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6.1 Name TWO rotating parts of a three-phase induction motor - NSC Electrical Technology Power Systems - Question 6 - 2024 - Paper 1

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6.1 Name TWO rotating parts of a three-phase induction motor. 6.2 FIGURE 6.2 below shows how the coils of a three-phase induction motor are placed in a stator. Answ... show full transcript

Worked Solution & Example Answer:6.1 Name TWO rotating parts of a three-phase induction motor - NSC Electrical Technology Power Systems - Question 6 - 2024 - Paper 1

Step 1

6.2.1 Determine the angle between coil U and coil V.

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Answer

The angle between coil U and coil V is 120°, as each coil is spaced 120° apart in a three-phase system.

Step 2

6.2.2 Explain how a rotating magnetic field is created when a three-phase supply is connected to the stator windings.

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Answer

When a three-phase supply is connected to the stator windings, the alternating current flowing through each of the coils creates a magnetic field. These coils are spaced 120° apart, causing each coil to be magnetised at different intervals. The combination of these magnetic fields rotates at a frequency corresponding to the supply frequency. For example, with a frequency of 50 Hz, the rotating magnetic field is produced, which is essential for the operation of induction motors.

Step 3

6.2.3 State why a squirrel cage is less of a fire hazard (explosion proof).

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Answer

A squirrel cage rotor does not use brushes or slip rings, which are components that can create sparks during operation. This lack of sparking minimizes the fire hazard, making the squirrel cage rotor safer in explosive environments.

Step 4

6.3.1 Synchronous speed

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Answer

The synchronous speed, nsn_s, can be calculated using the formula:

n_s = rac{60 imes f}{p} Substituting the known values: n_s = rac{60 imes 50}{2} = 1500 ext{ RPM}

Step 5

6.3.2 Rotor speed

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Answer

To find the rotor speed, nrn_r, we can use the slip formula:

nr=nsimes(1extslip)n_r = n_s imes (1 - ext{slip}) Considering a slip of 5%: nr=1500imes(10.05)=1500imes0.95=1425extRPMn_r = 1500 imes (1 - 0.05) = 1500 imes 0.95 = 1425 ext{ RPM}

Step 6

6.4.1 Efficiency

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Answer

The efficiency, extEfficiency ext{Efficiency}, can be calculated with the formula:

ext{Efficiency} = rac{P_{out}}{P_{in}} imes 100 Where:

  • Pout=PinextlossesP_{out} = P_{in} - ext{losses}
  • Pin=11750extWP_{in} = 11750 ext{ W}
  • extlosses=1750extW ext{losses} = 1750 ext{ W} Thus: Pout=117501750=10000extWP_{out} = 11750 - 1750 = 10000 ext{ W} Therefore, $$ ext{Efficiency} = rac{10000}{11750} imes 100 ext{ approx } 85.11 ext{ %} $

Step 7

6.4.2 Apparent power

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Answer

The apparent power, SS, can be calculated using:

S=extVimesextIimesextsqrt(3)S = ext{V} imes ext{I} imes ext{sqrt}(3) Substituting the known values: S=400imes20imesextsqrt(3)S = 400 imes 20 imes ext{sqrt}(3) Thus, Sextapprox.13856.41VA S ext{ approx. } 13856.41 VA

Step 8

6.4.3 Power factor

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Answer

The power factor, extPF ext{PF}, can be computed using:

ext{PF} = rac{P_{out}}{S} Substituting: $$ ext{PF} = rac{11750}{13856.41} ext{ approx. } 0.85 $

Step 9

6.4.4 Output power of the motor.

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Answer

The output power can be calculated using:

Pout=extsqrt(3)imesextVimesextIimesextPF P_{out} = ext{sqrt}(3) imes ext{V} imes ext{I} imes ext{PF} Substituting the known values: Pout=extsqrt(3)imes400imes20imes0.85extapprox10001.25extW P_{out} = ext{sqrt}(3) imes 400 imes 20 imes 0.85 ext{ approx } 10001.25 ext{ W}

Step 10

6.5.1 Identify the motor starter control circuit in FIGURE 6.5 above.

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Answer

The motor starter control circuit identified in Figure 6.5 is the manual sequence starter circuit.

Step 11

6.5.2 Explain the function of MC1 with reference to the main (power) circuit.

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Answer

MC1 is a contactor that energizes and de-energizes motor 1 in the main power circuit whenever the start button is pressed.

Step 12

6.5.3 State the advantage of having two overload relays in the circuit.

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Answer

Having two overload relays ensures that each motor is protected by its own overload relay independently, enhancing safety and reliability in case one motor experiences overload.

Step 13

6.5.4 Explain the disadvantage of connecting the overloads as shown in the control circuit in FIGURE 6.5 when MC1 and MC2 are controlling the two motors on the conveyor belt system.

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Answer

If one motor is overloaded and stops, the circuit can cause the other motor to also shut down, which might interrupt the entire conveyor system and reduce operational efficiency.

Step 14

6.5.5 Explain how the control circuit can be altered to prevent MC2 from being energised until after MC1 is.

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Answer

The control circuit can be modified by connecting a normally open (N/O) contact of MC1 in place of the temporary connection. This alteration ensures that MC2 can only be activated after MC1 is energized, preventing operation until it is safe.

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