Question

- With the instruments located on the high-voltage side and the low-voltage side
short-circuited, the short-circuit test readings for the 50 kVA 2400/240 V transformer are 48 V,
20.8 A, and 617 W. An open-circuit test with the low-voltage side energized gives instrument
readings on that side of 240 V, 5.41 A, and 186 W. Determine the efficiency (in %) at full load,
0.80 power factor lagging. Considering the magnitude of the equivalent resistance of 1.42 Ω,
and the equivalent reactance of 1.82 Ω referred to the high-voltage side.

Answer 1

Answer:

Efficiency, E = 98.25%, Voltage regulation, Vr = 99.5%

Explanation:

Given parameters

Apparent power, S = 50 kVA or 50,000 VA

full load power factor, cos∅ = 0.9

short circuit

short circuit voltage, Vsc = 48 V

short circuit current, Isc = 20.8 A

short circuit power loss (copper loss), Psc= 617 W

open circuit

open circuit voltage, Voc = 240 V

open circuit current, Ioc = 5.41 A

open circuit power loss (core loss) Poc = 186 W

Efficiency,  

E = P out / p out + p sc + p oc  *100

Pout = Scos∅, where S is as defined above

Pout = 50 * 0.9

Pout = 45kW or 45,000W

E = (Pout/(Pout + Psc + Poc)) * 100%

E = (45000/(45000 + 617 + 186)) * 100%

E = 98.25%

Voltage Regulation, Vr = Er cos∅ + Ex sin∅

Er = Isc * Rpu * cos∅, where Rpu = per unit resistance

Ex = Isc * Xpu * sin∅, where Xpu = per unit reactance

from P = I²R where R = P/I²

Rpu = Rsc = Psc/(Isc)² = 617/(20.8)² = 1.4261 ohms

Zpu = Vsc/Isc = 48/20.8 = 2.3077 ohms

from Z² = R² + X²

X² = Z² - R²

X = √(Z² - R²)

Xpu = √((2.3077)² - (1.4261)²)

Xpu = 1.8143 ohms

cos ∅ = 0.9

∅ = arccos 0.9 = 25.84

sin ∅ = sin 25.84 = 0.44

Isc = 20.8 A

Voltage Regulation, Vr = (20.8 * 1.4261 * 0.9) + (20.8 * 1.8143 * 0.44)

Vr = 99.5 V