Question

7. With the instruments located on the high-voltage side and with 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 and voltage regulation of the transformer if it is supplying rated load at 0.9 power factor leading at rated voltage at its low-voltage terminals.

Answer 1

Given Information:

Apparent power = S = 50 kVA

Primary voltage = Vp = 2400 V

Secondary voltage = Vs = 240 V

Short circuit current = Isc = 20.8 A

Short circuit voltage = Vsc = 48 V

Short circuit power = Psc = 617 W

Open circuit current = Ioc = 5.41 A

Open circuit voltage = Voc = 240 V

Open circuit power = Poc = 186 W

Power factor = pf = 0.9 leading

Required Information:

Efficiency = η = ?

Voltage regulation = VR = ?

Answer:

Efficiency = 98.24%

Voltage regulation = 0.42%

Explanation:

Efficiency is given by

η = output power/input power

Where output power can be found by

output power = S*pf

output power = 50,000*0.9

output power = 45,000 W

Input power is the sum of output power and losses

input power = output power + Pcu + Pcore

The short circuit test is conducted to find out the copper losses

Pcu =  617 W

The open circuit test is conducted to find out the core losses

Pcore = 186 W

input power = 45,000 + 617 + 186

input power = 45,803 W

Therefore, the efficiency is

η = 45,000/45,803

η = 0.9824

η = 98.24%

Voltage regulation is given by

VR = (Vnl - Vfl)/Vfl * 100%

Where Vnl is the no load voltage and Vfl is the full load voltage

Vfl = 2400 V

The load current is given by

I = 50,000/2400 < cos⁻¹(0.90)

I = 20.83 < 25.84° A  (since the pf is leading, the angle is positive)

Vnl = Vfl + I*(Req + jXeq)

Req and Xeq can be determined from short circuit test parameters

Req = Psc/Isc²

Req = 617/20.8²

Req = 1.42 Ω

Zeq = Vsc/Isc

Zeq = 48/20.8

Zeq = 2.307 Ω

Xeq = √(Zeq² - Req²)

Xeq = √(2.307² - 1.42²)

Xeq = j1.818 Ω

Vnl = 2400 + (20.83 < 25.84)*(1.42 + j1.818)

Vnl = 2400 + (10.115 + j46.97)

Vnl = 2410.11 + j46.974 V

Vnl = 2410.57 < 1.11° V

VR = (Vnl - Vfl)/Vfl * 100%

VR = (2410.11 - 2400)/2400 * 100%

VR = 0.42 %

A lower value of voltage regulation is desired which signifies that the variation in voltage is low and the voltage at the load is very close to supply voltage.

Answer 2

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, $E = \frac{Pout}{Pout + Psc + Poc} * 100$

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