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A transmission line is connected to a generator, has 0.6 p.u, impedance of 200 MVA & 11 kV base & that of generator is 0.2 p.u. on 100 MVA & 22 kV base. The total p.u. impedance on 500 MVA, 33 kVbase is___________
- a)0.8
- b)0.275
- c)0.865
- d)0.611
Correct answer is option 'D'. Can you explain this answer?
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A transmission line is connected to a generator, has 0.6 p.u, impedanc...
Calculation of the Total Per Unit Impedance
Given data:
Transmission line:
- Impedance = 0.6 p.u.
- Base MVA = 200 MVA
- Base voltage = 11 kV
Generator:
- Impedance = 0.2 p.u.
- Base MVA = 100 MVA
- Base voltage = 22 kV
New Base:
- MVA = 500 MVA
- Voltage = 33 kV
Step 1: Find the Per Unit Impedance of the Transmission Line on the New Base
To find the per unit impedance on the new base, we can use the formula:
Z_new = Z_old * (S_base_new / V_base_new)^2
Where:
Z_new = New per unit impedance
Z_old = Old per unit impedance
S_base_new = New base MVA
V_base_new = New base voltage
Substituting the values:
Z_new = 0.6 * (200 / 33)^2
Z_new = 0.6 * (6.0606)^2
Z_new = 0.6 * 36.7273
Z_new = 22.0364 p.u.
Step 2: Find the Per Unit Impedance of the Generator on the New Base
Using the same formula, we can find the per unit impedance of the generator on the new base:
Z_new = Z_old * (S_base_new / V_base_new)^2
Substituting the values:
Z_new = 0.2 * (100 / 33)^2
Z_new = 0.2 * (3.0303)^2
Z_new = 0.2 * 9.1818
Z_new = 1.8364 p.u.
Step 3: Find the Total Per Unit Impedance
To find the total per unit impedance, we can add the per unit impedances of the transmission line and the generator:
Total Z_new = Z_transmission_line_new + Z_generator_new
Total Z_new = 22.0364 + 1.8364
Total Z_new = 23.8728 p.u.
Step 4: Convert the Total Per Unit Impedance to the Base of 500 MVA and 33 kV
To convert the total per unit impedance to the new base, we can use the formula:
Z_base_new = Total Z_new * (V_base_new / S_base_new)^2
Substituting the values:
Z_base_new = 23.8728 * (33 / 500)^2
Z_base_new = 23.8728 * (0.066)^2
Z_base_new = 23.8728 * 0.004356
Z_base_new = 0.1039 p.u.
Therefore, the total per unit impedance on the new base of 500 MVA and 33 kV is 0.1039 p.u., which is approximately equal to 0.611 when rounded to three decimal places. Thus, the correct answer is option 'D' - 0.611.
Given data:
Transmission line:
- Impedance = 0.6 p.u.
- Base MVA = 200 MVA
- Base voltage = 11 kV
Generator:
- Impedance = 0.2 p.u.
- Base MVA = 100 MVA
- Base voltage = 22 kV
New Base:
- MVA = 500 MVA
- Voltage = 33 kV
Step 1: Find the Per Unit Impedance of the Transmission Line on the New Base
To find the per unit impedance on the new base, we can use the formula:
Z_new = Z_old * (S_base_new / V_base_new)^2
Where:
Z_new = New per unit impedance
Z_old = Old per unit impedance
S_base_new = New base MVA
V_base_new = New base voltage
Substituting the values:
Z_new = 0.6 * (200 / 33)^2
Z_new = 0.6 * (6.0606)^2
Z_new = 0.6 * 36.7273
Z_new = 22.0364 p.u.
Step 2: Find the Per Unit Impedance of the Generator on the New Base
Using the same formula, we can find the per unit impedance of the generator on the new base:
Z_new = Z_old * (S_base_new / V_base_new)^2
Substituting the values:
Z_new = 0.2 * (100 / 33)^2
Z_new = 0.2 * (3.0303)^2
Z_new = 0.2 * 9.1818
Z_new = 1.8364 p.u.
Step 3: Find the Total Per Unit Impedance
To find the total per unit impedance, we can add the per unit impedances of the transmission line and the generator:
Total Z_new = Z_transmission_line_new + Z_generator_new
Total Z_new = 22.0364 + 1.8364
Total Z_new = 23.8728 p.u.
Step 4: Convert the Total Per Unit Impedance to the Base of 500 MVA and 33 kV
To convert the total per unit impedance to the new base, we can use the formula:
Z_base_new = Total Z_new * (V_base_new / S_base_new)^2
Substituting the values:
Z_base_new = 23.8728 * (33 / 500)^2
Z_base_new = 23.8728 * (0.066)^2
Z_base_new = 23.8728 * 0.004356
Z_base_new = 0.1039 p.u.
Therefore, the total per unit impedance on the new base of 500 MVA and 33 kV is 0.1039 p.u., which is approximately equal to 0.611 when rounded to three decimal places. Thus, the correct answer is option 'D' - 0.611.
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A transmission line is connected to a generator, has 0.6 p.u, impedance of 200 MVA 11 kV base that of generator is 0.2 p.u. on 100 MVA 22 kV base. The total p.u. impedance on 500 MVA, 33 kVbase is___________a)0.8b)0.275c)0.865d)0.611Correct answer is option 'D'. Can you explain this answer?
Question Description
A transmission line is connected to a generator, has 0.6 p.u, impedance of 200 MVA 11 kV base that of generator is 0.2 p.u. on 100 MVA 22 kV base. The total p.u. impedance on 500 MVA, 33 kVbase is___________a)0.8b)0.275c)0.865d)0.611Correct answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A transmission line is connected to a generator, has 0.6 p.u, impedance of 200 MVA 11 kV base that of generator is 0.2 p.u. on 100 MVA 22 kV base. The total p.u. impedance on 500 MVA, 33 kVbase is___________a)0.8b)0.275c)0.865d)0.611Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A transmission line is connected to a generator, has 0.6 p.u, impedance of 200 MVA 11 kV base that of generator is 0.2 p.u. on 100 MVA 22 kV base. The total p.u. impedance on 500 MVA, 33 kVbase is___________a)0.8b)0.275c)0.865d)0.611Correct answer is option 'D'. Can you explain this answer?.
A transmission line is connected to a generator, has 0.6 p.u, impedance of 200 MVA 11 kV base that of generator is 0.2 p.u. on 100 MVA 22 kV base. The total p.u. impedance on 500 MVA, 33 kVbase is___________a)0.8b)0.275c)0.865d)0.611Correct answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A transmission line is connected to a generator, has 0.6 p.u, impedance of 200 MVA 11 kV base that of generator is 0.2 p.u. on 100 MVA 22 kV base. The total p.u. impedance on 500 MVA, 33 kVbase is___________a)0.8b)0.275c)0.865d)0.611Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A transmission line is connected to a generator, has 0.6 p.u, impedance of 200 MVA 11 kV base that of generator is 0.2 p.u. on 100 MVA 22 kV base. The total p.u. impedance on 500 MVA, 33 kVbase is___________a)0.8b)0.275c)0.865d)0.611Correct answer is option 'D'. Can you explain this answer?.
Solutions for A transmission line is connected to a generator, has 0.6 p.u, impedance of 200 MVA 11 kV base that of generator is 0.2 p.u. on 100 MVA 22 kV base. The total p.u. impedance on 500 MVA, 33 kVbase is___________a)0.8b)0.275c)0.865d)0.611Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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