In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor?

Question

Answer

Problem:

Calculate the efficiency of a 25 KVA 3300/230V single-phase transformer at 1/2 full load, 0.8 power factor when the iron and full load copper losses are 350 and 400 W, respectively.

Solution:

Given:

Transformer rating (S) = 25 KVA

Primary voltage (V1) = 3300 V

Secondary voltage (V2) = 230 V

Iron losses (Pcore) = 350 W

Copper losses at full load (Pcu) = 400 W

Load at which efficiency is to be calculated (kVA) = 1/2 of full load = 12.5 KVA

Power factor (pf) = 0.8

Step 1: Calculate the primary and secondary currents at 1/2 full load

The transformer rating is given in KVA, and the load is given in kVA. Therefore, the load current can be calculated as:

Load current (I2) = Load (kVA) / Secondary voltage (V2) = 12.5 / 230 = 0.05435 A

The primary current can be calculated as:

Primary current (I1) = S / V1 = 25,000 / 3300 = 7.5758 A

The primary current at 1/2 full load can be calculated as:

I1' = I1 * (I2 / pf) = 7.5758 * (0.05435 / 0.8) = 0.5147 A

Step 2: Calculate the copper losses at 1/2 full load

The copper losses in a transformer are proportional to the square of the current. Therefore, the copper losses at 1/2 full load can be calculated as:

Pcu' = (I1')^2 * R = (0.5147)^2 * R

where R is the resistance of the transformer winding. As the resistance is not given, we assume that it is proportional to the full load copper losses. Therefore, we can write:

R' / R = (I1' / I1)^2 = (0.5147 / 7.5758)^2

R' = R * (0.0036)

Substituting this value of R' in the expression for Pcu', we get:

Pcu' = (0.5147)^2 * R' = (0.5147)^2 * R * (0.0036) = 0.0107 * Pcu = 4.28 W

Step 3: Calculate the total losses at 1/2 full load

The total losses in a transformer are the sum of the iron losses and the copper losses

Answer

Answer

Efficiency at half load=95.69%

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