking the light on the dashboard.
To sense the temperature level. (2)
6.9
(10)
6.10
6.10.1 Diode bridge rectifier converts an AC supply voltage✓ to a DC voltage. ✓ (2)
6.10.2 Filtering circuit smooths the AC ripple ✓to ensure a pure DC voltage on the DC rail.✓ (2)
6.10.3 The inverter inverts the DC voltage✓ back to an AC voltage✓ at a variable frequency✓ which will then control the speed of the motor as it depends on the frequency supplied by the VSD. (3)
6.11 The VSD is used in:
Water pumping systems✓
Heating system✓
Exhaust air system, dust extraction and fan system✓
Variable air volume air conditioning system (3)
6.12
6.12.1 Regenerative braking methods may be used in:
Lifts✓
Cranes✓
Electrical locomotives (2)
6.12.2 Regenerative energy is energy recovered from the motor✓ when it slows down by converting mechanical energy to electrical energy✓ which can be either used immediately or stored until needed.✓ (3)
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TOTAL: 200