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1. What is the advantage of winding...
Q8. The following data are obtained when open circuit and short circuit tests are performed on...
Q8. The following data are obtained when open circuit and short circuit tests are performed on a single phase, 50 kVA, 2400/240 Volt 50 HZ distribution transformer. VOLTAGE(volts)CURRENT (amperes) POWER (watts) 213 Measurement in low voltage side with high voltage winding open Measurement on high voltage side with low voltage winding shorted 240 5.97 60 20.8 750 (a) Neglecting the series impedance, determine the exciting admittance referred to the high voltage side. (b) Neglecting the exciting admittance, determine the equivalent...
A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution transformer has a 1-ohm equivalent leakage reactance and a 5000-ohm...
A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution transformer has a 1-ohm equivalent leakage reactance and a 5000-ohm magnetizing reactance referred to the high-voltage side. If rated voltage is applied to the high-voltage winding, calculate the open-circuit secondary voltage. Neglect 'R and G^V losses. Assume equal series leakage reactances for the primary and the referred secondary
The short-circuit test readings for a 50-kVA 2400:240-V transformer are 48 V, 20.8 A, and 617...
The short-circuit test readings for a 50-kVA 2400:240-V transformer are 48 V, 20.8 A, and 617 W, with the instruments located on the high-voltage side and the low-voltage side short- circuited. 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. 1. Calculate the equivalent parameters of the transformer as referred to the high-voltage side. 2. Draw the equivalent approximate circuit of that transformer referred to the high...
The S.C. test on a 1-phase transformer, with the primary winding short-circuited and 30 V applied...
The S.C. test on a 1-phase transformer, with the primary winding short-circuited and 30 V applied to the secondary gave a wattmeter reading of 60 W and secondary current of 10 A. II the normal applied primary voltage is 200 V, the transformation ratio 1:2 and the full-load secondary current 10 A, calculate the secondary p.d. at full-load current for (a) unity power factor (b) power factor 0.8 lagging. these are the answers [ 394 V; 377.6 V ]
A 20 kVA, 60 Hz, 2400/240 V single-phase distribution transformer has been tested to determine its...
A 20 kVA, 60 Hz, 2400/240 V single-phase distribution transformer has been tested to determine its equivalent circuit. The results of the test are shown below: Open circuit Test (on low voltage side) Voc=240V, Ioc=1.028A, Poc=122W. Short circuit test (On high voltage side) Vsc=61.3V, Isc=8.33A, Psc=257W. The transformer is supplying power to a load that draws 15 kVA at 240 V. The load power factor is 0.90 lagging. Calculate the voltage regulation for this operation condition.
a) Discuss the main role of per-unit systems. (CO3:P01 - 3 marks) b) A single phase...
a) Discuss the main role of per-unit systems. (CO3:P01 - 3 marks) b) A single phase two winding transformer rated at 20 kVA, 415/230 volts 50 Hz is subjected to open and short circuit test. Table Q3 (a) summarize the results of the test; Table Q3 (a) Open Circuit Test 12 = 12 A P2 = 200 W Short Circuit Test V1 = 35 V P1 = 300 W I. Based on the short circuit test, determine the equivalent series...
single phase transformer with open circuit test and short circuit test
A 20 kVA, 12,000/380 V, single-phase transformer has the following test results:open circuit (low voltage side): V = 380, I = 2, P = 250short circuit (high voltage side): V = 200, I = 2.5, P = 250a)Determine the approximate equivalent circuit parameters of the transformer in per-unit values.b) the transformer is working at 80% full load with 0.85 power factor lagging. what is the secondary current at this working condition in both actual and per-unit values.
QUESTION 1 125 MARKS] The data in the Table 1 were obtained when open and short...
QUESTION 1 125 MARKS] The data in the Table 1 were obtained when open and short circuit test were performed on a single phase 50 kVA, 2400/240 V, 60 Hz transformer. Table 1 Voltage Current Power (V) (A) (Watt) 240 4.85 173 52 20.83 650 Test A Test B (a) Determine tests A, B and the sides the tests were done. Calculate the non-ideal transformer parameters referred to the low voltage side. Draw the equivalent circuit by labeling the parameters...
QUESTION 1 [25 MARKS] The data in the Table 1 were obtained when open and short...
QUESTION 1 [25 MARKS] The data in the Table 1 were obtained when open and short circuit test were performed on a single phase 50 kVA, 2400/240 V, 60 Hz transformer. Table 1 Voltage Current Power (V) (A) (Watt) 240 4.85 173 52 20.83 650 Test A Test B (a) Determine tests A, B and the sides the tests were done. Calculate the non-ideal transformer parameters referred to the low voltage side. Draw the equivalent circuit by labeling the parameters...
Problem 1 A 60Hz transformer is rated 30 (kVA), 4000(V)/120(V). The open circuit test, with the...
Problem 1 A 60Hz transformer is rated 30 (kVA), 4000(V)/120(V). The open circuit test, with the high voltage side open, gives Po= 100 (W), loc = 1.1455 (A). The short circuit test, measured with the low voltage side shorted, gives Pse = 180 (W), Vse = 129.79 (V). a) Determine the equivalent circuit for the transformer (both open and short circuit) b) Calculate the circuit components. (Req, Xeq, Rc, Xm)
Q8. The following data are obtained when open circuit and short circuit tests are performed on a single phase, 50 kVA, 2400/240 Volt 50 HZ distribution transformer. VOLTAGE(volts)CURRENT (amperes) POWER (watts) 213 Measurement in low voltage side with high voltage winding open Measurement on high voltage side with low voltage winding shorted 240 5.97 60 20.8 750 (a) Neglecting the series impedance, determine the exciting admittance referred to the high voltage side. (b) Neglecting the exciting admittance, determine the equivalent...
A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution transformer has a 1-ohm equivalent leakage reactance and a 5000-ohm magnetizing reactance referred to the high-voltage side. If rated voltage is applied to the high-voltage winding, calculate the open-circuit secondary voltage. Neglect 'R and G^V losses. Assume equal series leakage reactances for the primary and the referred secondary
The short-circuit test readings for a 50-kVA 2400:240-V transformer are 48 V, 20.8 A, and 617 W, with the instruments located on the high-voltage side and the low-voltage side short- circuited. 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. 1. Calculate the equivalent parameters of the transformer as referred to the high-voltage side. 2. Draw the equivalent approximate circuit of that transformer referred to the high...
a) Discuss the main role of per-unit systems. (CO3:P01 - 3 marks) b) A single phase two winding transformer rated at 20 kVA, 415/230 volts 50 Hz is subjected to open and short circuit test. Table Q3 (a) summarize the results of the test; Table Q3 (a) Open Circuit Test 12 = 12 A P2 = 200 W Short Circuit Test V1 = 35 V P1 = 300 W I. Based on the short circuit test, determine the equivalent series...
QUESTION 1 125 MARKS] The data in the Table 1 were obtained when open and short circuit test were performed on a single phase 50 kVA, 2400/240 V, 60 Hz transformer. Table 1 Voltage Current Power (V) (A) (Watt) 240 4.85 173 52 20.83 650 Test A Test B (a) Determine tests A, B and the sides the tests were done. Calculate the non-ideal transformer parameters referred to the low voltage side. Draw the equivalent circuit by labeling the parameters...
QUESTION 1 [25 MARKS] The data in the Table 1 were obtained when open and short circuit test were performed on a single phase 50 kVA, 2400/240 V, 60 Hz transformer. Table 1 Voltage Current Power (V) (A) (Watt) 240 4.85 173 52 20.83 650 Test A Test B (a) Determine tests A, B and the sides the tests were done. Calculate the non-ideal transformer parameters referred to the low voltage side. Draw the equivalent circuit by labeling the parameters...
Problem 1 A 60Hz transformer is rated 30 (kVA), 4000(V)/120(V). The open circuit test, with the high voltage side open, gives Po= 100 (W), loc = 1.1455 (A). The short circuit test, measured with the low voltage side shorted, gives Pse = 180 (W), Vse = 129.79 (V). a) Determine the equivalent circuit for the transformer (both open and short circuit) b) Calculate the circuit components. (Req, Xeq, Rc, Xm)
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