4.1 Name TWO cooling methods used in a dry transformer. 4.2 State the main cause that contributes to heat generation in transformers. 4.3 State TWO safety precautions when working with transformers. 4.4 FIGURE 4.4 below is a diagrammatic representation of a three-phase transformer connection. Answer the questions that follow. 4.4.1 Identify the type of transformer connection in FIGURE 4.4. 4.4.2 Name TWO applications of the transformer in FIGURE 4.4. 4.4.3 State, with a reason, whether the transformer is a step-up or a step-down transformer. 4.5 Compare single-phase transformers with three-phase transformers when they supply the same three-phase load. Refer to the following: 4.5.1 Economic cost. 4.5.2 Efficiency. 4.6 Refer to FIGURE 4.6 below and describe how the Buchholz relay would protect a transformer under minor and major faulty conditions. 4.7 Refer to FIGURE 4.7 below and answer the questions that follow. Given: TR = 5 : 1 Vph1 = 2 000 V S = 50 kVA Pout = 45 kW Transformer losses = 500 W Calculate the: 4.7.1 Secondary phase voltage. 4.7.2 Efficiency of the transformer. 4.7.3 Power factor of the transformer. 4.7.4 Current drawn by the load.

NSC Electrical Technology Power Systems Transformer Principles: 4.1 Name TWO cooling methods use

4.1 Name TWO cooling methods used in a dry transformer - NSC Electrical Technology Power Systems - Question 4 - 2021 - Paper 1

Question 4

4.1 Name TWO cooling methods used in a dry transformer. 4.2 State the main cause that contributes to heat generation in transformers. 4.3 State TWO safety precauti... show full transcript

Worked Solution & Example Answer:4.1 Name TWO cooling methods used in a dry transformer - NSC Electrical Technology Power Systems - Question 4 - 2021 - Paper 1

Step 1

4.4.1 Identify the type of transformer connection in FIGURE 4.4.

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Answer

The type of transformer connection shown in FIGURE 4.4 is a Delta-Delta (Δ-Δ) connection. In this configuration, both the primary and secondary windings are connected in a delta formation, which allows for higher starting torque and better voltage regulation.

Step 2

4.4.2 Name TWO applications of the transformer in FIGURE 4.4.

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Answer

Industrial power distribution, where the transformer steps down the voltage for machinery and equipment use.
Commercial buildings, providing power at the required voltage levels for lighting and HVAC systems.

Step 3

4.4.3 State, with a reason, whether the transformer is a step-up or step-down transformer.

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Answer

The transformer in FIGURE 4.4 is a step-down transformer. This is indicated by the turns ratio of 5:1, meaning that the secondary voltage is lower than the primary voltage, which is typical for stepping down voltage in power applications.

Step 4

4.5.1 Economic cost.

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Answer

Single-phase transformers are generally less expensive to manufacture and install due to their simpler design. However, when a larger power capacity is needed, three-phase transformers are more economical in terms of installation and operational cost, as they can deliver more power with fewer copper losses.

Step 5

4.5.2 Efficiency.

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Answer

Three-phase transformers are typically more efficient than single-phase transformers. This is due to better utilization of material and lower losses in the core and windings, resulting in higher efficiency ratings even under varying load conditions.

Step 6

4.6 Describe how the Buchholz relay would protect a transformer under minor and major faulty conditions.

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Answer

The Buchholz relay functions as a protective device that detects gas accumulation due to overheating or internal faults. Under minor faults, it triggers alarms when bubbles of gas form in the oil, indicating insulation breakdown or overheating. In major faults, such as a severe internal short circuit, it activates the trip mechanism, shutting down the transformer to prevent catastrophic failures and potential explosions.

Step 7

4.7.1 Secondary phase voltage.

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Answer

The secondary phase voltage can be calculated using the turns ratio as follows: $V_{ph2} = rac{V_{ph1}}{TR} = rac{2000 V}{5} = 400 V$

Step 8

4.7.2 Efficiency of the transformer.

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Answer

The efficiency ( \eta) of the transformer can be determined using the formula: $\eta = \frac{P_{out}}{P_{out} + losses} \times 100$ Substituting in the values: $\eta = \frac{45000 W}{45000 W + 500 W} \times 100 = 98.9\%$

Step 9

4.7.3 Power factor of the transformer.

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Answer

The power factor can be calculated using the formula: $\text{Cos} \theta = \frac{P}{S}$ Using the provided values: $\text{Cos} \theta = \frac{45000 W}{50000 VA} = 0.9$

Step 10

4.7.4 Current drawn by the load.

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Answer

The current drawn by the load can be calculated using the formula: $I_{2} = \frac{S}{\sqrt{3} \times V_{2} \times \text{Cos}\theta}$ Substituting the relevant values gives: $I_{2} = \frac{50000 VA}{\sqrt{3} \times 400 V \times 0.9} = 72.17 A$

4.1 Name TWO cooling methods used in a dry transformer - NSC Electrical Technology Power Systems - Question 4 - 2021 - Paper 1

Worked Solution & Example Answer:4.1 Name TWO cooling methods used in a dry transformer - NSC Electrical Technology Power Systems - Question 4 - 2021 - Paper 1

4.4.1 Identify the type of transformer connection in FIGURE 4.4.

4.4.2 Name TWO applications of the transformer in FIGURE 4.4.

4.4.3 State, with a reason, whether the transformer is a step-up or step-down transformer.

4.5.1 Economic cost.

4.5.2 Efficiency.

4.6 Describe how the Buchholz relay would protect a transformer under minor and major faulty conditions.

4.7.1 Secondary phase voltage.

4.7.2 Efficiency of the transformer.

4.7.3 Power factor of the transformer.

4.7.4 Current drawn by the load.

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