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QUESTION 4: THREE-PHASE TRANSFORMERS 4.1 List THREE external conditions that may cause transformer failure - NSC Electrical Technology Power Systems - Question 4 - 2019 - Paper 1

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QUESTION 4: THREE-PHASE TRANSFORMERS 4.1 List THREE external conditions that may cause transformer failure. 4.2 Explain what would happen if an earth fault occurs ... show full transcript

Worked Solution & Example Answer:QUESTION 4: THREE-PHASE TRANSFORMERS 4.1 List THREE external conditions that may cause transformer failure - NSC Electrical Technology Power Systems - Question 4 - 2019 - Paper 1

Step 1

4.1 List THREE external conditions that may cause transformer failure.

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Answer

The following external conditions could lead to transformer failure:

  1. Overload: When the transformer is subjected to a load greater than its design capacity, it can lead to overheating and subsequent failure.
  2. Lightning: High-voltage surges caused by lightning strikes can damage the insulation and other components of the transformer.
  3. Poor ventilation: Inadequate cooling and airflow around the transformer can cause overheating, which can severely affect its operational reliability.

Step 2

4.2 Explain what would happen if an earth fault occurs in one of the three phases of a protected transformer.

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Answer

When an earth fault occurs in one of the phases, it results in an unbalanced phase condition. This disbalance causes:

  • Activation of protective relays, which generally isolate the transformer from the electrical supply to prevent damage.
  • An increase in the voltage difference across the remaining phases, potentially exceeding safe operational limits.
  • If not detected promptly, this can lead to overheating and a significant risk of equipment failure.

Step 3

4.3 Describe how an increase in the load would affect the magneto motive force in the primary windings.

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An increase in load results in higher current being drawn from the transformer’s secondary side. Consequently:

  • The current flowing through the primary windings will also increase, which in turn increases the magneto motive force (MMF) in the primary windings, as MMF is directly proportional to the current.
  • This increased MMF helps maintain the necessary magnetic flux within the core, balancing the input and output requirements, which helps prevent saturation and ensures efficient operation.

Step 4

4.4 Name TWO types of cooling methods for a dry transformer.

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Answer

Two types of cooling methods for a dry transformer are:

  1. Air Natural Cooling (AN): Utilizes natural airflow around the transformer for cooling without any mechanical assistance.
  2. Air Forced Cooling (AF): Involves the use of fans to force air circulation, enhancing the cooling effect beyond what can be achieved by natural convection.

Step 5

4.5 State why the output power of a transformer is slightly less than the input power.

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Answer

The output power of a transformer is slightly less than the input power primarily due to losses incurred during the transformation process, which include:

  • Copper losses: I²R losses in the windings due to the resistance.
  • Iron losses: Hysteresis and eddy current losses within the core.
  • These factors contribute to the disparity, whereby the efficiency is typically less than 100%.

Step 6

4.6 Describe the construction of a three-phase core-type transformer.

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Answer

A three-phase core-type transformer consists of:

  • Coils: Three sets of primary and secondary windings arranged on a common core.
  • Core: Made up of laminated steel sheets to reduce eddy current losses, it has three legs around which the windings are placed.
  • Vertical Axis: The axis of the core's windings is aligned vertically for structural integrity.
  • The core structure supports magnetic coupling between the windings, facilitating efficient energy transfer.

Step 7

4.7.1 Turns ratio

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Answer

The turns ratio (TR) can be calculated using the formula:

TR=NpNs=60080=7.5:1TR = \frac{N_p}{N_s} = \frac{600}{80} = 7.5:1

Step 8

4.7.2 Primary phase voltage

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The primary phase voltage can be calculated as follows:

Vph(P)=VL(P)3=60003=3464 V3.46×103 VV_{ph(P)} = \frac{V_{L(P)}}{\sqrt{3}} = \frac{6000}{\sqrt{3}} = 3464 \text{ V} \approx 3.46 \times 10^3 \text{ V}

Step 9

4.7.3 Secondary phase voltage

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Answer

The secondary phase voltage can be derived using:

Vph(S)=Vph(P)×NsNp=3464×80600=461.88 V461.88 VV_{ph(S)} = V_{ph(P)} \times \frac{N_s}{N_p} = 3464 \times \frac{80}{600} = 461.88 \text{ V} \approx 461.88 \text{ V}

Step 10

4.7.4 Secondary line voltage

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Answer

For the secondary line voltage, we use the relation:

VL(S)=Vph(S)×3=461.88×3461.88 VV_{L(S)} = V_{ph(S)} \times \sqrt{3} = 461.88 \times \sqrt{3} \approx 461.88 \text{ V}

Step 11

4.8 With reference to Faraday's law, describe how mutual induction occurs in a transformer.

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Answer

Mutual induction occurs in a transformer when the magnetic field created by the current flowing through the primary winding induces a voltage in the secondary winding. This phenomenon can be explained as follows:

  • According to Faraday’s law, the induced EMF in a coil is proportional to the rate of change of magnetic flux through it.
  • Consequently, as the magnetic field from the primary winding expands and collapses, it induces a corresponding voltage in the secondary coil, thus transferring energy from the primary to the secondary side.

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