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6.1 Name the TWO types of rotor construction used in induction motors - NSC Electrical Technology Power Systems - Question 6 - 2022 - Paper 1

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6.1 Name the TWO types of rotor construction used in induction motors. 6.2 Refer to the speed of induction motors and answer the questions that follow. 6.2.1 Expla... show full transcript

Worked Solution & Example Answer:6.1 Name the TWO types of rotor construction used in induction motors - NSC Electrical Technology Power Systems - Question 6 - 2022 - Paper 1

Step 1

6.1 Name the TWO types of rotor construction used in induction motors.

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Answer

The two types of rotor constructions used in induction motors are:

  1. Wound Rotor
  2. Cage Rotor

Step 2

6.2.1 Explain the difference between rated speed and full-load speed.

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Answer

Rated speed is defined as the maximum speed at which the motor effectively operates under no load conditions. Full-load speed, on the other hand, is the speed at which the motor operates when maximum load is applied.

Step 3

6.2.2 Calculate the synchronous speed of a three-phase motor with four pole pairs per phase that are connected to a 400 V/50 Hz supply.

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Answer

The synchronous speed ( s) of a motor can be calculated using the formula: ns = rac{60 imes f}{p} Where: f = frequency (50 Hz) p = number of poles = 2 × pole pairs = 8 (since there are 4 pole pairs) Thus: ns = rac{60 imes 50}{8} = 375 ext{ rpm}

Step 4

6.2.3 Calculate the percentage slip if the rotor turns at 725 r/min (rpm).

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Answer

Percentage slip can be calculated using the formula: ext{Slip} = rac{n_s - n_r}{n_s} imes 100 Where: n_s = synchronous speed = 750 rpm (from 6.2.2) n_r = rotor speed = 725 rpm Substituting the values: ext{Slip} = rac{750 - 725}{750} imes 100 = 3.33 ext{ \\%}

Step 5

6.3 Name TWO mechanical inspections that could be carried out on the rotor and bearings of a motor.

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Answer

  1. Check if there is any play on the axis of the rotor.
  2. Inspect if the bearings rotate smoothly when turned by hand.

Step 6

6.4.1 Input power

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Answer

The input power (P) can be calculated using the formula: P = rac{\sqrt{3} imes I_L imes V_L imes \cos \theta}{1000}
Substituting the given values: P=3×5×380×cos(20°)1000=3.09kWP = \frac{\sqrt{3} \times 5 \times 380 \times \cos(20°)}{1000} = 3.09 kW

Step 7

6.4.2 Reactive power

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Answer

The reactive power (Q) can be calculated using the formula: Q=3×IL×VL×sinθ1000Q = \frac{\sqrt{3} \times I_L \times V_L \times \sin \theta}{1000} Substituting the values: Q=3×5×380×sin(20°)1000=1.13kVARQ = \frac{\sqrt{3} \times 5 \times 380 \times \sin(20°)}{1000} = 1.13 kVAR

Step 8

6.4.3 Output power if the motor is 90% efficient.

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Answer

The output power (P_o) can be calculated as follows: Po=Pinput×Efficiency100P_o = P_{input} \times \frac{Efficiency}{100} Where: Efficiency = 90% Substituting the input power calculated from 6.4.1: Po=3.09kW×90100=2.78kWP_o = 3.09 kW \times \frac{90}{100} = 2.78 kW

Step 9

6.5 FIGURE 6.5 shows the terminal box of a three-phase motor. Redraw the terminal box in the ANSWER BOOK and indicate how the motor can be connected in delta.

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Answer

To connect the motor in delta, connect:

  • U1 to V2
  • W1 to U2
  • V1 to W2 This creates a triangular connection that allows the motor to operate effectively under delta configuration.

Step 10

6.6.1 Identify component A.

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Answer

Component A is identified as an Overload Normally Closed contact.

Step 11

6.6.2 Explain the purpose of MC1/N.O.

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Answer

MC1/N.O. serves as a hold-out contact ensuring that the timer relay (MC1) and the star contactor (MC2) do not energise until MC1 is energized.

Step 12

6.6.3 Explain the importance of interlocking in this circuit.

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Answer

Interlocking ensures that MC2 and MC3 do not energise simultaneously, which prevents any electrical faults and unsafe operating conditions.

Step 13

6.6.4 Describe the operation of the circuit after the timer has timed through.

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Answer

When the timer has completed its timing, the Normally Closed contact opens, de-energising MC2. This also closes MC2/N.C. which opens MC3/N.C., ensuring that MC2 remains de-energised and the motor transitions to delta operation.

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