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A 200-kVA transformer has an efficiency of 98% at full load. If the maximum efficiency occurs at three quarters of full-load, calculate the efficiency at half load. Assume negligible magnetizing current and p.f. 0.8 at all loads.?
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A 200-kVA transformer has an efficiency of 98% at full load. If the ma...
Solution:

Given, the transformer rating = 200 kVA
Efficiency at full load = 98%
Maximum efficiency occurs at three-quarters of full load.

To calculate the efficiency at half load, we need to follow the below steps.

Step 1: Calculation of Full Load Current

Full Load Current (I1) = Transformer rating (kVA) / Transformer voltage (kV)

The voltage rating of the transformer is not given in the problem statement. Hence, we need to assume a voltage rating for further calculations.

Assuming the transformer voltage rating as 11 kV.

Full Load Current (I1) = 200 kVA / 11 kV ≈ 18.18 A

Step 2: Calculation of Three-quarters Load Current

As the maximum efficiency occurs at three-quarters of full load, we need to calculate the three-quarters load current.

Three-quarters Load Current (I3/4) = Full Load Current (I1) * √(3/4)

I3/4 = 18.18 * √(3/4) ≈ 15.86 A

Step 3: Calculation of Half Load Current

Half Load Current (I1/2) = Full Load Current (I1) * √(1/2)

I1/2 = 18.18 * √(1/2) ≈ 12.86 A

Step 4: Calculation of Efficiency at Half Load

To calculate the efficiency at half load, we need to use the following formula.

Efficiency = Output Power / Input Power

At half load, the output power is given by,

Output Power = Transformer rating (kVA) * Power Factor * Load Current (I1/2)

Assuming the power factor as 0.8.

Output Power = 200 kVA * 0.8 * 12.86 A ≈ 2060.8 W

The input power can be calculated as follows:

Input Power = Output Power / Efficiency

Efficiency at three-quarters load = 98%

Efficiency at half load can be obtained by interpolation.

Let x be the efficiency at half load.

(98 - x) / (98 - 94) = (15.86 - 12.86) / (15.86 - 18.18)

Solving the above equation, we get,

x ≈ 97.38%

Input Power = 2060.8 W / 0.9738 ≈ 2118.1 W

Therefore, the efficiency at half load is given by,

Efficiency = Output Power / Input Power = 2060.8 W / 2118.1 W ≈ 97.37%

Hence, the efficiency at half load is approximately 97.37%.
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A 200-kVA transformer has an efficiency of 98% at full load. If the ma...
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A 200-kVA transformer has an efficiency of 98% at full load. If the maximum efficiency occurs at three quarters of full-load, calculate the efficiency at half load. Assume negligible magnetizing current and p.f. 0.8 at all loads.?
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A 200-kVA transformer has an efficiency of 98% at full load. If the maximum efficiency occurs at three quarters of full-load, calculate the efficiency at half load. Assume negligible magnetizing current and p.f. 0.8 at all loads.? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A 200-kVA transformer has an efficiency of 98% at full load. If the maximum efficiency occurs at three quarters of full-load, calculate the efficiency at half load. Assume negligible magnetizing current and p.f. 0.8 at all loads.? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200-kVA transformer has an efficiency of 98% at full load. If the maximum efficiency occurs at three quarters of full-load, calculate the efficiency at half load. Assume negligible magnetizing current and p.f. 0.8 at all loads.?.
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