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TABLE 5.1 below shows the name plate of a three-phase induction motor - NSC Electrical Technology Power Systems - Question 5 - 2018 - Paper 1

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TABLE 5.1 below shows the name plate of a three-phase induction motor. Answer the questions that follow. TABLE 5.1: NAME PLATE OF A THREE-PHASE INDUCTION MOTOR MOTO... show full transcript

Worked Solution & Example Answer:TABLE 5.1 below shows the name plate of a three-phase induction motor - NSC Electrical Technology Power Systems - Question 5 - 2018 - Paper 1

Step 1

5.1.1 State the amount of current the motor will draw from the supply at full load.

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Answer

The motor will draw 1.3 A from the supply at full load as indicated on the nameplate.

Step 2

5.1.2 Explain why the motor is suitable for use in South Africa.

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Answer

The motor is suitable for use in South Africa because it is designed to operate with a supply voltage of 380 V and a frequency of 50 Hz, which are the standard electrical conditions commonly found in the region.

Step 3

5.1.3 State what the 7,5 kW on the name plate indicates.

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Answer

The 7.5 kW on the nameplate indicates the rated output power that the motor can deliver to drive the load efficiently.

Step 4

5.1.4 Determine the total number of poles.

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Answer

To determine the total number of poles, we can use the formula:

p=60×fnsp = \frac{60 \times f}{n_s}

Where:

  • f=50f = 50 Hz (frequency)
  • ns=1500n_s = 1500 r/min (synchronous speed)

Calculating:

  • p=60×501500=2p = \frac{60 \times 50}{1500} = 2 poles per phase. Thus, the total number of poles is: Total Poles=2×2=4.\text{Total Poles} = 2 \times 2 = 4.

Step 5

5.1.5 Calculate the efficiency of the motor at full load if the total loss is 1,2 kW.

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Answer

The efficiency (ηη) can be calculated using the formula:

η=PoutPout+Ploss×100η = \frac{P_{out}}{P_{out} + P_{loss}} \times 100

Where:

  • Pout=7.5kWP_{out} = 7.5 kW (output power)
  • Ploss=1.2kWP_{loss} = 1.2 kW (total losses)

Calculating: η=7.57.5+1.2×100=7.58.7×10086.21%.η = \frac{7.5}{7.5 + 1.2} \times 100 = \frac{7.5}{8.7} \times 100 \approx 86.21\%.

Step 6

5.2 Explain the purpose of no-volt protection with reference to motor control circuits.

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Answer

No-volt protection is designed to prevent the motor from restarting automatically after a power loss. If the supply voltage drops below a prescribed level or fails, the no-volt protection circuit will interrupt the power supply, thereby safeguarding the motor and protecting the operator.

Step 7

5.3 Explain how the direction of rotation of a three-phase induction motor can be changed.

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Answer

The direction of rotation of a three-phase induction motor can be changed by swapping any two of the three supply lines. This alteration reverses the magnetic field direction, thus changing the direction in which the motor rotates.

Step 8

5.4 State TWO mechanical inspections that must be carried out on an induction motor before commissioning.

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Answer

  1. Check that the end plates are securely fastened to prevent vibrations and ensure structural integrity.
  2. Verify that the shaft turns freely without any binding, allowing for smooth operation.

Step 9

5.5.1 Identify the control circuit in FIGURE 5.5.

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Answer

The control circuit identified in FIGURE 5.5 is a forward/reverse control circuit.

Step 10

5.5.2 State ONE application of the control circuit.

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Answer

One application of the control circuit is in elevators, where it is used to change the direction of the motor based on operational requirements.

Step 11

5.5.3 State the purpose of the overload relay.

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Answer

The purpose of the overload relay is to disconnect the power supply from the motor when the current exceeds a pre-set value, thus protecting the motor from overheating and potential damage.

Step 12

5.5.4 Describe the operation of the control circuit.

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Answer

When the start button is pressed, current will flow in the circuit and MC1/FWD becomes energized. This action engages the motor, enabling it to run forward. Conversely, when the reverse start button is pressed and MC2/REV is energized, the motor will run in reverse. The control circuit contains interlocks ensuring that both forward and reverse cannot be engaged simultaneously, maintaining safe operation.

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