Refer to the losses that occur in transformers and answer the questions that follow - NSC Electrical Technology Power Systems - Question 4 - 2019 - Paper 1

Two factors that may contribute to the excessive heating of transformers are:

1. Insufficient Circulating Air: Proper ventilation is necessary to remove heat. Lack of airflow increases temperature levels.

2. Insufficient Oil: If a transformer is oil-immersed, inadequate oil can reduce its cooling efficiency, leading to excess heat production.

Electromotive force (EMF) is induced in the secondary windings of transformers through the principle of electromagnetic induction. When an alternating current flows through the primary windings, it creates a magnetic field. This alternating magnetic field links with the secondary windings:

• Step 1: AC voltage supplies current to the primary windings, generating magnetic flux.
• Step 2: This magnetic flux then links with the secondary winding, inducing a voltage (EMF) across it, proportional to the rate of change of the magnetic field.

Cooling methods are crucial to prevent overheating in transformers. Efficient cooling ensures:

• The device does not overheat, extending its lifespan.
• Regular operational temperatures are maintained, which prevents deterioration of materials.

Protective devices are vital for the safety and efficiency of transformers. They help to ensure that:

• The transformer is isolated from the power supply in case of faults, thereby preventing damage.
• Any internal issues do not cause additional damage or a short circuit, which could have severe consequences.

To calculate the secondary line current, we use the formula:

$I_L = \frac{S}{\sqrt{3} \times V_L}$

Substituting the given values:

$I_L = \frac{10,000}{\sqrt{3} \times 500} \approx 11.55 A$

To find the transformer ratio, we use:

$\text{Transformer Ratio} = \frac{V_{ph}}{V_{s}}$

Where:

• Primary voltage: $V_{LP} = 6000 V$
• Secondary voltage: $V_{LS} = 500 V$

Calculating:

$\text{Ratio} = \frac{V_{LP}}{V_{LS}} = \frac{6000}{500} = 12:1$

The input power can be calculated as:

$P_{in} = \sqrt{3} \times V_L \times I_L \times Cos \theta$

Substituting the values:

$P_{in} = \sqrt{3} \times 500 \times 11.55 \times 0.97 \approx 9.70 kW$

To calculate efficiency, we use:

$\eta = \frac{P_{in} - P_{loss}}{P_{in}}\times 100$

Given that total loss is 80 W:

$\eta = \frac{9700 - 80}{9700} \times 100 \approx 99.17\%$

The secondary line current is higher than the primary line current because:

• The power rating (kVA) of the secondary side remains the same. Given the relationship between power, voltage, and current in transformers, if the secondary voltage is lower, the secondary current must increase to maintain equivalent power, consistent with the principle of conservation of energy.