QUESTION 6: THREE-PHASE MOTORS AND STARTERS
6.1 Name TWO continuity tests to be performed on a three-phase motor - NSC Electrical Technology Power Systems - Question 6 - 2022 - Paper 1
Question 6
QUESTION 6: THREE-PHASE MOTORS AND STARTERS
6.1 Name TWO continuity tests to be performed on a three-phase motor.
6.2 Explain the term cogging with reference to in... show full transcript
Worked Solution & Example Answer:QUESTION 6: THREE-PHASE MOTORS AND STARTERS
6.1 Name TWO continuity tests to be performed on a three-phase motor - NSC Electrical Technology Power Systems - Question 6 - 2022 - Paper 1
Step 1
6.1 Name TWO continuity tests to be performed on a three-phase motor.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Continuity test between the ends of each coil.
Continuity test between the frame of the motor and earth.
Step 2
6.2 Explain the term cogging with reference to induction motors.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Cogging is the tendency of the rotor rod to remain stuck under a stator tooth due to the direct magnetic attraction between the two.
Step 3
6.3 State TWO advantages of cage-type induction motors over wound rotor-type motors with slip rings and brushes.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Cage-type induction motors require less maintenance because of the absence of slip rings and brushes.
They are explosion-proof due to the absence of brushes and slip rings that cause sparking.
Step 4
6.4 Name TWO applications of squirrel-cage induction motors where constant speed and torque is essential.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Cranes
Conveyor belts
Step 5
6.5 Label points A, B and C on the characteristic curve in FIGURE 6.5.
97%
117 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
A - Pull-up torque
B - Breakdown torque
C - Full-load torque
Step 6
6.6 Differentiate between synchronous speed and rotor speed.
97%
121 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Synchronous speed is the speed at which the magnetic field in the stator rotates, whereas rotor speed is the speed at which the rotor rotates in an attempt to reach the synchronous speed.
Step 7
6.7.1 Synchronous speed of the motor.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To calculate the synchronous speed:
ns=p60×f
Substituting the given values:
ns=360×50=1000 rpm
Step 8
6.7.2 Percentage slip.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To calculate the percentage slip:
Percentage slip=nsns−nr×100
Substituting the calculated synchronous speed and the given rotor speed:
Percentage slip=10001000−950×100=5%
Step 9
6.8.1 Name TWO types of losses other than copper losses that influence efficiency of the motor.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Iron losses
Mechanical losses
Step 10
6.8.2 Calculate the efficiency of the motor.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the efficiency:
η=PinPin−losses×100
Given:
Input power Pin=5000W
Losses = 600 W
Substituting these values:
η=50005000−600×100=88%
Step 11
6.8.3 Calculate the output power of the motor.
97%
117 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To calculate the output power:
Pout=Pin−losses
Substituting the values:
Pout=5000W−600W=4400W
Thus, the output power is 4.4 kW.
Step 12
6.9.1 Identify component T.
97%
121 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
T is a timer.
Step 13
6.9.2 Explain the purpose of having TWO overload units in the circuit.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Each overload unit monitors the current drawn by each motor independently.
Step 14
6.9.3 Identify the error in the circuit.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The error is that the MC1/N/O2 auxiliary contact (Hold contact) connected in parallel with the start button is omitted.
Step 15
6.9.4 Explain how this error affects the operation of the circuit.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The moment the start button is released, MC1 will de-energize, opening MC1/N/O2, disconnecting the parallel section of the control circuit, disabling timer T1, thus MC2 will not be energized.