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A 500 kVA, 3-phase transformer has iron losses of 300 W and full load copper losses of 600 W. The percentage load at which the transformer is expected to have maximum efficiency is (Answer up to one decimal place)
Correct answer is '70.7'. Can you explain this answer?
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Most Upvoted Answer
A 500 kVA, 3-phase transformer has iron losses of 300 W and full load...
Solution:
Efficiency of transformer is given as:
Efficiency (η) = Output Power / Input Power
where Output Power = VIcosφ and Input Power = VI
Iron losses = Hysteresis losses + Eddy current losses
Given, iron losses = 300 W
Full load copper losses = 600 W
Total losses = Iron losses + Full load copper losses
= 300 + 600
= 900 W
At maximum efficiency, copper losses are equal to iron losses.
Therefore, Copper losses = Iron losses
i.e., V2 R = Iron losses
where R is the resistance of the transformer.
V2 = Iron losses / R
At maximum efficiency, output power is equal to input power.
Therefore, V2 = V1
where V1 is the input voltage and V2 is the output voltage.
At maximum efficiency, full load copper losses are equal to iron losses.
Therefore, I2 R2 = Iron losses
where I2 is the full load current and R2 is the resistance of the transformer.
I2 = Output Power / V2
= VIcosφ / V2
R2 = V2 / I2
= V2 / (VIcosφ / V2)
= V22 / (VIcosφ)
Substituting the value of V2, we get:
R2 = (Iron losses / R)2 / (VIcosφ)
= (300 / R)2 / (VIcosφ)
Putting Copper losses = Iron losses, we get:
V2 R2 = V22 / (VIcosφ)
Substituting the value of V2 and R2, we get:
Iron losses / R = (Iron losses / R)2 / (VIcosφ)
cosφ = 1 / √2
φ = cos-1(1/√2)
φ = 45°
Therefore, at maximum efficiency, the transformer is expected to operate at a load of 70.7% of full load.
Hence, the answer is 70.7.
Efficiency of transformer is given as:
Efficiency (η) = Output Power / Input Power
where Output Power = VIcosφ and Input Power = VI
Iron losses = Hysteresis losses + Eddy current losses
Given, iron losses = 300 W
Full load copper losses = 600 W
Total losses = Iron losses + Full load copper losses
= 300 + 600
= 900 W
At maximum efficiency, copper losses are equal to iron losses.
Therefore, Copper losses = Iron losses
i.e., V2 R = Iron losses
where R is the resistance of the transformer.
V2 = Iron losses / R
At maximum efficiency, output power is equal to input power.
Therefore, V2 = V1
where V1 is the input voltage and V2 is the output voltage.
At maximum efficiency, full load copper losses are equal to iron losses.
Therefore, I2 R2 = Iron losses
where I2 is the full load current and R2 is the resistance of the transformer.
I2 = Output Power / V2
= VIcosφ / V2
R2 = V2 / I2
= V2 / (VIcosφ / V2)
= V22 / (VIcosφ)
Substituting the value of V2, we get:
R2 = (Iron losses / R)2 / (VIcosφ)
= (300 / R)2 / (VIcosφ)
Putting Copper losses = Iron losses, we get:
V2 R2 = V22 / (VIcosφ)
Substituting the value of V2 and R2, we get:
Iron losses / R = (Iron losses / R)2 / (VIcosφ)
cosφ = 1 / √2
φ = cos-1(1/√2)
φ = 45°
Therefore, at maximum efficiency, the transformer is expected to operate at a load of 70.7% of full load.
Hence, the answer is 70.7.
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Community Answer
A 500 kVA, 3-phase transformer has iron losses of 300 W and full load...
Given that transformer rating is 500 kVA,
Iron losses = 300 W
Full load copper losses = 600 W
Maximum efficiency condition is
Wi = X2Wc
So,
= √1/2
Efficiency % = 0.707 x 100 = 70.7%
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A 500 kVA, 3-phase transformer has iron losses of 300 W and full load copper losses of 600 W. The percentage load at which the transformer is expected to have maximum efficiency is (Answer up to one decimal place)Correct answer is '70.7'. Can you explain this answer?
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