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A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA and reactance of 0.4 pu is connected to an infinite bus through a double circuit line, each line of reactance 0.35 pu. The generator is delivering 0.8 pu power and the infinite bus voltage is 1.0 pu. Then:
- a)The maximum power transfer is 2.08 pu
- b)The steady-state operating angle is 30°
- c)The maximum power transfer is 1.6 pu
- d)The steady-state operating angle is 22.54°
Correct answer is option 'D'. Can you explain this answer?
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A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5...
Given,
A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA.
The single line diagram is,
A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA.
The single line diagram is,
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A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5...
To find the maximum power transfer and the steady-state operating angle, we can use the power equation for a synchronous generator connected to an infinite bus:
P = (E_infinity * V)/(X_s + X_l)
where P is the power being delivered by the generator, E_infinity is the infinite bus voltage, V is the internal voltage of the generator, X_s is the reactance of the generator, and X_l is the reactance of the transmission line.
Given:
E_infinity = 1.0 pu
V = 1.2 pu
X_s = 0.4 pu
X_l = 0.35 pu
P = 0.8 pu
Substituting these values into the power equation:
0.8 = (1.0 * 1.2)/(0.4 + 2 * 0.35)
Simplifying the denominator:
0.8 = (1.2)/(0.4 + 0.7)
0.8 = 1.2/1.1
0.8 * 1.1 = 1.2
0.88 = 1.2
Therefore, the maximum power transfer is 0.88 pu (a).
To find the steady-state operating angle, we can use the power-angle equation for a synchronous generator:
P = (E_infinity * V * sin(delta))/(X_s + X_l)
where delta is the operating angle.
Given:
E_infinity = 1.0 pu
V = 1.2 pu
X_s = 0.4 pu
X_l = 0.35 pu
P = 0.8 pu
Substituting these values into the power-angle equation:
0.8 = (1.0 * 1.2 * sin(delta))/(0.4 + 2 * 0.35)
Simplifying the denominator:
0.8 = (1.2 * sin(delta))/(0.4 + 0.7)
0.8 = (1.2 * sin(delta))/(1.1)
Multiplying both sides by 1.1:
0.88 = 1.2 * sin(delta)
sin(delta) = 0.88/1.2
sin(delta) = 0.7333
Taking the inverse sine of both sides:
delta = sin^(-1)(0.7333)
Using a calculator, we find:
delta ≈ 46.02 degrees
Therefore, the steady-state operating angle is approximately 46.02 degrees (b).
P = (E_infinity * V)/(X_s + X_l)
where P is the power being delivered by the generator, E_infinity is the infinite bus voltage, V is the internal voltage of the generator, X_s is the reactance of the generator, and X_l is the reactance of the transmission line.
Given:
E_infinity = 1.0 pu
V = 1.2 pu
X_s = 0.4 pu
X_l = 0.35 pu
P = 0.8 pu
Substituting these values into the power equation:
0.8 = (1.0 * 1.2)/(0.4 + 2 * 0.35)
Simplifying the denominator:
0.8 = (1.2)/(0.4 + 0.7)
0.8 = 1.2/1.1
0.8 * 1.1 = 1.2
0.88 = 1.2
Therefore, the maximum power transfer is 0.88 pu (a).
To find the steady-state operating angle, we can use the power-angle equation for a synchronous generator:
P = (E_infinity * V * sin(delta))/(X_s + X_l)
where delta is the operating angle.
Given:
E_infinity = 1.0 pu
V = 1.2 pu
X_s = 0.4 pu
X_l = 0.35 pu
P = 0.8 pu
Substituting these values into the power-angle equation:
0.8 = (1.0 * 1.2 * sin(delta))/(0.4 + 2 * 0.35)
Simplifying the denominator:
0.8 = (1.2 * sin(delta))/(0.4 + 0.7)
0.8 = (1.2 * sin(delta))/(1.1)
Multiplying both sides by 1.1:
0.88 = 1.2 * sin(delta)
sin(delta) = 0.88/1.2
sin(delta) = 0.7333
Taking the inverse sine of both sides:
delta = sin^(-1)(0.7333)
Using a calculator, we find:
delta ≈ 46.02 degrees
Therefore, the steady-state operating angle is approximately 46.02 degrees (b).
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A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA and reactance of 0.4 pu is connected to an infinite bus through a double circuit line, each line of reactance 0.35 pu. The generator is delivering 0.8 pu power and the infinite bus voltage is 1.0 pu. Then:a)The maximum power transfer is 2.08 pub)The steady-state operating angle is 30°c)The maximum power transfer is 1.6 pud)The steady-state operating angle is 22.54°Correct answer is option 'D'. Can you explain this answer?
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A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA and reactance of 0.4 pu is connected to an infinite bus through a double circuit line, each line of reactance 0.35 pu. The generator is delivering 0.8 pu power and the infinite bus voltage is 1.0 pu. Then:a)The maximum power transfer is 2.08 pub)The steady-state operating angle is 30°c)The maximum power transfer is 1.6 pud)The steady-state operating angle is 22.54°Correct answer is option 'D'. Can you explain this answer? for RRB NTPC/ASM/CA/TA 2024 is part of RRB NTPC/ASM/CA/TA preparation. The Question and answers have been prepared according to the RRB NTPC/ASM/CA/TA exam syllabus. Information about A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA and reactance of 0.4 pu is connected to an infinite bus through a double circuit line, each line of reactance 0.35 pu. The generator is delivering 0.8 pu power and the infinite bus voltage is 1.0 pu. Then:a)The maximum power transfer is 2.08 pub)The steady-state operating angle is 30°c)The maximum power transfer is 1.6 pud)The steady-state operating angle is 22.54°Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for RRB NTPC/ASM/CA/TA 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA and reactance of 0.4 pu is connected to an infinite bus through a double circuit line, each line of reactance 0.35 pu. The generator is delivering 0.8 pu power and the infinite bus voltage is 1.0 pu. Then:a)The maximum power transfer is 2.08 pub)The steady-state operating angle is 30°c)The maximum power transfer is 1.6 pud)The steady-state operating angle is 22.54°Correct answer is option 'D'. Can you explain this answer?.
A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA and reactance of 0.4 pu is connected to an infinite bus through a double circuit line, each line of reactance 0.35 pu. The generator is delivering 0.8 pu power and the infinite bus voltage is 1.0 pu. Then:a)The maximum power transfer is 2.08 pub)The steady-state operating angle is 30°c)The maximum power transfer is 1.6 pud)The steady-state operating angle is 22.54°Correct answer is option 'D'. Can you explain this answer? for RRB NTPC/ASM/CA/TA 2024 is part of RRB NTPC/ASM/CA/TA preparation. The Question and answers have been prepared according to the RRB NTPC/ASM/CA/TA exam syllabus. Information about A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA and reactance of 0.4 pu is connected to an infinite bus through a double circuit line, each line of reactance 0.35 pu. The generator is delivering 0.8 pu power and the infinite bus voltage is 1.0 pu. Then:a)The maximum power transfer is 2.08 pub)The steady-state operating angle is 30°c)The maximum power transfer is 1.6 pud)The steady-state operating angle is 22.54°Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for RRB NTPC/ASM/CA/TA 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA and reactance of 0.4 pu is connected to an infinite bus through a double circuit line, each line of reactance 0.35 pu. The generator is delivering 0.8 pu power and the infinite bus voltage is 1.0 pu. Then:a)The maximum power transfer is 2.08 pub)The steady-state operating angle is 30°c)The maximum power transfer is 1.6 pud)The steady-state operating angle is 22.54°Correct answer is option 'D'. Can you explain this answer?.
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ample number of questions to practice A 50 Hz synchronous generator having an internal voltage 1.2 pu, H = 5.2 MJ/MVA and reactance of 0.4 pu is connected to an infinite bus through a double circuit line, each line of reactance 0.35 pu. The generator is delivering 0.8 pu power and the infinite bus voltage is 1.0 pu. Then:a)The maximum power transfer is 2.08 pub)The steady-state operating angle is 30°c)The maximum power transfer is 1.6 pud)The steady-state operating angle is 22.54°Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice RRB NTPC/ASM/CA/TA tests.
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