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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
  2. Shell type These constructions determine the configuration and efficiency of the transformer.

Step 2

5.2 Explain why dielectric oil is used inside a transformer.

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Answer

Dielectric oil serves two primary functions in transformers:

  • It acts as a non-conductor of electricity, providing electrical insulation between the internal components of the transformer.
  • It helps in cooling the transformer by facilitating the dissipation of heat generated during operation, thus preventing overheating.

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 located in-line between the conservator and the transformer housing. This placement allows it to detect gas and oil movement, signaling potential faults within the transformer.

Step 4

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

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Answer

The three-phase delta-star step-down transformer can be depicted as follows:

  • In the delta configuration, the three primary windings are connected in a loop.
  • The secondary windings are connected in a star configuration with a common neutral point. This arrangement provides the necessary voltage transformation from high to low.

Step 5

5.5.1 Output power

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Answer

To calculate the output power (P_out) of the transformer, use the formula:

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

Thus, the output power is 8,000 W.

Step 6

5.5.2 Efficiency

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To find the efficiency (η) of the transformer:

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

Substituting the values:

  • Losses = Copper loss + Core loss = 300 W + 50 W = 350 W

Therefore,

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

The efficiency of the transformer is 95.8%.

Step 7

5.6.1 Rating of the transformer (apparent power)

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Answer

To calculate the apparent power (S) of the transformer:

S(Papp)=3×VL×IL=3×6,000×2S(P_{app}) = \sqrt{3} \times V_{L} \times I_{L} = \sqrt{3} \times 6,000 \times 2

= 20,784.61 VA or approximately 20.78 kVA.

Step 8

5.6.2 Power factor of the load

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Answer

To find the power factor (cos φ) of the load:

cosφ=PS=18,00020,784.61=0.87cos φ = \frac{P}{S} = \frac{18,000}{20,784.61} = 0.87

Therefore, the power factor of the load is 0.87.

Step 9

5.6.3 Primary phase voltage

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Answer

For a delta connected transformer:

VL=VPH3V_{L} = V_{PH} \sqrt{3}

Thus, primary phase voltage:

VPH=VL3=6,0003=3,464.1VV_{PH} = \frac{V_{L}}{\sqrt{3}} = \frac{6,000}{\sqrt{3}} = 3,464.1 V

Step 10

5.6.4 Turns ratio

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Answer

To calculate the turns ratio (TR) between primary and secondary:

TR=VPH(1)VPH(2)=6,000240=25TR = \frac{V_{PH(1)}}{V_{PH(2)}} = \frac{6,000}{240} = 25

Hence, the turns ratio is 25.

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