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5.1 Name TWO types of transformer core constructions used in three-phase transformers - NSC Electrical Technology Power Systems - Question 5 - 2022 - Paper 1

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5.1 Name TWO types of transformer core constructions used in three-phase transformers. 5.2 Explain why dielectric oil is used inside a transformer. 5.3 State where... show full transcript

Worked Solution & Example Answer:5.1 Name TWO types of transformer core constructions used in three-phase transformers - NSC Electrical Technology Power Systems - Question 5 - 2022 - Paper 1

Step 1

5.1 Name TWO types of transformer core constructions used in three-phase transformers.

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Answer

The two types of transformer core constructions used in three-phase transformers are:

  1. Core-type: This construction features a core that encircles the windings of the transformer, providing efficient magnetic coupling.
  2. Shell type: In this design, the windings are located inside a shell-like core, which offers better leakage flux control.

Step 2

5.2 Explain why dielectric oil is used inside a transformer.

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Answer

Dielectric oil is used in transformers primarily for its electric insulating properties. It ensures that the high-voltage parts of the transformer are insulated from each other, preventing electrical discharge. Additionally, the oil acts as a coolant by transferring heat away from the components, maintaining operational efficiency and safety.

Step 3

5.3 State where the Buchholz relay is situated in an oil-cooled transformer.

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Answer

The Buchholz relay is typically positioned in-line between the conservator and the transformer housing. Its location allows it to monitor gas accumulation and oil flow, providing critical protection against faults.

Step 4

5.4 Draw a three-phase delta-step-down transformer unit by using three identical single-phase transformers.

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Answer

The drawing should depict three single-phase transformers connected in a delta configuration on the primary side, with appropriate connections indicating that the secondary side is also set to deliver a lower voltage. The arrangement should represent the step-down functionality clearly, showing primary and secondary line voltages and load connections.

Step 5

5.5.1 Output power

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Answer

To calculate the output power, use the formula:

Pout=Simesextp.f.P_{out} = S imes ext{p.f.}

Substituting the values:

Pout=10,000imes0.8=8,000extWP_{out} = 10,000 imes 0.8 = 8,000 ext{ W}

Step 6

5.5.2 Efficiency

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The efficiency can be calculated using the formula:

extEfficiency(extn)=PoutPout+extlosses×100 ext{Efficiency} ( ext{n}) = \frac{P_{out}}{P_{out} + ext{losses}} \times 100

Where the losses are the sum of copper loss and core loss:

losses=300+50=350extW\text{losses} = 300 + 50 = 350 ext{ W}

Thus:

n=8,0008,000+350×100=8,0008,350×10095.8%n = \frac{8,000}{8,000 + 350} \times 100 = \frac{8,000}{8,350} \times 100 \approx 95.8\%

Step 7

5.6.1 Rating of the transformer (apparent kV)

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Answer

To find the apparent power rating, use:

S(Papp)=3×VL×ILS(P_{app}) = \sqrt{3} \times V_L \times I_L

Substituting the values:

S(Papp)=3×6,000×220,784.61extVAS(P_{app}) = \sqrt{3} \times 6,000 \times 2 \approx 20,784.61 ext{ VA}

Step 8

5.6.2 Power factor of the load

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Answer

The power factor can be calculated using:

extCosθ=PS ext{Cos } \theta = \frac{P}{S}

Substituting the values:

cosθ=18,00020,784.610.87\cos \theta = \frac{18,000}{20,784.61} \approx 0.87

Step 9

5.6.3 Primary phase voltage

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Answer

For a delta connection, the phase voltage ( VphV_{ph}) is given by:

Vph=VLV_{ph} = V_L For the primary side:

VL=6,000extVV_L = 6,000 ext{ V}

Step 10

5.6.4 Turns ratio

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Answer

The turns ratio (TR) can be determined by the relation:

TR=Vph(1)Vph(2)TR = \frac{V_{ph(1)}}{V_{ph(2)}}

Where:

TR=6,000240=25TR = \frac{6,000}{240} = 25 Thus, the turns ratio is 25:1.

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