4.10 Refer to the circuit diagram in FIGURE 4.10 below and answer the questions that follow. Given: S = 200 kVA V_{L(1)} = 11 kV V_{L(3)} = 380 V I_{L(1)} = 30 A P_{in} = 171,8 kW Calculate the: 4.10.1 Efficiency of the transformer if it operates at a power factor of 0,85 lagging. 4.10.2 Turns ratio. 4.10.3 Secondary line current.

NSC Electrical Technology Power Systems Transformer Losses: 4.10 Refer to the circuit diagra

4.10 Refer to the circuit diagram in FIGURE 4.10 below and answer the questions that follow - NSC Electrical Technology Power Systems - Question 4 - 2020 - Paper 1

Question 4

4.10 Refer to the circuit diagram in FIGURE 4.10 below and answer the questions that follow. Given: S = 200 kVA V_{L(1)} = 11 kV V_{L(3)} = 380 V I_{L(1)} = 30 A P... show full transcript

Worked Solution & Example Answer:4.10 Refer to the circuit diagram in FIGURE 4.10 below and answer the questions that follow - NSC Electrical Technology Power Systems - Question 4 - 2020 - Paper 1

Step 1

4.10.1 Efficiency of the transformer if it operates at a power factor of 0,85 lagging.

96%

114 rated

Answer

To calculate the efficiency ( $\eta$ ) of the transformer, we first find the output power ( $P_{out}$ ) using the formula:

$P_{out} = S \times \text{power factor} = 200 \, kVA \times 0.85 = 170 \, kW$

Next, we apply the efficiency formula:

$\eta = \frac{P_{out}}{P_{in}} \times 100$

Substituting the values:

$\eta = \frac{170000 \, W}{171800 \, W} \times 100 = 98.95\%$

Thus, the efficiency of the transformer is approximately 98.95%.

Step 2

4.10.2 Turns ratio.

99%

104 rated

Answer

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

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

Where:

$V_{ph(2)} = 380 \, V$ (secondary voltage)
$V_{ph(1)} = 11000 \, V$ (primary voltage)

Substituting the values:

$TR = \frac{380}{11000} = \frac{1}{29} = 1:29$

Therefore, the turns ratio is 1:29.

Step 3

4.10.3 Secondary line current.

96%

101 rated

Answer

To find the secondary line current ( $I_{L(3)}$ ), we can use the following formula:

$I_{L(3)} = \frac{S}{V_{L(3)}}$

Substituting the values:

$I_{L(3)} = \frac{200000 \, VA}{380 \, V} \approx 526.32 \, A$

Thus, the secondary line current is approximately 526.32 A.

4.10 Refer to the circuit diagram in FIGURE 4.10 below and answer the questions that follow - NSC Electrical Technology Power Systems - Question 4 - 2020 - Paper 1

Worked Solution & Example Answer:4.10 Refer to the circuit diagram in FIGURE 4.10 below and answer the questions that follow - NSC Electrical Technology Power Systems - Question 4 - 2020 - Paper 1

4.10.1 Efficiency of the transformer if it operates at a power factor of 0,85 lagging.

4.10.2 Turns ratio.

4.10.3 Secondary line current.

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