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A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.
- a)4°
- b)2.6°
- c)1.8°
- d)3°
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an ine...
GH = 30 × 10 = 300MJ
Pm = 26800hp = 26800 × 0.746 = 20,000KW
Pe = 16000KW
Pa = Pm − Pe = 20,000 − 16,000 = 4000KW
Pa = Md2δ/dt2
M = GH/180f = 300/(180 × 50) = 1/30
4 = 1/30(d2δ/d1)
d2δ/dt2 = 120%sec2 = 120 × π/180 = 2.09rad/sec2
Multiply by 2(dδ/dt). dt on both sides 
Integrating the above equation 
Further integration
constant for 10 cycle which means 10 cycles = 0.2 sec δ = 1.045(0.2)2 = 0.0418rad = 1.8∘
Most Upvoted Answer
A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an ine...
Given data:
- Frequency (f) = 50 cycles
- Number of poles (p) = 4
- Rated power (S) = 30 MVA
- Rated voltage (V) = 11.2 kV
- Inertia constant (H) = 10 kW sec/KVA
- Acceleration computed for 10 cycles (Δt) = 10 cycles
- Input less rotational losses (P_in) = 26800
- Electric power developed (P_gen) = 16 MW
To find: Change in torque angle (Δδ) in that period
Formula:
Acceleration (Δω) = (2πfΔt) / p
Change in torque angle (Δδ) = (Δω * H) / (2πf)
Calculation:
1. Convert rated power to VA:
30 MVA = 30,000,000 VA
2. Convert rated voltage to V:
11.2 kV = 11,200 V
3. Convert electric power developed to VA:
16 MW = 16,000,000 VA
4. Calculate the change in angular velocity:
Δω = (2π * 50 * 10) / 4
= 785.4 rad/sec
5. Calculate the change in torque angle:
Δδ = (785.4 * 10 * 10^3) / (2π * 50)
= 785.4 * 10^3 / 314.16
≈ 2500 rad
6. Convert the change in torque angle from radians to electrical degrees:
1 rad = 180/π electrical degrees
2500 rad ≈ (2500 * 180) / π electrical degrees
≈ 40181.47 electrical degrees
7. Calculate the change in torque angle in electrical degrees:
Δδ = 40181.47 electrical degrees
Therefore, the change in torque angle in that period is approximately 40181.47 electrical degrees, which is equivalent to 1.8° (option c).
- Frequency (f) = 50 cycles
- Number of poles (p) = 4
- Rated power (S) = 30 MVA
- Rated voltage (V) = 11.2 kV
- Inertia constant (H) = 10 kW sec/KVA
- Acceleration computed for 10 cycles (Δt) = 10 cycles
- Input less rotational losses (P_in) = 26800
- Electric power developed (P_gen) = 16 MW
To find: Change in torque angle (Δδ) in that period
Formula:
Acceleration (Δω) = (2πfΔt) / p
Change in torque angle (Δδ) = (Δω * H) / (2πf)
Calculation:
1. Convert rated power to VA:
30 MVA = 30,000,000 VA
2. Convert rated voltage to V:
11.2 kV = 11,200 V
3. Convert electric power developed to VA:
16 MW = 16,000,000 VA
4. Calculate the change in angular velocity:
Δω = (2π * 50 * 10) / 4
= 785.4 rad/sec
5. Calculate the change in torque angle:
Δδ = (785.4 * 10 * 10^3) / (2π * 50)
= 785.4 * 10^3 / 314.16
≈ 2500 rad
6. Convert the change in torque angle from radians to electrical degrees:
1 rad = 180/π electrical degrees
2500 rad ≈ (2500 * 180) / π electrical degrees
≈ 40181.47 electrical degrees
7. Calculate the change in torque angle in electrical degrees:
Δδ = 40181.47 electrical degrees
Therefore, the change in torque angle in that period is approximately 40181.47 electrical degrees, which is equivalent to 1.8° (option c).
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A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. Can you explain this answer?
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A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. 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 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. 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 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. Can you explain this answer?.
A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. 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 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. 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 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. Can you explain this answer?.
Solutions for A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. Can you explain this answer?, a detailed solution for A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.a)4°b)2.6°c)1.8°d)3°Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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